The behaviour of gases-Igori wallace

The behaviour of gases 2016

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CHAPTER ONE

MATTER

1.0 Introduction Long before the science of chemistry was established, materials

were described as existing in one of three physical states. There are either rigid, solid objects, having a definite volume and a fixed shape, nonrigid liquids, having no fixed shape other than that of their containers but having definite volumes or gases, which have neither fixed shape nor fixed volume.

The techniques used for handling various materials depend on their physical states as well as their chemical properties. While it is comparatively easy to handle liquids and solids, it is not as convenient to measure out a quantity of a gas. Fortunately, except under rather extreme conditions, all gases have similar physical properties, and the chemical identity of the substance does not influence those properties. For example, all gases expand when they are heated in a nonrigid container and contract when they are cooled or subjected to increased pressure. They readily diffuse through other gases. Any quantity of gas will occupy the entire volume of its container, regardless of the size of the container.

1.1. States of Matter

Matter is anything that has mass and occupies space. All the material things in the universe are composed of matter, including anything we can touch as well as the planets in the solar system and all the stars in the sky. It is composed of tiny particles such as atoms, molecules, or ions and can exist in three physical states- solid, liquid and gas. Solid State

In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between


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the particles. As a result, solids have a definite shape, volume and are incompressible. Most solids are hard, but some (like waxes) are relatively soft. Some large crystals look the way they do because of the regular arrangement of atoms (ions) in their crystal structure. Solids usually have their constituent particles arranged in a regular, three-dimensional array of alternating positive and negative ions called a crystal. Some solids, especially those composed of large molecules, cannot easily organize their particles in such regular crystals and exist as amorphous (literally, ―without form‖) solids. Glass is one example of an amorphous solid.

Liquid State A liquid is a nearly incompressible fluid that conforms to the

shape of its container but retains a (nearly) constant volume independent of pressure. The volume is definite if the temperature and pressure are constant. The molecules have enough energy to move relative to each other and the structure is mobile. Gaseous State

Gases consist of tiny particles widely spaced (Figure 1.1). Under typical conditions, the average distance between gas particles is about ten times their diameter. Because of these large distances, the volume occupied by the particles themselves is very small compared to the volume of the empty space around them. For a gas at room temperature and pressure, the gas particles themselves occupy about 0.1% of the total volume. The other 99.9% of the total volume is empty space (whereas in liquids and solids, about 70% of the volume is occupied by particles). Because of the large distances between gas particles, the attractions or repulsions among them are weak.

The particles in a gas are in rapid and continuous motion. For example, the average velocity of nitrogen molecules, N2, at 20 °C is about 500 m/s. As the temperature of a gas increases, the particles‘ velocity increases. The average velocity of nitrogen molecules at 100 °C is about 575 m/s.

http://en.wikipedia.org/wiki/Fluid

http://en.wikipedia.org/wiki/Temperature

http://en.wikipedia.org/wiki/Pressure


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The particles in a gas are constantly colliding with the walls of the container and with each other. Because of these collisions, the gas particles are constantly changing their direction of motion and their velocity. In a typical situation, a gas particle moves a very short distance between collisions. For example, oxygen, O2, molecules at normal temperatures and pressures move an average of 10-7 m between collisions.

Fig.1.1. A Representation of the Solid, Liquid, and Gas States The various characteristics or properties of the states of matter discussed above are summarized in table 1.1 below.

Table 1.1. Characteristics of the Three States of Matter Characteristic Solid Liquid Gas

Shape Definite conforms to the shape of its container

Indefinite

Volume Definite Definite Indefinite

Relative intermolecular

interaction strength

Strong Moderate Weak

Relative particle positions

in contact and fixed in place

in contact but not fixed

not in contact, random

positions

Compressibility incompressible incompressible Compressible fluid


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1.2. Phase Transition Phase transition is a term used to describe a state of change of

matter from one state to another. The state or phase of a given set of matter can change depending on pressure and temperature conditions, transitioning to other phases as these conditions change to favour their existence; for example, solid transitions to liquid with an increase in temperature. Near absolute zero, a substance exists as a solid. As heat is added to this substance it melts into a liquid at its melting point, boils into a gas at its boiling point, and if heated high enough would enter a plasma state in which the electrons are so energized that they leave their parent atoms. 1.2.1. Melting point

This is the temperature at which the solid and liquid forms of a pure substance can exist at equilibrium. As heat is applied to a solid, its temperature will increase until the melting point is reached. More heat then will convert the solid into a liquid with no temperature change. When the entire solid has melted, additional heat will raise the temperature of the liquid. The melting temperature of crystalline solids is a characteristic figure and is used to identify pure compounds and elements. Most mixtures and amorphous solids melt over a range of temperatures.

The melting temperature of a solid is generally considered to be the same as the freezing point of the corresponding liquid; because a liquid may freeze in different crystal systems and because impurities lower the freezing point, however, the actual freezing point may not be the same as the melting point. Thus, for characterizing a substance, the melting point is preferred. A typical example is the change of solid ice to liquid water as shown below. H2O(s) → H2O(l) (melting, fusion) Ice, snow liquid water

http://en.wikipedia.org/wiki/Pressure


http://en.wikipedia.org/wiki/Absolute_zero

http://en.wikipedia.org/wiki/Solid

http://en.wikipedia.org/wiki/Liquid

http://en.wikipedia.org/wiki/Melting_point

http://en.wikipedia.org/wiki/Boiling_point

http://en.wikipedia.org/wiki/Plasma_%28physics%29

http://en.wikipedia.org/wiki/Electrons

http://www.britannica.com/EBchecked/topic/586581/temperature

http://www.britannica.com/EBchecked/topic/553257/solid

http://www.britannica.com/EBchecked/topic/343026/liquid

http://www.britannica.com/EBchecked/topic/374164/melt

http://www.britannica.com/EBchecked/topic/218744/freezing-point

http://www.britannica.com/EBchecked/topic/218744/freezing-point

http://www.britannica.com/EBchecked/topic/218717/freezing


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1.2.2. Freezing point This is the temperature at which a liquid becomes a solid. As

with the melting point, increased pressure usually raises the freezing point. The freezing point is lower than the melting point in the case of mixtures and for certain organic compounds such as fats. As a mixture freezes, the solid that forms first usually has a composition different from that of the liquid, and formation of the solid changes the composition of the remaining liquid, usually in a way that steadily lowers the freezing point. This principle is used in purifying mixtures, successive melting and freezing gradually separating the components. The heat of fusion (heat that must be applied to melt a solid), must be removed from the liquid to freeze it. Some liquids can be supercooled i.e., cooled below the freezing point without solid crystals forming. Putting a seed crystal into a supercooled liquid triggers freezing, whereupon the release of the heat of fusion raises the temperature rapidly to the freezing point. Freezing of liquid water to ice is a common example. H2O(l) → H2O(s) (freezing) liquid water Ice 1.2.3. Condensation

This is change of a gas to either liquid or solid state, generally upon a surface that is cooler than the adjacent gas. The change of vapour to solid is sometimes called deposition. A substance condenses when the pressure exerted by its vapour exceeds the vapour pressure of the liquid or solid phase of the substance at the temperature of the surface where condensation occurs. Heat is released when a vapour condenses. Unless this heat is removed, the surface temperature will increase until it is equal to that of the surrounding vapour. In the atmosphere, however, there is an abundant supply of aerosols, which serve as nuclei, called condensation nuclei, on which water vapour may condense. Some are hygroscopic (moisture-attracting), and condensation begins on them when the relative humidity is less than


http://www.britannica.com/EBchecked/topic/374185/melting-point

http://www.britannica.com/EBchecked/topic/218731/freezing

http://www.britannica.com/EBchecked/topic/258697/heat-of-fusion

http://www.britannica.com/EBchecked/topic/455270/phase

http://www.britannica.com/EBchecked/topic/131435/condensation-nucleus


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100 percent, but other nuclei require some supersaturation before condensation begins. Condensation accounts for the formation of dew (liquid water formed by condensation of water vapour from the atmosphere), and Frost (solid water formed by direct condensation of water vapour from the atmosphere without first forming liquid water). H2O(g) → H2O(l) (condensation) Water vapour dew

H2O(g) → H2O(s) (condensation, deposition) Water vapour frost, snow 1.2.4. Vapourization

This refers to the conversion of a substance from the liquid or solid phase into the gaseous (vapour) phase. Heat must be supplied to a solid or liquid to effect vaporization. If the surroundings do not supply enough heat, it may come from the system itself as a reduction in temperature. The atoms or molecules of a liquid or solid are held together by cohesive forces, and these forces must be overcome in separating the atoms or molecules to form the vapour; the heat of vaporization is a direct measure of these cohesive forces. H2O(l) → H2O(g) (vaporization) Liquid water water vapour 1.2.5. Sublimation

The change of a solid directly to the vapour without its becoming liquid is specifically referred to as sublimation. Although the vapor pressure of many solids is quite low, some (usually molecular solids) have appreciable vapor pressure. Ice, for instance, has a vapour pressure of 4.7 mmHg at 0oC. For this reason, a pile of snow slowly disappears in winter even though the temperature is too low for it to melt. The snow is being changed directly to water vapour.


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H2O(s) → H2O(g) (sublimation) Ice, snow Water vapour

Sublimation can be used to purify solids such as impure iodine that readily vaporize. Impure iodine is heated in a beaker so that it vaporizes, leaving nonvolatile impurities behind. The vapour crystallizes on the bottom surface of a dish containing ice that rests on top of the beaker. Freeze-drying of foods is a commercial application of sublimation. Brewed coffee, for example, is frozen and placed in a vacuum to remove water vapour. The ice continues to sublime until it is all gone, leaving freeze-dried coffee. Most freeze-dried foods are easily reconstituted by adding water. The following diagram summarizes these phase transitions.

Fig.1.2. Diagram showing the nomenclature for the different phase transitions.

1.3. Heat of Phase Transition Any change of state involves the addition or removal of energy

as heat to or from the substance. A simple experiment shows that this is the case. Suppose you add heat at a constant rate to a beaker containing ice at -20oC. In Figure 1.3 below, we have plotted the temperature of the different phases of water as heat is added. The


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temperature of the ice begins to rise from -20oC, as you would expect; the addition of heat normally raises the temperature of a substance. At 0oC, the ice begins to melt, so that you get a beaker of ice in water. Note the flat region in the curve, labeled ice and water. Why is this region flat? It means that heat is being added to the system without a change in temperature; the temperature remains at 0oC. This temperature, of course, is the melting point of ice. The heat being added is energy required to melt ice to water at the same temperature. The intermolecular forces binding water molecules to specific sites in the solid phase must be partially broken to allow water molecules the ability to slide over one another easily, as happens in the liquid state. Note the flat regions for each of the phase transitions. Because heat is being added at a constant rate, the length of each flat region is proportional to the heat of phase transition.

Fig. 1.3. Heating curve for water: Heat is being added at a constant rate to a system containing water. Note the flat regions of the curve. When heat is added during a phase transition, the temperature does not change.

The heat needed for the melting of a solid is called the heat of

fusion (or enthalpy of fusion) and is denoted ∆Hfus. For ice, the heat of fusion is 6.01 kJ per mole. H2O(s) → H2O(l); ∆Hfus = 6.01 kJ/mol


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The heat needed for the vaporization of a liquid is called the heat of vaporization (or enthalpy of vaporization) and is denoted ∆Hvap. At 100oC, the heat of vaporization of water is 40.7 kJ per mole. H2O(l) → H2O(g); ∆Hvap = 40.7 kJ/mol Note that much more heat is required for vaporization than for melting. Melting needs only enough energy for the molecules to escape from their sites in the solid. For vaporization, enough energy must be supplied to break most of the intermolecular attractions. A refrigerator relies on the cooling effect accompanying vaporization. The mechanism contains an enclosed gas that can be liquefied under pressure, such as ammonia or 1,1,1,2-tetrafluoroethane, CH2FCF3. As the liquid is allowed to evaporate, it absorbs heat and thus cools its surroundings (the interior space of the refrigerator). Gas from the evaporation is recycled to a compressor and then to a condenser, where it is liquefied again. Heat leaves the condenser, going into the surrounding air. 1.4. Pressure of Gases

The molecules of a gas, being in continuous motion, frequently strike the inner walls of their container. As they do so, they immediately bounce off without loss of kinetic energy, but the reversal of direction (acceleration) imparts a force to the container walls. This force, divided by the total surface area on which it acts, is the pressure of the gas.

The pressure of a gas is observed by measuring the pressure that must be applied externally in order to keep the gas from expanding or contracting. To visualize this, imagine some gas trapped in a cylinder having one end enclosed by a freely moving piston. In order to keep the gas in the container, a certain amount of weight (more precisely, a force, f) must be placed on the piston so as to exactly balance the force exerted by the gas on the bottom of the piston, and tending to push it up. The pressure of the gas (P) is simply the quotient f/A, where A is the cross-section area of the piston.


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Example 1.1. If a force of 16N is pressed against an area of 2.44 m2, what is the pressure in pascals?

Solution Given force, F = 16N, area, A = 2.44 m2

Apply the relationship,

𝑃 = 𝐹

𝐴

𝑃 = 16𝑁

2.44m2

= 6.57𝑁𝑚−2

1.4.1. Pressure Units The unit of pressure in the SI system is the pascal (Pa), defined

as a force of one newton per square metre (1 Nm–2 = 1 kg m–1 s–2 ). In chemistry, it is more common to express pressures in units of atmospheres or torr:

1 atm = 101325 Pa = 760 torr.

The older unit millimetre of mercury (mm Hg) is almost the same as the torr; it is defined as one mm of level difference in a mercury barometer at 0°C. In meteorology, the pressure unit most commonly used is the bar:

1 bar = 106 N m–2 = 0.987 atm. For conversion purposes,

1 atm = 760 torr =760 mmHg = 1.01325 × 105 Nm-2

Example 1.2. How many atmospheres are in 1547mmHg

Solution Use the conversion factor; 1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔 ∴ 𝑥 𝑎𝑡𝑚 = 1547 𝑚𝑚𝐻𝑔 Cross multiplying and making 𝑥 the subject gives


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𝑥 = 1 𝑎𝑡𝑚 ×1547 𝑚𝑚𝐻𝑔

760 𝑚𝑚𝐻𝑔

𝑥 = 2.04 𝑎𝑡𝑚

Example 1.3. Write the conversion factor to determine how many mmHg are in 9.65 atm.

Solution Use the same conversion factor as in example 1.2 above 1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔 ∴ 9.65 𝑎𝑡𝑚 = 𝑥 𝑚𝑚𝐻𝑔 Cross multiplying and making 𝑥 the subject give

𝑥 = 9.65 𝑎𝑡𝑚 ×760 𝑚𝑚𝐻𝑔

1 𝑎𝑡𝑚

𝑥 = 7334 𝑚𝑚𝐻𝑔

Example 1.4. How many torr are in 1.56 atm

Solution

Use the conversion factor;

1 𝑎𝑡𝑚 = 760 𝑡𝑜𝑟𝑟

∴ 1.56 𝑎𝑡𝑚 = 𝑥 𝑡𝑜𝑟𝑟

Cross multiplying and making 𝑥 the subject give

𝑥 = 1.56𝑎𝑡𝑚 ×760 𝑡𝑜𝑟𝑟

1 𝑎𝑡𝑚

𝑥 = 1190 𝑡𝑜𝑟𝑟

Example 1.5. Blood pressures are expressed in mmHg. What would be the blood pressure in atm if a patient‘s systolic and diastolic blood pressures are 120 mmHg and 82 mmHg respectively? (In medicine,


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such a blood pressure would be reported as ―120/82‖, spoken as ―one hundred twenty over eighty-two‖ ).

Solution Use the same conversion factor as in example one above 1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔 ∴ 9.65 𝑎𝑡𝑚 = 𝑥 𝑚𝑚𝐻𝑔

Cross multiplying and making 𝑥 the subject give

Use the same conversion factor as in example one above

1 𝑎𝑡𝑚 = 760 𝑚𝑚𝐻𝑔

∴ 120 𝑚𝑚𝐻𝑔 = 120 𝑚𝑚𝐻𝑔 ×1 𝑎𝑡𝑚


= 0.157 atm

82 𝑚𝑚𝐻𝑔 = 82 𝑚𝑚𝐻𝑔 ×1 𝑎𝑡𝑚


= 0.107 atm

∴ 120 𝑚𝑚𝐻𝑔

82 𝑚𝑚𝐻𝑔= 0.157 𝑎𝑡𝑚: 0.107 𝑎𝑡𝑚

1.4.2. Atmospheric Pressure This is defined as the force per unit area exerted against a surface by the weight of the air above that surface. In most circumstances atmospheric pressure is closely approximated by the hydrostatic pressure caused by the weight of air above the measurement point. On a given plane, low-pressure areas have less atmospheric mass above their location, whereas high-pressure areas have more atmospheric mass above their location. Likewise, as elevation (altitude) increases, there is less overlying atmospheric mass, so that atmospheric pressure decreases with increasing elevation.

http://en.wikipedia.org/wiki/Fluid_pressure

http://en.wikipedia.org/wiki/Weight

http://en.wikipedia.org/wiki/Earth%27s_atmosphere

http://en.wikipedia.org/wiki/Elevation


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1.4.3. Measurement of Gas Pressure A barometer is piece of lab equipment specifically designed to

measure the atmospheric pressure. Invented in the early 17th century by the Italian EVANGELISTA TORRICELLI. The barometer consists of a vertical glass tube closed at the top and evacuated, and open at the bottom, where it is immersed in a dish of a liquid. The atmospheric pressure acting on this liquid will force it up into the evacuated tube until the weight of the liquid column exactly balances the atmospheric pressure. If the liquid is mercury, the height supported will be about 760 cm; this height corresponds to standard atmospheric pressure.

Fig. 1.4. A simple barometer

The formula for this pressure in the atmosphere is derived as shown below: 𝒇𝒐𝒓𝒄𝒆 = 𝒎𝒂𝒔𝒔 × 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏

or 𝑭 = 𝒎𝒂 or mg


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Earth's acceleration of objects is based on its gravitational field and equals approximately 9.80665 m s-2. Additionally, since pressure is the force per the unit area being measured, then

𝑷 =𝑭

𝑨

=𝒎𝒈

𝑨

Since mass (m) = density (d) × volume (v)

𝑷 = 𝒈 ×𝒅 ×𝑽

𝑨

Since 𝑉𝑜𝑙𝑢𝑚𝑒 (𝑋3)

𝐴𝑟𝑒𝑎 (𝑋2)= 𝑕𝑖𝑒𝑔𝑕𝑡 (𝑋)

𝑷 = 𝑔 × 𝑑 × 𝑕 Where d = density, g = gravity and h = height of the liquid or gas. Example 1.6. Mercury has a density of 13.6 g/cm3 and water has a density of 1.00 g/cm3. If a column of mercury has a height of 755 mm, how high would a corresponding column of water be in feet? Solution: Let us begin by setting the pressures equal:

Pmercury = Pwater Since 𝑷 = 𝑔 × 𝑑 × 𝑕 We can write:

𝑕𝑤𝑎𝑡𝑒𝑟 = 𝑑𝐻𝑔 × 𝑕𝐻𝑔

𝑑𝑤𝑎𝑡𝑒𝑟

=13.6𝑔𝑐𝑚 −3 ×755 𝑚𝑚

1.00𝑔𝑐𝑚 −3

= 10268 𝑚𝑚 = 33.7 𝑓𝑡 1.4.4. The Manometer

A modification of the barometer, the U-tube manometer, provides a simple device for measuring the pressure of any gas in a container. There are a variety of manometer designs. A simple,


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common design is to seal a length of glass tubing and bend the glass tube into a U-shape. The glass tube is then filled with a liquid, typically mercury, so that all trapped air is removed from the sealed end of the tube. The glass tube is then positioned with the curved region at the bottom. The mercury settles to the bottom.

After the mercury settles to the bottom of the manometer, a vacuum is produced in the sealed tube. The open tube is connected to the system whose pressure is being measured. In the sealed tube, there is no gas to exert a force on the mercury (except for some mercury vapor). In the tube connected to the system, the gas in the system exerts a force on the mercury. The net result is that the column of mercury in the sealed tube is higher than that in the unsealed tube. The difference in the heights of the columns of mercury is a measure of the pressure of gas in the system.

In the open-tube manometer, the pressure of the gas is given by h (the difference in mercury levels) in units of torr or mmHg. Atmospheric pressure pushes on the mercury from one direction, and the gas in the container pushes from the other direction. In a manometer, since the gas in the bulb is pushing more than the atmospheric pressure, you add the atmospheric pressure to the height difference:

Pgas > Patm

Gas pressure = atmospheric pressure + h (height of the mercury) Pgas < Patm Gas pressure = atmospheric pressure - h (height of the mercury) The closed-tube manometer look similar to regular manometers except that the end that is open to the atmospheric pressure in a regular manometer is sealed and contains a vacuum. In these systems, the difference in mercury levels (in mmHg) is equal to the pressure in torr.


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Fig.1.5. The Manometer

Example 1.7. Find the pressures using the manometer set up below.

Solution

since Pgas > Patm

Pgas= Patm + h

Pgas= (755 + 24 )mmHg

=779mmHg

since Pgas < Patm Pgas= Patm ‒ h

Pgas= (763 ‒35)g

Pgas= 728 mmHg


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Example 1.8. Suppose you want to construct a closed-end manometer to measure gas pressures in the range 0.000–0.200 atm. Because of the toxicity of mercury, you decide to use water rather than mercury. How tall a column of water do you need? (The density of water is 1.00 g/cm3; the density of mercury is 13.6 g/cm3). Solution Given: pressure range and densities of water and mercury, column height unknown. Strategy: Step 1. Calculate the height of a column of mercury corresponding to 0.200 atm in millimeters of mercury. This is the height needed for a mercury-filled column. Step 2. From the given densities, use a proportion to compute the height needed for a water-filled column. In millimeters of mercury, a gas pressure of 0.200 atm 1atm = 760mmHg

∴ 0.200 atm will be 0.200 𝑎𝑡𝑚 ×760𝑚𝑚𝐻𝑔

1 𝑎𝑡𝑚

= 152 𝑚𝑚𝐻𝑔 Using a mercury manometer, you would need a mercury column of at least 152 mm high.

Because water is less dense than mercury, you need a taller column of water to achieve the same pressure as a given column of mercury. The height needed for a water-filled column corresponding to a pressure of 0.200 atm is proportional to the ratio of the density of mercury to the density of water; Using 𝑷 = 𝑔 × 𝑑 × 𝑕 Where d = density, g = gravity and h = height of the liquid or gas. Let us begin by setting the pressures equal:

Pmercury = Pwater We can then write: 𝑔 × 𝑑𝐻𝑔 × 𝑕𝐻𝑔 = 𝑔 × 𝑑𝑤𝑎𝑡𝑒𝑟 × 𝑕𝑤𝑎𝑡𝑒𝑟


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𝑕𝑤𝑎𝑡𝑒𝑟 = 𝑑𝐻𝑔 × 𝑕𝐻𝑔

𝑑𝑤𝑎𝑡𝑒𝑟

=13.6𝑔𝑐𝑚 −3 × 152 𝑚𝑚

1.00𝑔𝑐𝑚 −3

= 2070 𝑚𝑚 Comment: it takes a taller column of a less dense liquid to achieve the same pressure. 1.4.5. Effect of Pressure on the volume of gases

For a gas whose volume is not fixed, increasing the pressure will cause the gas to contract (reducing the volume), and decreasing the pressure will cause the gas to expand (increasing the volume). If the volume is fixed, then increasing the pressure will increase the temperature, and decreasing the pressure will decrease the temperature. 1.4.6. Simple Pressure Related Applications

• Drinking straw: A drinking straw is used by creating a

suction with your mouth. Actually this causes a decrease in air pressure on the inside of the straw. Since the atmospheric pressure is greater on the outside of the straw, liquid is forced into and up the straw.

• Siphon: With a siphon water can be made to flow "uphill". A siphon can be started by filling the tube with water (perhaps by suction). Once started, atmospheric pressure upon the surface of the upper container forces water up the short tube to replace water flowing out of the long tube. 1.5. Density of a Gas

This is defined as mass divided by the volume of a gas

𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑑 = 𝑚𝑎𝑠𝑠 (𝑔)

𝑣𝑜𝑙𝑢𝑚𝑒 (𝐿)


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The density of a gas in grams/L can be obtained from ideal gas equation as follows: 𝑃𝑉 = 𝑛𝑅𝑇

Number of mole of a gas (n) = 𝑚𝑎𝑠𝑠 (𝑚)

𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)

Substituting ―n‖ into ideal gas equation above

𝑃𝑉 = 𝑚

𝑀× 𝑅𝑇

Cross multiplying we have 𝑀 × 𝑃𝑉 = 𝑚 × 𝑅𝑇 Divide both side by V gives

𝑃 × 𝑀 = 𝑚

𝑉× 𝑅𝑇

Lastly divide both by RT gives density 𝑚

𝑉=

𝑃 × 𝑀

𝑅𝑇

𝑑 = 𝑃 × 𝑀

𝑅𝑇

Example 1.9. What is the density of oxygen at STP? [R= 0.8206L atm mol-1K-1] Solution Data collection S.t = 273K S.p = 1 atm R= 0.8206L atm mol-1K-1

Molecular weight, M of oxygen = 32.0gmol-1

Using 𝑑 = 𝑚

𝑉=

𝑃 × 𝑀

𝑅𝑇

=1 𝑎𝑡𝑚 × 32.0 𝑔𝑚𝑜𝑙−1

0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 273.15𝐾

= 1.428𝑔/𝐿 Example 1.10. A 0.0125g sample of a gas with an empirical formula of CHF2 is placed in a 165-mL flask. It has a pressure of 13.7 mm Hg at 22.5 °C. What is the molecular formula of the compound? [R= 0.8206L atm mol-1K-1]


20

Solution Collect the available data and convert as necessary to agree with the given unit of R then find the value of density from which the molecular weight of the

gas can be determined using the relation: 𝑑 = 𝑃 × 𝑀

𝑅𝑇

Mass of gas sample = 0.0125g Volume = 165 mL = 0.156 L Temperature, T = 22.5°C = 295.7K

Pressure, P = 13.7 mm Hg = 1 atm × 13.7 𝑚𝑚𝐻𝑔

760 𝑚𝑚𝐻𝑔= 0.0180 𝑎𝑡𝑚𝑎𝑡𝑚

Now density, 𝑑= 𝑚

𝑉 𝑑 =

𝑚

𝑉=

0.0125𝑔 𝑔

0.156 𝐿𝐿

= 0.0758 𝑔𝑔/𝐿𝐿 To find molecular weight of gas, we use

𝑑= 𝑃 × 𝑀

𝑅𝑇

Making molecular weight, M the subject and substituting

M = 𝑑𝑅𝑇

𝑃

= 0.075𝑔𝐿−1 × 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 273.15𝐾

0.0180 𝑎𝑡𝑚

M = 102𝑔𝑚𝑜𝑙−1 The molecular formula is (CHF2)2 or C2H2F4. Example 1.11. If 0.11 g of H2O2 decomposes in a 2.50 L flask at 25 oC, what is the pressure of O2 & H2O?[0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 , H =1, 0 = 16, ] Solution Step 1: Write the balanced chemical reaction. Step 2: Calculate the moles of each product. Step 3: Find the pressure of each via PV = nRT Equation of reaction : 2H2O2(l) → 2H2O (g) + O2 (g)

From the equation of reaction, 2 mol of 2H2O2 produce 2 mol of H2O and a mol of O2.


21

Therefore mol of H2O2 = 𝑚𝑎𝑠𝑠

𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠

= 0.11𝑔

34𝑔/𝑚𝑜𝑙

=0.0032 mol

mol of O2 = 1

2 × 0.0032 mol of H2O2

=0.0016 mol mol of H2O = 1 × 0.0032 mol of H2O2

=0.0032 mol Using PV = nRT to calculate the pressure of the gases

P(O2) = 𝑛𝑅𝑇

𝑉

= 0.0016 𝑚𝑜𝑙 × 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 298𝐾

2.5 𝐿

= 0.016 atm

P(H2O) = 𝑛𝑅𝑇

𝑉

= 0.0032 𝑚𝑜𝑙 × 0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 298𝐾

2.5 𝐿

= 0.032 atm Example 1.12. A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 8.14 g/L at 47°C and 3.15 atm. Calculate the molar mass of the compound and determine its molecular formula. Solution We can calculate the molar mass of a gas if we know its density, temperature, and pressure. The molecular formula of the compound must be consistent with its molar mass. What temperature unit should we use? Data provided density = 8.14 g/L T = 47°C = 320 K P = 3.15 atm

Using the relationship, 𝑀 =𝑑𝑅𝑇

𝑃 to solve for molar mass,


22

=8.14 𝑔𝐿−1 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾−1 ×320 𝐾

3.15 𝑎𝑡𝑚

= 67.9 𝑔𝑚𝑜𝑙−1 We can determine the molecular formula of the compound by trial and error, using only the knowledge of the molar masses of chlorine (35.45 g) and oxygen (16.00 g). We know that a compound containing one Cl atom and one O atom would have a molar mass of 51.45 g, which is too low, while the molar mass of a compound made up of two Cl atoms and one O atom is 86.90 g, which is too high. Thus, the compound must contain one Cl atom and two O atoms and have the formula ClO2 , which has a molar mass of 67.45 g. Example 1.13. The density of a gaseous organic compound is 3.38 g/L at 40°C and 1.97 atm. What is its molar mass? Solution Data provided d = 3.38 g/L T = 40°C = 313 K P = 1.97 atm

Using the relationship 𝑀 =𝑑𝑅𝑇

𝑃

=3.38 𝑔𝐿−1 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾−1 ×313 𝐾

1.97 𝑎𝑡𝑚

= 44.0 𝑔𝑚𝑜𝑙−1 1.5.1. The effects of temperature on density

The density of a gas depends quite strongly on its temperature, so hot air has a smaller density than does cold air; colder air is more dense than hot air. From everyday experience, we know that something is dense if it tries to drop, which is why a stone drops to the bottom of a pond and a coin sinks to the bottom of a pan of water. This relative motion occurs because both the stone and the coin have higher densities than does water, so they drop. Similarly, we are more dense than air and will drop if we fall off a roof. Just like the coin in water, cold air sinks because it is denser than warmer air. We sometimes see


23

this situation stated as warm air ‗displaces‘ the cold air, which subsequently takes its place. Alternatively, we say ‗warm air rises‘, which explains why we place our clothes above a radiator to dry them, rather than below it.

Light entering the room above the radiator passes through these pockets of warm air as they rise through colder air, and therefore passes through regions of different density. The rays of light bend in transit as they pass from region to region, much in the same way as light twists when it passes through a glass of water. We say the light is refracted. The eye responds to light, and interprets these refractions and twists as different intensities.

So we see swirling eddy (or ‗convective‘) patterns above a radiator because the density of air is a function of temperature. If all the air had the same temperature, then no such difference in density would exist, and hence we would see no refraction and no eddy currents – which is the case in the summer when the radiator is switched off. Then again, we can sometimes see a ‗heat haze‘ above a hot road, which is caused by exactly the same phenomenon. 1.6. Temperature ` This is the numerical measure of the degree of hotness or coldness of a body. It is an important property of any gas. If two bodies are at different temperatures, heat will flow from the warmer to the cooler one until their temperatures are the same. This is the principle on which thermometry is based; the temperature of an object is measured indirectly by placing a calibrated device known as a thermometer in contact with it. When thermal equilibrium is obtained, the temperature of the thermometer is the same as the temperature of the object. 1.6.1. Temperature Scale

A thermometer makes use of some temperature-dependent quantity, such as the density of a liquid, to allow the temperature to be found indirectly through some easily measured quantity such as the

http://www.grc.nasa.gov/WWW/k-12/airplane/gasprop.html


24

length of a mercury column. The resulting scale of temperature is entirely arbitrary; it is defined by locating its zero point, and the size of the degree unit.

Celsius temperature scale locates the zero point at the freezing temperature of water; the Celsius degree (C °) is defined as 1/100 of the difference between the freezing and boiling temperatures of water at 1 atm pressure. The older Fahrenheit scale placed the zero point at the coldest temperature it was possible to obtain at the time (by mixing salt and ice.) The 100° point was set with body temperature (later found to be 98.6°F.) On this scale, water freezes at 32°F and boils at 212°F. The Fahrenheit scale is a finer one than the Celsius scale; there are 180 Fahrenheit degrees in the same temperature interval that contains 100 Celsius degrees, so 1F° = 9/5 C . Since the zero points are also different by 32F, conversion between temperatures expressed on the two scales requires the addition or subtraction of this offset, as well as multiplication by the ratio of the degree size. These selections allow us to write the following relations.

t(oF) = 9

5 t(oC) + 32

t(oC) = 9

5 t(oF) – 32

Where

t(oF) is the temperature in degree Fahrenheit and

t(oC) is the temperature in degree Celsius.

1.6.2. Absolute temperature In 1787 the French mathematician and physicist JACQUES

CHARLES discovered that for each Celsius degree that the temperature of a gas is lowered, the volume of the gas will diminish by 1/273 of its volume at 0°C. The obvious implication of this is that if the temperature could be reduced to –273°C, the volume of the gas would contract to zero. Of course, all real gases condense to liquids before this


25

happens, but at sufficiently low pressures their volumes are linear functions of the temperature (Charles' Law), and extrapolation of a plot of volume as a function of temperature predicts zero volume at -273°C. This temperature, known as absolute zero, corresponds to the total absence of thermal energy.

Because the Kelvin scale is based on an absolute, rather than on an arbitrary zero of temperature, it plays a special significance in scientific calculations; most fundamental physical relations involving temperature are expressed mathematically in terms of absolute temperature. The diagram below compares the different temperature scales with respect to boiling and freezing point of water.

Fig. 1.6. Comparison of Temperature Scales (Schematic) 1.6.3. Conversion between Celsius and Kelvin Scale

In order to covert temperature in degree Celsius to temperature in Kelvin, the expression below is used. toC = (273 + t )K = T (K)


26

Where t is the temperature on the Celsius scale, T is the temperature on the Kelvin scale. Example 1.14. Covert the following temperatures to Kelvin scale:

a. 27oC; b. -10oC. Solutions

a. Using the relationship

toC = (273 + t )K = T (K)

27oC = (273 + 27)K = 300K

b. toC = (273 + t )K = T (K)

-10oC = (273- 10)K = 263K

In order to convert absolute temperature T K to degree Celsius, 273 is simply subtracted from the value. Example 1.15. Covert the following temperatures to degree Celsius:

a. 298K b. 25K

Solutions a. Using the relationship

toC = (273 + t )K = T (K)

toC = (298 ‒ 273) oC = 25oC

b. Using the relationship

toC = (273 + t )K = T (K)

toC = (25 ‒ 273) oC = ‒ 248 oC

1.7. The Volume of Gas The volume of a gas is simply the space in which the molecules

of the gas are free to move. If we have a mixture of gases, such as air, the various gases will coexist within the same volume. In these


27

respects, gases are very different from liquids and solids, the two condensed states of matter. The volume of a gas can be measured by trapping it above mercury in a calibrated tube known as a gas burette (fig. 1.7). The SI unit of volume is the cubic meter, but in chemistry the liter and the milliliter (mL) are commonly used.

Fig. 1.7. Gas burette

It is important to bear in mind, however, that the volume of a gas varies with both the temperature and the pressure, so reporting the volume alone is not very useful. A common practice is to measure the volume of the gas under the ambient temperature and atmospheric pressure, and then to correct the observed volume to what it would be at standard atmospheric pressure and some fixed temperature, usually 0° C or 25°C. The table below shows some commonly used volume measurement units and their conversion factor.


28


29

1.8. Effect of Temperature on the volume of gases If the volume of the container is not fixed, increasing the

temperature will cause a gas to expand (increase the volume), and contract when cooled (decreasing the volume). This would be the case for a gas inside a piston, or inside a rubber balloon. If the volume is fixed, then increasing the temperature will increase the pressure, and decreasing the temperature will decrease the pressure. This would be the case for a gas in a closed solid container, like a canister or sealed metal box. Why does thunder accompany lightning?

Lightning is one of the most impressive and yet frightening manifestations of nature. It reminds us just how powerful nature can be. Lightning is quite a simple phenomenon. Just before a storm breaks, perhaps following a period of hot, fine weather, we often note how the air feels ‗tense‘. In fact, we are expressing an experiential truth: the air contains a great number of ions – charged particles. The existence of a large charge on the Earth is mirrored by a large charge in the upper atmosphere. The only difference between these two charges is that the Earth bears a positive charge and the atmosphere bears a negative charge.

Accumulation of a charge difference between the Earth and the upper atmosphere cannot proceed indefinitely. The charges must eventually equalize somehow: in practice, negative charge in the upper atmosphere passes through the air to neutralize the positive charge on the Earth. The way we see this charge conducted between the Earth and the sky is lightning: in effect, air is ionized to make it a conductor, allowing electrons in the clouds and upper atmosphere to conduct through the air to the Earth‘s surface. This movement of electrical charge is a current, which we see as lightning. Incidentally, ionized air emits light, which explains why we see lightning. Lightning comprises a massive amount of energy, so the local air through which it conducts tends to heat up to as much as a few thousand degrees centigrade. And we have already seen how air expands when warmed, e.g. as described


30

mathematically by Charles‘s law. In fact, the air through which the lightning passes increases in volume to an almost unbelievable extent because of its rise in temperature. And the expansion is very rapid. 1.9. Standard Temperature and Pressure, s.t.p.

Suppose two scientists work on the same research project, but one resides in the far north of the Arctic Circle and the other lives near the equator. Even if everything else is the same – such as the air pressure, the source of the chemicals and the manufacturers of the equipment – the difference between the temperatures in the two laboratories will cause their results to differ widely. For example, the ‗room energy‘ RT will differ. One scientist will not be able to repeat the experiments of the other, which is always bad science.

An experiment should always be performed at known temperature. Furthermore, the temperature should be constant throughout the course of the experiment, and should be noted in the laboratory notebook. But to enable complete consistency, sets of universally accepted arbitrary standards were devised and are called a set of standard conditions. ‗Standard pressure‘ was set as 1 atm and ‗Standard temperature‘ has the value of 0oC (273 K). If both the pressure and the temperature are maintained at these standard conditions, then we say the measurement was performed at ‗standard temperature and pressure‘, which is universally abbreviated to ‗s.t.p.‘ If the scientists at the equator and the Arctic Circle perform their work in thermostatically controlled rooms, both at s.t.p., then the results of their experiments will be identical. If we know the volume of a sample of a gas at any condition, we can easily calculate the volume it would have as an ideal gas at STP by employing the combined gas law. 1.10. Molar volume of a gas

The volume occupied by one mole of a gas under any conditions of temperature and pressure is called the molar volume, Vm. The molar volume of an ideal gas depends on the conditions of temperature and pressure; at s.t.p. it is 22.4 L (or 22400 cm3).


31

How did we arrive at this value? It is simply the volume of 1.00 mol of gas at STP At s.t.p, pressure (P) = 1atm, temperature (T) = 27K, for one mole of gas, n = 1, R= 0.0821 L atm mol-1K-1 Using ideal gas equation to calculate the volume

PV = nRT

V = 𝑛𝑅𝑇

𝑃

= 1.00 𝑚𝑜𝑙 ×0.0821 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙 −1 𝐾−1 ×273 𝐾

1.00 𝑎𝑡𝑚

= 22.4 𝐿

1.11. Molecular weight and density of a gas

The molar volumes of all gases are the same when measured at the same temperature and pressure. But the molar masses of different gases will vary. This means that different gases will have different densities (different masses per unit volume). If we know the molecular weight of a gas, we can calculate its density.

More importantly, if we can measure the density of an unknown gas, we have a convenient means of estimating its molecular weight. This is one of many important examples of how a macroscopic measurement (one made on bulk matter) can yield microscopic information (that is, about molecular-scale objects).

Determination of the molecular weight of a gas from its density is known as the Dumas method, after the French chemist JEAN DUMAS (1800-1840) who developed it. One simply measures the weight of a known volume of gas and converts this volume to its STP equivalent, using Boyle's and Charles' laws. The weight of the gas divided by its STP volume yields the density of the gas, and the density multiplied by 22.4Lmol–1 gives the molecular weight. Pay careful attention to the examples of gas density calculations shown below.


32

Example 1.16. Calculate the approximate molar mass of a gas whose measured density is 3.33 g/L at 30oC and 780 torr. Solution. Data provided

Molar mass?

Density = 3.33 g/L

Volume = 1L

Temperature,T = 30oC = (30 +273)K Pressure, P = 780 torr = (780/760) atm From the ideal gas equation, the number of moles contained in one litre of the gas is

𝑛 = 𝑃𝑉

𝑅𝑇

=

780760

𝑎𝑡𝑚 × (1.00 𝐿)

0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 393𝐾

= 0.413 𝑚𝑜𝑙

Now density, 𝑑 = 𝑚𝑎𝑠𝑠 (𝑚)

𝑉𝑜𝑙𝑢𝑚𝑒 (𝑉)

Therefore, 𝑚 = 𝑑𝑣 But mass (m) = number of mole (n) × molar mass (M)

Therefore 𝑑𝑣 = 𝑛𝑚

M= 𝑑 ×𝑣

𝑛

Substituting gives

M = 33𝑔 𝐿−1 ×1.0 𝐿

0.0413 𝑚𝑜𝑙

= 80.6gmol-1

Example 1.17. The density of air at 15OC and 1.00 atm is 1.23g/L. What is the molar mass of the air?


33

Solution First calculate the mole of air from which the molar mass can be gotten. Data provided Density = 1.23 g/L Volume = 1L Temperature,T = 15oC = (15 +273)K = 288K Pressure, P = 1 atm Molar mass? From the ideal gas equation, the number of moles contained in one litre of the air is

𝑛 = 𝑃𝑉

𝑅𝑇

= 1 𝑎𝑡𝑚 × (1.00 𝐿)

0.8206 𝐿 𝑎𝑡𝑚 𝑚𝑜𝑙−1 𝐾−1 × 288𝐾

= 0.0423 𝑚𝑜𝑙

M= 𝑑 ×𝑣

𝑛

Substituting gives

M = 1.23𝑔 𝐿−1 ×1.0 𝐿

0.0423 𝑚𝑜𝑙

= 29.1gmol-1


34

CHAPTER TWO

THE GAS LAWS

2.1. Introduction Experience has shown that several properties of a gas can be

related to each other under certain conditions. The properties are pressure (P), volume (V), temperature (T, in kelvins), and amount of material expressed in moles (n). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur. These properties and other variables such as rate of diffusion of any gaseous substance bear a simple mathematical relationship to each other. These are collectively called gas laws. 2.2. Pressure – Volume Relationship

Robert Boyle (1627–1691), an Irish physical scientist, discovered that the volume of a given sample of a gas at a constant temperature is inversely proportional to its pressure. This generalization, known as Boyle’s law, applies approximately to any gas, no matter what its composition. (It does not apply to liquids or solids.)

Inverse proportionality occurs when one variable gets larger by the same factor as another gets smaller. For example, average speed and the time required to travel a certain distance are inversely proportional. If we double our speed, the time it takes us to complete the trip is halved. Similarly, if the pressure on a given sample of gas at a given temperature is doubled (increased by a factor of 2), its volume is halved (decreased by a factor of 2).

Boyle might have observed the following data on volume and pressure for a given sample of gas at a given temperature, under four different sets of conditions:

Volume (L) Pressure (atm) 1 4.00 1.00 2 2.00 2.00 3 1.00 4.00


35

4 0.500 8.00 Note that tabulating data is very helpful when two or more variables are being considered. The units are usually included in the column headings in such a table. The data in the table show that the product of the volume (V) and the pressure (P) is a constant. The table may be expanded to show this relationship:

Volume (L) Pressure (atm) Volume × Pressure (L. atm) 1 4.00 1.00 4.00 2 2.00 2.00 4.00 3 1.00 4.00 4.00 4 0.500 8.00 4.00

Mathematically expression of the law;

[V α 1

𝑃]T

[V = 𝐾

𝑃 ]T

PV = K (Where K = constant of proportionality). A more useful form of the law can be written as: P1V1= P2V2

Where V1 and P1 refer to the original volume and pressure, V2 and P2 refer to the volume and pressure under the new or changed conditions.

If we place the values of P on the horizontal axis and the values of V on the vertical axis, plot the preceding tabulated values for P and V, and smoothly connect the points, we get a curve that can tell us what the volume will be at any intermediate pressure (Figure 2.1a). We can also plot 1/V versus P and get a straight line through the origin (Figure 2.1b).

V (L) 1/V (1/L) P (atm) 1 4.00 0.250 1.00 2 2.00 0.500 2.00 3 1.00 1.00 4.00 4 0.500 2.00 8.00


36

(a) (b)

Fig. 2.1. Graphical illustration of Boyle’s law: (a) Plot of V versus P. (b) Plot of 1/V versus P. 2.3. Kinetic Theory and Boyle’s Law

The pressure of gas is due to continuous collision of the gaseous molecules with the walls of the container. At constant temperature, the average kinetic energy of the gas molecules is constant. If the size of the container is reduced to a half (volume reduces), the frequency of collision of the gas molecules with the walls of the container will be doubled. This is due to the fact that the distance to the walls has been reduced to a half. Therefore, the gas pressures will double the initial value.

On the other hand if the volume of the container (size) is doubled, the frequency of collision of the gas molecules with the walls of the container will become reduced by a half, since the distance between the molecules before colliding with the walls has been doubled. Hence the pressure will be half of the initial value. Example 2.1. A certain mass of a gas occupies 400cm3 at 1.0 × 105 Nm-2. Calculate its volume when the pressure is 4.0 × 105 Nm-2 at constant temperature.


37

Solution In trying to solve this kind of problem, it is always good to collect the given information together so as to easily detect the variable you are asked to find. Data provided; P1 = 1.0 × 105 Nm-2, V1 =400cm3, P2 = 2.0 × 105 Nm-2, V2 = ? According to boyle‘s law, P1V1= P2V2

Making V2 the subject,

V2 = P1V1

P2

On substituting, V2 = 1 × 105 ×400

2 × 105 = 200cm3

Example 2.2. If 4 Liters of methane gas has a pressure of 1.0 atm, what will be the pressure of the gas if we squish it down so it has a volume of 2.5 L? Solution Data provided; P1 = 1.0 atm V1 = 4.0L P2 = ? V2 = 2.5 L According to boyle‘s law, P1V1= P2V2

Making P2 the subject,

P2 = p1v1

v2

P2 = 1.0 × 4

2.5 = 1.6 𝑎𝑡𝑚

Example 2.3. A 3.50-L sample of gas has a pressure of 0.750 atm. Calculate the volume after its pressure is increased to 1.50 atm at constant temperature. Solution Alternatively, data collection can be in the form of table as shown below

Pressure Volume 1 0.750 atm 3.50 L


38

2 1.50 atm ? Using P1V1= P2V2


V2 = p1v1

P2

Substitution of the values into the equation yields

V2 = 0.750 ×3.50

1.50= 1.75 𝐿

Note that multiplying the pressure by 2 causes the volume to be reduced to half. Example 2.4. A sample of gas initially occupies 35.0 mL at 1.50 atm. Calculate the pressure required to reduce its volume to 20.5 mL at constant temperature. Solution Data collection

Pressure Volume 1 1.50 atm 35.0 mL 2 ? 20.5 mL Using P1V1= P2V2


P2 = p1v1

V2


P2 = 1.50 ×35.0

20.5= 2.56 𝑎𝑡𝑚

Note that the units of pressure and volume must be the same on each side of the equation P1V1= P2V2 . If the units given in a problem are not the same, one or more of the units must be converted. Example 2.5. A 1.45-L sample of gas has a pressure of 0.950 atm. Calculate the volume after its pressure is increased to 787 torr at constant temperature. Solution Because the pressures are given in two different units, one of them must be changed.


39

Pressure Volume 1 0.950 atm 1.45 L

2 787 𝑡𝑜𝑟𝑟 1 𝑎𝑡𝑚

760 𝑡𝑜𝑟𝑟 = 1.036 𝑎𝑡𝑚 ?

Using P1V1= P2V2


V2 = p1v1

P2


V2 = 0.950 ×1.45

1.036

= 1.33 𝐿 Alternatively, we can change 0.950 atm to torr and still arrive at the same answer. (722 torr) (1.45 L) = (787 torr)V2

V2 = 1.33 𝐿 Note: 1 atm = 760 torr Example 2.6. Calculate the initial volume of a sample of gas at 1.20 atm if its volume is changed to 70.4 mL as its pressure is changed to 744 torr at constant temperature Solution Data collection

Pressure Volume 1 1.20 atm ?

2 744 𝑡𝑜𝑟𝑟 1 𝑎𝑡𝑚

760 𝑡𝑜𝑟𝑟 = 0.979 𝑎𝑡𝑚 70.4 L

Using P1V1= P2V2


V1 = p2v2

P1


V1 = 0.979 ×70.4

1.2

= 57.4 𝐿


40

Example 2.7. Calculate the pressure required to change a 3.38-L sample of gas initially at 1.15 atm to 925 mL, at constant temperature. Solution Collect the data and convert 925 mL to L (mL ≡ cm3, 1000mL = 1L)

Pressure Volume 1 1.15 atm 3.38 L 2 ? 925 mL = 0.925 L Using P1V1= P2V2


P2 = p1v1

V2


P2 = 1.15 ×3.38

0.925

= 4.20 𝑎𝑡𝑚 The pressure must be raised to 4.20 atm. Practice questions

1. State Boyle‘s law (i) in words (ii) mathematically 2. Explain Boyle‘s law in terms of kinetic theory. 3. Fill the following gaps: (Measurements at constant

temperatures).

Initial pressure Initial volume Final pressure Final volume

1.0 × 105 Nm-2 300cm3 1.5 × 105 Nm-2 -

1.0 × 105 Nm-2 225cm3 - 900cm3

- 3.50dm3 760 mmHg 700 cm3

800 mm Hg 300cm3 650 mmHg -

4. 30dm3 of oxygen at 10 atmospheres is placed in a 20dm3

container. Calculate the new pressure if temperature is kept constant.


41

5. Calculate the initial pressure of a 485-mL sample of gas that has been changed at constant temperature to 1.16 L and 1.18 atm.

2.4. Temperature – Volume Relationship

In 1787, 125 years after Boyle published the law that bears his name, J. A. C. Charles (1746–1823) discovered a law relating the volume of a given sample of gas to its absolute temperature. It took more than a century to discover this law because of the requirement that the temperature be absolute.

The volume of a sample of gas varies with the temperature, as shown in Table 2.1 and plotted in Figure 2.2(a) for a particular sample. Although the volume changes with the Celsius temperature, the relationship is not a direct proportionality. That is, when the Celsius temperature doubles, the volume does not double, all other factors being held constant. On the graph, the plotted points form a straight line, but the line does not pass through the origin. For a direct proportionality to exist, the straight line must pass through the origin. If the straight line corresponding to the points in Table 12.1 is extended until the volume reaches 0 L, the Celsius temperature is -273K (Figure 2.2b). Charles defined a new temperature scale in which the lowest possible temperature is absolute, corresponding to -273K. This temperature is called absolute zero. Table 2.1 Temperature and Volume Data for a Particular Sample of Gas at a Given Pressure

Temperature(°C) Volume(L)

1 0 0.400

2 100 0.548

3 200 0.692

4 300 0.840


42

(a) (b)

Fig. 2.2. Dependence of Volume on Temperature at Constant Pressure (a) Plot of the data given in Table 2.1. (b) Extension of the line in part (a) to absolute zero, with the Kelvin scale added to the horizontal axis. We can state Charles‘ findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature. This means an increase in the temperature of a fixed mass of a gas leads to a corresponding increase in the volume of the gas by the same proportion, and vice versa, with the proviso that pressure remains the same. Mathematically expression of the law; [ V∝ T ]P

[ V= KT ] P

[ V/𝑇 = K ] P

(Where K = constant of proportionality).

A more useful form of the law can be express as:

𝑉1

𝑇1 =

𝑉2

𝑇2

Where V1 and T1 refer to the original volume and pressure, V2 and T2 refer to the volume and pressure under the new or changed conditions.


43

2.5. Kinetic Theory and Charles’ Law As the temperature of the gas molecules increase, the average

kinetic energy is equally raised, and hence, the average velocity of gas molecules. The gas molecules move more rapidly colliding with one another and more frequently with the walls of the container. For gas pressure to remain constant, the volume of the container must be increased with an increase in temperature. Example 2.8. Assume that the volume of a balloon filled with H2 is 1.00 L at 25°C. Calculate the volume of the balloon when it is cooled to -78°C in a low-temperature bath made by adding dry ice to acetone. Solution Collect the given information and convert as necessary Data provided; V1 = 1.00L,

T1 = 250C = (25 + 273)K = 298K

T2 = -780C = (273 - 78)K =195K

V2 = ?

Applying Charles‘ law,

𝑉1

𝑇1 =

𝑉2

𝑇2

V2 = 195 ×1.00

293

= 0.65L

Example 2.9. The volume of a fixed mass of gas measured at

atmospheric pressure and 260C is 3.0 dm3. Calculate the volume at

1270C and at the same pressure.


44

Solution

Data provided;

V1 = 3.0 dm3,

T1 = 260C = (25 + 273)K = 299K

T2 = 1270C = (273 + )K =400 K

V2 = ?

Applying Charles‘ law,

𝑉1

𝑇1 =

𝑉2

𝑇2

V2 = 400 ×3.00

299

= 4.0 dm3

Example 2.10. If 250cm3 of a gas at s.t.p. is heated to 270C at constant

pressure, calculate its new volume.

Solution

Data provided;

V1 = 250 cm3,

T1 = s.t = 273K

T2 = 270C = (273 +27 )K =300 K

V2 = ? Applying Charles‘ law,

V1/ T1 = V2/ T2

V2 = V1 × T2/ T1

V2 = 250 × 300

273


45

= 274.7 cm3

Example 2.11. Show that the data in Table below prove (a) that the

Celsius temperature is not directly proportional to volume and (b) that

the Kelvin temperature is directly proportional to volume.

Temperature and Volume data for a particular Sample of gas at a given pressure

Solution

As the absolute temperature 273 K is increased to 373 K or 473 K, the

volume increases to 373/273 = 1.37 or 473/273 = 1.37 times the original

volume. The ratio of V to T is constant (see Table above). The volume is

directly proportional to absolute temperature.

Example 2.12. Calculate the Celsius temperature to which a 678-mL

sample of gas at 0oC must be heated at constant pressure for the

volume to change to 0.896 L.

Solution

Data provided

V1 = 678 mL = 0.678 L

T1 = 0oC = 273K

V2 = 0.896 L


46

T2 = ?

Using the relationship

𝑉1

𝑇1 =

𝑉2

𝑇2

T2 = 273 ×0.896

0.678

= 361 𝐾𝐾

= (361 − 273)oC

= 88oC

Note: 1000 mL = 1L

Example 2.13. Calculate the original temperature of a 0.456-mL gas

sample if it is expanded at constant pressure to 1.75 L at 55°C.

Solution

Data provided

V1 = 0.456 mL = 0.000456 L

T1 = ?

V2 = 1.75 L

T2 = 55 OC = (273 + 55)K = 238K

Using the relationship below and making T1 the subject;

𝑉1

𝑇1 =

𝑉2

𝑇2

T1 = 238 ×0.000456

1.75

= 0.1 𝐾K

= (0.1 − 273) oC

= −272.9 oC


47

Example 2.14. A plastic bag of peanuts is laid on a windowsill in the

sun, where its temperature increases from 20OC to 30OC. If the original

volume is 100.0 cm3, what is the final volume after warming?

Solution

Data collection

V1 = 100 cm3 T1 = 20OC = 293 K

V2 = ? T2 = 30 oC = 303 K Using the relationship below and making V2 the subject and substituting;

𝑉1

𝑇1 =

𝑉2

𝑇2

V2= 303 ×100

293

= 103.4 cm3

Example 2.15. The temperature of a 4.00 L sample of gas is changed

from 10.0 °C to 20.0 °C. What will the volume of this gas be at the new

temperature if the pressure is held constant?

Solution

Data collection

V1 = 4.00L

T1 = 10OC = 283 K

V2 = ?

T2 = 20 OC = 293 K


48

Using the relationship below and making V2 the subject and

substituting;

𝑉1

𝑇1 =

𝑉2

𝑇2

𝑉2 = 𝑇2𝑉1

𝑇1

= 293 ×4.00

283

= 4.1 𝐿

Example 2.16. Carbon dioxide is usually formed when gasoline is

burned. If 30.0 L of CO2 is produced at a temperature of 1.00 x103 °C

and allowed to reach room temperature (25.0 °C) without any pressure

changes, what is the new volume of the carbon dioxide?

Solution Data collection

V1 = 30.0L T1 = 1.00 x103 °C = (273 + 1000)K = 1273 K

V2 = ? T2 = 25 OC = 298 K Using the relationship below and making V2 the subject and

substituting;

𝑉1

𝑇1 =

𝑉2

𝑇2

𝑉2 = 𝑇2𝑉1

𝑇1

= 298 × 30.00

1273

= 7.0 𝐿


49

Example 2.17. The volume of a gas syringe which contains 56.05

milliliters was raised to 67.7 milliliters at 107.5 oC. Determine the initial

temperature of the gas?

Solution

Data collection

V1 = 56.05 mm = 0.05605L

T1 =

V2 = 67.7 mm = 0.068L

T2 = 107.5 OC = 380.5 K

Using the relationship below and making T1 the subject and

substituting;

𝑉1

𝑇1 =

𝑉2

𝑇2

𝑇1 = 𝑇2𝑉1

𝑉2

= 380.5 × 0.05605

0.068

= 313.6 𝐾

= (313.6 − 273) = 40.6 oC

Example 2.18. If 15.0 liters of neon at 25.0 °C is allowed to expand to

45.0 liters, what is the new temperature?

Solution

Data provided

V1 = 15.0L


50

T1 = 25 °C = (273 + 25)K = 298 K

V2 = 45.0 L

T2 = ?

Using the relationship below and making T2 the subject and

substituting;

𝑉1

𝑇1 =

𝑉2

𝑇2

𝑇2 = 𝑇1𝑉2

𝑉1

= 298 × 45.00

15

= 294 𝐾

Example 2.19. A balloon has a volume of 2500.0 mL on a day when the

temperature is 30.0 °C. If the temperature at night falls to 10.0 °C, what

will be the volume of the balloon if the pressure remains constant?

Solution

Data collection

V1 = 2500 mL

T1 = 30OC = 303 K

V2 = ?

T2 = 10 OC = 283 K

Using the relationship below and making V2 the subject and

substituting;

𝑉1

𝑇1 =

𝑉2

𝑇2


51

𝑉2 = 𝑇2𝑉1

𝑇1

= 283 ×2500.00

303

= 2335 𝑚𝐿

2.6. Temperature-Pressure Relationship

Boyle‘s Law is the relationship between Pressure and Volume but does not address temperature. How does temperature change affect the properties of a sample of gas? Recall that temperature is a measure of the average kinetic energy of particles. As the particles of a substance move faster, the substance‘s temperature increases. The particles bump into each other and the sides of the container more often.

How would this affect a system where the volume is closed and constant? This observation was first made by Gay-Lussac. He observed that pressure has a direct proportional link with temperature of a sample of gas in a closed container (volume constant). Properly put, this law states that at constant volume, the pressure of a fixed mass of a gas is directly proportional to its absolute temperature. The law is expressed mathematically as follows:

𝑃 ∝ 𝑇 (Constant volume)

𝑃

𝑇= 𝑘

A more useful form of the law can be express as:

𝑃1

𝑇1 =

𝑃2

𝑇2

Where P1 and T1 refer to the original pressure and temperature, P2 and T2 refer to the pressure and temperature under the new or changed conditions. Note: in solving or addressing mathematical problems with this law, the temperature must be expressed in Kelvin and the pressure in a standard uint.


52

Example 2.20. 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What

would be the required temperature (in Celsius) to change the pressure

to standard pressure?

Solution Data provided P1 = 97.0 kPa T1 = 25.0°C = ( 25 + 273)K = 298.0 K P2 = s.p. = 101.325kPa T2 = ?

Applying 𝑃1

𝑇1 =

𝑃2

𝑇2 and making T2 the subject

T2 = 𝑇1𝑃2

𝑃1

= 298.0 𝐾 ×101.325 𝑘𝑃𝑎

97.0 𝑘𝑃𝑎

= 311K

Converting to degree in Celsius;

311K = (311 ‒ 273) °C

= 38°C

Example 2.21. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be? Solution Data provided P1 = 15 atm


53

T1 = 25.0°C = ( 25 + 273)K = 298.0 K P2 = 16 atm T2 = ?

Applying 𝑃1

𝑇1 =

𝑃2

𝑇2 and making T2 the subject

T2 = 𝑇1𝑃2

𝑃1

= 298.0 𝐾 ×16 𝑎𝑡𝑚

15 𝑎𝑡𝑚

= 317 K

Example 2.22. A 30.0 L sample of nitrogen inside a metal container at 20.0 °C was placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was 3.00 atm. What is the pressure of the nitrogen after its temperature was increased?

Solution

Collect data and convert temperatures to Kevin

P1 = 3.00 atm

T1 = 25.0°C = ( 20 + 273)K = 293.0 K

P2 = ?

T2 = 50.0°C = ( 50 + 273)K = 323.0 K

Applying 𝑃1

𝑇1 =

𝑃2

𝑇2 and making P2 the subject

P2 = 𝑇2𝑃1

𝑇1


54

= 323.0 𝐾 ×3.00 𝑎𝑡𝑚

293 𝑎𝑡𝑚

= 3.3 atm

Example 2.23. The temperature of a sample of gas in a steel container at 30.0 kPa is increased from ‒100.0 °C to 1.00 x 103 °C. What is the final pressure inside the tank?

Solution

Collect data and convert temperatures to Kevin

P1 = 30 kPa

T1 = ‒100.0 °C = (‒100.0 + 273)K = 173.0 K

P2 = ?

T2 = 1.00 x 103 °C = (1.00 x 103 + 273)K = 1273.0 K

Applying 𝑃1

𝑇1 =

𝑃2

𝑇2 and making P2 the subject

P2 = 𝑇2𝑃1

𝑇1

= 1273 𝐾 ×30 𝑘𝑃𝑎

173 𝐾

= 220 kPa

2.7. The Combined Gas Law Boyle‘s and Charles‘ laws may be merged into one law, called the

combined gas law, expressed in equation form as derived below: From Boyle‘ law: V∝ 1/𝑃 (T constant)


55

From Charles‘ law: V ∝ 𝑇 (P constant) V ∝ 1/𝑃 ∝ T 𝑉 = 𝑘𝑇/𝑃 𝑃𝑉

𝑇 = k

That is, for a given sample of a gas, PV/T remains constant, and therefore

𝑃1 𝑉1

𝑇1=

𝑃2 𝑉2

𝑇2 (a given sample of a gas)

This expression is a mathematical statement of the combined (or general) gas law. In words, the volume of given sample of a gas is inversely proportional to its pressure and directly proportional to its absolute temperature.

Note that if the temperature is constant, T1 = T2, then the expression reduces to the equation for Boyle‘s law, P1V1 = P2V2. Alternatively, if the pressure is constant, P1 = P2, the expression is equivalent to Charles‘ law, V1/T1 = V2/T2. When the initial volume V1 of a gas at temperature T1 and pressure P1 is subjected to changes in temperature to T2 and pressure to P2, its new volume V2 is obtained from the equation.

To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically.

Example 2.24. A certain mass of a gas occupies 330 cm3 at 27oC and 9.0

× 104 Nm-2 pressure. Calculate its volume at s.t.p. (s.p = 1.0 × 105 Nm-2).

Solution

Write the given data down, convert as variable to appropriate units

and substitute into the form to find the unknown.


56

Data provided:

V1 = 330 cm3

P1 = 9.0 × 104 Nm-2

T1 = 27oC = (27 + 273)K = 300K

T2 = s.t. = 273K

P2 = s.p. = 1.0 × 105Nm-2

V2 = ?

Using the gas equation:

𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2

Making V2 the subject of the formula:

V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 9.0 × 104 × 330 × 273

1.0 × 105 × 300

= 270 cm3

Example 2.25. Calculate the volume of a sample of gas originally

occupying 908 mL at 717 torr and 20OC after its temperature and

pressure are changed to 72OC and 1.07 atm.

Solution

In attempting this problem, the volume can be stated in millilitres in

both states. The pressure can be stated in atmospheres in both but the

temperature must be in kelvins in both states.

Data provided

V1 = 908 mL


57

P1 = 717

760 𝑎𝑡𝑚 = 0.94 𝑎𝑡𝑚

T1 = 20oC = (20 + 273)K = 293K

T2 = 72 oC = 345K

P2 = 1.07 atm

V2 = ?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 0.94 × 908 × 345

1.07 × 293

= 943 mL

Example 2.27. Calculate the original volume of a sample of gas that is

at 700 torr and 22 oC before its volume, temperature, and pressure are

changed to 998 mL, 82°C, and 2.07 atm

Solution

Data provided

V1 = ?

P1 = 700

760 𝑎𝑡𝑚 = 0.92 𝑎𝑡𝑚

T1 = 22oC = (22 + 273)K = 295K

T2 = 82 oC = 355K

P2 = 2.07 atm


58

V2 = 998 mL


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V1 = 𝑃2𝑉2𝑇1

𝑃1𝑇2

= 2.07 × 998 × 298

0.92 × 355

= 1884 mL

Example 2.28. 17.3-mL sample of gas originally at standard

temperature and pressure is changed to 10.9 mL at 678 torr. Calculate

its final temperature in degrees

Celsius.

Solution

Data provided

V1 = 17.3 mL

P1 = s.p. = 760 torr

T1 = s.t. = 273 K

T2 = ?

P2 = 678 torr

V2 =10.9 mL


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2

Making T2 the subject of the formula:


59

T2 = 𝑃2𝑉2𝑇1

𝑃1𝑉1

= 678 × 10.9 × 273

760 × 17.3

= 153 K

Converting to degree Celsius

= (153 − 273) oC

= 120 oC

Example 2.29. Calculate the volume at standard temperature and

pressure of a sample of gas that has a volume of 49.7 mL at 52°C and

811 torr.

Solution

Data provided

V1 = 49.7 mL

P1 = 811

760 𝑎𝑡𝑚 = 1.07 𝑎𝑡𝑚

T1 = 52°C = 325 K

T2 = s.t. = 273 K

P2 = s.p. = 1 atm

V2 =?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1


60

= 1.07 × 49.7 × 273

1.0 × 325

= 45 mL

Example 2.30. Calculate the new volume after a 2.00-L sample of gas at

1.50 atm and 127oC is changed to 27oC at 3.50 atm.

Solution

Data provided

V1 = 2.00 L

P1 =1.50 𝑎𝑡𝑚

T1 = 127°C = 400 K

T2 = 27oC =300 K

P2 = 3.50 atm

V2 =?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 1.5 × 2.00 × 300

3.50 × 400

= 0.64 L

Example 2.31. 500.0 liters of a gas are prepared at 700.0 mmHg and

200.0 °C. The gas is placed into a tank under high pressure. When the

tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the

volume of the gas?


61

Solution

Data provided

V1 = 500 L

P1 = 700.0 𝑚𝑚𝐻𝑔 = 700760 𝑎𝑡𝑚 = 0.92 𝑎𝑡𝑚

T1 = 200°C = 473 K

T2 = 20oC =293 K

P2 = 30.0 atm

V2 =?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 0.92 × 500 × 293

30 × 473

= 9.5 L

Example 2.32. A gas balloon has a volume of 106.0 liters when the

temperature is 45.0 °C and the pressure is 740.0 mm of mercury. What

will its volume be at 20.0 °C and 780 .0 mm of mercury pressure?

Solution

Data provided

V1 = 106 L

P1 = 740.0 𝑚𝑚𝐻𝑔

T1 = 45°C = 318 K


62

T2 = 20oC =293 K

P2 = 780.0 𝑚𝑚𝐻𝑔

V2 =?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 740 × 106 × 293

780 × 318

= 92.7 L

Example 2.33. The volume of a gas originally at standard temperature

and pressure was recorded as 488.8 mL. What volume would the same

gas occupy when subjected to a pressure of 100.0 atm and temperature

of -245.0 °C?

Solution

Data provided

V1 = 488.8 L

P1 = 𝑠. 𝑝. = 1.0 𝑎𝑡𝑚

T1 = s.t. = 273 K

T2 = ‒245oC =28 K

P2 = 100 𝑎𝑡𝑚

V2 =?



63

𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 1.0 × 488.8 × 28

100 × 273

= 0.5 L

Example 2.34. A gas is heated from 263.0 K to 298.0 K and the volume

is increased from 24.0 liters to 35.0 liters by moving a large piston

within a cylinder. If the original pressure was 1.00 atm, what would the

final pressure be?

Solution

Data provided

V1 = 24.0 L

P1 = 1.0 𝑎𝑡𝑚

T1 = 263.0 K

T2 =298.0 K

P2 = ?

V2 = 35.0 L


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2

Making P2 the subject of the formula:

P2 = 𝑃1𝑉1𝑇2

𝑉2𝑇1


64

= 1.0 × 24.0 × 298

35 × 263

= 0.78 atm

Example 2.35. The pressure of a gas is reduced from 1200.0 mmHg to

850.0 mmHg as the volume of its container is increased by moving a

piston from 85.0 mL to 350.0 mL. What would the final temperature be

if the original temperature was 90.0 °C?

Solution

Data provided

V1 = 85.0 mL

P1 = 1200 𝑚𝑚𝐻𝑔

T1 = 90.0 °C = 363 K

T2 =?

P2 = 850 𝑚𝑚𝐻𝑔

V2 = 350.0 mL


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


T2 = 𝑃2𝑉2𝑇1

𝑉1𝑃1

= 850 × 350 × 363

85 × 1200

= 1059 K


= (1059 − 273) oC


65

= 786 oC

Example 2.36. If a gas is heated from 298.0 K to 398.0 K and the

pressure is increased from 2.230 x 103 mmHg to 4.560 x 103 mmHg

what final volume would result if the volume is allowed to change

from an initial volume of 60.0 liters?

Solution

Data provided

V1 = 60.0 L

P1 = 2.230 × 103 𝑚𝑚𝐻𝑔

T1 = 298.0 K

T2 =398.0 K

P2 = 4.560 × 103 𝑚𝑚𝐻𝑔

V2 =?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 2.230 × 103 × 60.0 × 398

4.560 × 103 × 298

= 39.2 L

Example 2.37. A balloon containing a sample of gas has a temperature

of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The

balloon has a volume of 1,070 mL. The balloon is transported by plane


66

to Denver, where the temperature is 11°C and the pressure is 655 torr.

What is the new volume of the balloon?

Solution

Data provided

V1 = 1070 mL

P1 = 1.09 𝑎𝑡𝑚

T1 = 22°C = 295 K

T2 =11°C = 284 K

P2 = 655 𝑡𝑜𝑟𝑟 = 655760 𝑎𝑡𝑚 = 0.86 𝑎𝑡𝑚

V2 =?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 1.09 × 1070 × 284

0.86 × 295

= 1306 mL

Example 2.38. A balloon used to lift weather instruments into the

atmosphere contains gas having a volume of 1,150 L on the ground,

where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this

gas has a pressure of 6.88 torr and a temperature of −15°C. What is the

new volume of the gas?


67

Solution

Data provided

V1 = 1150 L

P1 = 0.977 𝑎𝑡𝑚

T1 = 18°C = 291 K

T2 = −15°C = 258 K

P2 = 6.88 𝑡𝑜𝑟𝑟 = 6.88760 𝑎𝑡𝑚 = 0.0091 𝑎𝑡𝑚

V2 =?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1

= 0.977 × 1150 × 258

0.0091 × 291

= 109465 L

2.8. Relationship between Amount and Volume 2.8.1. Gay-Lussac's Law of Combining Volumes

In the same 1808 article in which Gay-Lussac published his observations on the thermal expansion of gases, he pointed out that when two gases react, they do so in volume ratios that can always be expressed as small whole numbers. This came to be known as the Law of combining volumes. Example 2.39. Ammonium carbonate decomposes when heated to yield carbon dioxide, ammonia, and water vapour. Calculate the ratio


68

of the (separate) volume of ammonia to that of water vapour, each at 450°C and 1.00 atm. Solution

The mole ratio of the gases, given in the balanced equation, is 2 mol NH3: 1 mol CO2: 1 molH2O

The ammonia and water vapour are separated and measured at the given temperature and pressure. The ratio of their volumes can be calculated as follows:

𝑉𝑁𝐻 3

𝑉𝐻2𝑂=

𝑛𝑁𝐻 3 𝑅𝑇 /𝑃

𝑛𝐻2𝑂 𝑅𝑇 /𝑃

Because R is a constant and both T and P are the same for the two gases, this equation reduces to

𝑉𝑁𝐻 3

𝑉𝐻2𝑂

= 𝑛𝑁𝐻 3

𝑛𝐻2𝑂

The right side of this equation is the ratio of the numbers of moles—the ratio given by the balanced chemical equation. The left side of the equation is the ratio of the volumes, so the ratio given by the balanced chemical equation is equal to the volume ratio under these conditions. The ratio is 2: 1. Example 2.40. If 2.00 L H2 of and 1.00 L of both at standard temperature and pressure, are allowed to react, will the water vapor they form at 250°C and 1.00 atm occupy 2.00 L? Solution 2H2 (g) + O2 (g) → 2H2O (g)

The volumes of H2 and O2 that react are in the ratio given in the balanced equation because the two gases have the same temperature and pressure. The volume of water vapour formed is not in that ratio, however, because its temperature is different. Its volume will be much greater than 2 L. 2.8.2. Avogadro's Law

The work of the Italian scientist Amedeo Avogadro complemented the studies of Boyle, Charles, and Gay-Lussac. In 1811


69

he published a hypothesis stating that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules (or atoms if the gas is monatomic). This law states that equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecule. Mathematically: V ∝ 𝑛 (at constant T and P) 𝑉 = 𝑘𝑛

𝑉

𝑛= 𝑘

Where V is the volume of gas, n is the number of molecules and 𝑘 is the proportionality constant.

This law relates the volume of a fixed mass of a gas to the number of molecules it contains. It shows that the volume occupied by a gas depends on the number of molecules it contains, at a given temperature and pressure. An increase in the number of gas molecules leads to an increase in gas volume, and vice versa.

According to Avogadro‘s law we see that when two gases react with each other, their reacting volumes have a simple ratio to each other. If the product is a gas, its volume is related to the volume of the reactants by a simple ratio (a fact demonstrated earlier by Gay-Lussac). For example, consider the synthesis of ammonia from molecular hydrogen and molecular nitrogen:

3H2(g) + N2(g) → 2NH3(g) 3 mol 1 mol 2 mol

Because, at the same temperature and pressure, the volumes of gases are directly proportional to the number of moles of the gases present, we can now write

3H2(g) + N2(g) → 2NH3(g) 3 volume 1 volume 2 volume

The volume ratio of molecular hydrogen to molecular nitrogen is 3:1, and that of ammonia (the product) to molecular hydrogen and molecular nitrogen combined (the reactants) is 2:4, or 1:2.


70

Example 2.41. 50 cm3 of sulphur (IV) oxide were produced at s.t.p. when some quantity of powdered sulphur was burnt in excess oxygen. (a) Write a balanced chemical equation for the reaction. (b) Calculate the volume of oxygen used up during the reaction. (c) Which of the laws is applicable? State the law. Solution (a). S(g) + O2(g) → SO2(g)

(b). From the balanced chemical equation in (a) above; At s.t.p: 22400 cm3 of SO2 required 22400 cm3 of O2 Hence 1 cm3 of SO2 will require 1 cm3 of O2

∴ 50 cm3 of SO2 will use 50 cm3 of O2

(c). Avogadro‘s law is applicable in (b) above and it state that at the same temperature and pressure equal volume of gases contain the same number of molecules. 2.9. The Ideal Gas Law

So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law. The general ideal gas equation is a combination of Boyle‘s, Charles‘ and Avogadro‘s laws involving the four gas variables: pressure (P), volume (V), number of mole of gas (n), and temperature (T).

From Boyle‘ law: V∝ 1/𝑃 (T constant) From Charles‘ law: V ∝ 𝑇 (P constant) From Avogadro‘s law: V ∝ 𝑛 (P,T constant) V ∝ 1/𝑃 ∝ T ∝ 𝑛

V = R × 1/𝑃 × T × 𝑛

PV = nRT


71

In this equation, P is pressure, V is volume, n is amount of moles, and T is temperature. R is called the ideal gas law constant and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts i and f to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under any conditions.

2.10. Evaluation of the Gas Constant, R The gas constant can be expressed in various units, all having

the dimension of energy per degree per mol. From the general equation PV = nRT we get:

𝑅 = 𝑃𝑉

𝑛𝑇

Where P is pressure, V is volume, n is amount, and T is temperature. R is most easily calculated from the fact that the hypothetical volume of an ideal gas is 22.4L at STP (273.K and 1 atm).

i. If volume is expressed in liters and pressure in atmospheres, then the proper value of R is as follows:

𝑅 = 𝑃𝑉

𝑛𝑇

= 1.0 𝑎𝑡𝑚 ×22.414 𝐿

1.0 𝑚𝑜𝑙 ×273.15 𝐾

R= 0.08206 atm L mol-1K-1

ii. if pressure is in Nm-2 and volume in m3 then the proper value of R is thus:

𝑅 = 𝑃𝑉

𝑛𝑇

Where P = 101325 Nm-2, V = 22.4/1000 = 0.0224 m3

R = 101325 𝑁𝑚−2 ×0.0224 𝑚3

1.0 𝑚𝑜𝑙 ×273.15 𝐾

= 8.314N m mol-1K-1

iii. if pressure is in atm and volume in cm3 then



72

𝑅 = 𝑃𝑉

𝑛𝑇

= 1.0 𝑎𝑡𝑚 ×22414 𝑐𝑚 3

1.0 𝑚𝑜𝑙 ×273.15 𝐾

= 82.06 atm cm3 mol-1K-1

[22.414 L =22400 cm3]

iv. if pressure is in Pa and volume in liter

𝑅 = 𝑃𝑉

𝑛𝑇

[1 atm = 1.01325× 105 Pa; 1 L= 10−3m3]

= 1.01325 ×105 𝑃𝑎 ×22.414 × 10−3 𝑚3

1.0 𝑚𝑜𝑙 ×273.15 𝐾

= 8.314 Pa m3 mol-1K-1

v. In JK−1 mol−1,

R = 8.314 kgm2s−2 = 8.314 JK−1 mol−1 [1 Pa = 1 kgm−1 s−2]

vi. In cal K−1 mol−1 (1 cal = 4.184 J), R = 1.987 calK−1mol−1

Example 2.42. 50.0 g of N2 (M = 28.0 g) occupies a volume of 750mL at

298.15 K. Assuming the gas behaves ideally, calculate the pressure of

the gas in kPa.


73

Solution

Collect the data, convert volume to liter and find the number of mole

of nitrogen before substituting into the ideal gas equation to find

pressure

Data given

Mass of N2 = 50g

Molar mass of N2 = 28g/mol

Volume , V = 750mL = 750

1000 𝐿 = 0.750 𝐿

Temperature, T = 298.15 K

Number of mole of nitrogen gas (n) = 𝑚𝑎𝑠𝑠


= 50 𝑔

28𝑔𝑚𝑜𝑙 −1

=1.79 mol

Using PV = nRT

𝑃 = 𝑛𝑅𝑇

𝑉

= 1.79 𝑚𝑜𝑙 ×0.08206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙 −1 𝐾−1 ×298.15 𝐾

0.750 𝐿

= 58.39 atm

Converting to kPa

= (58.39 × 101325 )

1000

= 5.916 × 103 𝑘𝑃𝑎

Example 2.43. Calculate the volume occupied by 2.5 moles of an ideal

gas at -23 oC, and 4.0 atmospheres. [R = 0.082 atm dm3 K-1 mol-1]


74

Solution

Data provided:

P = 4.0 atm

T = -23 + 273 = 250K

n = 2.5 moles

R = 0.082 atm dm3 K-1 mol-1

V = ?

Applying PV = nRT

Making V the subject, and substituting:

V = 𝑛𝑅𝑇

𝑃

= 2.5 ×0.082 ×250

4

=12.8 dm3

Example 2.44. Calculate the volume of 1.63 mol of carbon dioxide gas

at 295 K and 1.14 atm.

Solution

Data provided:

P = 1.14 atm

T = 295K

n = 1.63 moles

R = 0.082 atm L K-1 mol-1

V = ?

Applying PV = nRT



75

V = 𝑛𝑅𝑇

𝑃

= 1.63 ×0.082 ×295

1.14

=34.6 L

Example 2.45. Calculate the volume of 0.898 mol of methane gas, CH4,

at 292 K and 1.06 atm.

Solution

Data provided:

P = 1.06 atm

T = 292 K

n =0 .898 moles

R = 0.082 atm L K-1 mol-1

V = ?

Applying PV = nRT


V = 𝑛𝑅𝑇

𝑃

= 0.898 mol ×0.082 atm L mol −1 K−1 ×292 K

1.06 atm

= 20.3 L

Example 2.46. Calculate the volume of 42.6 g of oxygen gas at 35oC and

792 torr

Solution

1. First convert temperature to Kelvin and pressure to atm.

2. Find the number of mole of oxygen


76

3. Plug data into idea gas equation to find the volume

Data provided:

P = 792

760 𝑎𝑡𝑚 = 1.04 atm

T = 35oC = 308 K

R = 0.082 atm L K-1 mol-1

n = mass/molar mass = 42.6g/32gmol-1 = 1.33 mol

V = ?

Applying PV = nRT


V = 𝑛𝑅𝑇

𝑃

= 1.33 mol ×0.082 atm L mol −1 K−1 ×308 K

1.04 atm

=32.3 L

Students sometimes wonder “How do I decide when to use the combined gas

law and when to use the ideal gas law?” The answer depends on the problem,

naturally. If moles are involved, the combined gas law cannot be used.

Example 2.47. Decide which gas law should be used to solve each of

the following:

(a) Calculate the final volume of a sample of gas that has an initial

volume of 7.10 L at STP if the temperature and pressure are changed to

33oC and 696 torr.

(b) Calculate the volume of 0.977 mol of gas at 33oC and 792 torr.


77

Solution

(a) The combined gas law can be used because it does not involve

number of moles and initial and final conditions are involved.

(b) This problem involves moles and must be solved with the ideal gas

law.

Example 2.48. Calculate the pressure of 0.0789 mol of chlorine gas that

occupies 891 mL at ‒15°C.

Solution

The quantities given are converted to the units generally used with the

ideal gas law equation. Note that the nature of the gas is immaterial as

long as the number of moles is known.

Data provided:

T = ‒15°C = (‒15 +273) K = 258 K

n =0 .0789 moles

R = 0.082 atm L K-1 mol-1

V = 891 mL = (891/1000) L = 0.891 L

P = ?

Applying PV = nRT

Making P the subject, and substituting:

P = 𝑛𝑅𝑇

𝑉

= 0.0789 mol ×0.082 atm L mol −1 K−1 ×258 K

0.891 L

=1.87 atm


78

Example 2.49. Calculate the pressure of 0.0855 mol of neon gas that

occupies 66.1 mL at 25°C.

Solution

Data provided:

T = 25°C = (25 +273) K = 298 K

n =0 .0855 moles

R = 0.082 atm L K-1 mol-1

V = 66.1 mL = (66.1/1000) L = 0.0661 L

P = ?

Applying PV = nRT


P = 𝑛𝑅𝑇

𝑉

= 0.0855 mol × 0.082 atm L mol −1 K−1 ×298 K

0.0661 L

= 31.6 atm

Example 2.50. Calculate the number of moles of oxygen gas in a 2.60-L

container at 19°C and

755 torr.

Solution

Data provided:

T = 19°C = (19 +273) K = 292 K

R = 0.082 atm L K-1 mol-1

V = 2.60 L


79

P = 755 torr = 755

760 𝑎𝑡𝑚 = 0.99 atm

n =?

Applying PV = nRT

Making n the subject, and substituting:

n = 𝑃𝑉

𝑅𝑇

= 0.99 atm ×2.60 L

0.082 atm L mol −1 K−1 ×292 K

= 0.12 mol

5.0g of neon is at 256 mm Hg and at a temperature of 35º C. What is the

volume?

Solution

Step 1: Write down your given information:

P = 256 mmHg

V = ?

m = 5.0 g

R = 0.082 L atm mol-1K-1

T = 35oC

Step 2: Convert as necessary:

T = 35oC = (35 + 273)K = 308 K

P = 256 mmHg = (256/760) atm = 0.34 atm

n = mass/molar mass = 5.0g/20.1797 gmol-1 = 2.5 mol

Applying PV = nRT



80

V = 𝑛𝑅𝑇

𝑃

= 2.5 mol ×0.082 atm L mol −1 K−1 ×308 K

0.34 atm

=186 L

Example 2.51. What is a gas‘s temperature in Celsius when it has a

volume of 25 L, 203 mol, 143.5 atm?

Solution

Data provided:

R = 0.082 atm L K-1 mol-1

V = 25 L

P = 143.5 atm

n = 203 mol

T = ?

Applying PV = nRT

Making T the subject, and substituting:

T = 𝑃𝑉

𝑛𝑅

= 143.5 atm × 25 L

0.082 atm L mol −1 K−1 ×203 mol

= 215.5 K


= (215.5 − 273) oC

= −57.5 oC


81

Example 2.52. Sodium azide (NaN3) is used in some automobile air

bags. The impact of a collision triggers the decomposition of NaN3 as

follows:

2NaN3(s) →2Na(s) + 3N2(g)

The nitrogen gas produced quickly inflates the bag between the driver

and the windshield and dashboard. Calculate the volume of N2

generated at 85°C and 812 mmHg by the decomposition of 50.0 g of

NaN3.

Strategy From the balanced equation we see that 2 mol NaN3 gives 3

mol N2 so the conversion factor between NaN3 and N2 is

3 mol N2

2 𝑚𝑜𝑙 𝑁𝑎𝑁3

Because the mass of NaN3 is given, we can calculate the number of

moles of NaN3 and hence the number of moles of N2 produced. Finally,

we can calculate the volume of N2 using the ideal gas equation.

Solution

The sequence of conversions is as follows:

grams of NaN3 → moles of NaN3 →moles of N2 →volume of N2

First, we calculate the number of moles of N2 produced by 50.0 g of

NaN3:

mole of NaN3 = mass/ molar mass

= 50g/65.02gmol-1

= 0.769 mol


82

Mole of N2 = 0.769 mol × 3 mol N2

2 𝑚𝑜𝑙 𝑁𝑎𝑁3

= 1.15 mol N2

The volume of 1.15 mol of N2 can be obtained by using the ideal gas

equation:

PV = nRT


V = 𝑛𝑅𝑇

𝑃

= 1.15 mol ×0.082 atm L mol −1 K−1 ×(85+273) K

(812

760) atm

= 33.7594 𝐿

1.068

=31.6 L

Example 2.53. The equation for the metabolic breakdown of glucose

(C6H12O6) is the same as the equation for the combustion of glucose in

air:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)

Calculate the volume of CO2 produced at 37°C and 1.00 atm when 5.60

g of glucose is used up in the reaction. [C=12, O=16,H=1]

Solution

Collect given data and convert as necessary

R = 0.082 atm L K-1 mol-1

V = ?

P = 1.00 atm


83

n = ?

T = 37°C = 310 K

Mass of glucose = 5.6g

Strategy

1. First calculate the molar mass of glucose

2. Calculate the mole of C6H12O6 from which the mole of CO2 can

be gotten since 1 mole of C6H12O6 produced 6 moles of CO2

from the equation.

3. Use ideal gas equation to find the volume of CO2 produced.

C6H12O6 =[(12× 6 )+ (1 × 12) + (16× 6)] = 179gmol-1

Mole of C6H12O6 = 5.6g/179gmol-1

= 0.03 mol

From the balanced equation we see that 1 mol C6H12O6 gives 6 mol CO2

so the conversion factor between C6H12O6 and CO2 is

6 mol CO2

1 𝑚𝑜𝑙 𝐶6𝐻12𝑂6

Mole of CO2 = 0.03 mol × 6 mol CO 2

1 𝑚𝑜𝑙 𝐶6𝐻12𝑂6

=0.18 mol CO2

The volume of 0.18 mol of CO2 can be obtained by using the ideal gas

equation:

PV = nRT


V = 𝑛𝑅𝑇

𝑃


84

= 0.18 mol ×0.082 atm L mol −1 K−1 ×310 K

1.0 atm

= 4.6 𝐿

Example 2.54. Assuming ideal behaviour, which of the following

samples of gases will have the greatest volume at STP? Which of these

gases will have the greatest density at STP? (a) 0.82 mole of He. (b) 24 g

of N2. (c) 5.0 × 1023 molecules of Cl2

Solution

At STP, [T = 273K, P=1.0atm, R= 0.082 atm L mol−1K−1 ]

(a) n = 0.82 mole He

Using PV = nRT


V = 𝑛𝑅𝑇

𝑃

= 0.82 mol ×0.082 atm L mol −1 K−1 × 273 K

1.0 atm

= 18.4 L

Density, d of He = 𝑚

𝑉

But mass, 𝑚 = 𝑛𝑚 = 0.82 × 4.003

= 0.33𝑔

∴ dendity, d = 0.88 𝑔

18.4 𝐿

=0.018g/L

(b) 24 g of N2

No of mole of nitrogen 𝑛 = 𝑚𝑎𝑠𝑠

𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠


85

= 24g

14𝑔𝑚𝑜𝑙 −1

= 1.71 𝑚𝑜𝑙

Using PV = nRT to calculate the volume of N2 at STP


V = 𝑛𝑅𝑇

𝑃

= 1.71 mol ×0.082 atm L mol −1 K−1 × 273 K

1.0 atm

= 38.3 L

Density, d of N2 = 𝑚

𝑉

= 24 𝑔

38.3 𝐿

= 0.627 g/L

(c) 5.0 × 1023 molecules of Cl2

Using PV = nRT to calculate the volume of chlorine molecule at STP,

Make V the subject, and substite:

V = 𝑛𝑅𝑇

𝑃

= 5.0 × 1023 mol ×0.082 atm L mol −1 K−1 × 273 K

1.0 atm

= 1.1193 × 1025 L

Density, d of Cl2 = 𝑚

𝑉

= 35.5 𝑔

1.1193 ×1025 𝐿

= 3.17 × 10−24 g/L


86

Summary of the results

He gas N2 gas Cl2 gas

Volume at STP (L) 18.4 38.3 1.1193 × 1025

Density at STP (g/L) 0.018 0.627 3.17 × 10−24

Results from the calculations showed chlorine gas has the highest

volume at STP 1.1193 × 1025 L while nitrogen gas has the greatest

density of 0.627 g/L.

Example 2.55. How many moles of O2 are present in a 0.500-L sample

at 25oC and 1.09 atm?

Solution

Collect the given data and convert as necessary

T = 25°C = (25 +273) K = 298 K

R = 0.082 atm L K-1 mol-1

V = 0.500 L

P = 1.09 atm

n =?

Applying PV = nRT to find n of O2;


n = 𝑃𝑉

𝑅𝑇

= 1.09 atm ×0.500 L

0.082 atm L mol −1 K−1 × 298 K


87

=0.022 mol of O2

Example 2.56. What is the volume of 1.00 mol of gas at STP?

Solution

Data provided:

P = s.p. = 1.0 atm

T = s.t. = 273 K

R = 0.082 atm L K-1 mol-1

n = 1.0 mol

V = ?

Applying PV = nRT


V = 𝑛𝑅𝑇

𝑃

= 1.0mol ×0.082 atm L mol −1 K−1 ×273 K

1.0 atm

= 22.4 L

Note that the volume of 1.00 mol of gas at STP is called the molar volume of a

gas. This value should be memorized.

Example 2.57. How many moles of SO2 are present in a 765-mL sample

at 37oC and 775 torr?

Solution

Since R is defined in terms of liters and atmospheres, the pressure and

volume are first converted to those units.


88

Collect the given data

T = 37°C = (37 +273) K = 310 K

R = 0.082 atm L K-1 mol-1

V = 765 mL = (765/1000)L = 0.765 L

P = 775 torr = (775/760) atm = 1.02 atm

n =?

Applying PV = nRT to find n of SO2;


n = 𝑃𝑉

𝑅𝑇

= 1.02 atm ×0.765 L

0.082 atm L mol −1 K−1 × 310 K

= 0.03 mol of SO2

Example 2.57. At what temperature will 0.0750 mol of CO2 occupy 2.75

L at 1.11 atm?

Solution


V = 2.75 L

P = 1.11 atm

n = 0.0750 mol

T = ?

R = 0.082 atm L K-1 mol-1


89

Applying PV = nRT and Making T the subject, and substituting:

T = 𝑃𝑉

𝑛𝑅

= 1.11 atm × 2.75 L

0.0750 mol × 0.082 atm L mol −1 K−1

= 496 K

Example 2.58. What volume will 7.00 g of Cl2 occupy at STP?

Solution

The value of n is not given explicitly in the problem, but the mass is

given, from which we can calculate the number of moles:

Data provided

V = ?

P = s.p. = 1.0 atm

T = s.t. = 273 K

R = 0.082 atm L K-1 mol-1

Number of mole of Cl2 = 𝑚𝑎𝑠𝑠

𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑕𝑙𝑜𝑟𝑖𝑛𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒

= 7.00 𝑔

35.5 × 2 𝑔/𝑚𝑜𝑙

= 0.0986 mol of Cl2

Applying PV = nRT to find the volume of Cl2


V = 𝑛𝑅𝑇

𝑃

= 0.0986 mol × 0.082 atm L mol −1 K−1 273 K

1.0 atm

= 2.23 L


90

Example 2.59. If 4.58 g of a gas occupies 3.33 L at 27oC and 808 torr,

what is the molar mass of the gas?

Solution

If you do not see at first how to solve this problem to completion, at

least you can recognize that P, V, and T data are given. First calculate

the number of moles of gas present from which you can get the molar

mass.


T = 27°C = (27 +273) K = 300 K

R = 0.082 atm L K-1 mol-1

V = 3.33 L

P = 808 torr = (808/760) atm = 1.06 atm

Mass (m) = 4.58g

n =?

Applying PV = nRT to find n of gas;


n = 𝑃𝑉

𝑅𝑇

= 1.06 atm × 3.33 L

0.082 atm L mol −1 K−1 × 300 K

= 0.143 mol of gas

Recall, number of mole (n) = 𝑚𝑎𝑠𝑠 (𝑚)

𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀)

Therefore molar mass of gas (M) = 𝑚𝑎𝑠𝑠 (𝑚)

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒 (𝑛)

= 4.58 𝑔

0.143 𝑚𝑜𝑙


91

= 32.0g/mol

Example 2.60. What volume is occupied by the oxygen liberated by

heating 0.250 g of KClO3 until it completely decomposes to KCl and

oxygen? The gas is collected at STP.

Solution

From the balanced equation below, we see that 2 mol KClO3 gives 3

mol O2 so the conversion factor between KClO3 and O2 is

3 mol O2

2 𝑚𝑜𝑙 KClO3

2 KClO3 → 2KCl + 3O2

Because the mass of KClO3 is given, we can calculate the number of

moles of KClO3 and hence the number of moles of O2 produced.

Finally, we can calculate the volume of O2 using the ideal gas equation.


T = s.t. = 273 K

R = 0.082 atm L K-1 mol-1

V = ?

P = 𝑠. 𝑝. = 1.0 atm

Mass (m) = 0.250 g of KClO3

n =?

Molar mass of KClO3 [39.10 +35.5 +(16× 3)] = 122.6g/mol

mole of KClO3 = mass/ molar mass

= 0.250g/122.6gmol-1


92

= 0.002 mol KClO3

Mole of O2 = 0.002 mol × 3 mol O2

2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3

= 0.003 mol O2

The volume of 0.003 mol O2 can be obtained by using the ideal gas

equation:

PV = nRT


V = 𝑛𝑅𝑇

𝑃

= 0.003 mol ×0.082 atm L mol −1 K−1 × 273 K

1.0 atm

= 0.067 L

5.0 moles of oxygen gas are contained in a 1.13 dm3 container at 127oC.

wha is the pressure of the system in Nm-2? [R = 8.314NmK-1mol-1]

Solution

Collect information provided and convert as necessary

T = 127oC = 400 K

n = 0.32 mol

R = 8.314NmK-1mol-1

V = 1.13 dm3 = (1.13/1000) m3 = 1.13 × 10−3𝑚3

[1000 dm3 = 1 m3]

P =?

Use PV = nRT to find P



93

P= 𝑛𝑅𝑇

𝑉

= 0.32 mol ×8.314 Nm mol −1 K−1 ×400 K

0.00113 m3

= 941762.8 𝑁𝑚−2

= 9.43 × 105 𝑁𝑚−2

Example 2.61. A vessel contains 2.5 dm3 of oxygen gas at 29oC under

2.1 atmospheres. Estimate the amount of the gas at STP.

Solution

We will use the general gas equation to get the volume of oxygen gas

at STP then apply ideal gas law to get the amount of the gas.

Collect information provided and convert as necessary

T1 = 29oC = (29 + 273) K = 302 K

P1 = 2.1 atm

V1 = 2.5 dm3

T2 = s.t. = 273 K

P2 = s.p. = 1.0 atm

V2 = ?

n = ?


𝑃1 𝑉1

𝑇1 =

𝑃2 𝑉2

𝑇2


V2 = 𝑃1𝑉1𝑇2

𝑃2𝑇1


94

= 2.1 atm × 2.5 dm 3 × 273 K

1.0 atm × 302 K

= 4.75 dm3

Now we can apply ideal gas equation in order to find the amount of

oxygen gas at s.t.p;

PV = nRT


n = 𝑃𝑉

𝑅𝑇

But P = 1 atm = 101325 𝑁𝑚−2

Volume, V = 4.75 dm3 = (4.75/1000) m3 = 4.75 × 10−3 𝑚3

Temperature, T = 273 K

R = 8.314NmK-1mol-1

n = 101325 Nm −2 × 4.75 × 10−3 m3

8.314 NmK −1mol −1 × 273 K

= 0.212 mol of O2 gas

Example 2.62.Calculate the volume occupied by 40g of carbon dioxide

(CO) at 4.58 × 104 𝑁𝑚−2 and 50oC , assuming ideal gas law is obeyed.

[O = 16, C =12]

Solution

First we calculate the molar mass of CO and find its number f mole

then apply ideal gas equation to get the volume.


T = 50°C = (50 +273) K = 323 K


95

R = 8.314NmK-1mol-1

V = ?

P = 4.58 × 104 𝑁𝑚−2

Mass of CO(m) = 40g

n =?

Molar mass of CO = 12 +16 = 28g/mol



= 40 𝑔

28𝑔/ 𝑚𝑜𝑙

= 1.43 mol

Applying PV = nRT to find V of CO gas;


V = 𝑛𝑅𝑇

𝑃

= 1.43 mol 8.314 Nm mol −1 K−1 × 323 K

4.58 × 104 𝑁𝑚−2

= 0.082 m3 of CO gas

= 82.0 dm3 of CO gas

[1000dm3 = 1m3]

Example 2.63. How many moles of a gas are contained in 890.0 mL at

21.0 °C and 750.0 mmHg pressure?

Solution

Collect the given data and convert to appropriate units

T = 21°C = (21 +273) K = 294 K

R = 0.082 atm L K-1 mol-1


96

V = 890 mL = (890/1000)L = 0.890 L

P = 750.0 mmHg = (750/760) atm = 0.99 atm

n =?



n = 𝑃𝑉

𝑅𝑇

= 0.99 atm × 0.890 L

0.082 atm L mol −1 K−1 × 294 K

= 0.037 mol of gas

Example 2.64. Calculate the molecular weight of a gas if 35.44 g of the

gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at a constant

temperature of 35.5 °C.

Solution

We first find the number of mole of the gas using ideal gas equation

and then relate the mole and molar mass.


T = 27°C = (35.5 +273) K = 308.5 K

R = 0.082 atm L K-1 mol-1

V = 7.50 L

P = 60.0 atm

Mass (m) = 20.44g

n =?


97



n = 𝑃𝑉

𝑅𝑇

= 60.0 atm × 8.5 L

0.082 atm L mol −1 K−1 × 308.5 K

= 20.1 mol of gas



Therefore molar mass of gas (M) = 𝑚𝑎𝑠𝑠 (𝑚)

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒 (𝑛)

= 20.44 𝑔

20.1 𝑚𝑜𝑙

= 1.0 g/mol

2.11. Dalton’s Law of Partial Pressure

Thus far we have concentrated on the behaviour of pure

gaseous substances, but experimental studies very often involve

mixtures of gases. For example, for a study of air pollution, we may be

interested in the pressure-volume-temperature relationship of a sample

of air, which contains several gases. In this case, and all cases involving

mixtures of gases, the total gas pressure is related to partial pressures,

that is, the pressures of individual gas components in the mixture. In 1801

Dalton formulated a law, now known as Dalton’s law of partial

pressures, which states that the total pressure of a mixture of gases is just

the sum of the pressures that each gas would exert if it were present alone.

Figure 2.3 illustrates Dalton‘s law.


98

Figure 2.3. Schematic illustration of Dalton’s law of partial pressures. Consider a case in which two gases, A and B, are in a container of volume V. The pressure exerted by gas A, according to the ideal gas equation, is

𝑃𝐴 =𝑛𝐴𝑅𝑇

𝑉

Where nA is the number of moles of A present. Similarly, the pressure exerted by gas B is

𝑃𝐵 =𝑛𝐵𝑅𝑇

𝑉

In a mixture of gases A and B, the total pressure PT is the result of the collisions of both types of molecules, A and B, with the walls of the container. Thus, according to Dalton‘s law, 𝑃𝑇 = 𝑃𝐴 + 𝑃𝐵

= 𝑛𝐴𝑅𝑇

𝑉 +

𝑛𝐵𝑅𝑇

𝑉

= 𝑅𝑇

𝑉 (𝑛𝐴 + 𝑛𝐵)

= 𝑛𝑅𝑇

𝑉


99

Where n, the total number of moles of gases present, is given by n = nA + nB, and PA and PB are the partial pressures of gases A and B, respectively. For a mixture of gases, then, PT depends only on the total number of moles of gas present, not on the nature of the gas molecules. In general, the total pressure of a mixture of gases is given by 𝑃𝑇 = 𝑃1 + 𝑃2 + 𝑃3 + − − − − Where P1, P2, P3, . . . are the partial pressures of components 1, 2, 3, . . . . To see how each partial pressure is related to the total pressure, consider again the case of a mixture of two gases A and B. Dividing PA by PT, we obtain

𝑃𝐴

𝑃𝑇 =

𝑛𝐴𝑅𝑇𝑉

(𝑛𝐴 +𝑛𝐵) 𝑅𝑇𝑉

= 𝑛𝐴

𝑛𝐴 +𝑛𝐵

= 𝑋𝐴

Where XA is called the mole fraction of A. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of one component to the number of moles of all components present. In general, the mole fraction of component i in a mixture is given by

𝑋𝑖 = 𝑛 𝑖

𝑛𝑇

Where ni and nT are the number of moles of component i and the total number of moles present, respectively. The mole fraction is always smaller than 1. We can now express the partial pressure of A as 𝑃𝐴 = 𝑋𝐴𝑃𝑇 Similarly, 𝑃𝐵 = 𝑋𝐵𝑃𝑇 Note that the sum of the mole fractions for a mixture of gases must be unity. If only two components are present, then

𝑋𝐴 + 𝑋𝐵 = 𝑛𝐴

𝑛𝐴 + 𝑛𝐵 +

𝑛𝐵

𝑛𝐴 + 𝑛𝐵= 1

If a system contains more than two gases, then the partial pressure of the ith component is related to the total pressure by 𝑃𝑖 = 𝑋𝑖𝑃𝑇


100

How are partial pressures determined? A manometer can measure only the total pressure of a gaseous mixture. To obtain the partial pressures, we need to know the mole fractions of the components, which would involve elaborate chemical analyses. The most direct method of measuring partial pressures is using a mass spectrometer. The relative intensities of the peaks in a mass spectrum are directly proportional to the amounts, and hence to the mole fractions, of the gases present. Example 2.65. A mixture of gases contains 3.85 moles of neon (Ne), 0.92 mole of argon (Ar), and 2.59 moles of xenon (Xe). Calculate the partial pressures of the gases if the total pressure is 2.50 atm at a certain temperature. Solution The partial pressure of a gas is equal to the product of its mole fraction and the total pressure (PT) Given data Mole of Ne = 3.85 moles Mole of Ar = 0.92 moles Mole of Xe = 2.59 moles Total pressure, PT = 2.5

Mole fraction of neon (𝑋𝑁𝑒 ) = 𝑛𝑁𝑒

𝑛𝑁𝑒 +𝑛𝐴𝑟 +𝑛𝑋𝑒

= 3.85 𝑚𝑜𝑙

3.85 𝑚𝑜𝑙 +0.92 𝑚𝑜𝑙 +2.59 𝑚𝑜𝑙

= 0.523 Therefore 𝑃𝑁𝑒 = 𝑋𝑁𝑒𝑃𝑇 = 0.523 × 2.50 = 1.31 𝑎𝑡𝑚 Similarly, we can calculate the mole fraction of argon and its partial pressure:

Mole fraction of Argon (𝑋𝐴𝑟 ) = 𝑛𝐴𝑟

𝑛𝐴𝑟 + 𝑛𝑁𝑒 +𝑛𝑋𝑒

= 0.92 𝑚𝑜𝑙

0.92 𝑚𝑜𝑙 + 3.85 𝑚𝑜𝑙 +2.59 𝑚𝑜𝑙


101

= 0.125 Therefore 𝑃𝐴𝑟 = 𝑋𝐴𝑟𝑃𝑇 = 0.125 × 2.50 = 0.313 𝑎𝑡𝑚 Finally, we calculate the mole fraction of xenon and its partial pressure:

Mole fraction of Xenon (𝑋𝑋𝑒 ) = 𝑛𝑋𝑒

𝑛𝑋𝑒 +𝑛𝐴𝑟 +𝑛𝑁𝑒

= 2.59 𝑚𝑜𝑙

2.59 𝑚𝑜𝑙 +0.92 𝑚𝑜𝑙 +2.59 𝑚𝑜𝑙

= 0.352 Therefore 𝑃𝑋𝑒 = 𝑋𝑋𝑒𝑃𝑇 = 0.352 × 2.50 = 0.88 𝑎𝑡𝑚 Check: The individual partial pressures must be less than the total pressure and make sure that the sum of the partial pressures is equal to the total pressure; that is, (1.31 + 0.313 + 0.880) atm = 2.50 atm. Example 2.66. A sample of natural gas contains 8.24 moles of methane (CH4), 0.421 mole of ethane (C2H6), and 0.116 mole of propane (C3H8). If the total pressure of the gases is 1.37 atm, what are the partial pressures of the gases? Solution Data provided

Mole of methane = 8.24 moles

Moles of ethane = 0.421 moles

Mole of propane = 0.116 moles

Total pressure = 1.37 atm

Mole fraction of neon (𝑋𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 ) = 𝑛𝑚𝑒𝑡 𝑕𝑎𝑛𝑒

𝑛𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 +𝑛𝑒𝑡 𝑕𝑎𝑛𝑒 +𝑛𝑝𝑟𝑜𝑝𝑎𝑛𝑒

= 8.24 𝑚𝑜𝑙

8.24 𝑚𝑜𝑙 +0.421 𝑚𝑜𝑙 +0.116 𝑚𝑜𝑙


102

= 0.94

Therefore

𝑃𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 = 𝑋𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 𝑃𝑇

= 0.94 × 1.37

= 1.29 𝑎𝑡𝑚

Similarly, we can calculate the mole fraction of ethane and its partial

pressure:

Mole fraction of Ethane (𝑋𝑒𝑡𝑕𝑎𝑛𝑒 ) = 𝑛𝑒𝑡 𝑕𝑎𝑛𝑒

𝑛𝑒𝑡𝑕𝑎𝑛𝑒 + 𝑛𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 +𝑛𝑝𝑟𝑜𝑝𝑎𝑛𝑒

= 0.421 𝑚𝑜𝑙

0.92 𝑚𝑜𝑙 + 8.24 𝑚𝑜𝑙 +0.116 𝑚𝑜𝑙

= 0.05

Therefore

𝑃𝑒𝑡𝑕𝑎𝑛𝑒 = 𝑋𝑒𝑡𝑕𝑎𝑛𝑒 𝑃𝑇

= 0.05 × 1.37

= 0.069 𝑎𝑡𝑚

Finally, we calculate the mole fraction of propane and its partial

pressure:

Mole fraction of Propane (𝑋𝑝𝑟𝑜𝑝𝑎𝑛𝑒 )

= 𝑛𝑝𝑟𝑜𝑝𝑎𝑛𝑒

𝑛𝑝𝑟𝑜𝑝𝑎𝑛𝑒 +𝑛𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 +𝑛𝑒𝑡 𝑕𝑎𝑛𝑒

= 0.116 𝑚𝑜𝑙

0.116 𝑚𝑜𝑙 +8.24 𝑚𝑜𝑙 +0.92 𝑚𝑜𝑙

= 0.01

Therefore

𝑃𝑝𝑟𝑜𝑝𝑎𝑛𝑒 = 𝑋𝑝𝑟𝑜𝑝𝑎𝑛𝑒 𝑃𝑇


103

= 0.01 × 1.37

= 0.014 𝑎𝑡𝑚

2.12. Gases Collected Over Water

Whenever a gas is collected over water, it becomes wet by water vapour. Since water vapour is a gas, it exerts its own pressure, and a mixture of gases is obtained. The pressure exerted is, therefore, the sum of the partial pressures of the gas and that of the water vapour at that temperature, i.e. PTotal = Pgas +Pwater vapour Pgas = PTotal − Pwater vapour Dalton‘s law of partial pressures is useful for calculating volumes of gases collected over water. For example, when potassium chlorate (KClO3) is heated, it decomposes to KCl and O2:

2KClO3(s) → 2KCl(s) + 3O2(g) The oxygen gas can be collected over water, as shown in Figure 2.4. Initially, the inverted bottle is completely filled with water. As oxygen gas is generated, the gas bubbles rise to the top and displace water from the bottle. This method of collecting a gas is based on the assumptions that the gas does not react with water and that it is not appreciably soluble in it. These assumptions are valid for oxygen gas, but not for gases such as NH3, which dissolves readily in water. The oxygen gas collected in this way is not pure, however, because water vapour is also present in the bottle. The total gas pressure is equal to the sum of the pressures exerted by the oxygen gas and the water vapour: PT = P 𝑜2 + P𝐻2O Consequently, we must allow for the pressure caused by the presence of water vapour when we calculate the amount of O2 generated. Table 2.2 shows the pressure of water vapour at various temperatures.


104

Table 2.2. pressure of water vapour at various temperatures Temperature Water vapour pressure

(oC ) (mmHg) 0 4.58 10 9.21 30 31.82 50 92.51 70 233.7 80 355.1 90 525.76 100 760.00


105

Figure 2.4: An apparatus for collecting gas over water. The oxygen generated by heating potassium chlorate (KClO3) in the presence of a small amount of manganese dioxide (MnO2), which speeds up the reaction, is bubbled through water and collected in a bottle as shown. Water originally present in the bottle is pushed into the trough by the oxygen gas. Example 2.67. Oxygen gas generated by the decomposition of potassium chlorate is collected as shown in Figure 2.4. The volume of oxygen collected at 26°C and atmospheric pressure of 771 mmHg is 141 mL. Calculate the mass (in grams) of oxygen gas obtained. The pressure of the water vapour at 26°C is 25.2 mmHg. [O = 16] Solution To solve for the mass of O2 generated, we must first calculate the partial pressure of O2 in the mixture. What gas law do we need? How do we convert pressure of O2 gas to mass of O2 in grams? Data provided PT = 771 mmHg P𝐻2O = 25.2 mmHg P 𝑜2 = ? V = 141mL = (141/1000)L = 0.141 L Molar mass (M) = 32.0 g/mol T = 26°C = (26 + 273)K = 299 K R = 0.082 atm L K-1 mol-1 From Dalton‘s law of partial pressures we know that PT = P 𝑜2 + P𝐻2O P 𝑜2 = PT ‒ P𝐻2O = 771 mmHg ‒ 25.2 mmHg

= 746 mmHg = (746/760) atm = 0.98 atm

From the ideal gas equation we write 𝑃𝑉 = 𝑛𝑅𝑇


106

But 𝑛 = 𝑚

𝑀

Therefore 𝑃𝑉 = 𝑚

𝑀. 𝑅𝑇

and 𝑃𝑉𝑀 = 𝑚𝑅𝑇

𝑚 = 𝑃𝑉𝑀

𝑅𝑇

= 0.98 𝑎𝑡𝑚 × 0.141 𝐿 × 32.0 𝑔/𝑚𝑜𝑙

0.0821 𝑎𝑡𝑚 𝐿 𝐾−1 𝑚𝑜𝑙 −1 ×299 𝐾

= 0.180 g Example 2.68. Hydrogen gas generated when calcium metal reacts with water is collected as shown in Figure 5.14. The volume of gas collected at 30°C and pressure of 988 mmHg is 641 mL. What is the mass (in grams) of the hydrogen gas obtained? The pressure of water vapor at 30°C is 31.82 mmHg. [H =1]. Solution Data provided PT = 988 mmHg P𝐻2O = 31.82 mmHg P 𝐻2 = ? V = 641mL = (641/1000)L = 0.641 L Molar mass (M) = 2.0 g/mol T = 30°C = (30 + 273)K = 303 K R = 0.082 atm L K-1 mol-1 From Dalton‘s law of partial pressures we know that PT = P 𝑜2 + P𝐻2O P 𝐻2 = PT ‒ P𝐻2O = 988 mmHg ‒ 31.82 mmHg

= 956.18 mmHg = (956.18/760) atm = 1.26 atm

From the ideal gas equation we write 𝑃𝑉 = 𝑛𝑅𝑇

But 𝑛 = 𝑚

𝑀


107

Therefore 𝑃𝑉 = 𝑚

𝑀. 𝑅𝑇

and 𝑃𝑉𝑀 = 𝑚𝑅𝑇

𝑚 = 𝑃𝑉𝑀

𝑅𝑇

= 1.26 𝑎𝑡𝑚 × 0.641 𝐿 × 2.0 𝑔/𝑚𝑜𝑙

0.0821 𝑎𝑡𝑚 𝐿 𝐾−1 𝑚𝑜𝑙 −1 ×303 𝐾

= 0.065 g Example 2.69. 20 dm3 of hydrogen were collected over water at 17oC and 79.7KNm-2 pressure. Calculate (i) pressure of dry hydrogen at this temperature. (ii) volume of dry hydrogen gas at s.t.p. (vapour pressure of water is 1.9 KNm-2 at 17oC; 1atm = 101.3 KNm-2 ). Solution Data given:

PTotal = 79.7KNm-2

Pwater vapour = 1.9 KNm-2 at 17oC

Pgas = ?

(i) According to Dalton‘s law:

Pgas = PTotal − Pwater vapour

p(H2) = (79.7 − 1.9 ) KNm-2

= 77.8 KNm-2

(ii) To find the volume of dry hydrogen gas at s.t.p.

Data provided.

P1= 77.8 KNm-2

V1= 20 dm3

T1 = 17oC = (17 + 273)K = 290K

P2 = s.p. = 101.3 KNm-2

T2 = s.t. = 273K


108

V2 = ? Using the general gas equation and substituting, P1V1/T1 = P2V2/T2

V2 = 77.8 ×20 ×273

101.3 ×290 = 14.5 dm3

Example 2.70. (i). If I try to put a 1.00-L sample of O2 at 300 K and 1.00 atm plus a 1.00-L sample of N2 at 300 K and 1.00 atm into a rigid 1.00-L container at 300 K, will they fit? (ii) If so, what will be their total volume and total pressure? Solution (i) The gases will fit; gases expand or contract to fill their containers. (b) The total volume is the volume of the container—1.00 L. The temperature is 300 K, given in the problem. The total pressure is the sum of the two partial pressures. Partial pressure is the pressure of each gas (as if the other were not present). The oxygen pressure is 1.00 atm. The oxygen has been moved from a 1.00-L container at 300 K to another 1.00-L container at 300 K, and so its pressure does not change. The nitrogen pressure is 1.00 atm for the same reason. The total pressure is 1.00 atm + 1.00 atm = 2.00 atm. Example 2.71. A 1.00-L sample of O2 at 300 K and 1.00 atm plus a 0.500-L sample of N2 at 300 K and 1.00 atm are put into a rigid 1.00-L container at 300 K. What will be their total volume, temperature, and total pressure? Solution The total volume is the volume of the container—1.00 L. The temperature is 300 K, given in the problem. The total pressure is the sum of the two partial pressures. The oxygen pressure is 1.00 atm. The nitrogen pressure is 0.500 atm, since it was moved from 0.500 L at 1.00 atm to 1.00 L at the same temperature (Boyle‘s law). The total pressure is 1.00 atm + 0.500 atm = 1.50 atm


109

Example 272. A 1.00-L sample of O2 at 300 K and 1.00 atm plus a 0.500-L sample of N2 at 300 K and 1.00 atm are put into a rigid 1.00-L container at 300 K. What will be their total volume, temperature, and total pressure? Calculate the number of moles of O2 both before and after mixing. Solution Data provided P = 1.0 atm T = 300 K V = 1.0 L R = 0.082 atm L K-1 mol-1 n = ? Applying the ideal gas equation to find n before mixing 𝑃𝑉 = 𝑛𝑅𝑇


n = 𝑃𝑉

𝑅𝑇

= 1 atm × 1.0 L

0.082 atm L mol −1 K−1 × 300 K


After mixing

n𝑜2 = 𝑃𝑜2𝑉

𝑅𝑇

= 1 atm × 1.0 L

0.082 atm L mol −1 K−1 × 300 K


There is no change in the number of moles of oxygen gas before and

after mixing.


110

Example 2.73. Calculate the volume of 1.00 mol of H2O at 1.00-atm pressure and a temperature of 25oC. Solution Water (H2O) is not a gas under these conditions, and so the equation PV = nRT does not apply. (The ideal gas law can be used for water vapour, e.g., water over 100 oC at 1 atm or water at lower temperatures mixed with air). At 1-atm pressure and 25 oC, water is a liquid with a density of about 1.00 g/mL. 1 mol of H2O contain 18g

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑉 = 𝑚𝑎𝑠𝑠

𝑑𝑒𝑛𝑠𝑖𝑡𝑦

= 18 𝑔

1.00 𝑔/𝑚𝐿

= 18 mL Example 2.74. A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container? Solution

Data provided

Poxygen = 2.00 atm

Pcarbon dioxide = 3.00 atm

Phelium = 4.00 atm

PT = ?

From Dalton‘s law of partial pressure,

PT = sum of the individual partial pressures in the reaction vessel

= Poxygen + Pcarbon dioxide + Phelium

= (2.0 + 3.0 +4.0) atm = 9.0 atm


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Example 2.75. A tank contains 480.0 grams of oxygen and 80.00 grams of helium at a total pressure of 7.00 atmospheres. Calculate the following. a) How many moles of O2 are in the tank? b) How many moles of He are in the tank? c) Total moles of gas in tank. d) Mole fraction of O2.

e) Mole fraction of He.

f) Partial pressure of O2.

g) Partial pressure of He.

Solution

Let us first find the mole of the gases with the simple relation (n = m/M)

Data given:

Mass of O2 = 480.0 g

Mass of helium = 80g

Total pressure (PT) = 7.00 atm

a). Mole of O2 in the tank (n) = 𝑚𝑎𝑠𝑠 (𝑚)


= 480 𝑔

32 𝑔/𝑚𝑜𝑙

= 15 mol of O2

b). Mole of He in the tank (n) = 𝑚𝑎𝑠𝑠 (𝑚)


= 80 𝑔

4.0 𝑔/𝑚𝑜𝑙


112

= 20 mol of He

c). Total mole of gas in the tank = mole of O2 + mole of He

= (15 + 20)mole

= 35 moles

d). Mole fraction of O2 (𝑋𝑂2) =

𝑛𝑂2

𝑛𝑂2 + 𝑛𝐻𝑒

= 15 𝑚𝑜𝑙

15 𝑚𝑜𝑙 +20 𝑚𝑜𝑙

= 0.4286

e). Mole fraction of He (𝑋𝐻𝑒 ) = 𝑛𝐻𝑒

𝑛𝑂2+ 𝑛𝐻𝑒

= 20 𝑚𝑜𝑙

15 𝑚𝑜𝑙 +20 𝑚𝑜𝑙

= 0.5714 f). Partial pressure of O2 (𝑃𝑂2

) = 𝑋𝑂2𝑃𝑇

= 0.4286 × 7.00 = 3.0 atm g). Partial pressure of He (𝑃𝐻𝑒) = 𝑋𝐻𝑒𝑃𝑇 = 0.5714× 7.00 = 3.99 atm Example 2.76. A tank contains 5.00 moles of O2, 3.00 moles of neon, 6.00 moles of H2S, and 4.00 moles of argon at a total pressure of 1620.0 mmHg. Calculate the following. a) Total moles of gas in tank b) Mole fraction of gases c) Partial pressure of gases d) Pressure fraction of gases

Solution

Data given:

Mole of O2 = 5.00 moles


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Mole of Ne = 3.00 moles

Mole of H2S = 6.00 moles

Mole of Ar = 4.00 moles

Total pressure (PT) = 1620.0 mmHg = (1620/760) atm = 2.1 atm

a). Total mole of gas in the tank = mole of O2 + mole of Ne + mole of H2S + mole of Ar

= (5+3+6 + 4) mole

= 18 moles

bi). Mole fraction of O2 (𝑋𝑂2) =

𝑛𝑂2

𝑛𝑂2 + 𝑛𝑁𝑒 + 𝑛𝐻2𝑆+ 𝑛𝐴𝑟

= 5 𝑚𝑜𝑙

5 𝑚𝑜𝑙 +3 𝑚𝑜𝑙 +6 𝑚𝑜𝑙 +4 𝑚𝑜𝑙

= 0.28

bii). Mole fraction of Ne (𝑋𝑁𝑒 ) = 𝑛𝑁𝑒


= 3 𝑚𝑜𝑙


= 0.17

biii). Mole fraction of 𝐻2𝑆 (𝑋𝐻2𝑆) = 𝑛𝐻2𝑆

𝑛𝑂2+ 𝑛𝑁𝑒 + 𝑛𝐻2𝑆+ 𝑛𝐴𝑟

= 6 𝑚𝑜𝑙


= 0.33

biv). Mole fraction of Ar (𝑋𝐴𝑟 ) = 𝑛𝐴𝑟


= 4 𝑚𝑜𝑙


= 0.22 ci). Partial pressure of O2 (𝑃𝑂2

) = 𝑋𝑂2𝑃𝑇

= 0.28 × 2.1


114

= 0.588 atm cii). Partial pressure of Ne (𝑃𝑁𝑒) = 𝑋𝑁𝑒𝑃𝑇 = 0.17 × 2.1 = 0.357 atm ciii). Partial pressure of 𝐻2𝑆(𝑃𝐻2𝑆) = 𝑋𝐻2𝑆𝑃𝑇

= 0.33 × 2.1 = 0.693 atm civ). Partial pressure of Ar (𝑃𝐴𝑟 ) = 𝑋𝐴𝑟𝑃𝑇 = 0.22 × 2.1 = 0.462 atm

di). pressure fraction of O2 (𝑃𝑂2) =

𝑃𝑂2

𝑃𝑂2 + 𝑃𝑁𝑒 + 𝑃𝐻2𝑆+ 𝑃𝐴𝑟

= 0.588 𝑎𝑡𝑚

0.588 𝑎𝑡𝑚 +0.357 𝑎𝑡𝑚 +0.693 𝑎𝑡𝑚 +0.462 𝑎𝑡𝑚

= 0.28 atm

dii). Pressure fraction of Ne (𝑃𝑁𝑒 ) = 𝑃𝑁𝑒

𝑃𝑂2+ 𝑃𝑁𝑒 + 𝑃𝐻2𝑆+ 𝑃𝐴𝑟

= 0.357 𝑎𝑡𝑚

0.588 𝑎𝑡𝑚 +0.357 𝑎𝑡𝑚 +0.693 𝑎𝑡𝑚 +0.462 𝑎𝑡𝑚

= 0.17 atm

diii). Pressure fraction of 𝐻2𝑆 (𝑃𝐻2𝑆) = 𝑃𝐻2𝑆


= 0.693 𝑎𝑡𝑚

0.588 𝑎𝑡𝑚 +0.357 𝑎𝑡𝑚 +0.693 𝑎𝑡𝑚 +0.462 𝑎𝑡𝑚

= 0.33 atm

div). Pressure fraction of Ar (𝑃𝐴𝑟 ) = 𝑃𝐴𝑟


= 0.462 𝑎𝑡𝑚

0.588 𝑎𝑡𝑚 +0.357 𝑎𝑡𝑚 +0.693 𝑎𝑡𝑚 +0.462 𝑎𝑡𝑚

= 0.22 atm Example 2.77. A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder. [H = 1, C = 12]. Solution Let us first find the mole of the gases with the simple relation (n = m/M) then apply ideal gas equation to find the partial pressure of the gases.


115

Data given:

Mass of methane = 1813 g

Mass of ethane = 336g

T = 22.0°C = 295 K

V = 20.0 L

Mole of methane in the cylinder (n) = 𝑚𝑎𝑠𝑠 (𝑚)


But molar mass of methane (CH4) = [12 + (1× 4)] = 16 g/mol

n = 1813 𝑔


= 113 mol of methane

Mole of ethane in the cylinder (n) = 𝑚𝑎𝑠𝑠 (𝑚)


But molar mass of methane (C2H6) = [(12 ×2) + (1× 6)] = 30 g/mol

n = 338 𝑔


= 11.3 mol of ethane

Partial pressure of methane (𝑃) =𝑛𝑚𝑒𝑡 𝑕𝑎𝑛𝑒 × 𝑅𝑇

𝑉

=113 𝑚𝑜𝑙 × 0.0821 𝑎𝑡𝑚 𝐿 𝐾−1 𝑚𝑜𝑙 −1 ×295 𝐾

20 𝐿

= 136.8 atm

Partial pressure of methane (𝑃) =𝑛𝑒𝑡 𝑕𝑎𝑛𝑒 × 𝑅𝑇

𝑉


116

=11.3 𝑚𝑜𝑙 × 0.0821 𝑎𝑡𝑚 𝐿 𝐾−1 𝑚𝑜𝑙 −1 ×295 𝐾

20 𝐿

= 13.6 atm

Total pressure in the cylinder (PT) = Pmethane + Pethane

= (136 + 13.6)atm

= 149.4 𝑎𝑡𝑚 2.13. Gas Diffusion and Effusion 2.13.1. Gas Diffusion

A direct demonstration of random motion is provided by diffusion, the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. Despite the fact that molecular speeds are very great, the diffusion process takes a relatively long time to complete. For example, when a bottle of concentrated ammonia solution is opened at one end of a laboratory bench, it takes some time before a person at the other end of the bench can smell it. The reason is that a molecule experiences numerous collisions while moving from one end of the bench to the other. Thus, diffusion of gases always happens gradually, and not instantly as molecular speeds seem to suggest. Furthermore, because the root-mean square speed of a light gas is greater than that of a heavier gas, a lighter gas will diffuse through a certain space more quickly than will a heavier gas. Figure 2.5 illustrates gaseous diffusion.

In 1832 the Scottish chemist Thomas Graham found that under the same conditions of temperature and pressure, rates of diffusion for gases are inversely proportional to the square roots of their molar masses. This statement, now known as Graham’s law of diffusion, is expressed mathematically as

𝑟1

𝑟2=

𝑀2

𝑀1


117

Where r1 and r2 are the diffusion rates of gases 1 and 2, and M1 and M2 are their molar masses, respectively.

Figure 2.5: A demonstration of gas diffusion. NH3 gas (from a bottle containing aqueous ammonia) combines with HCl gas (from a bottle containing hydrochloric acid) to form solid NH4Cl. Because NH3 is lighter and therefore diffuses faster, solid NH4Cl first appears nearer the HCl bottle (on the right). 2.13.2. Gas Effusion

Whereas diffusion is a process by which one gas gradually mixes with another, effusion is the process by which a gas under pressure escapes from one compartment of a container to another by passing through a small opening. Figure 2.6 shows the effusion of a gas into a vacuum. Although effusion differs from diffusion in nature, the rate of effusion of a gas has the same form as Graham‘s law of diffusion [see Equation for diffusion of a gas]. A helium-filled rubber balloon deflates faster than an air-filled one because the rate of effusion through the pores of the rubber is faster for the lighter helium atoms than for the air molecules. Industrially, gas effusion is used to separate uranium isotopes in the forms of gaseous 235UF6 and 238UF6. By subjecting the gases to many stages of effusion, scientists were able to obtain highly enriched 235U isotope, which was used in the construction of atomic bombs during World War II.


118

Figure 2.6: Gas effusion. Gas molecules move from a high-pressure region (left) to a low-pressure one through a pinhole. Example 2.78. A flammable gas made up only of carbon and hydrogen is found to effuse through a porous barrier in 3.50 min. Under the same conditions of temperature and pressure, it takes an equal volume of chlorine gas 7.34 min to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what this gas might be. Solution Data given Time for effusion of chlorine = 7.34 min Time of effusion of unknown gas X = 3.5 min Molar mass of chlorine gas = 70.90 g/mol Molar mass of X = ? Strategy We find the rate of effusion of both gases and the use graham‘s law to find the molar mass of gas X

Rate of effusion of Cl2 = 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑐𝑕𝑙𝑜𝑟𝑖𝑛𝑒

7.34 𝑚𝑖𝑛

Rate of effusion of X = 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑋

3.5 𝑚𝑖𝑛


119

𝑟𝑎𝑟𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐶𝑙2

𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑋=

𝑀𝑋

𝑀𝐶𝑙2

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝐶𝑙2 7.34 𝑚𝑖𝑛

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑓 𝑋 3.5 𝑚𝑖𝑛

= 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝐶𝑙2

7.34 𝑚𝑖𝑛 ×

3.5 𝑚𝑖𝑛

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑋

= 3.5 𝑚𝑖𝑛

7.34 𝑚𝑖𝑛

From the molar mass of Cl2, we write

3.5 𝑚𝑖𝑛

7.34 𝑚𝑖𝑛 =

𝑀𝑋

70.90 𝑔/𝑚𝑜𝑙

𝑀𝑋 = (3.5 𝑚𝑖𝑛

7.34 𝑚𝑖𝑛 )2 × 70.90 g/mol

𝑀𝑋 = 16.1 𝑔/𝑚𝑜𝑙 Because the molar mass of carbon is 12.01 g and that of hydrogen is 1.008 g, the gas is methane (CH4). Note: Because lighter gases effuse faster than heavier gases, the molar mass of the unknown gas must be smaller than that of chlorine gas. Indeed, the molar mass of methane (16.04 g) is less than the molar mass of chlorine gas (70.90 g). Example 2.79. Methane effuses through a small opening in the side of a container at rate of 1.3 mols -1. An unknown gas X effuses through the same opening at the rate of 5.42 mols-1 when maintained at the same temperature and pressure as methane. Determine the molar mass of the unknown gas. [H= 1, C= 12]

Solution

Data provided

Rate of effusion of methane = 1.3 mols-1

Rate of effusion of gas X = 5.42 mols-1

Molar mass of methane = 16.0g/mol


120

Molar mass of gas X = ?

Use graham‘s law to find molar mass of gas X

𝑟𝑎𝑟𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐶𝐻4


𝑀𝑋

𝑀𝐶𝐻4

1.3 mols−1

5.42 mols−1=

𝑀𝑋

16.0

(2.3985)2 = 𝑀𝑋

16

𝑀𝑋 = 16 × 5.7528 = 92. 045 Therefore molar mass of gas X is 92.045g/mol Example 2.80. The time required for a volume of O2 to diffuse through opening is 40seconds. Calculate the molar mass of gas which requires 50seconds for the same volume to diffuse through the same opening under the same conditions [O = 16]. Solution Data provided Rate of diffusion of oxygen = 40s

Rate of diffusion of gas X = 50s

Molar mass of oxygen = 32.0g/mol

Molar mass of gas X = ?

Rate of diffusion of O2 = 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛

35𝑠

Rate of diffusion of X = 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑋

50𝑠

𝑟𝑎𝑟𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑂2


𝑀𝑋

𝑀𝑂2


121

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑂2 40𝑠

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑓 𝑋 50𝑠

= 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑂2

40𝑠 ×

50𝑠

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑋 =

50𝑠

40𝑠

From the molar mass of O2, we write

50𝑠

40𝑠 =

𝑀𝑋


𝑀𝑋 = (50𝑠

40𝑠 )2 × 32.0 g/mol

𝑀𝑋 = (1.25)2 × 32.0 𝑔/𝑚𝑜𝑙 𝑀𝑋 = 1.5625 × 32.0 𝑔/𝑚𝑜𝑙

𝑀𝑋 = 50.0 𝑔/𝑚𝑜𝑙

Example 2.81. If equal amounts of hydrogen and argon are placed in a porous container and allowed to escape, which gas will escape faster and why? Solution If equal amounts of hydrogen and argon are placed in a porous container and allowed to escape, hydrogen gas will effuse faster than argon because hydrogen is a lighter gas (with molecular mass of 2.0g/mol) than argon (with molecular mass of 39.95g/mol) and the rate of effusion varies inversely with molecular weight of a gas. Example 2.82. The time required for a volume of gas y to effuse through a small hole was 112.2seconds. The time required for the same volume of oxygen was 84.7seconds. Calculate the molecular weight of gas y. Solution Data provided Rate of effusion of oxygen = 48.7s

Rate of effusion of gas y = 112.2s


122

Molar mass of oxygen = 32.0g/mol

Molar mass of gas y = ?

Rate of effusion of O2 = 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛

48.5𝑠

Rate of effusion of y = 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑌

112.2𝑠

𝑟𝑎𝑟𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑂2

𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑌=

𝑀𝑌

𝑀𝑂2

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑂2 48.5𝑠

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑜𝑓 𝑌 112.2𝑠

= 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑂2

48.5𝑠 ×

112.2𝑠

𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑜𝑓 𝑌 =

112.2𝑠

48.5𝑠

From the molar mass of O2, we write

112.2𝑠

48.5𝑠 =

𝑀𝑌


𝑀𝑌 = (112.2𝑠

48.5𝑠 )2 × 32.0 g/mol

= (2.31)2 × 32.0 𝑔/𝑚𝑜𝑙 = 5.35 × 32.0 𝑔/𝑚𝑜𝑙

= 171.2 𝑔/𝑚𝑜𝑙

Example 2.83. Consider the reaction represented by the following equation:

State what would happen to the vapour density of N2O4 as the temperature of the system is increased. If the system is cooled, would


123

the gases become lighter or darker in colour? Explain your answer in each case. Solution The density becomes lighter as the temperature of the system increases. If the system is cooled, the product becomes lighter in colour. At low temperature, dinitrogen (IV) oxide (N2O4) predominates. Hence, the gases become lighter while at high temperature, N2O4 dissolves to nitrogen (IV) oxide (NO2) molecules and the gases becomes darker in colour and lighter in density.


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CHAPTER THREE

THE KINETIC MOLECULAR THEORY OF GASES

3.1 Introduction

The gas laws help us to predict the behaviour of gases, but they do not explain what happens at the molecular level to cause the changes we observe in the macroscopic world. For example, why does a gas expand on heating?

This section introduces the kinetic molecular theory of gases, which explains the gas laws and when extended, also explains some properties of liquids and solids.

Five postulates explain why gases behave as they do: 1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. 2. The volume of the molecules is insignificant compared with the volume occupied by the gas. 3. Forces between the molecules are negligible, except when the molecules collide with one another. 4. Molecular collisions are perfectly elastic; that is, no energy is lost when the molecules collide. 5. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy. The average kinetic energy of a molecule is given by

𝐾𝐸 = 12 𝑚𝑢2

Where m is the mass of the molecule and u is its speed. The horizontal

bar denotes an average value. The quantity 𝑢2 is called mean square speed; it is the average of the square of the speeds of all the molecules:

𝑢2 = 𝑢12 + 𝑢2

2 + − − − 𝑢𝑁2

Where N is the number of molecules. Assumption 5 enables us to write 𝐾𝐸 𝛼 𝑇

12 𝑚𝑢2 𝛼 𝑇


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𝐾𝐸 = 12 𝑚𝑢2 = 𝐶𝑇

Where C is the proportionality constant and T is the absolute temperature.

According to the kinetic molecular theory, gas pressure is the result of collisions between molecules and the walls of their container. It depends on the frequency of collision per unit area and on how ―hard‖ the molecules strike the wall. The theory also provides a molecular interpretation of temperature. According to Equation above, the absolute temperature of a gas is a measure of the average kinetic energy of the molecules. In other words, the absolute temperature is a measure of the random motion of the molecules—the higher the temperature, the more energetic the molecules. Because it is related to the temperature of the gas sample, random molecular motion is sometimes referred to as thermal motion. 3.2. Application to the Gas Laws

Although the kinetic theory of gases is based on a rather simple model, the mathematical details involved are very complex. However, on a qualitative basis, it is possible to use the theory to account for the general properties of substances in the gaseous state. The following examples illustrate the range of its utility: • Compressibility of Gases. Because molecules in the gas phase are separated by large distances (assumption 1), gases can be compressed easily to occupy less volume. • Boyle’s Law. The pressure exerted by a gas results from the impact of its molecules on the walls of the container. The collision rate, or the number of molecular collisions with the walls per second, is proportional to the number density (that is, number of molecules per unit volume) of the gas. Decreasing the volume of a given amount of gas increases its number density and hence its collision rate. For this reason, the pressure of a gas is inversely proportional to the volume it occupies; as volume decreases, pressure increases and vice versa. • Charles’s Law. Because the average kinetic energy of gas molecules is proportional to the sample‘s absolute temperature (assumption 5),


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raising the temperature increases the average kinetic energy. Consequently, molecules will collide with the walls of the container more frequently and with greater impact if the gas is heated, and thus the pressure increases. The volume of gas will expand until the gas pressure is balanced by the constant external pressure. • Avogadro’s Law. We have shown that the pressure of a gas is directly proportional to both the density and the temperature of the gas. Because the mass of the gas is directly proportional to the number of moles (n) of the gas, we can represent density by n/V. Therefore,

𝑃 ∝ 𝑛

𝑉 𝑇

For two gases, 1 and 2, we write

𝑃1 ∝ 𝑛1𝑇1

𝑉1 = 𝐶

𝑛1𝑇1

𝑉1

𝑃2 ∝ 𝑛2𝑇2

𝑉2 = 𝐶

𝑛2𝑇2

𝑉2

Where C is the proportionality constant. Thus, for two gases under the same conditions of pressure, volume, and temperature (that is, when P1 = P2, T1 = T2, and V1 = V2), it follows that n1 = n2, which is a mathematical expression of Avogadro‘s law. • Dalton’s Law of Partial Pressures. If molecules do not attract or repel one another (assumption 3), then the pressure exerted by one type of molecule is unaffected by the presence of another gas. Consequently, the total pressure is given by the sum of individual gas pressures. 3.3. Distribution of Molecular Speeds

The kinetic theory of gases enables us to investigate molecular motion in more detail. Suppose we have a large number of gas molecules, say, 1 mole, in a container. As long as we hold the temperature constant, the average kinetic energy and the mean square speed will remain unchanged as time passes. As you might expect, the motion of the molecules is totally random and unpredictable. At a given instant, how many molecules are moving at a particular speed? To answer this question Maxwell analyzed the behaviour of gas molecules at different temperatures.


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Figure 3.1(a) shows typical Maxwell speed distribution curves for nitrogen gas at three different temperatures. At a given temperature, the distribution curve tells us the number of molecules moving at a certain speed. The peak of each curve represents the most probable speed, that is, the speed of the largest number of molecules. Note that the most probable speed increases as temperature increases (the peak shifts toward the right). Furthermore, the curve also begins to flatten out with increasing temperature, indicating that larger numbers of molecules are moving at greater speed. Figure 3.1(b) shows the speed distributions of three gases at the same temperature. The difference in the curves can be explained by noting that lighter molecules move faster, on average, than heavier ones.

(a)


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(b) Figure 3.1 (a) The distribution of speeds for nitrogen gas at three different temperatures. At the higher temperatures, more molecules are moving at faster speeds. (b) The distribution of speeds for three gases at 300 K. At a given temperature, the lighter molecules are moving faster, on the average. 3.4. Root-Mean-Square Speed

How fast does a molecule move, on the average, at any temperature T? One way to estimate molecular speed is to calculate the root-mean-square (rms) speed (urms), which is an average molecular speed. One of the results of the kinetic theory of gases is that the total kinetic

energy of a mole of any gas equals 3

2𝑅𝑇. Earlier we saw that the

average kinetic energy of one molecule is 1 2 𝑚𝑢2 and so we can write

𝑁𝐴 12 𝑚𝑢2 =

3

2𝑅𝑇

Where NA is Avogadro‘s number and m is the mass of a single molecule. Because NAm = M, where M is the molar mass, this equation can be rearranged to give

𝑢2 = 3𝑅𝑇

𝑀

Taking the square root of both sides gives


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𝑢2 = 𝑢𝑟𝑚𝑠 = 3𝑅𝑇

𝑀 (3.1)

Equation (3.1) shows that the root-mean-square speed of a gas increases with the square root of its temperature (in kelvins). Because M appears in the denominator, it follows that the heavier the gas, the more slowly its molecules move. If we substitute 8.314 J/K-1 mol-1 for R and convert the molar mass to kg/mol, then urms will be calculated in meters per second (m/s). Example 3.1. Calculate the root-mean-square speeds of helium atoms and nitrogen molecules in m/s at 25°C. Strategy

To calculate the root-mean-square speed we use 𝑢𝑟𝑚𝑠 = 3𝑅𝑇

𝑀

expressed in m/s and convert temperature to kelvin Solution To calculate urms, the units of R should be 8.314 J/K-1 mol-1 and, because 1 J= 1 kg m2s-2, the molar mass must be in kg/mol. The molar mass of He is 4.003 g/mol, or 4.003 × 10-23 kg/mol. 25°C = (25 + 273)K = 298 K

From 𝑢𝑟𝑚𝑠 = 3𝑅𝑇

𝑀

= 3 8.314 𝐽𝐾−1𝑚𝑜𝑙 −1 × 298 𝐾

4.003 × 10−3 𝑘𝑔/𝑚𝑜𝑙

= 1.86 × 106 𝐽/𝑘𝑔 Use the conversion factor 1 J= 1 kg m2s-2 we get

= 1.86 × 106 m2/s2 = 1.36 × 103 m/s

The procedure is the same for N2, the molar mass of which is 28.02 g/mol, or 2.802 × 10-2 kg/mol so that we write

𝑢𝑟𝑚𝑠 = 3 8.314 𝐽𝐾−1𝑚𝑜𝑙 −1 × 298 𝐾

2.802 × 10−2 𝑘𝑔/𝑚𝑜𝑙


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= 2.65 × 105 m2/s2 = 515 m/s Example 3.2. Calculate the root-mean-square speed of molecular chlorine in m/s at 20°C. Solution

To calculate the root-mean-square speed we use 𝑢𝑟𝑚𝑠 = 3𝑅𝑇

𝑀

expressed in m/s and convert temperature to kelvin 20°C = (20 + 273)K = 293 K Molar mass of Cl2 = 71g/mol = 7.1 × 10-2 kg/mol

From 𝑢𝑟𝑚𝑠 = 3𝑅𝑇

𝑀

= 3 8.314 𝐽𝐾−1𝑚𝑜𝑙 −1 × 293 𝐾

7.1 × 10−2 𝑘𝑔/𝑚𝑜𝑙

= 1.03 × 105 m2/s2 = 320.8 m/s

The calculation in Example 3.1 has an interesting relationship to the composition of Earth‘s atmosphere. Unlike Jupiter, Earth does not have appreciable amounts of hydrogen or helium in its atmosphere. Why is this the case? A smaller planet than Jupiter, Earth has a weaker gravitational attraction for these lighter molecules. A fairly straightforward calculation shows that to escape Earth‘s gravitational field, a molecule must possess an escape velocity equal to or greater than 1.1 × 104 m/s. Because the average speed of helium is considerably greater than that of molecular nitrogen or molecular oxygen, more helium atoms escape from Earth‘s atmosphere into outer space. Consequently, only a trace amount of helium is present in our atmosphere. On the other hand, Jupiter, with a mass about 320 times greater than that of Earth, retains both heavy and light gases in its atmosphere.


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3.5. Ideal and Real Gases Any gas that obeys the gas laws at all temperatures and pressures is called an ideal or perfect gas. Hence, the gas the equation: PV = nRT is applicable to ideal gases only. Real gases only obey gas laws under normal conditions of temperature and pressure. 3.6. Differences between ideal gas and real gases

1. An ideal gas obeys the gas laws at all temperatures and pressure, while a real gas obeys the gas laws under normal conditions of temperature and pressure.

2. The actual volume of the molecules of an ideal gas is negligible compared with the volume of the container, while the actual volume of the molecules of a real gas is not negligible i.e. molecules of a real gas occupy space.

3. In an ideal gas there are no intermolecular attractions at all temperatures and pressures, while intermolecular attraction is strong and appreciable in a real gas at high pressure and low temperature.

3.7. Deviation from Ideal Behaviour The gas laws and the kinetic molecular theory assume that

molecules in the gaseous state do not exert any force, either attractive or repulsive, on one another. The other assumption is that the volume of the molecules is negligibly small compared with that of the container. A gas that satisfies these two conditions is said to exhibit ideal behaviour.

Although we can assume that real gases behave like an ideal gas, we cannot expect them to do so under all conditions. For example, without intermolecular forces, gases could not condense to form liquids. The important question is: Under what conditions will gases most likely exhibit nonideal behaviour?


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Figure 3.2 shows PV/RT plotted against P for three real gases and an ideal gas at a given temperature. This graph provides a test of ideal gas behaviour. According to the ideal gas equation (for 1 mole of gas), PV/RT equals 1, regardless of the actual gas pressure. (When n = 1, PV = nRT becomes PV = RT, or PV/RT = 1.) For real gases, this is true only at moderately low pressures (≤ 5 atm); significant deviations occur as pressure increases. Attractive forces operate among molecules at relatively short distances. At atmospheric pressure, the molecules in a gas are far apart and the attractive forces are negligible. At high pressures, the density of the gas increases; the molecules are much closer to one another. Intermolecular forces can then be significant enough to affect the motion of the molecules, and the gas will not behave ideally.

Another way to observe the nonideal behaviour of gases is to lower the temperature. Cooling a gas decreases the molecules‘ average kinetic energy, which in a sense deprives molecules of the drive they need to break from their mutual attraction.

To study real gases accurately, then, we need to modify the ideal gas equation, taking into account intermolecular forces and finite molecular volumes. Such an analysis was first made by the Dutch physicist J. D. van der Waals in 1873. Besides being mathematically simple, van der Waals‘s treatment provides us with an interpretation of real gas behaviour at the molecular level.


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Figure 3.2: Plot of PV/RT versus P of 1 mole of a gas at 0°C. For 1 mole of an ideal gas, PV/RT is equal to 1, no matter what the pressure of the gas is. For real gases, we observe various deviations from ideality at high pressures. At very low pressures, all gases exhibit ideal behavior; that is, their PV/RT values all converge to 1 as P approaches zero

Consider the approach of a particular molecule toward the wall of a container. The intermolecular attractions exerted by its neighbours tend to soften the impact made by this molecule against the wall. The overall effect is a lower gas pressure exerted by an ideal gas, Pideal, is related to the experimentally measured; that is, observed pressure, Pobs, by the equation


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Where a is a constant and n and V are the number of moles and volume of the gas, respectively. The correction term for pressure (an2/V2) can be understood as follows. The intermolecular interaction that gives rise to nonideal behavior depends on how frequently any two molecules approach each other closely. The number of such ―encounters‖ increases with the square of the number of molecules per unit volume, (n/V)2, because the presence of each of the two molecules in a particular region is proportional to n/V and so a is just a proportionality constant. The quantity Pideal is the pressure we would measure if there were no intermolecular attractions.

Another correction concerns the volume occupied by the gas molecules. In the ideal gas equation, V represents the volume of the container. However, each molecule does occupy a finite, although small, intrinsic volume, so the effective volume of the gas becomes (V ‒ nb), where n is the number of moles of the gas and b is a constant. The term nb represents the volume occupied by n moles of the gas.

Having taken into account the corrections for pressure and volume, we can rewrite the ideal gas equation as follows:

Equation (3.2), relating P, V, T, and n for a nonideal gas, is known as the van der Waals equation. The van der Waals constants a and b are selected to give the best possible agreement between Equation (3.2) and observed behaviour of a particular gas. Table 3.1 lists the values of a and b for a number of gases. The value of a indicates how strongly molecules of a given type of gas attract one another. We see that helium atoms have the weakest attraction for one another, because


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helium has the smallest a value. There is also a rough correlation between molecular size and b. Generally, the larger the molecule (or atom), the greater b is, but the relationship between b and molecular (or atomic) size is not a simple one.

Example 3.3. Given that 2.75 moles of CO2 occupy 4.70 L at 53°C, calculate the pressure of the gas (in atm) using (a) the ideal gas equation and (b) the van der Waals equation. Solution Strategy To calculate the pressure of CO2 using the ideal gas equation and van der Waals equation, we proceed by applying the ideal gas equation and then find the correction terms in van der Waals equation and substitute in the equation to get the pressure. (a).


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Data provided: V = 4.70 L T = 53°C = (53 + 273) K = 326 K n = 2.75 mol R = 0.0821 L atm/K -1 mol-1 Substituting these values in the ideal gas equation, we write 𝑃𝑉 = 𝑛𝑅𝑇

𝑃 =𝑛𝑅𝑇

𝑉

= 2.75 𝑚𝑜𝑙 ×0.0821𝐿 𝑎𝑡𝑚 𝐾−1𝑚𝑜𝑙 −1 ×326 𝐾

4.70 𝐿

= 15.7 atm (b). Applying equation 3.2, It is convenient to first calculate the correction terms in Equation separately. From Table 3.1, we have

a = 3.59 atm L2/mol2 b = 0.0427 L/mol

so that the correction terms for pressure and volume are

𝑎𝑛2

𝑉2=

3.59 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙−2 × (2.75 𝑚𝑜𝑙)2

(4.70 𝐿)2

= 1.23 atm 𝑛𝑏 = 2.75 𝑚𝑜𝑙 × (0.0427 L/mol)

= 0.117 L Finally, substituting these values in the van der Waals equation, we have

(P + 1.23 atm) (4.70 L ‒ 0.117 L)= (2.75 mol) (0.0821 L atm/K -1 mol-1) (326 K) (P + 1.23 atm) (4.688 L) =73.6 L 4.688P + 5.77 = 73.6 4.688P = 73.6 ‒ 5.77 4.688P = 67.83


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P = 14.5 atm Example 3.4. Using the data shown in Table 3.1, calculate the pressure exerted by 4.37 moles of nitrogen gas confined in a volume of 2.45 L at 38°C using van der Waals equation. Solution Data provided V = 2.45 L T =38°C = (38 + 273) K = 311 K n = 4.37 mol R = 0.0821 L atm/K -1 mol-1 Now let us apply equation 3.2, to find the pressure. From Table 3.1, we have

a = 1.39 atm L2/mol2 b = 0.0913 L/mol


𝑎𝑛2

𝑉2=

1.39 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙−2 × (4.37 𝑚𝑜𝑙)2

(2.45 𝐿)2

= 4.4 atm 𝑛𝑏 = 4.37 𝑚𝑜𝑙 × (0.0913 L/mol)


(P + 4.4 atm) (2.45 L ‒ 0.399 L)= (4.37 mol) (0.0821 L atm/K -1 mol-1) (311 K) P = 50. 0 atm


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Example 3.5. Using (a) the ideal gas law equation and (b) Van der Waal‘s eqauation, cal culate the pressure exerted by 50.0g of carbon (IV) oxide in 1.00 L vessel at 25oC. [Find values of a and b in table 3.1] Solution Convert the mass of CO2 to mole and temperature to Kelvin. Data provided a = 3.592 atm L2/mol2 b = 0.043 L/mol V = 1.00 L T =25°C = (25 + 273) K = 298 K R = 0.0821 L atm/K -1 mol-1

Mass of CO2 = 50.0g

Number of mole of CO2 (n) = 𝑚𝑎𝑠𝑠


But molar mass of CO2 = 12 + (16 × 2) = 44 g/mol

∴ (n) = 50 𝑔

44 𝑔/𝑚𝑜𝑙= 1.136 𝑚𝑜𝑙

(a). Let us substitute these values in the ideal gas equation, we write 𝑃𝑉 = 𝑛𝑅𝑇

𝑃 =𝑛𝑅𝑇

𝑉

= 1.136 𝑚𝑜𝑙 ×0.0821𝐿 𝑎𝑡𝑚 𝐾−1𝑚𝑜𝑙 −1 ×298 𝐾

1.00 𝐿

= 27.78 atm (b). Let us apply equation 3.2, to find the pressure.


𝑎𝑛2

𝑉2=

3.592 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙−2 × (1.136 𝑚𝑜𝑙)2

(1.00 𝐿)2

= 4.635 atm 𝑛𝑏 = 1.136 𝑚𝑜𝑙 × (0.043 L/mol)



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(P + 4.635 atm) (1.00 L ‒ 0.0488 L) = (1.136 mol) (0.0821 L atm/K -1 mol-1) (298 K) P = 24.61 atm Example 3.6. Oxygen is supplied to hospital and chemical laboratories in large steel cylinders. Typically, such a cylinder has an internal volume of 28.0 litres and contains 6.80kg of oxygen. Use Van der Waal‘s equation to estimate the pressure inside such cylinder at 20oC. Solution Data provided a = 1.36 atm L2/mol2 b = 0.032 L/mol V = 28.0 L T =20°C = (20 + 273) K = 293 K R = 0.0821 L atm/K -1 mol-1

Mass of O2 = 6.8kg = 6800g

Number of mole of O2 (n) = 𝑚𝑎𝑠𝑠


But molar mass of O2 = (16 × 2) = 32 g/mol

∴ (n) = 6800 𝑔

32 𝑔/𝑚𝑜𝑙= 212.5 𝑚𝑜𝑙

Let us apply Van der Wall‘s equation to find the pressure.


𝑎𝑛2

𝑉2=

1.36 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙−2 × (212.5 𝑚𝑜𝑙)2

(28.00 𝐿)2

= 78.3 atm 𝑛𝑏 = 212.5 𝑚𝑜𝑙 × (0.032 L/mol)

= 6.8 L


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Finally, substituting these values in the van der Waals equation, we have

(P + 78.3 atm) (28 L ‒ 6.8 L)= (212.5 mol) (0.0821 L atm/K -1 mol-1) (293 K) P = 162.82 atm Example 3.7. Use Van der Waal‘s equation, calculate the pressure exerted by 1 mol of ammonia at 0OC in a volume of

(a) 1 .0 litre and (b) 0.05 litre

Solution Use Van der Waal‘s equation. Convert temperatures to kelvins. Data provided a = 4.17 atm L2/mol2 b = 0.037 L/mol R = 0.0821 L atm/K -1 mol-1

(a) V = 1.0 L T =0°C = (0 + 273) K = 273 K n = 1 mol Applying Van der Waal‘s equation to find the pressure;

But for 1 mol, the equation is reduced to

𝑃 + 𝑎

𝑉2 𝑉 − 𝑏 = 𝑅𝑇

Substituting data into above equation;

𝑃 + 4.17 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙 −2

(1.00 𝐿)2 1.00 𝐿 − 0.037 𝐿/𝑚𝑜𝑙 =

0.0821 L atmK−1 mol−1 × 273 K 𝑃 + 4.17 𝑎𝑡𝑚𝑚𝑜𝑙−1 0.963 𝐿/𝑚𝑜𝑙 = 22.4 L atm


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0.963𝑃 𝐿/𝑚𝑜𝑙 + 4. 𝐿 𝑎𝑡𝑚 = 22.4 L atm 0.963𝑃𝐿/𝑚𝑜𝑙 = (22.4 − 4 )𝐿𝑎𝑡𝑚 0.963𝑃𝐿/𝑚𝑜𝑙 = 18.4 𝐿𝑎𝑡𝑚

𝑃 = 18 𝐿𝑎𝑡𝑚

0.963 𝐿/𝑚𝑜𝑙

= 19 𝑎𝑡𝑚 𝑚𝑜𝑙−1 (b) V = 0.05 L T =0°C = (0 + 273) K = 273 K n = 1.0 mol Applying Van der Waal‘s equation to find the pressure;

But for 1 mol, the equation is reduced to

𝑃 + 𝑎

𝑉2 𝑉 − 𝑏 = 𝑅𝑇

Substituting data into above equation;

𝑃 + 4.17 𝑎𝑡𝑚 𝐿2 𝑚𝑜𝑙 −2

(0.05 𝐿)2 0.05 𝐿 − 0.037 𝐿/𝑚𝑜𝑙 =

0.0821 L atmK−1 mol−1 × 273 K 𝑃 + 1668 𝑎𝑡𝑚𝑚𝑜𝑙−1 0.013 𝐿/𝑚𝑜𝑙 = 22.4 L atm 0.013𝑃 𝐿/𝑚𝑜𝑙 + 21.68 𝐿 𝑎𝑡𝑚 = 22.4 L atm 0.013𝑃 𝐿/𝑚𝑜𝑙 = (22.4 − 21.68 )𝐿𝑎𝑡𝑚 0.013𝑃 𝐿/𝑚𝑜𝑙 = 0.716 𝐿𝑎𝑡𝑚

𝑃 = 0.716 𝐿𝑎𝑡𝑚

0.013𝑃 𝐿/𝑚𝑜𝑙𝑙

= 55.0 𝑎𝑡𝑚 𝑚𝑜𝑙−1 3.8. Intermolecular Forces Intermolecular forces are attractive forces between molecules. Intermolecular forces are responsible for the nonideal behaviour of gases. They exert even more influence in the condensed phases of matter—liquids and solids. As the temperature of a gas drops, the average kinetic energy of its molecules decreases. Eventually, at a sufficiently low temperature, the molecules no longer have enough energy to break away from the attraction of neighbouring molecules.


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At this point, the molecules aggregate to form small drops of liquid. This transition from the gaseous to the liquid phase is known as condensation.

In contrast to intermolecular forces, intramolecular forces hold atoms together in a molecule. Intramolecular forces stabilize individual molecules, whereas intermolecular forces are primarily responsible for the bulk properties of matter (for example, melting point and boiling point).

Generally, intermolecular forces are much weaker than intramolecular forces. Much less energy is usually required to evaporate a liquid than to break the bonds in the molecules of the liquid. For example, it takes about 41 kJ of energy to vaporize 1 mole of water at its boiling point; but about 930 kJ of energy are necessary to break the two O‒H bonds in 1 mole of water molecules. The boiling points of substances often reflect the strength of the intermolecular forces operating among the molecules.

At the boiling point, enough energy must be supplied to overcome the attractive forces among molecules before they can enter the vapour phase. If it takes more energy to separate molecules of substance A than of substance B because A molecules are held together by stronger intermolecular forces, then the boiling point of A is higher than that of B. The same principle applies also to the melting points of the substances. In general, the melting points of substances increase with the strength of the intermolecular forces.

To discuss the properties of condensed matter, we must understand the different types of intermolecular forces. Dipole-dipole, dipole-induced dipole, and dispersion forces make up what chemists commonly refer to as van der Waals forces, after the Dutch physicist Johannes van der Waals. Ions and dipoles are attracted to one another by electrostatic forces called ion-dipole forces, which are not van der Waals forces. Hydrogen bonding is a particularly strong type of dipole-dipole interaction. Because only a few elements can participate in hydrogen bond formation, it is treated as a separate category. Depending on the phase of a substance, the nature of chemical bonds,


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and the types of elements present, more than one type of interaction may contribute to the total attraction between molecules, as we will see below. 3.8.1. Dipole-Dipole Forces

Dipole-dipole forces are attractive forces between polar molecules, that is, between molecules that possess dipole moments. Their origin is electrostatic, and they can be understood in terms of Coulomb‘s law. The larger the dipole moment, the greater the force. 3.8.2. Ion-Dipole Forces

Coulomb‘s law also explains ion-dipole forces, which attract an ion (either a cation or an anion) and a polar molecule to each other. The strength of this interaction depends on the charge and size of the ion and on the magnitude of the dipole moment and size of the molecule. The charges on cations are generally more concentrated, because cations are usually smaller than anions. Therefore, a cation interacts more strongly with dipoles than does an anion having a charge of the same magnitude. 3.8.3. Dispersion Forces What attractive interaction occurs in nonpolar substances? To answer this question, let us consider the arrangement shown in Figure 3.2 below.


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Figure 3.2: (a) Spherical charge distribution in a helium atom. (b) Distortion caused by the approach of a cation. (c) Distortion caused by the approach of a dipole. If we place an ion or a polar molecule near an atom (or a nonpolar molecule), the electron distribution of the atom (or molecule) is distorted by the force exerted by the ion or the polar molecule, resulting in a kind of dipole. The dipole in the atom (or nonpolar molecule) is said to be an induced dipole because the separation of positive and negative charges in the atom (or nonpolar molecule) is due to the proximity of an ion or a polar molecule. The attractive interaction between an ion and the induced dipole is called ion-induced dipole interaction, and the attractive interaction between a polar molecule and the induced dipole is called dipole-induced dipole interaction. The likelihood of a dipole moment being induced depends not only on the charge on the ion or the strength of the dipole but also on the polarizability of the atom or molecule—that is, the ease with which the electron distribution in the atom (or molecule) can be distorted. Generally, the larger the number of electrons and the more diffuse the electron cloud in the atom or molecule, the greater its polarizability. By diffuse cloud we mean an electron cloud that is spread over an appreciable volume, so that the electrons are not held tightly by the nucleus.

Figure 3.3: Induced dipoles interacting with each other. Such patterns exist only momentarily; new arrangements are formed in the next instant. This type of interaction is responsible for the condensation of nonpolar gases.

Polarizability allows gases containing atoms or nonpolar molecules (for example, He and N2) to condense. In a helium atom, the electrons are moving at some distance from the nucleus. At any instant


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it is likely that the atom has a dipole moment created by the specific positions of the electrons. This dipole moment is called an instantaneous dipole because it lasts for just a tiny fraction of a second. In the next instant, the electrons are in different locations and the atom has a new instantaneous dipole, and so on. Averaged over time (that is, the time it takes to make a dipole moment measurement), however, the atom has no dipole moment because the instantaneous dipoles all cancel one another. In a collection of He atoms, an instantaneous dipole of one He atom can induce a dipole in each of its nearest neighbors (Figure 3.3). At the next moment, a different instantaneous dipole can create temporary dipoles in the surrounding He atoms. The important point is that this kind of interaction produces dispersion forces, attractive forces that arise as a result of temporary dipoles induced in atoms or molecules. At very low temperatures (and reduced atomic speeds), dispersion forces are strong enough to hold He atoms together, causing the gas to condense. The attraction between nonpolar molecules can be explained similarly.

A quantum mechanical interpretation of temporary dipoles was provided by the German physicist Fritz London in 1930. London showed that the magnitude of this attractive interaction is directly proportional to the polarizability of the atom or molecule. As we might expect, dispersion forces may be quite weak. This is certainly true for helium, which has a boiling point of only 4.2 K, or 2269°C. (Note that helium has only two electrons, which are tightly held in the 1s orbital. Therefore, the helium atom has a low polarizability.)

Dispersion forces, which are also called London forces, usually increase with molar mass because molecules with larger molar mass tend to have more electrons, and dispersion forces increase in strength with the number of electrons. Furthermore, larger molar mass often means a bigger atom whose electron distribution is more easily disturbed because the outer electrons are less tightly held by the nuclei. Table 3.2 compares the melting points of similar substances that consist of nonpolar molecules. As expected, the melting point increases as the number of electrons in the molecule increases. Because these are all


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nonpolar molecules, the only attractive intermolecular forces present are the dispersion forces.

In many cases, dispersion forces are comparable to or even greater than the dipole-dipole forces between polar molecules. For a dramatic illustration, let us compare the boiling points of CH3F (278.4°C) and CCl4 (76.5°C). Although CH3F has a dipole moment of 1.8 D, it boils at a much lower temperature than CCl4, a nonpolar molecule. CCl4 boils at a higher temperature simply because it contains more electrons. As a result, the dispersion forces between CCl4 molecules are stronger than the dispersion forces plus the dipole-dipole forces between CH3F molecules. (Keep in mind that dispersion forces exist among species of all types, whether they are neutral or bear a net charge and whether they are polar or nonpolar.)

Table 3.2 melting points of similar Nonpolar compounds Compound Melting point (OC) CH4 ‒182.5 CF4 ‒150.0 CCl4 ‒23.0 CBr4 90.0 Cl4 171.0 Example 3.8. What type(s) of intermolecular forces exist between the following pairs: (a) HBr and H2S, (b) Cl2 and CBr4, (c) I2 and NO3

‒, (d) NH3 and C6H6? Strategy Classify the species into three categories: ionic, polar (possessing a dipole moment), and nonpolar. Keep in mind that dispersion forces exist between all species.


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Solution (a) Both HBr and H2S are polar molecules.

Therefore, the intermolecular forces present are dipole-dipole forces, as well as dispersion forces.

(b) Both Cl2 and CBr4 are nonpolar, so there are only dispersion forces between these molecules.

(c) I2 is a homonuclear diatomic molecule and therefore

nonpolar, so the forces between it and the ion NO3‒ are ion-

induced dipole forces and dispersion forces.

(d) NH3 is polar, and C6H6 is nonpolar. The forces are dipole-induced dipole forces and dispersion forces.

3.9. The Hydrogen Bond

Normally, the boiling points of a series of similar compounds containing elements in the same periodic group increase with increasing molar mass. This increase in boiling point is due to the


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increase in dispersion forces for molecules with more electrons. Hydrogen compounds of Group 4A follow this trend, as Figure 3.4 shows. The lightest compound, CH4, has the lowest boiling point, and the heaviest compound, SnH4, has the highest boiling point. However, hydrogen compounds of the elements in Groups 5A, 6A, and 7A do not follow this trend. In each of these series, the lightest compound (NH3, H2O, and HF) has the highest boiling point, contrary to our expectations based on molar mass. This observation must mean that there are stronger intermolecular attractions in NH3, H2O, and HF, compared to other molecules in the same groups. In fact, this particularly strong type of intermolecular attraction is called the hydrogen bond, which is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as N‒ H, O‒ H, or F‒ H, and an electronegative O, N, or F atom. The interaction is written

A‒H •••• B or A‒H ••• A A and B represent O, N, or F; A‒H is one molecule or part of a molecule and B is a part of another molecule; and the dotted line represents the hydrogen bond. The three atoms usually lie in a straight line, but the angle AHB (or AHA) can deviate as much as 30° from linearity. Note that the O, N, and F atoms all possess at least one lone pair that can interact with the hydrogen atom in hydrogen bonding.

The average energy of a hydrogen bond is quite large for a dipole-dipole interaction (up to 40 kJ/mol). Thus, hydrogen bonds have a powerful effect on the structures and properties of many compounds. Figure 3.5 shows several examples of hydrogen bonding.


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Figure 3.4: Boiling points of the hydrogen compounds of Groups 4A, 5A, 6A, and 7A elements. Although normally we expect the boiling point to increase as we move down a group, we see that three compounds (NH3, H2O, and HF) behave differently. The anomaly can be explained in terms of intermolecular hydrogen bonding.


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Figure 3.5: Hydrogen bonding in water, ammonia, and hydrogen fluoride. Solid lines represent covalent bonds, and dotted lines represent hydrogen bonds.

The strength of a hydrogen bond is determined by the coulombic interaction between the lone-pair electrons of the electronegative atom and the hydrogen nucleus. For example, fluorine is more electronegative than oxygen, and so we would expect a stronger hydrogen bond to exist in liquid HF than in H2O. In the liquid phase, the HF molecules form zigzag chains:

The boiling point of HF is lower than that of water because each H2O takes part in four intermolecular hydrogen bonds. Therefore, the forces holding the molecules together are stronger in H2O than in HF. Example 3.9. Which of the following can form hydrogen bonds with water? CH3OCH3, CH4, F

‒, HCOOH, Na+.


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Strategy A species can form hydrogen bonds with water if it contains one of the three electronegative elements (F, O, or N) or it has an H atom bonded to one of these three elements. Solution There are no electronegative elements (F, O, or N) in either CH4 or Na+. Therefore, only CH3OCH3, F‒ and, HCOOH can form hydrogen bonds with water.

The intermolecular forces discussed so far are all attractive in

nature. Keep in mind, though, that molecules also exert repulsive forces on one another. Thus, when two molecules approach each other, the repulsion between the electrons and between the nuclei in the molecules comes into play. The magnitude of the repulsive force rises very steeply as the distance separating the molecules in a condensed phase decreases. This is the reason that liquids and solids are so hard to compress. In these phases, the molecules are already in close contact with one another, and so they greatly resist being compressed further. Example 3.10. What intermolecular forces besides dispersion forces, if any, exist in each substance? Are any of these substances solids at room temperature? 1. potassium chloride (KCl) 2. ethanol (C2H5OH) 3. bromine (Br2)


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Solution 1. Potassium chloride is composed of ions, so the intermolecular interaction in potassium chloride is ionic forces. Because ionic interactions are strong, it might be expected that potassium chloride is a solid at room temperature. 2. Ethanol has a hydrogen atom attached to an oxygen atom, so it would experience hydrogen bonding. If the hydrogen bonding is strong enough, ethanol might be a solid at room temperature, but it is difficult to know for certain. (Ethanol is actually a liquid at room temperature.) 3. Elemental bromine has two bromine atoms covalently bonded to each other. Because the atoms on either side of the covalent bond are the same, the electrons in the covalent bond are shared equally, and the bond is a nonpolar covalent bond. Thus, diatomic bromine does not have any intermolecular forces other than dispersion forces. It is unlikely to be a solid at room temperature unless the dispersion forces are strong enough. Bromine is a liquid at room temperature. 3.10. Intermolecular Forces at Low Temperature, High Molecular Weight, and High Pressure

At low temperatures where the gas molecules have lower kinetic energies, the contribution for attractive forces increases, which the ideal gas law does not account for.

The universal attractive force, or London dispersion force, also generally increases with molecular weight. The London dispersion force is caused by correlated movements of the electrons in interacting molecules. Electrons that belong to different molecules start "fleeing" and avoiding each other at the short intermolecular distances, which is frequently described as formation of "instantaneous dipoles" that attract each other.

Finally, as a gas is compressed and pressure increases, repulsive forces from the gas molecules oppose the decrease in


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volume. The frequency of collisions also increases at higher pressure, thereby increasing the contribution of these intermolecular forces.

3.11. The Mean Free Path of Gas Molecules

The motion of a molecule in a gas is complicated. Besides colliding with the walls of the confinement vessel, the molecules collide with each other. A useful parameter to describe this motion is the mean free path. The mean free path is the average distance traversed by a molecule between collisions. The mean free path of a molecule is related to its size; the larger its size the shorter its mean free path.

Suppose the gas molecules are spherical and have a diameter d. Two gas molecules will collide if their centers are separated by less than 2d. Suppose the average time between collisions is ∆t. During this time, the molecule travels a distance v . ∆t, and sweeps a volume equal to see diagrams below


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If on average it experiences one collision, the number of molecules in the volume V must be 1. If N is the number of molecules per unit volume, this means that

or

The time interval ∆t defined in this manner is the mean time between collisions, and the mean free path is given by

Here we have assumed that only one molecule is moving while all others are stationary. If we carry out the calculation correctly (all molecules moving), the following relation is obtained for the mean free path:

The Number Density

For N1 stationary particles, the number of molecules per unit volume

V

N

V

NNd

1


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Collision Frequency

Z1 = 𝜋𝑑2𝑁𝑑

The Mean Collision Time

The mean collision time is average time elapsed between successive collisions.

coll 1

𝑍1 =

1

𝜋𝑑2𝑁𝑑

Factors affecting mean free path

1. Density: As gas density increases, the molecules become closer to each other. Therefore, they are more likely to run into each other, so the mean free path decreases.

2. Radius of molecule: increasing the radius of the molecules will decrease the space between them, causing them to run into each other thereby decreasing the mean free path.

3. Pressure, temperature, and other factors that affect density can indirectly affect mean free path.


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CHAPTER FOUR

THE MOLE CONCEPT

4.1 Introduction Most chemical experiments involve enormous numbers of

atoms or molecules. In order to estimate the quantities of various chemical substances, Chemists adopted a convenient concept involving collection of elementary units such as atoms, molecules, ions, or electrons of a chemical substance. This is concept is ―the mole‖.

This quantity is sometimes referred to as the chemical amount. In Latin mole means a "massive heap" of material. It is convenient to think of a chemical mole as such. Visualizing a mole as a pile of particles, however, is just one way to understand this concept. A sample of a substance has a mass, volume (generally used with gases), and number of particles that is proportional to the chemical amount (measured in moles) of the sample. For example, one mole of oxygen gas (O2 ) occupies a volume of 22.4 L at standard temperature and pressure (STP; 0°C and 1 atm), has a mass of 31.998 grams, and contains about 6.022 × 10 23 molecules of oxygen. Measuring one of these quantities allows the calculation of the others and this is frequently done in stoichiometry.

The mole is the amount of a chemical substance which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12; its symbol is "mol." When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

The term "mole" commonly is used to represent the number of molecules (or atoms) in a quantity of material; that is, one mole of molecules = Avogadro's number of molecules. In this sense, a mole is a dimensionless number, just as a dozen means 12. For example, you could talk about a mole of caterpillars, meaning 6.023 x 1023

http://www.chemistryexplained.com/knowledge/Kilogram.html


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caterpillars. However, the number is seldom useful except in talking about molecules or atoms. The metric prefixes commonly are used to give such units as millimoles, nanomoles, or picomoles. 4.2. Avogadro's Number

We have noted that one mole of a substance always contains a certain number of molecules (or atoms), regardless of the substance involved. This number is called Avogadro’s number (in honor of the scientist who first suggested the concept, long before the value of the number could be determined). The number represented as NA is 6.023 x 1023. We know that the number of particles (atoms or molecules) in 1 mole of a substance is 6.022×1023 atoms or molecules. We can therefore say: 1 mole of carbon atoms weighs 12.0 g and contains 6.023 x 1023 atoms 1 mole of sodium atoms weighs 23.0 g and contains 6.023 x 1023 atoms 4.3. Calculations Using Mole Concept

Example 4.1. How many molecules are there in 20.0 g of benzene, C6H6? Solution:

First find how many moles of C6H6 there are in 20.0 g, then use

Avogadro's number to find the number of molecules.

Molar mass of C6H6 = (6 x 12.0) + (6 x 1.0)

= 78.0 g/mol

Moles of C6H6 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶6𝐻6

𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶6𝐻6

= 20 𝑔


= 0.256 𝑚𝑜𝑙𝑒𝑠

Number of molecules = moles × Avogadro‘s number

=(0.256 𝑚𝑜𝑙𝑒𝑠)(6.023 x 1023 molecules/mole)


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= 1.54 × 1023 molecules

Example 4.2. How many atoms are represented by 3.00 moles of calcium (Ca) ? Solution:

3.00 mol Ca × 6.023 x 1023

1 .00 mol Ca

= 1.81 x 1024 atoms Ca

Example 4.3. How many atoms are represented by 3 g of calcium (Ca) ?

[Ca = 40, Avogadro constant = 6.023 x 1023]

Solution

40 g (1 mol) of calcium contains 6.023 x 1023 atoms

1 g of calcium will contain 6.023 x 1023

40 atoms

∴ 3 g will contain 6.023 x 1023 × 3

40 atoms

= 4.5 × 1022 atoms

Example 4.4. What is the mass of 6.02 × 1024 atoms of magnesium many atoms are represented by 3 g of calcium (Ca) ? [Mg = 24, Avogadro constant = 6.023 x 1023] Solution 6.023 x 1023 atoms of magnesium weigh 24 g (1 mol)

∴ 6.02 × 1024 atoms weigh 6.02 x 1024 ×24

6.023 x 1023

= (24 × 10)g

= 240 g


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Example 4.5. How many moles are in 9.03 × 1023 atoms of sodium? [Na = 23, Avogadro constant = 6.023 x 1023] Solution Let y represent the amount of Na 6.023 x 1023 atoms ≡ 1 mole 9.03 × 1023 atoms ≡ y mole

y = 9.03 x 1023 × 1.0

6.023 x 1023

= 1.50 mol Example 4.6. Calculate the mass of sodium, which would contain the same number of atoms as 9.0 g of carbon? [Na = 23, C = 12] Solution Since equal amount of two or more elements contain the same number of atoms, then: 1 mole (12.0 g) of carbon contains the same number of atoms as 1 mole (23.0 g) of sodium. i.e. 12.0 g of carbon ≡ 23.0 g of sodium

∴ 9.0 g of carbon ≡ 9.0 × 23.0

12

= 30.0 g of sodium


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CHAPTER FIVE

GASES IN CHEMICAL REACTION 5.1. Introduction

Gases that are involved in chemical reactions obey the same laws of stoichiometry that apply to substances in any other state therefore, the ideal gas law can be used to calculate the quantities of gaseous substances involved in a reaction and then those results used to find the quantities of other substances. Figure 5.1 presents the conversions allowed by the ideal gas law to determine the number of moles of a gaseous reactant or product.

Figure 5.1: Mole Conversions, Including Application of the Ideal Gas Law to Determine the Number of Moles of a Gaseous Reactant or Product. Example 5.1. How many liters of oxygen gas at 21oC and 1.13 atm can be prepared by thermal decomposition of 0.950 g of KClO3? [K = 39, Cl = 35.5, O = 16]


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Solution We first find the number of mole of KClO3 from which the number of mole of oxygen can be determine through the balanced equation and finally apply ideal gas equation to find the volume of oxygen that can be prepared.

Solution

Data provided and necessary conversion

T = 21oC = (21 +273)K = 294K

P = 1.13 atm

Mass of KClO3 = 0.950 g

V = ?

Molar mass of KClO3 = [39 + 35.5 + (16× 3)] = 122.5 g/mol

Mole of KClO3 (n) = 𝑚𝑎𝑠𝑠


= 0.950 g

122.5 𝑔/𝑚𝑜𝑙

= 0.007787 mol of KClO3

The number of moles of O2 produced is

0.007787 mol of KClO3 (3 𝑚𝑜𝑙 𝑂2

2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂3) = 0.01168 mol O2

We can now use the ideal gas law equation:

𝑉 = 𝑛𝑅𝑇

𝑃 =

0.01168 mol 0.0821 𝐿 𝑎𝑡𝑚 𝐾−1𝑚𝑜𝑙 −1 (294 𝐾)

1.13 𝑎𝑡𝑚

= 0.249 L


162

Example 5.2. A chemist decomposes 1.06 g of Hg2O in a sealed system. The oxygen produced has a pressure of 0.514 atm and a volume of 62.5 mL at 35°C Calculate the value of R from these data. Solution Data provided and necessary conversion T = 35oC = (35 +273)K = 308K P = 0.514 atm Mass of Hg2O = 1.06 g V = 62.5 mL = (62.5/1000) L = 0.0625 L Molar mass of Hg2O = [ (200.5 × 2 )+ 16] = 417 g/mol

Mole of Hg2O (n) = 𝑚𝑎𝑠𝑠


= 1.06 g


= 0.00254 mol of Hg2O

The number of moles of O2 produced is

0.00254 mol of Hg2O (1 𝑚𝑜𝑙 𝑂2

2 𝑚𝑜𝑙 𝐻𝑔2𝑂) = 0.00127 mol O2

𝑅 = 𝑃𝑉

𝑛𝑇 =

0.514 atm (0.0625 L)

0.00127 𝑚𝑜𝑙 (308 𝐾)

= 0.0821 L atm K-1mol-1

Example 5.3. As N2 and H2 react to form NH3 in a large cylinder at 500°C, what happens to (a) the total number of atoms? (b) the total number of molecules? (c) the total pressure? Solution (a) The number of atoms stays the same, as is true for all reactions. That is the basis for the balanced chemical equation.


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(b) N2 (g) + 3H2 (g) → 2NH3 (g) The total number of moles of gas decreases as this reaction proceeds, so the number of molecules also decreases. (c) The total pressure decreases as the total number of moles of gas decreases. Example 5.4. How many litres of CO2 at STP can be prepared by the complete thermal decomposition of 0.150 mol of Ca(HCO3)2. The products are CaO, CO2 and H2O. Solution Data provided T = 273K P = 1.0 atm R = = 0.0821 L atm K-1mol-1 Mole of Ca(HCO3)2 = 0.150 mol V = ?

The number of moles of CO2 produced is

0.150 mol of Ca(HCO3)2 (2 𝑚𝑜𝑙 𝑂2

1 𝑚𝑜𝑙 𝐶𝑎(𝐻𝐶𝑂3)2) = 0.30 mol CO2

We can now use the ideal gas law equation:

𝑉 = 𝑛𝑅𝑇

𝑃 =

0.30 mol 0.0821 𝐿 𝑎𝑡𝑚 𝐾−1𝑚𝑜𝑙 −1 (273 𝐾)

1.0 𝑎𝑡𝑚

= 6.7249 L of CO2


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CHAPTER SIX

CHEMICAL EQUILIBRIUM IN GASES

6.1. Introduction Few chemical reactions proceed in only one direction. Most are,

at least to some extent, reversible. At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process—that is, the formation of reactant molecules from product molecules—begins to take place. When the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products no longer change with time, chemical equilibrium is reached. Chemical equilibrium is a dynamic process. As such, the rate of product formation and conversion back to reactant molecules is constant. Note that a chemical equilibrium reaction involves different substances as reactants and products. Equilibrium between two phases of the same substance is called physical equilibrium because the changes that occur are physical processes. The vaporization of water in a closed container at a given temperature is an example of physical equilibrium. In this instance, the number of H2O molecules leaving and the number returning to the liquid phase are equal: H2O(l) ⇌ H2O(g)

(The double arrow means that the reaction is reversible.) The study of physical equilibrium yields useful information,

such as the equilibrium vapour pressure. However, chemists are particularly interested in chemical equilibrium processes, such as the reversible reaction involving nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4). The progress of the reaction

N2O4(g) ⇌ 2NO2(g) can be monitored easily because N2O4 is a colorless gas, whereas NO2 has a darkbrown colour that makes it sometimes visible in polluted air. Suppose that a known amount of N2O4 is injected into an evacuated flask. Some brown colour appears immediately, indicating the


165

formation of NO2 molecules. The colour intensifies as the dissociation of N2O4 continues until eventually equilibrium is reached. Beyond that point, no further change in colour is observed. By experiment we find that we can also reach the equilibrium state by starting with pure NO2 or with a mixture of NO2 and N2O4. In each case, we observe an initial change in colour, caused either by the formation of NO2 (if the colour intensifies) or by the depletion of NO2 (if the colour fades), and then the final state in which the colour of NO2 no longer changes. Depending on the temperature of the reacting system and on the initial amounts of NO2 and N2O4, the concentrations of NO2 and N2O4 at equilibrium differ from system to system (Figure 6.1).

Figure 6.1: Change in the concentrations of NO2 and N2O4 with time, in three situations. (a) Initially only NO2 is present. (b) Initially only N2O4 is present. (c) Initially a mixture of NO2 and N2O4 is present. In each case, equilibrium is established to the right of the vertical line. The NO2–N2O4 System at 6.2. The Equilibrium Constant

Let us consider the following reversible reaction: aA + bB ⇌cC + d D

in which a, b, c, and d are the stoichiometric coefficients for the reacting species A, B, C, and D. The equilibrium constant for the reaction at a particular temperature is


166

𝐾 = [𝐶]𝑐 [𝐷]𝑑

[𝐴]𝑎 [𝐵]𝑏 ( 6.1)

Equation (6.1) is the mathematical form of the law of mass action. It relates the concentrations of reactants and products at equilibrium in terms of a quantity called the equilibrium constant. The equilibrium constant is defined by a quotient. The numerator is obtained by multiplying together the equilibrium concentrations of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation. The same procedure is applied to the equilibrium concentrations of reactants to obtain the denominator. This formulation is based on purely empirical evidence, such as the study of reactions like NO2–N2O4.

The equilibrium constant has its origin in thermodynamics, however, we can gain some insight into K by considering the kinetics of chemical reactions. Let us suppose that this reversible reaction occurs via a mechanism of a single elementary step in both the forward and reverse directions:

in which kf and kr are the rate constants for the forward and reverse directions, respectively. At equilibrium, when no net changes occur, the two rates must be equal:

ratef = rater Or

kf[A][B]2 = kr[AB2]


167

𝐾𝑓

𝐾𝑟=

[𝐴𝐵2]

𝐴 [𝐵]2

Because both kf and kr are constants at a given temperature, their ratio is also a constant, which is equal to the equilibrium constant Kc.

𝐾𝑓

𝐾𝑟= 𝐾𝑐 =

[𝐴𝐵2]

𝐴 [𝐵]2

So Kc is a constant regardless of the equilibrium concentrations of the reacting species because it is always equal to kf/kr, the quotient of two quantities that are themselves constant at a given temperature. Because rate constants are temperature-dependent, it follows that the equilibrium constant must also change with temperature.

Finally, we note that if the equilibrium constant is much greater than 1 (that is, K > 1), the equilibrium will lie to the right of the reaction arrows and favour the products. Conversely, if the equilibrium constant is much smaller than 1 (that is, K ˂ 1), the equilibrium will lie to the left and favour the reactants (Figure 6.2).

Figure 6.2: (a) At equilibrium, there are more products than reactants, and the equilibrium is said to lie to the right. (b) In the opposite situation, when there are more reactants than products, the equilibrium is said to lie to the left.


168

6.3. Ways of Expressing Equilibrium Constants To use equilibrium constants, we must express them in terms of

the reactant and product concentrations. Our only guidance is the law of mass action [Equation (6.1)]. However, because the concentrations of the reactants and products can be expressed in different units and because the reacting species are not always in the same phase, there may be more than one way to express the equilibrium constant for the same reaction. To begin with, we will consider reactions in which the reactants and products are in the same phase. 6.4. Homogeneous Equilibria

The term homogeneous equilibrium applies to reactions in which all reacting species are in the same phase. An example of homogeneous gas-phase equilibrium is the dissociation of N2O4. The equilibrium constant is

𝐾𝑐 = [𝑁𝑂2]2

[𝑁2𝑂4]

Note that the subscript in Kc denotes that the concentrations of the reacting species are expressed in moles per liter. The concentrations of reactants and products in gaseous reactions can also be expressed in terms of their partial pressures. At constant temperature the pressure P of a gas is directly related to the concentration in moles per liter of the gas; that is, P = (n/V)RT. Thus, for the equilibrium process

N2O4(g) ⇌ 2NO2(g)

We can write

𝐾𝑃 = 𝑃2𝑁𝑂2

𝑃𝑁2𝑂4

in which P𝑁𝑂2 and 𝑃𝑁2𝑂4 are the equilibrium partial pressures (in atmospheres) of NO2 and N2O4, respectively. The subscript in KP tells us that equilibrium concentrations are expressed in terms of pressure.

In general, Kc is not equal to Kp, because the partial pressures of reactants and products are not equal to their concentrations expressed


169

in moles per liter. A simple relationship between KP and Kc can be derived as follows. Let us consider this equilibrium in the gas phase:

aA(g) ⇌ bB(g)

in which a and b are stoichiometric coefficients. The equilibrium constant Kc is

𝐾𝑐 = [𝐵]𝑏

[𝐴]𝑎

and the expression for KP is

𝐾𝑝 = 𝑃𝑏𝐵

𝑃𝑎𝐴

in which PA and PB are the partial pressures of A and B. Assuming ideal gas behaviour, 𝑃𝐴𝑉 = 𝑛𝐴𝑅𝑇

𝑃𝐴 = 𝑛𝐴𝑅𝑇

𝑉

in which V is the volume of the container in liters. Also, 𝑃𝐵𝑉 = 𝑛𝐵𝑅𝑇

𝑃𝐵 = 𝑛𝐵𝑅𝑇

𝑉

Substituting these relations into the expression for KP, we obtain

Now both nA/V and nB/V have the units of moles per liter and can be replaced by [A] and [B], so that

in which


170

∆𝑛 = 𝑏 − 𝑎 = moles of gaseous products ‒ moles of gaseous reactants

Because pressure is usually expressed in atmospheres, the gas constant R is given by 0.0821 L atm/K mol, and we can write the relationship between KP and Kc as

𝐾𝑃 = 𝐾𝐶(0.0821T)∆𝑛 (6.2) In general, KP ≠ Kc except in the special case when ∆n = 0. In that case, Equation (5.2) can be written as 𝐾𝑃 = 𝐾𝐶(0.0821T)0 𝐾𝑃 = 𝐾𝐶 6.5. Equilibrium Constant and Units

Note that it is general practice not to include units for the equilibrium constant. In thermodynamics, the equilibrium constant is defined in terms of activities rather than concentrations. For an ideal system, the activity of a substance is the ratio of its concentration (or partial pressure) to a standard value, which is 1 M (or 1 atm). This procedure eliminates all units but does not alter the numerical parts of the concentration or pressure. Consequently, K has no units. Example 6.1. Write expressions for KP if applicable, for the following reversible reactions at equilibrium: (a). HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq)

(b). 2NO (g) + O2(g) ⇌ 2NO2 (g)

Strategy Keep in mind the following facts: (1) the KP expression applies only to gaseous reactions and (2) the concentration of solvent (usually water) does not appear in the equilibrium constant expression. Solution

(a) Because there are no gases present, KP does not apply.

(b) 𝐾𝑃 = 𝑃2𝑁𝑂2

𝑃2𝑁𝑂 𝑃𝑂2


171

Example 6.2 The equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine PCl5(g) ⇌ PCl3(g) + Cl2(g)

is found to be 1.05 at 250°C. If the equilibrium partial pressures of PCl5 and PCl3 are 0.973 atm and 0.548 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250°C. Strategy The concentrations of the reacting gases are given in atm, so we can express the equilibrium constant in KP. From the known KP value and the equilibrium pressures of PCl3 and PCl5, we can solve for P𝐶𝑙2. Solution Data given Partial pressures of PCl5 = 0.973 atm Partial pressures of PCl3 = 0.548 atm 𝐾𝑃 = 1.05 Partial pressures of Cl2 = ? First, we write KP in terms of the partial pressures of the reacting species

𝐾𝑃 = 𝑃𝑃𝐶𝑙 3 𝑃𝐶𝑙2

𝑃𝐶𝑙5

Knowing the partial pressures, we write

1.05 = 0.548 ( 𝑃𝐶𝑙2 )

(0.973)

𝑃𝐶𝑙2=

1.05 ×0.973

0.548

= 1.86 atm Example 6.3 The equilibrium constant KP for the reaction 2NO2 ⇌ 2NO + O2 (g)

is 158 at 1000 K. Calculate PO2 if PNO2 = 0.400 atm and PNO = 0.270 atm. Solution Data given


172

Partial pressures of NO2 = 0.400 atm Partial pressures of NO = 0.270 atm 𝐾𝑃 = 158 Partial pressures of O2 =?

𝐾𝑃 = 𝑃2

𝑁𝑂 𝑃𝑂2

𝑃2𝑁𝑂 2

158 = (0.270)2 × 𝑃𝑂2

(0.400)2

𝑃𝑂2 =

0.16 (158)

0.073

= 346 atm Example 6.4 Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) ⇌ CH3OH(g)

The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of KP at this temperature? Strategy Apply the relationship between Kc and KP is given by Equation (6.2). What is the change in the number of moles of gases from reactants to product? Recall that ∆n = moles of gaseous products ‒ moles of gaseous reactants. Convert temperature to kelvins Solution Data given Kc = 10.5

T = 220oC = (220 + 273)K = 493 K

∆n = (1‒3) = ‒2

Applying 𝐾𝑃 = 𝐾𝐶(0.0821T)∆𝑛

𝐾𝑃 = (10.5)(0.0821 × 493)-2

= 6.41× 10-3

Example 6.5 For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)


173

KP is 4.3 × 10-4 at 375°C. Calculate Kc for the reaction. Solution

Data given

Kp = 4.3 × 10-4

T = 375oC = (375 + 273)K = 648 K

∆n = (2‒4) = ‒2


= 𝐾𝐶 (0.0821 × 648)-2

4.3 × 10-4 = 𝐾𝐶 (3.5 × 10-4)

𝐾𝐶 = 4.3 × 10−4

3.5 × 10−4

= 1.2

6.6. Heterogeneous Equilibria A reversible reaction involving reactants and products that are in

different phases leads to a heterogeneous equilibrium. For example, when calcium carbonate is heated in a closed vessel, this equilibrium is attained:

CaCO3(s) ⇌ CaO(s) + CO2(g)

The two solids and one gas constitute three separate phases. At equilibrium, we might write the equilibrium constant in terms of partial pressure for gases as 𝐾𝑃 = 𝑃𝐶𝑂2

The equilibrium constant in this case is numerically equal to the pressure of CO2 gas, an easily measurable quantity. Example 6.6 Consider the following heterogeneous equilibrium:

CaCO3(s) ⇌ CaO(s) + CO2(g)

At 800°C, the pressure of CO2 is 0.236 atm. Calculate (a) KP and (b) Kc for the reaction at this temperature.


174

Strategy Remember that pure solids do not appear in the equilibrium constant expression. The relationship between KP and Kc is given by Equation (6.2). Solution (a) Using 𝐾𝑃 = 𝑃𝐶𝑂2

we write

𝐾𝑃 = 𝑃𝐶𝑂2

= 0.236

Applying 𝐾𝑃 = 𝐾𝐶(0.0821T)∆𝑛 Where T = 800°C = (800 + 273)K = 1073 K ∆n = 1 𝐾𝑃 = 0.236 0.236 = 𝐾𝐶 (0.0821 × 1073)1 0.236 = 𝐾𝐶 (88.09)

𝐾𝐶 = 0.236

88.09

= 2.68 × 10-3 Example 6.7 Consider the following equilibrium at 395 K:

NH4HS(s) ⇌ NH3(g) + H2S(g) The partial pressure of each gas is 0.265 atm. Calculate KP and Kc for the reaction. Solution

Data given

T = 395oC = (395 + 273)K = 670 K

∆n = 2

𝑃𝑁𝐻3= 0.265

𝑃𝐻2𝑆 = 0.265

Kp = ?

Kc = ?


175

𝐾𝑃 = 𝑃𝑁𝐻3 𝑃𝐻2𝑆

= 0.265 (0.265)

= 0.07


= 𝐾𝐶 (0.0821 × 670)2

0.07 = 𝐾𝐶 (3025)

𝐾𝐶 = 0.07

3025

= 2.3 × 10-5

Example 6.8 Starting with a 3: 1 mixture of H2 and N2 at 450.0°C, the equilibrium mixture is found to be 9.6% NH3, 22.6% N2, and 67.8% H2 by volume. The total pressure is 50.0 atm. Calculate KP and Kc.

The reaction is N2 + 3H2 ⇌ 2NH3. Solution: According to Dalton's law of partial pressures, the partial pressure of a gas in a mixture is given by the product of its volume fraction and the total pressure. Therefore the equilibrium pressure of each gas is 𝑃𝑁𝐻3

= (0.096)(50.0 atm) = 4.8 atm

𝑃𝑁2 = (0.226)(50.0 atm) =11.3 atm

𝑃𝐻2 = (0.678)(50.0 atm) = 33.9 atm

Total pressure = 50.0 atm T = 450.0°C = (450 + 273)K = 723 K ∆n = (2‒4) = ‒2 The equilibrium expression for the reaction is

𝐾𝑃 = 𝑃2

𝑁𝐻 3

𝑃3𝐻2 𝑃𝑁2

Substituting the data gives

= (4.80)2

(33.9)3 ×(11.3)

= 23.04

(440227 )


176

= 5.2 × 10-5

Applying 𝐾𝑃 = 𝐾𝐶(0.0821T)∆𝑛 to find KC

= 𝐾𝐶 (0.0821 × 723)-2

5.2 × 10-5 = 𝐾𝐶 (2.8 × 10-4)

𝐾𝐶 = 5.2 × 10−5

2.8 × 10−4

= 0.184 6.7. The Effect of Change in Partial Pressure of One Gas on Kp and Equilibrium Position

It often is important to know the yield of a chemical reaction—that is, the percentage of reactants converted to products. The following example shows how this yield may be calculated, and how conditions may be altered to increase the yield. Example 6.9 Kp = 54.4 at 355.0°C for the reaction H2 + I2 ⇌ 2HI. What percentage of I2 will be converted to HI if 0.20 mole each of H2 and I2 are mixed and allowed to come to equilibrium at 355.0°C and a total pressure of 0.50 atm? Solution: Assume that X moles each of H2 and I2 are used up in reaching equilibrium to give 2X moles of HI, in accordance with the chemical equation, leaving 0.20 - X moles each of H2 and I2. The partial pressure of each gas is given by the product of its mole fraction and the total pressure.

𝑃𝐻𝐼 =

2𝑋

0.4 (0.50 𝑎𝑡𝑚)


177

𝑃𝐻2= 𝑃𝐼2

= 0.2 −𝑋

0.4 (0.50 𝑎𝑡𝑚)

𝐾𝑃 = 𝑃2

𝐻𝐼

𝑃𝐻2 𝑃𝐼2

= [

2𝑋

0.4 (0.50 𝑎𝑡𝑚 )]2

[ 0.2 −𝑋

0.4 (0.50 𝑎𝑡𝑚 )]2

54.4 = (2𝑋

0.2 −𝑋)2

Taking the square root of each side, we obtain

7.4 = 2𝑋

0.2 −𝑋

2X = 7.4 (0.2 ‒ X) 2X = 1.48 ‒ 7.4X 9.4X = 1.48 X = 0.157 = moles of H2 and I2 used up

Percentage conversion (yield) = 0.157

0.200 × 100 = 78.5%

Example 6.10 What percentage of I2 will be converted to HI at equilibrium at 355.0°C, if 0.200 mole of I2 is mixed with 2.00 moles of H2 at total pressure of 0.50 atm? Solution: In this problem, it is advantageous first to assume that the large excess of H2 will use almost the entire amount of I2, leaving only X moles of it unused. In general, it is always advantageous to let X represent the smallest unknown entity because it often simplifies the mathematical solution. If X moles of I2 are not used, then 0.20 ‒ X moles are used. For every mole of I2 used up, one of H2 is used up, and two of HI are formed. Proceeding as in the last problem, the number of moles of each component at equilibrium is


178

The partial pressure of each component will be the mole fraction of each times the total pressure, as follows.

𝑃𝐻2=

1.80+ 𝑋

2.20 (0.50 𝑎𝑡𝑚)

𝑃𝐼2 =

𝑋

2.20 (0.50 𝑎𝑡𝑚)

𝑃𝐻𝐼 = 0.40 − 2𝑋

2.20 (0.50 𝑎𝑡𝑚)

When we substitute these partial pressures into the expression for Kp, we get an expression that will be tedious to solve unless we make a reasonable approximation: we assume that X is negligible in comparison with 0.20 and 1.80.

𝐾𝑃 = 54.4 = [

0.40 − 2𝑋

2.20 (0.50 )]2

1.80+ 𝑋

2.20 (0.50 ) [

𝑋

2.20 (0.50 )]

≅ (0.40)2

𝐼.80 (𝑋)

X = (0.40)2

𝐼.80 (54.4)

= 0.0016 moles of I2 not used

0.200 ‒ 0.0016 = 0.1984 moles of I2 used

Percentage of I2 used = 0.1984

0.200 × 100 = 99.2%

Note that the wise decision to let X = the amount of I2 not used instead of the amount of I2 that was used really did simplify the solution by making it possible to neglect X when added to or subtracted from larger numbers. If we had solved the quadratic equation instead, we


179

would have found that 99.197% of the I2 had been used up. This is a common method of simplifying a mathematical problem, and at the end you can always check to see whether your answer really is negligible compared to what you said it was. Many chemists say that if X is less than 10.0% of what it is added to or subtracted from, it is okay to neglect it. The preceding problem illustrates the fact that, although the value of Kp does not change with changes in concentration, the equilibrium position will change to use up part of the excess of any one reagent that has been added. In this problem, the large excess of H2 shifts the equilibrium position to the right, causing more of the I2 to be used up (99.2% compared to 78.5%) than when H2 and I2 are mixed in equal proportions. Advantage may be taken of this principle by using a large excess of a cheap chemical to convert the maximum amount of an expensive chemical to a desired product. In this case I2, the more expensive chemical, is made to yield more HI by using more of the cheaper H2. 6.8. The Percentage Decomposition of Gases

Many gases decompose into simpler ones at elevated temperatures, and it often is important to know the extent to which decomposition takes place. Example 6.11 Kp = 1.78 atm at 250.0°C for the decomposition reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). Calculate the percentage of PC15 that dissociates if 0.0500 mole of PC15 is placed in a closed vessel at 250.0°C and 2.00 atm pressure. Solution: Although you are told that you are starting with 0.0500 mole PCl5, this piece of information is not needed to find the percentage dissociation at the given pressure and temperature. If you were asked for the volume of the reaction vessel, then you would need to know the actual number of moles; otherwise not. To answer the question that is asked, it is simpler to just start with one mole (don't worry about the volume)


180

and assume that X moles of PCl5 dissociate to give X moles each of PCl3 and Cl2 and 1 - X moles of PCl5 at equilibrium.

The partial pressures are given by the mole fractions times the total pressure, and are substituted into the Kp expression, to give

𝐾𝑝 = 1.78 =

𝑋

1+𝑋 2.0 𝑎𝑡𝑚

𝑋

1+𝑋 2.0 𝑎𝑡𝑚

1 −𝑋

1+𝑋 0.50 𝑎𝑡𝑚

= 2𝑋2

1−𝑋 (1+𝑋)

1.78 = 2𝑋2

1− 𝑋2

1.78 ‒ 1.78X2 = 2X2

𝑋2 = 1.78

3.78= 0.478

𝑋 = 0.478

= 0.686 moles PCl5 dissociate

Percentage of PCl5 dissociated = 0.686

1.00 × 100 = 68.6%

This was not a difficult quadratic equation to solve but, even if it had been, it would not be possible to neglect X compared to 1.00; it is too large. If we had neglected X, we would have obtained the extremely erroneous answer of 94.3% dissociated. If Kp is very large (or very small), it means that the equilibrium position lies far to the right (or to


181

the left). In either of these cases it is possible to choose X so that it will be very small and amenable to a simplified math solution. The value of Kp for the PCl5 equilibrium is neither very large nor very small, and hence it never will be possible to neglect X.


182

CHAPTER SEVEN

WORK OF EXPANSION AND COMPRESSION OF GAS

7.1. Introduction

We have seen that work can be defined as force F multiplied by distance d: w = Fd

In thermodynamics, work has a broader meaning that includes mechanical work (for example, a crane lifting a steel beam), electrical work (a battery supplying electrons to light the bulb of a flashlight), and surface work (blowing up a soap bubble). In this section we will concentrate on mechanical work.

One way to illustrate mechanical work is to study the expansion or compression of a gas. Many chemical and biological processes involve gas volume changes. Breathing and exhaling air involves the expansion and contraction of the tiny sacs called alveoli in the lungs. Another example is the internal combustion engine of the automobile. The successive expansion and compression of the cylinders due to the combustion of the gasoline-air mixture provide power to the vehicle. Figure 6.1 shows a gas in a cylinder fitted with a weightless, frictionless movable piston at a certain temperature, pressure, and volume. As it expands, the gas pushes the piston upward against a constant opposing external atmospheric pressure P. The work done by the gas on the surroundings is

Work done = 𝑃 × 𝐴 × 𝑕

But 𝐴 × 𝑕 = 𝑉

Therefore 𝑤 = 𝑃∆𝑉

w = ‒P∆V (7.1)


183

Where ∆V, the change in volume, is given by Vf ‒ Vi. The minus sign in Equation (7.1) takes care of the sign convention for w. For gas expansion (work done by the system), ∆V > 0, so ‒P∆V is a negative quantity. For gas compression (work done on the system), ∆V ˂ 0, and ‒P∆V is a positive quantity. Note that ―‒P∆V‖ is often referred to as ―P-V‖ work.

Figure 7.1: The expansion of a gas against a constant external pressure (such as atmospheric pressure). The gas is in a cylinder fitted with a weightless movable piston. The work done is given by ‒P∆V. According to Equation (6.1), the units for work done by or on a gas are liters atmospheres. To express the work done in the more familiar unit of joules, we use the conversion factor 1 L . atm = 101.3 J Example 7.1. A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 1.2 atm. Strategy A simple sketch of the situation is helpful here:


184

The work done in gas expansion is equal to the product of the external, opposing pressure and the change in volume. Solution (a) Because the external pressure is zero, no work is done in the expansion:

w = ‒P∆V = ‒ (0)(6.0 ‒ 2.0) L = 0 (b) The external, opposing pressure is 1.2 atm, so

w = ‒P∆V = ‒ (1.2 atm) (6.0‒ 2.0) L

= ‒ 4.8 L . atm To convert the answer to joules, we write

w = ‒ 4.8 L . atm × 101.3𝐽

1 𝐿.𝑎𝑡𝑚

= ‒ 4.9 × 102 J Because this is gas expansion (work is done by the system on the surroundings), the work done has a negative sign. Example 7.2. A gas expands from 264 mL to 971 mL at constant temperature. Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 4.00 atm. Solution This is similar to above example. Convert volume to litre.


185

(a) Because the external pressure is zero, no work is done in the expansion:

w = ‒P∆V

= ‒ (0)(971 ‒ 264) mL

= ‒ (0)(527/1000) L

= 0

(b) The external, opposing pressure is 4.0 atm, so

w = ‒P∆V

= ‒ (4.0 atm) (971 ‒ 264) Ml

=‒ (4.0 atm) (971 ‒ 264)

=‒ (4.0 atm)(527/1000) L

= ‒ (4.0 atm)(0.527) L

= ‒ 2.108 L. atm

To convert the answer to joules, we write

w = ‒ 2.108 L . atm × 101.3𝐽

1 𝐿.𝑎𝑡𝑚

= ‒ 2.13 × 102 J Example 7.3. The work done when a gas is compressed in a cylinder is 387 J. During this process, there is a heat transfer of 152 J from the gas to the surroundings. Calculate the energy change for this process. Strategy Compression is work done on the gas, so what is the sign for w? Heat is released by the gas to the surroundings. Is this an endothermic or exothermic process? What is the sign for q? Solution To calculate the energy change of the gas, we need ∆U = q + w. Work of compression is positive and because heat is released by the gas, q is negative. Therefore, we have

∆U = q + w


186

= ‒152 J + 387 J = 235 J

As a result, the energy of the gas increases by 235 J. Example 7.4. A gas expands and does P-V work on the surroundings equal to 279 J. At the same time, it absorbs 216 J of heat from the surroundings. What is the change in energy of the system? Expansion is work done by system on surrounding, so what is the sign for w? Heat is absorbed from the surroundings. Is this an endothermic or exothermic process? What is the sign for q? Solution To calculate the energy change of the gas, we use ∆U = q + w. Work of expansion is negative and because heat is absorbed from the surroundings, q is positive. Therefore, we have ∆U = q + w

= 216 J ‒ 279 J = ‒ 63 J

7.2. Enthalpy

Internal energy of a system is a state function, and this property is used in the discussion of any change in the heat content of the chemical reaction. Consider the reaction:

2CO(g) +O2(g) → 2CO2(g)

The reactants CO and oxygen are the initial states of the atoms and the product CO2 gives the final state. Since internal energy is a state function, the energy change ∆E associated with the reaction depends only on the initial and final states but not on the path taken by the reaction.

The energy change, ∆E of each chemical reaction run at constant temperature is a measure of the relative bond strength of reactants and products. The expression for pressure – volume (P-V) work shows than ∆E can be measured using the equation


187

∆𝐸 = 𝑞 – 𝑤

= q – 𝑝𝑒𝑥𝑑𝑣𝑣2

𝑣1

For reaction at constant volume

V1=V2,

∆V = 0

Therefore ∆E = q – 0 (constant volume)

So ∆E = 𝑞𝑉

(Change in internal energy is equal to the heat absorbed by the system when the process occurs at constants volume).

Change in internal energy can easily be measured using closed vessel such as bomb calorimeter but most often than not, chemical reactions are run at constant pressure like using open vessels, instead of constant volume, such that heat absorbed by the system is neither equal to 𝑞𝑣 or ∆E.

We shall therefore develop another state function called the enthalpy which takes care of both the internal energy and work done by the system.

𝜕𝑈 = 𝜕𝑞 − 𝑃𝜕𝑉

∆𝑈 = 𝑞𝑃 –𝑃∆𝑉

∆𝑈 = 𝑈2 ‒ 𝑈1

∆𝑉 = 𝑉2 ‒ 𝑉1

Hence 𝑈2 ‒ 𝑈1 = 𝑞𝑝 –𝑃 (𝑉2 ‒ 𝑉1)

𝑈2 ‒ 𝑈1 = 𝑞𝑝 –𝑃𝑉2 + 𝑃𝑉1


188

𝑈2 + 𝑃𝑉2 = 𝑞𝑝 + 𝑈1 + 𝑃𝑉1

𝑈, 𝑃 𝑎𝑛𝑑 𝑉 are all state functions therefore 𝑈 + 𝑃𝑉 (internal energy plus work of expansion) will therefore give us another state function represented by 𝐻 (𝑈 + 𝑃𝑉) - enthalpy.

Therefore at constant pressure,

𝜕𝑞 = 𝑈2 + 𝑃𝑉2 − (𝑈1 + 𝑃𝑉1)

= 𝐻2 − 𝐻1

𝑞𝑝 = ∆𝐻

Addition of heat at constant pressure result to increase in enthalpy. The enthalpy change is equal to the heat absorbed only when the process is carried out at constant pressure. Because of equality between ∆H and 𝑞𝑝 , the enthalpy is often called heat content of a system.

For a process at constant pressure in which heat is evolved by the system to the surroundings, ∆H and 𝑞𝑝 are both negative. This is

because enthalpy of the final state is lower than the enthalpy of the initial state.

∆H = 𝐻𝑓 − 𝐻𝑖 , negative

This process is exothermic. There is rise in temperature, on the other hand when the system absorbs heat from the surroundings both ∆H and 𝑞𝑝 are the positive, the process is endothermic. There is fall in

temperature.

Our results so far show that

∆E = 𝑞𝑉

∆H = 𝑞𝑝

∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)


189

= ∆𝐸 + 𝑃2𝑉2 − 𝑃1𝑉1

∆H and ∆E differ only by the difference in the PV products of the final and initial states.

If gases are produced or consumed in a chemical reaction, ∆(𝑃𝑉) may be quite appreciable and ∆H and ∆E differ significantly.

For ideal gases reacting at constant temperature producing gases as the only products,

aA (g) + bB (g) → cC (g) + dD (g)

𝑃𝑉 (products) = (𝑐 + 𝑑)𝑅𝑇

𝑃𝑉 (reactants) = (𝑎 + 𝑏)𝑅𝑇

Therefore ∆(𝑃𝑉) is given by the expression

∆(𝑃𝑉) = [(𝑐 + 𝑑)𝑅𝑇 − (𝑎 + 𝑏)𝑅𝑇 ]

= ∆𝑛𝑅𝑇

The total number of moles of gaseous products minus the total number of moles of gaseous reactants is defined as ∆𝑛, change in number of moles

In general then we can say

∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)

= ∆𝐸 + ∆𝑛𝑅𝑇

Example 6.7. For decomposition of MgCO3(s) by the reaction,

MgCO3(s) → MgO(s) + CO2(g)

∆H = 108.78kJ at 900K and 1atm pressure. If the molar volume of MgCO3(s) is 0.028dm3 and that of MgO(g) is 0.011dm3, find ∆E


190

Solution:

The ∆(𝑃𝑉) is divided into two terms, that due to volume change in solids and that due to production of gas.

∆(𝑃𝑉) = ∆(𝑃𝑉) solids + ∆(𝑃𝑉)gas

∆(𝑃𝑉)s = p (0.011− 0.028) = − p × 0.017 negligible

Since 1mole of gas appear in the products, ∆n = 1

∆(𝑃𝑉)g = ∆nRT = 1×8.314×900 = 7.483kJ

∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)

∆𝐸 = ∆𝐻 − ∆(𝑃𝑉)

= (108.78 – 7.483) kJ

= 101.297 kJ

= 101.30 kJ

Example 7.8. Liquid water is vaporized at 100oC and 1.013bar. The heat of vaporization is 40.69 kJ mol-1. What are the values of

(i) Wrev per mole (ii) q per mole (iii) ∆E and (iv) ∆H?

Solution

(i) Assuming that water vapour is an ideal gas and that the volume of liquid water is negligible compared to steam.

𝑊𝑟𝑒𝑣 = + 𝑃∆𝑉 = + 𝑅𝑇

= + (8.314×10-3 kJ mol-1K-1)(373.15K)


191

= + 3.10 kJ mol-1

(ii) The heat of vapourization is 40.69kJmol-1 and since heat is absorbed, q has a positive value, q = 40.69 kJ mol-1

(iii) ∆𝐸 = 𝑞 − 𝑤 = (40.69 – 3.10) kJ mol-1 = 37.59 kJ mol-1

(iv) ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉) = (37.59+3.10) kJ mol-1 =40.69 kJ mol-1

Example 7.9. Calculate the work done on the surrounding when one mole of water is vaporized at 100oC and 1atm. Giving molar volume of liquid water as 18cm3 mol-1 and molar volume of steam as 24dm3 mol-1

at 1atm

Solution: Work done on the surrounding is = 𝑃∆𝑉 Molar volume of liquid water is 18cm3 mol-1 Molar volume of steam = 24dm3 mol-1

Therefore the volume of 1 mole of liquid water is negligible

1 atm = 1.01325× 105𝑁𝑚−2

24dm3 = 24× 10−3m3

Work done = 1.01325× 105𝑁𝑚−2 × 24× 10−3m3 mol-1

= 24.319 × 102 Nm mol-1

= 2.43 kJ mol-1

Example 7.10. If the enthalpy change for the process in question 3 above is 40.70 kJ mol-1, what is the change in internal energy ∆E?


192

Solution

From ∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)

∆𝐸 = ∆𝐻 − ∆(𝑃𝑉)

Where ∆H = 40.70 kJ mol-1

∆E = 40.70 kJ mol-1 – 2. 43 kJ mol-1

= 38.27 kJ mol-1

Example 7.11. For the combustion of benzene according to the equation

C6H6 (l ) + 71

2 O2 (g) → 6CO2 (g) + 3H2O (l)

If the heat of reaction at constant pressure is

∆H25oC = −3267.62 kJ mol-1, find ∆E

Solution:

There is contraction of gaseous volume from 7.5 to 6 moles of gas.

Hence ∆n = 6 − 7.5 = −1.5

∆𝐻 = ∆𝐸 + ∆(𝑃𝑉)

= ∆𝐸 + ∆𝑛𝑅𝑇

∆𝐸 = ∆𝐻 − ∆𝑛𝑅𝑇

= ∆H – (– 1.5× 8.314 × 298.2)J

= (– 3267.62 + 3.72) kJ

= – 3263.90 kJ


193

Example 7.12. To vaporize 100.0g of CCl4 at its normal boiling point, 349.0K and 1atm; 19.5kJ of heat is required. Calculate ∆Hvap for CCl4 and compare it with ∆E for the same process.

Solution:

Molar mass of CCl4 = 154.0g mol-1

Number of moles of CCl4 in 100g = 100𝑔

154𝑔 𝑚𝑜𝑙 −1

= 0.65 mol

For 0.65mol, enthalpy change = 19.5 kJ

For 1mol, enthalpy change will be 1 𝑚𝑜𝑙 𝐶𝐶𝑙4

0.65 𝑚𝑜𝑙× 19.5 kJ

= 30.0 kJ

∴ ∆𝐻vap = 30.0K

∆𝐻 = ∆𝐸 + ∆ 𝑃𝑉

∆𝐸 = ∆𝐻 − ∆(𝑃𝑉)

= ∆𝐻vap − ∆𝑛𝑅𝑇

∆n = 1 because CCl4 (l) → CCl4 (g)

Volume of CCl4 (l) is negligible compared to CCl4 (g)

For ∆n = 1 at 349.0K,

∆E = 30.0KJ – 1mol × 8.314JK-1mol-1) (349K)


194

= 30.0 kJ – 2.90 kJ

= 27.1 kJ

Thus of the 30.0KJ of energy transferred from the surroundings in the form of heat, 27.1KJ is used to increase the internal energy of the molecules ∆E and 2.9KJ is used to expand the resulting vapour, ∆(𝑃𝑉).

7.3. Heat Capacities of Gases The amount of heat expressed in joules, necessary to produce a standard change of 1oC in one gram of material is called specific heat. The product of the specific heat and the molar mass of a substance is the heat required to raise the temperature of one mole of that substance by 1oC. This is called the molar heat capacity and is a positive number which has the unit of joules per mole – degree (J.mol-1 K-1). Since the heat capacity C is the amount of heat needed to produce a temperature change of 1oC, it would appear that the quantity of heat required to produce a total temperature change, ∆T, is

q = C∆T = C(T2 – T1).

Heat is not a state function therefore it does not depend on the initial and final states only but on how the process is carried out. The equation above says nothing about the process of producing the temperature change. We can remove this vagueness by defining two molar heat capacities for gases, one Cp for processes at constant pressure the other Cv for processes at constant volume.

Thus Cp = 𝑞𝑝

∆𝑇 and Cv = 𝑞𝑣

∆𝑇

The equivalent definitions in terms of infinitesimal changes are:

Cp = 𝑑𝑞𝑝

𝑑𝑇 =

𝑑𝐻

𝑑𝑇 and


195

Cv = 𝑑𝑞𝑣

𝑑𝑇 =

𝑑𝐸

𝑑𝑇

From the above, we can find that the heat required to change the temperature of n moles of material from T1 to T2 is:

qp = 𝑛𝐶𝑝𝑑𝑇𝑇2

𝑇1 for process at constant pressure and

qv = 𝑛𝐶𝑣𝑑𝑇𝑇2

𝑇1 for process at constant volume

Near room temperature, 𝐶𝑝 and 𝐶𝑣 are constants independent of

temperature therefore:

qp = nCp (T2 − T1)

qv = nCv (T2 − T1)

For an ideal gas, there is a relationship between Cp and Cv. This relationship is shown easily by combing the definition of enthalpy with definitions of Cp and Cv. Thus for one mole of an ideal gas,

𝐻 = 𝐸 + 𝑃𝑉

Differentiating the above equation gives

𝑑𝐻 = 𝑑𝐸 + 𝑑(𝑃𝑉)

Dividing through by 𝑑𝑇

𝑑𝐻

𝑑𝑇 =

𝑑𝐸

𝑑𝑇 +

𝑑(𝑃𝑉)

𝑑𝑇

Cp = Cv + 𝑑(𝑃𝑉)

𝑑𝑇 (Cp =

𝑑𝐻

𝑑𝑇 , Cv =

𝑑𝐸

𝑑𝑇 )

Cp = Cv + 𝑑(𝑅𝑇)

𝑑𝑇 (PV = RT)

Cp = Cv + 𝑅𝑑𝑇

𝑑𝑇 (R is gas constant)


196

Cp = Cv + R

The heat capacity at constant pressure Cp is always larger than the heat capacity at constant volume Cv because PV – work is done when gas is heated at constant pressure. The relationship is visualized quite easily for an ideal gas. When one mole of ideal gas is heated at constant pressure, the work done in pushing back the piston is P∆V = R∆T. For a 1oC change in temperature, the amount of work done is equal to R, and this is just the extra energy required to heat a mole of ideal gas at constant pressure over that required to heat it through 1oC at constant volume.

Example 6.13. How much heat is required to raise the temperature of 10g of argon through 10OC at (a) constant volume (b) constant pressure?

[ Cv = 12.468J, Cp = 20.794J, Ar = 40]

Solution:

Atomic mass of Ar, a mono atomic gas = 40gmol-1

No. of moles in 10g = 10𝑔

40𝑔𝑚𝑜𝑙 −1 = 0.25mol

(a). qv = nCv∆T

= 0.25 mol× 12.468𝐽𝐾−1 mol−1 × 10𝐾

= 31.17J = 31J

(b). qp = nCp∆T

= 0.25mol× 20.794JK-1 mol-1× 10𝐾

= 51.985J = 51.99J = 52J


197

Example 6.14. Suppose that 1.00kJ of heat is transferred to 2.00mol of argon at 298.0K, 1atm. What will the final temperature Tf be if the heat is transferred at (a) Constant volume (b) Constant pressure? Calculate the energy change ∆E in each case. [ Cv = 12.468 JK-1 mol-1, Cp = 20.794 JK-1 mol-1]

Solution:

(a). At constant volume

First calculate the rise in temperature

qv = nCv∆T

∆T = 𝑞𝑣

𝑛𝐶𝑣

= 1000 𝐽

2𝑚𝑜𝑙 (12.468𝐽𝐾−1𝑚𝑜 𝑙−1)

= 40.10K

∴ 𝑇𝑓 = 298K + 40.10K = 338.1K

Use the rise in temperature to calculate ∆E

∆E = nCv∆T

= (2 mol)( 12.468 JK-1 mol-1)(40.10K)

=1000J

(b). At constant pressure

First calculate the rise in temperature

qp = nCp∆T

∆T = 𝑞𝑝

𝑛𝐶𝑝


198

= 1000𝐽

2𝑚𝑜𝑙 (20.794𝐽𝐾−1𝑚𝑜 𝑙−1)

= 24.05K

∴ 𝑇𝑓 = 298K + 24.05K = 322.05K

Use this rise in temperature to calculate ∆E

∆E = nCp∆T

= (2 mol)( 12.468 JK-1 mol-1)(24.05K)

=600J

Note: the expression for ∆E involves Cv even though the process is carried out at constant pressure. The difference of 400J between the input qp and ∆E is the work done by the gas as it expands.

7.4. Reversible Isothermal and Adiabatic Processes

Before discussing adiabatic processes, let us look at reversible isothermal expansion of gases. Isothermal expansion of gas is the expansion carried out at constant temperature. The maximum work that is obtainable from the isothermal expansion of an ideal gas is easily calculated using the ideal equation of state,

𝑃𝑉 = 𝑛𝑅𝑇

P = 𝑛𝑅𝑇𝑉

In reversible isothermal expansion,

W = − 𝑃𝑑𝑉𝑉2

𝑉1

Substituting for p,

Wrev = − 𝑛𝑅𝑇

𝑉

𝑉2

𝑉1𝑑𝑉


199

= − nRT 𝑑𝑉

𝑉

𝑣2

𝑣1

= − 𝑛𝑅𝑇𝐼𝑛𝑉2

𝑉1

= − 2.303𝑛𝑅𝑇 𝑙𝑜𝑔𝑉2

𝑉1

Since V2 > V1, and the logarithm is positive then Wrev < 0. A negative W indicates that work is done by the system on the surroundings.

In compression, the final volume V2 is less than V1 so Wrev is positive. The positive value means that work is done on the gas.

Example 7.15. A mole of CH4 expands reversibly and isothermally from 1dm3 to 50 dm3 at 250C. Calculate the work by the gas in joules assuming the gas is ideal.

Solution:

𝑊𝑟𝑒𝑣 = − 𝑛𝑅𝑇𝑙𝑛 𝑉2

𝑉1

= −2.303nRTlog𝑉2

𝑉1

= − 2.303× 1𝑚𝑜𝑙 8.314𝐽𝐾−1𝑚𝑜𝑙−1 (298.1𝐾 log50

1)

= − 2.303× 1 × 8.314𝐽 × 298.15 log 50

= −2.303× 1 × 8.314 × 298.15 × 1.6990

= − 9698.9J

= − 9.70KJ

7.5. Expression of Wrev in terms of Pressure Let us consider a situation in which a change in pressure causes an infinitesimal change in volume of the gas. This means there are


200

changes in both pressure and volume, so work done by the system will be W = ‒ 𝑑(𝑃𝑉). Since it is assumed that the change in pressure is more pronounced, above equation can be expressed as W = ‒ 𝑉𝑑𝑃. But V is not constant, rather a function of pressure, so using general gas equation of state

PV = nRT

V = 𝑛𝑅𝑇

𝑃

Substituting for V in the above equation W =‒ 𝑛𝑅𝑇

𝑝 𝑑𝑝

For isothermal process, T is constant

∴ 𝑤 = 𝑛𝑅𝑇 𝑑𝑃

𝑃

𝑃2

𝑃1

= ‒𝑛𝑅𝑇𝑙𝑛𝑃2

𝑃1

∴ 𝑊𝑟𝑒𝑣 =‒𝑛𝑅𝑇𝑙𝑛𝑃2

𝑃1

Assuming Boyles‘ law P1V1 = P2V2

𝑃2

𝑃1=

𝑉1

𝑉2 𝑜𝑟

𝑉2

𝑉1 =

𝑃1

𝑉2

From the equation Wrev = ‒nRTln𝑉2

𝑉1 = nRTln

𝑃1

𝑃2

‒ nRTln𝑉2

𝑉1 = 𝑛𝑅𝑇𝑙𝑛

𝑃1

𝑃2

𝑊𝑟𝑒𝑣 = +𝑛𝑅𝑇𝑙𝑛𝑃2

𝑃1

When P2 is greater than P1, work is done ON the system by the surroundings.


201

Example 7.16. One mole of nitrogen at 250C and 1.01325× 105Nm-2 is expanded reversibly and isothermally to a pressure of 1.32× 104 Nm-2 (a) what is the value of Wrev? (b) What is the value of Wrev if the gas is expanded against a constant pressure of 1.32× 104 Nm-2 ?

Solution:

(a). For one mole of gas

Wrev = RTln𝑃2

𝑃1

= (8.314JK-1mol-1)(298.15K)ln 1.32×104

1.01325 ×105

= − 5.05kJmol-1

(b). P1V1 = RT,

V1 = 𝑅𝑇

𝑃1

= 8.314𝐽𝐾−1𝑚𝑜 𝑙−1 (298.15𝐾)

1.01325 × 105𝑁𝑚−2

= 2.446× 10−2𝑚3𝑚𝑜𝑙−1

V2 = (8.314𝑁𝑚𝐾−1𝑚𝑜 𝑙−1)(298.15𝐾)

1.32×104𝑁𝑚−2

= 0.1878m3 mol-1

𝑤 = 𝑃∆𝑉

= 1.32× 104(0.1878 − 0.02446)


202

= 1.32× 104 × 0.1633𝐽

= 2.156KJ

Example 7.17. An ideal gas expanded reversibly and isothermally from 10 bar to 1 bar at 250C. What are the values of

(a) W per mole?

(b) q per mole

(c) ∆E and

(d) ∆H

Solution:

(a). For one mole,

W = RTln𝑃2

𝑃1

= (8.314JK-1mol-1) (298.15K)ln1

10

= - 5707.67J

= - 5.71kJ

(b). For isothermal expansion,

q = − w = − (− 5.71kJ) = 5.71kJ

(c). ∆E = 0, since this is an ideal gas

(d) 𝑑𝐻 = 𝑑𝐸 + 𝑑(𝑃𝑉) = 0 (Expansion is not at constant pressure).


203

7.6. Adiabatic Expansion An adiabatic process is one in which there is neither gain nor loss of heat, that is one in which the system under investigation is thermally isolated from its environment such that q = 0. In an adiabatic expansion, work is done at the expense of the internal energy of the gas, resulting in drop of temperature. When a gas expands adiabatically to a larger volume and a lower pressure, the volume is usually smaller than it would be for an isothermal expansion to the same pressure. The work done by isothermal reversible expansion of a gas is always larger than the work done by the adiabatic expansion. The energy for doing the additional work in isothermal expansion is provided by heat absorbed from constant temperature reservoir. The energy for doing work in the adiabatic expansion comes only from the cooling of the gas itself.

Now let us consider the reversible adiabatic expansion of one mole of ideal gas. Since for an adiabatic process, dq = 0, then by the first law of thermodynamics,

𝑑𝐸 = − 𝑑𝑤 (𝑑𝐸 = 𝑑𝑞 – 𝑑𝑤)

dw = − dE

From the equation dE = CvdT

dw = − CvdT

∴ 𝑤𝑟𝑒𝑣 = − 𝐶𝑣𝑑𝑇𝑇2

𝑇1

Since work, PV, is done at the expense of internal energy

dE = − PdV

CvdT = − PdV


204

For one mole of gas,

PV = RT

P = 𝑅𝑇

𝑉

CvdT = − 𝑅𝑇

𝑉 dV

𝐶𝑣𝑑𝑇

𝑇 = − R .

𝑑𝑉

𝑉

If V1 is the volume of the gas at T1 and V2 at T2 and Cv is independent of temperature,

𝐶𝑣𝑑𝑇

𝑇

𝑇2

𝑇1 = − 𝑅

𝑑𝑉

𝑉

𝑉2

𝑉1

Within these limits, on integration,

Cvln𝑇2

𝑇1 = − Rln

𝑉2

𝑉1

For ideal gas during expansion V2>V1 so T2<T1, there is temperature drop in adiabatic expansion. In other words the gas cools. Conversely, adiabatic compression of gas produces an increase in temperature.

Example 7.18. Calculate the temperature increase of helium if a mole of it is compressed adiabatically and reversibly from 44.8dm3 at 10C to 22.4dm3

[Cv for He = 12.55JK-1 mol-1, R = 8.314JK-1 mol-1]

Solution:

V1 = 44.8dm3, V2 = 22.4dm3

T1 = 273.15K , T2 = ?


205

Cv = 12.55K-1 mol-1 R = 8.314JK-1 mol-1

Use equation Cvln𝑇2

𝑇1 = − Rln

𝑉2

𝑉1

=12.55JK-1 mol-1 ln(𝑇2

273.15) = − 8.314JK-1mol-1ln

22.4

44.8

= ln 𝑇2

273.15𝐾 = −

8.314

12.55 ln

22.4

44.8

lnT2 − ln273.15 = − 0.6625ln(1

2)

lnT2 = − 0.6625 × − 0.6931 + 𝑙𝑛273.15

lnT2 = 0.4592 + 5.6100

= 6.0692

T2 = 432.34K

Temperature increase = 432.34 – 273.15

= 159.19oC

≃ 159oC

Above calculation shows that the compression was carried out so

rapidly that there was no heat transfer to the container but sufficiently

slow to make it reversible.

To find the change in pressure, we use the equation,


206

Cv ln𝑇2

𝑇1 = − Rln

𝑉2

𝑉1

ln𝑇2

𝑇1 = −

𝑅

𝐶𝑣 In

𝑉2

𝑉1

= 𝑅

𝐶𝑣 In

𝑉1

𝑉2

Taking the antilog of both sides reduces the equation to

𝑇2

𝑇1 = (

𝑉1

𝑉2) R/Cv

From Cp = Cv + R,

R = Cp − Cv

∴ 𝑇2

𝑇1 =

𝑉1

𝑉2

𝐶𝑝− 𝐶𝑉

𝐶𝑉

= 𝑉1

𝑉2 γ −1

Where γ = 𝐶𝑝

𝐶𝑣

Example 7.19. A mole of argon is allowed to expand adiabatically and

reversibly from a pressure of 10 bar (106 Pa) and 298.15K to 1 bar (105

Pa). What is the final temperature and how much work is done by the

gas? [Cp = 5

2 R, Cv =

3

2 R ]

Solution:


207

From Cp ln 𝑇2

𝑇1 = 𝑅 𝑙𝑛

𝑃2

𝑃1

5

2 R ln

𝑇2

𝑇1= 𝑅 𝑙𝑛

𝑃2

𝑃1

2.5 × log𝑇2

𝑇1 = log

𝑃2

𝑃1

log𝑇2

𝑇1 =

1

2.5 𝑙𝑜𝑔 (0.1)

log 𝑇2 − log 𝑇1 = 1

2.5 log 0.1

log 𝑇2 = log(298.15 ) − 0.4

= 2.4744 – 0.4

= 2.0744

T2 = 102.0744

∴ 𝑇2 = 118.70𝐾

Alternatively,

5

2𝐼𝑛

𝑇2

𝑇1 = 𝐼𝑛

𝑃2

𝑃1


208

2. 5 𝐼𝑛 𝑇2

298.15 = 𝐼𝑛 (

105𝑝𝑎

106𝑝𝑎)

= 𝐼𝑛𝑇2

298.15 =

1

2.5 𝐼𝑛

1

10

ln T2 – In 298.15 = –0.9210

𝑙𝑛 T2 = 𝐼𝑛 298.15 – 0.9210

= 5.6979 – 0.9210

= 4.7766

= 118.70K

(b). W = 𝐶𝑉𝑑𝑇𝑇2

𝑇1= 𝐶𝑉 𝑇2 − 𝑇1

= 3

2R 𝑇2−𝑇1

= 3

2 ( 8.31455JK-1mol-1)(118.7−298.15𝐾)

=−2.238kJmol-1

The maximum work that can be done by the gas is 2.24kJmol-1

Example 7.20. An ideal monatomic gas at 298.15K and 105Nm-2 is expanded in reversible adiabatic process to a final pressure of 5×


209

104𝑁𝑚−2. Calculate: (a) final temperature (b) q per mole (c) W per

mole, and (d) ∆E. [Cp = 5

2𝑅, 𝑅 = 8.3145𝐽𝐾−1𝑚𝑜𝑙−1, 𝐶𝑉 =

3

2R]

Solution:

From Cp ln 𝑇2


𝑃2

𝑃1

5

2 R ln

𝑇2

𝑇1= 𝑅 𝑙𝑛

𝑃2

𝑃1

2.5 ln𝑇2

𝑇1= 𝑙𝑛

𝑃2

𝑃1

ln 𝑇2

298.15𝐾 =

1

2.5 In

5×104𝑁𝑚−2

105𝑁𝑚−2

ln 𝑇2 − 𝑙𝑛 298.15 =1

2.5𝑙𝑛 0.5

ln 𝑇2 = 𝑙𝑛 298.15 +1

2.5𝑙𝑛 0.5

= 5.6976 – 0.2773

= 5.4203

T2 = e5.4203

= 225.95K

= 226.0K

(b). For adiabatic process, no heat gain or less

∴ 𝑞 = 0

(C). W = − Cv (𝑇2 − 𝑇1)

= − 3

2𝑅 226.0 − 298.15 𝐽𝑚𝑜𝑙−1


210

= − 1.5(8.3145JK-1)(-72.15K)

= + 899.84Jmol-1

PdV = W = −∆E

∆E = q – W

= 0 – W

= − W

= − 899.84Jmol-1

Note: for infinitesimal increase in volume dV at the pressure P, work done by the gas is PdV. Since this work is accomplished at the expense of the internal energy of the gas, the internal energy must decrease by an amount dE.

Example 7.21. A tank contains 20dm3 of compressed N2 at 10 bar (106 Pa) and 250C. Calculate w when the gas is allowed to expand reversibly to 1 bar (105Pa) pressure (a) isothermally (b) adiabatically.

Solution:

For the isothermal expansion of ideal gas,

𝑊rev = RT ln 𝑃2

𝑃1

= (8.3145JK-1mol-1)(298.15K)(ln 105𝑝𝑎

106 𝑝𝑎)

= (2478.9582) (−2.3026)Jmol-1

= − 5708.0122 = −5.71kJmol-1

However there are n moles present


211

From 𝑃𝑉 = 𝑛𝑅𝑇

n = 𝑃𝑉

𝑅𝑇

= 106𝑁𝑚−2 20×10−3𝑚3

8.3145𝑁𝑚𝐾−1𝑚𝑜 𝑙−1 298.15𝐾

= 8.068

= 8.07moles

𝑊 = 𝑊rev per mole × 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠

= − 5.71× 8.07𝑘𝐽

= − 46.08 𝑘𝐽

= − 46.1 𝑘𝐽

(b). For adiabatic expansion, Cp = 29.125JKmol-1, Cv = 20.811JK-1mol-1, R = 8.314JK-1mol-1.

Cp ln 𝑇2


𝑃2

𝑃1

29.125 ln T2

298.15K = 8.314ln

8.314

29.125 ln (

105 Nm −2

106 Nm −2)

ln T2 − ln 298.15 = 8.314

29.125 ln (

1

10)

= − 0.6573

ln T2 = ln 298.15 – 0.6573

= 5.6976 – 0.6573

= 5.0403

∴ 𝑇2 = 154.5𝐾


212

W = 𝐶𝑉𝑑𝑇𝑇2

𝑇1 = Cv (T2-T1)

= (20.811JK-1mol-1)(154.5 – 298.15K) = (20.811)(−143.65)Jmol-1 = −2.99kJmol-1 For 8.07moles W = −2.99kJmol-1× 8.07𝑚𝑜𝑙𝑒𝑠

= − 24.12kJ 7.7. Degrees of freedom and equipartition of energy

For each atom in a solid or gas phase, three coordinates have to be specified to describe the atom‘s position – a single atom has 3 degrees of freedom for its motion. A solid or a molecule composed of N atoms has 3N degrees of freedom. We can also think about the number of degrees of freedom as the number of ways to absorb energy. The theorem of equipartition of energy (classical mechanics) states that in thermal equilibrium the same average energy is associated with each independent degree of freedom and that the energy is ½kBT. For the interacting atoms, e.g. liquid or solid, for each atom we have

½ kBT for kinetic energy and ½ kBT for potential energy - equality of kinetic and potential energy in harmonic approximation is addressed by the virial theorem of classical mechanics. Based on equipartition principle, we can calculate heat capacity of the ideal gas of atoms - each atom has 3 degrees of freedom and internal energy of 3/2kBT. The molar internal energy U=3/2NAkBT=3/2RT and the molar heat capacity under conditions of constant volume is Cv =[dU/dT]V=3/2R In an ideal gas of molecules only internal vibrational degrees of freedom have potential energy associated with them. For example, a diatomic molecule has 3 translational + 2 rotational + 1 vibrational = 6 total degrees of freedom. Potential energy contributes ½ kBT only to the energy of the vibrational degree of freedom, and Umolecule = 7/2kBT if all degrees of freedom are “fully” excited.


213

CHAPTER EIGHT

CHEMICAL KINETICS

We have learnt on several occasions that a balanced chemical equation

is a chemical statement that gives the mole ratios of reactants and

products as well as the ratios of formula units. A balanced chemical

equation as ordinarily written provides valuable chemical information

as to the masses, or volumes (if gases are involved) and is therefore an

essential quantitative tool for calculating product yields from amounts

of reacting substances. However, a balanced chemical equation tells us

nothing about how fast or quickly chemical changes occur, or what

energy changes are associated with the molecular interaction in a given

chemical reaction. Knowing how quickly a chemical reaction occurs is

a crucial factor in how the reaction affects its surroundings. Therefore,

knowing the rate of a chemical reaction and the energy changes

associated with the molecular interaction during the reaction are

integral to understanding the reaction.

The questions of ―how fast does the reaction go‘? and ‗what

conditions or factors bring about variations in speed‘ in a given

chemical reaction are the subject of this chapter.

The concept of rate applies to a number of phenomenon in our

daily life. For example the change in distance by an athlete over time is

the running rate of the athlete. The number of soap bars that are

produced in a given time is the rate of production of soap etc. We


214

apply the same principle in chemical reaction. This time as products

are formed reactants are used up and rate (speed) of a chemical

reaction can be expressed as the ratio of the change in the

concentration of a reactant (or product) to a change in time. The study

that deals with the movement/motion-the speeds, or rates of chemical

reactions is known as chemical kinetics.

We know that any reaction can be represented by the general equation

Reactants → products

This equation tells us that, during the course of a reaction, reactant

molecules are consumed while product molecules are formed. As a

result, we can follow the progress

of a reaction by monitoring either the decrease in concentration of the

reactants or the increase in concentration of the products.

Let us consider a simple reaction in which A molecules are converted

to B molecules (for example, the conversion of cis-1,2-dichloroethylene

to trans-1,2-dichloroethylene):

A → B

The decrease in the number of A molecules leads to increase in the

number of B molecules with time. In general, it is more convenient to

express the rate in terms of change in concentration with time. Thus,

for the preceding reaction we can express the rate as

𝑅𝑎𝑡𝑒 = − ∆[𝐴]

∆𝑡 𝑜𝑟 𝑅𝑎𝑡𝑒 =

∆[𝐵]

∆𝑡


215

in which Δ[A] and Δ[B] are the changes in concentration (in molarity)

over a period Δt. Because the concentration of A decreases during the

time interval, Δ[A] is a negative quantity. The rate of a reaction is a

positive quantity, so a minus sign is needed in the rate expression to

make the rate positive. On the other hand, the rate of product

formation does not require a minus sign because Δ[B] is a positive

quantity (the concentration of B increases with time).

For more complex reactions, we must be careful in writing the rate

expression.

Consider, for example, the reaction

2A → B

Two moles of A disappear for each mole of B that forms—that is, the

rate at which B forms is one half the rate at which A disappears. We

write the rate as either

𝑟𝑎𝑡𝑒 = − 1

2 ∆[𝐴]

∆𝑡 𝑜𝑟 𝑟𝑎𝑡𝑒 =

∆[𝐵]

∆𝑡

Consider the following hypothetical reaction between reactants

A and B to form products C and D

aA + bB → cC + dD


216

( a, b, c and d are the stoichiometric coefficients of A, B, C, and D

respectively). The rate of this reaction is the speed at which A or B is

consumed or, alternatively, the speed at which C or D is formed.

Mathematically this is given by

𝑟𝑎𝑡𝑒 = −1

𝑎 ∆[𝐴]

∆𝑡=

−1

𝑏 ∆[𝐵]

∆𝑡=

1

𝑐 ∆[𝐶]

∆𝑡=

1

𝑑 ∆[𝐷]

∆𝑡

Figure 8.1: The rate of reaction A →B, represented as the decrease of A molecules with time and as the increase of B molecules with time. Example 8.1

Write the rate expressions for the following reactions in terms of the

disappearance of the reactants and the appearance of the products:

(a). I-(aq) + OCl-(aq) → Cl-(aq) + OI- (aq)

(b). 3O2 (g) → 2O3 (g)

(c). 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O (g)

Solution


217

(a). Because each of the stoichiometric coefficients equals 1

𝑟𝑎𝑡𝑒 = − ∆[𝐼−]

∆𝑡=

∆[𝑂𝐶𝐼−]

∆𝑡=

∆[𝐶𝑙−]

∆𝑡=

∆[𝑂𝐼−]

∆𝑡

(b). Here the coefficients are 3 and 2, so


3 ∆[𝑂2]

∆𝑡=

1

2 ∆[𝑂3]

∆𝑡

(c) In this reaction


4 ∆[𝑁𝐻3]

∆𝑡= −

1

5 ∆[𝐶𝑂2]

∆𝑡=

1

4 ∆[𝑁𝑂]

∆𝑡=

1

6 ∆[𝐻2𝑂]

∆𝑡

Practice Exercise

Write the rate expression for the following reaction:

Example 8.2

Consider the reaction

4NO2 (g) + O2 (g) → 2N2O5 (g)

Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of 0.037 M/s. (a) At what rate is N2O5 being formed? (b) At what rate is NO2 reacting?


218

Strategy To calculate the rate of formation of N2O5 and disappearance of NO2, we need to express the rate of the reaction in terms of the stoichiometric coefficients as in Example 8.1:


4 ∆[𝑁𝑂2]

∆𝑡 = −

∆[𝑂2]

∆𝑡 =

1

2 ∆[𝑁2𝑂5]

∆𝑡

We are given

∆[𝑂2]

∆𝑡 = 0.037 M/s

Where the minus sign shows that the concentration of O2 is decreasing

with time.

Solution

(a) From the preceding rate expression, we have

− ∆[𝑂2]

∆𝑡 =

1

2 ∆[𝑁2𝑂5]

∆𝑡

Therefore,

∆[𝑁2𝑂5]

∆𝑡= −2(− 0.037 𝑀𝑠−1) = 0.074 M/s

(b) Here we have

= − 1

4 ∆[𝑁𝑂2]

∆𝑡 = −

∆[𝑂2]

∆𝑡

So

∆[𝑁𝑂2]

∆𝑡 = 4 − 0.037 𝑀𝑠−1 = −0.15𝑀/𝑠

Practice Exercise

Consider the reaction


219

Suppose that, at a particular moment during the reaction, molecular

hydrogen is being

formed at the rate of 0.078 M/s. (a) At what rate is P4 being formed? (b)

At what rate is

PH3 reacting?

Example 8.3

Hydrogen gas is used for fuel aboard the Space Shuttle and may be

used in

Earth-bound engines of the future:

a) Express the rate of this reaction in terms of changes in [H2], [O2], and

[H2O] with time.

b) When [O2] is decreasing at 0.23 mol/L.s, at what rate is [H2O]

increasing?

Solution


220

Example 8.4

Consider the reaction of nitrogen dioxide with fluorine to give

nitrylfluoride, NO2F.

How is the rate of formation of NO2F related to the rate of reaction of

fluorine?

Problem Strategy

We need to express the rate of this reaction in terms of concentration

changes with time of the product, NO2F, and reactant, F2, and then

relate these two rates. The rate of disappearance of reactants is

expressed as a negative quantity of concentration change

per some time interval as we all know. The rate of formation of

products is expressed as a positive quantity of concentration change

per some time interval. In order to equate rate expressions, we need to


221

divide each by the coefficient of the corresponding substance in the

chemical equation.

Solution

You write

Rate of formation of NO2F = ∆[𝑁𝑂2𝐹]

∆𝑡

and

Rate of reaction of F2 = − ∆[𝐹2]

∆𝑡

You divide each rate by the corresponding coefficient (if applicable)

and then equate them:

1

2 ∆[𝑁𝑂2𝐹]

∆𝑡 = −

∆[𝐹2]

∆𝑡

Example 8.5

Calculate the average rate of decomposition of N2O5, - ∆[N2O5]/∆t, by

the reaction

during the time interval from t = 600s to t =1200s (regard all time

figures as significant).

Use the following data:


222

Problem Strategy

An average reaction rate is the change in concentration of a reactant or

product over a time interval; in this case it‘s the rate of decomposition

of the reactant N2O5, (- ∆[N2O5]/∆t). The ∆[ N2O5] in the equation is the

change in concentration of N2O5 (final value minus initial value). The

∆t is the time interval (final minus initial) over which the concentration

change occurred.

Solution

Average rate of decomposition of N2O5 = − ∆[𝑁2𝑂5]

∆𝑡=

Practice Exercise

For the reaction given in Example 8.3, how is the rate of formation of

NO2F related to the rate of reaction of NO2?


223

Example 8.6

Would a wet piece of iron metal rust faster in air or pure oxygen?

Explain (Hint: consider the effect of concentration on reaction rate)

Solution

Iron exposed to moist air will react with oxygen to form iron oxide.

This oxidation process is called rusting.

4Fe(s) + 3O2(g) → 2Fe2O3(s)

Oxygen is more concentrated in its pure form than mixed with other

gases in air. And we know that, as concentration increases the rate of a

reaction will also increase. Therefore, a wet piece of iron metal will rust

faster in pure oxygen than in air.

Factors Affecting Reaction Rates

At the beginning of this unit it was pointed out that various reactions

or changes that occur in nature or otherwise take place at a variety of

speeds depending on the conditions. Why do these times for different

changes turn out as they do? As the speed of an athlete depends on

several factors such as temperature surroundings, wind direction,

health condition etc. so is the speed of a chemical reaction affected by

several factors. In general, the rate at which a given chemical reaction

takes place depends upon a number of factors. The rates of chemical

reactions can be affected by:


224

Reactant concentrations

Often the rate of reaction increases when the concentration of a

reactant is increased. For a chemical reaction to have noticeable rate,

there should be noticeable number of molecules with the energy equal

or greater than the activation energy. With an increase in concentration

the number of molecules with the minimum required energy for a

reaction to take place will increase and thereby the rate of the reaction

increases. Suppose that at any one time 1 in a million particles have

enough energy to equal or exceed the activation energy. If you had 100

million particles, 100 of them would react. If you had 200 million

particles in the same volume, 200 of them would now react. The rate of

reaction has doubled by doubling the concentration. A piece of steel

wool burns with some difficulty in air (20% O2) but bursts into a

dazzling white flame in pure oxygen. The rate of burning increases

with the concentration of O2. In some reactions, however, the rate is

unaffected by the concentration of a particular reactant, as long as it is

present at some concentration. Another example is the reaction of zinc

and hydrochloric acid in the lab, zinc granules react fairly slowly with

dilute hydrochloric acid, but much faster if the acid is concentrated.

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)


225

Concentration of catalyst.

A catalyst is a substance that increases the rate of reaction without being

consumed in the overall reaction. Because the catalyst is not consumed by

the reaction, it does not appear in the balanced chemical equation

(although its presence may be indicated by writing its formula over the

arrow). A solution of pure hydrogen peroxide, H2O2, is stable, but

when hydrobromic acid, HBr(aq), is added, H2O2 decomposes rapidly

into H2O and O2.

Here HBr acts as a catalyst to speed decomposition.

Temperature at which the reaction occurs

Usually reactions speed up when the temperature increases. It takes

less time to boil an egg at sea level than on a mountaintop, where

water boils at a lower temperature. Reactions during cooking go faster

at higher temperature.

Surface area of a solid reactant

If a reaction involves a solid with a gas or liquid, the surface area of

the solid affects the reaction rate. Because the reaction occurs at the

surface of the solid, the rate increases with increasing surface area. A


226

wood fire burns faster if the logs are chopped into smaller pieces.

Similarly, the surface area of a solid catalyst is important to the rate of

reaction. The greater the surface area per unit volume, the faster the

reaction.

Pressure

Increasing the pressure on a reaction involving reacting gases increases

the rate of reaction. Increasing the pressure of a gas is exactly the same

as increasing its concentration. If you have a given mass of gas, the

way you increase its pressure is to squeeze it into a smaller volume. If

you have the same mass in a smaller volume, then its concentration is

higher. Thus the effect is the same as the concentration effect. In the

manufacture of ammonia by the Haber Process, the rate of reaction

between the hydrogen and the nitrogen is increased by the use of very

high pressures.

N2(g) + 3H2(g) ↔ 2NH3(g) + heat (ΔH = -92kJmol-1)

In fact, the main reason for using high pressures is to improve the

percentage of ammonia in the equilibrium mixture, but there is a useful

effect on rate of reaction as well. Changing the pressure on a reaction

which involves only solids or liquids has no effect on the rate.


227

Nature of reactants

Substances differ markedly in the rates at which they undergo

chemical change. For example, hydrogen and fluorine molecules react

explosively, even at room temperature, producing hydrogen fluoride

molecules:

H2 + F2 → 2HF (very fast at room temp.)

Under similar conditions, hydrogen and oxygen molecules react so

slowly that no chemical change is apparent:

2H2 + O2 → 2H2O (Very slow at room temp.)

The differences in reactivity between reactions may be attributed to the

different structures of the atoms and molecules of the reacting

materials (for instance whether the substances are in solution or in the

solid state-include) If a reaction involves two species of molecules with

atoms that are already joined by strong covalent bonds (for example

quartz (SiO2) and water (H2O)) collisions between these molecules at

ordinary temperatures may not provide enough energy to break these

bonds unlike the collisions which take place between molecules whose

atoms are joined by weak covalent bonds. Therefore, reactions between

molecules whose atoms are bound by weak covalent bonds take place

at a faster rate than reactions between molecules whose atoms are

bound by strong covalent bonds. For example, when methane gas is

mixed with chlorine gas and exposed to sunlight an explosive reaction


228

takes place in which chlorinated methane products are produced along

with hydrogen chloride.

CH4 + Cl2 + energy → CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl

Concentration dependence of Rate: Order of a reaction

We just discussed how the rate of a chemical reaction depends on

several factors such as temperature, catalyst, surface area of reactants,

presence or absence of a catalyst, nature of reactants and concentration.

We have seen qualitatively that the rates of most chemical reactions

increase when the concentrations of the reactants increase. In this

section, you will explore the quantitative relationships between the

rate of a reaction and the concentrations of the reactants.

Now consider the general reaction

aA + bB → products (occuring at a constant temperature)

where A and B represent the reactant formulae and, a and b represent

the stoichiometric coefficients. In this section, you will study reaction

rates that are not affected by the concentrations of the products.

Therefore, you do not need to use symbols for the products.

In general, the rate of a reaction is proportional to the

concentration of each reactant raised to some power, where the power

on a given reactant is called, the order of the reaction with respect to

that reactant. The overall order of a reaction is the sum of all the

exponents of the concentration terms in the rate equation.


229

The rate for the above reaction can be given as

Rate α [A]m[B]n ………………………………….………………8.1

where m and n are the rate law exponents and indicate the order of the

reaction with respect to the corresponding reactants. The values of m

and n for a given reaction must be determined experimentally and do

not change with temperature.

This relationship given by equation 8.1 can be expressed in a general

equation given below, called the rate law equation.

Rate = k [A]m [B]n ………………………………….………….8.2

The rate law equation expresses the relationship between the

concentrations of the reactants and the rate of the reaction. The letter k

represents a proportionality constant called the rate constant and it

indicates how fast or slow a reaction is proceeding. A small rate

constant indicates a slow reaction and a large rate constant indicates a

fast reaction. The value of k for a given reaction is temperature

dependent and is constant under constant temperature and pressure

conditions.

The exponents m and n do not necessarily correspond to the

stoichiometric coefficients of their reactants. Usually the value of a rate

law exponent is 1 or 2. But, seldom values of 0, 3; and even fractions

can occur. If the exponent for a given reactant is 1, then the reaction is

said to be first order with respect to that reactant. Similarly, if the

exponent of a reactant is 2, the reaction is said to be second order in


230

this reactant. For example, the rate law equation below represents a

reaction that is first order in A, second order in B, and third order (1 +

2) overall.

Rate = k[A]1[B]2

For example, the reaction between nitric oxide and ozone

NO(g) + O3(g) → NO2(g) + O2(g)

is first order in nitric oxide and first order in ozone. The rate law

equation for this reaction is:

Rate = k{NO]1[O3]1

The overall order of the reaction is 1 + 1 = 2.

Determining reaction Orders

The order of a reaction with respect to its reactants can be determined

by running a series of experiments each of which starts with a different

set of reactant concentrations and the initial rate is obtained. The

experiments are designed to change one reactant concentration while

keeping the other constant. This method of determining order of a

reaction is known as the initial rate method.

Example 8.7

For the reaction A + B → products, the following rate data were

obtained in three separate experiments:


231

a) What is the order of the reaction with respect to A and B?

b) What is the rate constant, k?

Solution

a) The general rate expression for this reaction is Rate = k[A]x [B]y and

the values of x and y must be determined from the above data.

The data obtained shows that the rate doubles in experiment number 2

than it was in experiment number 1 when the concentration of B is

doubled keeping A constant. In such a condition, when the rate of a

reaction doubles by doubling the concentration of a given reactant,

order of the reaction with respect to that reactant is 1.

In experiment number 3 the rate is found not to change, than it was in

experiment number 2, when the concentration of A is doubled. This

indicates that the rate does not depend on the concentration of A. This

means, order of the reaction with respect to A is zero.

The experimentally determined rate equation will then be Rate =

k[A]o[B] = k[B] and the overall order = 1

The order with respect to each reactant can also be determined by

calculation as follows:

Rate1 = 2 x 10-5 M min-1 = k[2]x [1]y


232

Rate2 = 4 x 10-5 M min-1 = k[2]x [2]y

Taking the ratio of the two rates we have

The order of the reaction with respect to the reactant A can be

calculated by taking the ratio of the rate expressions of experiment

number 2 and 3.

Thus, the experimentally determined rate equation is given by

Rate = k[A]x[B]y = K[A]o[B] = k[B]

b) Once we get the order of the reaction the rate constant can be

calculated by taking the data obtained in any of the three experiments.

Consider experiment number 1


233

Example 8.8

For the reaction X + 2Y + 2Z → products, the following rate data were

obtained:

a) What is the order of each reactant in the system?

b) What is the value of the rate constant?

c) What is the rate of disappearance of X in experiment 4?

Solution

a) In the first experiment the rate of the reaction was determined to be

1.0 x 10-6 M/min when the concentrations of all reactants were kept at

0.1 M.

In the second experiment the rate was found to be tripled when the

concentrations of X and Y were tripled and that of Z was kept constant.

This indicates that the rate of the reaction depends on the

concentration on either X or Y or on both of them.


234

In the third experiment, the rate remained the same as in experiment

number 1 when the concentration of X was quadrupled and those of Y

and Z were kept constant. This fact indicates that, the rate of the

reaction does not depend on the concentration of X.

Therefore, Rate α [X]0

Going back to experiment number 2, the rate was tripled when the

concentrations of X and Y were each tripled. But we have decided that

the rate does not depend on the concentration of X. Therefore, the rate

was tripled when the concentration of Y was tripled.

Thus, Rate α [Y]1

When we compare experiments 2 and 4, neglecting X; the

concentration of Y is kept constant while that of Z is tripled. As a result

the rate was found to increase by a factor of 9.

Thus, Rate α [Z]2 since 9 = 32

b) k can be evaluated from any one of the 4 data sets, once orders are

known; e.g., expt. 1.

Rate = k[X]0[Y]1[Z]2

c) When [Z] = 0.15, [Y] = 0.15, and [X] = 0.125, from balanced equation,

Rate = k[Y] [Z]2

= 10-3 M-2 / min (0.15) (0.15)2 = 3.4 x 10-6 M/min


235

Relation between Reactant Concentrations and Time Rate laws enable us to calculate the rate of a reaction from the rate

constant and reactant concentrations. They can also be converted into

equations that enable us to determine the concentrations of reactants at

any time during the course of a reaction. We will illustrate this

application by considering first one of the simplest kind of rate

laws—that applying to reactions that are first order overall.

First-Order Reactions

A first-order reaction is a reaction whose rate depends on the reactant

concentration raised to the first power. In a first-order reaction of the type

A→ product

the rate is

From the rate law, we also know that

Thus,

We can determine the units of the first-order rate constant k by

transposing:


236

Because the unit for ∆[A] and [A] is M and that for ∆t is s, the unit for k

is

Derivation of first order reaction

The preceding first order rate in differential form becomes

Half-Life

The half-life of a reaction, t1/2, is the time required for the concentration of

a reactant to decrease to half of its initial concentration.


237

By the definition of half-life,

The equation above tells us that the half-life of a first-order reaction is

independent of the initial concentration of the reactant. Measuring the

half-life of a reaction is one way to determine the rate constant of a

first-order reaction.

Example 8.9

The decomposition of ethane (C2H6) to methyl radicals is a first-order

reaction with a rate constant of 5.36 × 10-4 s-1 at 700°C:

Calculate the half-life of the reaction in minutes.

Strategy

To calculate the half-life of a first-order reaction, we use half life

equation above. A

conversion is needed to express the half-life in minutes.

Solution


238

For a first-order reaction, we only need the rate constant to calculate

the half-life of the reaction.

Using

Second-Order Reactions

A second-order reaction is a reaction whose rate depends on the

concentration of one reactant raised to the second power or on the

concentrations of two different reactants, each raised to the first power. The

simpler type involves only one kind of reactant molecule:

A → product


239

Another type of second-order reaction is

A + B → product

and the rate law is given by

rate = k[A][B]

The reaction is first order in A and first order in B, so it has an overall

reaction order of 2.

Using calculus, we can obtain the following expressions for ―A→

product‖ second-order reactions:


240

The equation above is a result of

Zero-Order Reactions

First- and second-order reactions are the most common reaction types.

Reactions whose order is zero are rare. For a zero-order reaction

A → product

the rate law is given by rate = k[A]0

= k Thus, the rate of a zero-order reaction is a constant, independent of

reactant concentration.

Using calculus, we can show that

The Equation above has the form of a linear equation. A plot of [A]t

versus t gives a straight line with slope = -k and y intercept = [A]0. To

calculate the half-life of a zero-order reaction, we set [A]t = [A]0/2 in

the equation above equation and obtain


241

Many of the known zero-order reactions take place on a metal surface.

An example is the decomposition of nitrous oxide (N2O) to nitrogen

and oxygen in the presence of platinum (Pt):

2N2O(g) → 2N2(g) + O2(g)

When all the binding sites on Pt are occupied, the rate becomes

constant regardless of the amount of N2O present in the gas phase.

Third-order and higher order reactions are quite complex; they are not

presented in this book. Table 8.1 summarizes the kinetics of zero-order,

first-order, and second order reactions.

Table 8.1: Summary of the Kinetics of zero-order, first order, and second-order

reactions


242

Activation Energy and Temperature Dependence of Rate Constants

With very few exceptions, reaction rates increase with increasing

temperature. For example, much less time is required to hard-boil an

egg at 100°C (about 10 min) than at 80°C (about 30 min). Conversely,

an effective way to preserve foods is to store them at subzero

temperatures, thereby slowing the rate of bacterial decay. Figure 8.2

shows a typical example of the relationship between the rate constant

of a reaction and temperature. To explain this behavior, we must ask

how reactions get started in the first place.

Figure 8.2: Dependence of rate constant on temperature. The rate constants of most reactions increase with increasing temperature.


243

The Collision Theory of Chemical Kinetics

The kinetic molecular theory of gases states that gas molecules

frequently collide with one another. Therefore it seems logical to

assume—and it is generally true—that chemical reactions occur as a

result of collisions between reacting molecules.

In terms of the collision theory of chemical kinetics, then, we

expect the rate of a reaction to be directly proportional to the number

of molecular collisions per second, or to the frequency of molecular

collisions:

This simple relationship explains the dependence of reaction rate on

concentration.

Consider the reaction of A molecules with B molecules to form

some product. Suppose that each product molecule is formed by the

direct combination of an A molecule and a B molecule. If we doubled

the concentration of A, say, then the number of A-B collisions would

also double, because, in any given volume, there would be twice as

many A molecules that could collide with B molecules. Consequently,

the rate would increase by a factor of 2. Similarly, doubling the

concentration of B molecules would increase the rate twofold. Thus, we

can express the rate law as

Rate = k[A][B]


244

The reaction is first order in both A and B and obeys second-order

kinetics.

The collision theory is intuitively appealing, but the

relationship between rate and molecular collisions is more complicated

than you might expect. The implication of the collision theory is that a

reaction always occurs when an A and a B molecule collide. However,

not all collisions lead to reactions. Calculations based on the kinetic

molecular theory show that, at ordinary pressures (say, 1 atm) and

temperatures (say, 298 K), there are about 1 × 1027 binary collisions

(collisions between two molecules) in 1 mL of volume every second, in

the gas phase. Even more collisions per second occur in liquids. If

every binary collision led to a product, then most reactions would be

complete almost instantaneously. In practice, we find that the rates of

reactions differ greatly. This means that, in many cases, collisions alone

do not guarantee that a reaction will take place.

Any molecule in motion possesses kinetic energy; the faster it moves,

the greater the kinetic energy. When molecules collide, part of their

kinetic energy is converted to vibrational energy. If the initial kinetic

energies are large, then the colliding molecules

will vibrate so strongly as to break some of the chemical bonds. This

bond fracture is the first step toward product formation. If the initial

kinetic energies are small, the molecules will merely bounce off each


245

other intact. Energetically speaking, there is some minimum collision

energy below which no reaction occurs.

We postulate that, to react, the colliding molecules must have a

total kinetic energy equal to or greater than the activation energy (Ea),

which is the minimum amount of energy required to initiate a chemical

reaction. Lacking this energy, the molecules remain intact, and no

change results from the collision. The species temporarily formed by the

reactant molecules as a result of the collision before they form the product is

called the activated complex (also called the transition state).


246

Figure 8.3: Potential energy profiles for (a) exothermic and (b) endothermic reactions. These plots show the change in potential energy as reactants A and B are converted to products C and D. The transition state is a highly unstable species with a high potential energy. The activation energy is defined for the forward reaction in both (a) and (b). Note that the products C and D are more stable than the reactants in (a) and less stable than those in (b). Figure 8.3 shows two different potential energy profiles for the

reaction

A + B → C + D

If the products are more stable than the reactants, then the reaction will

be accompanied


247

by a release of heat; that is, the reaction is exothermic [Figure 8.3(a)].

On the other hand, if the products are less stable than the reactants,

then heat will be absorbed by the reacting mixture from the

surroundings and we have an endothermic reaction [Figure 8.3(b)]. In

both cases, we plot the potential energy of the reacting system versus

the progress of the reaction. Qualitatively, these plots show the

potential energy changes as reactants are converted to products.

We can think of activation energy as a barrier that prevents less

energetic molecules from reacting. Because the number of reactant

molecules in an ordinary reaction is very large, the speeds, and hence

also the kinetic energies of the molecules, vary greatly. Normally, only

a small fraction of the colliding molecules—the fastest-moving ones—

have enough kinetic energy to exceed the activation energy. These

molecules can therefore take part in the reaction. The increase in the

rate (or the rate constant) with temperature can now be explained: The

speeds of the molecules obey the Maxwell distributions shown in

earlier chapters. Compare the speed distributions at two different

temperatures. Because more high-energy molecules are present at the

higher temperature, the rate of product formation is also greater at the

higher temperature.


248

The Arrhenius Equation

The dependence of the rate constant of a reaction on temperature can

be expressed by this equation, now known as the Arrhenius equation:

in which Ea is the activation energy of the reaction (in kilojoules per

mole), R is the gas constant (8.314 J/K . mol), T is the absolute

temperature, and e is the base of the natural logarithm scale. The

quantity A represents the collision frequency and is called the frequency

factor. It can be treated as a constant for a given reacting system over a

fairly wide temperature range. The equation above shows that the rate

constant is directly proportional to A and, therefore, to the collision

frequency. Further, because of the minus sign associated with the

exponent Ea/RT, the rate constant decreases with increasing activation

energy and increases with increasing temperature. This equation can

be expressed in a more useful form by taking the natural logarithm of

both sides:

Equation above can take the form of a linear equation:


249

Thus, a plot of ln k versus 1/T gives a straight line whose slope m is

equal to –Ea/R and whose intercept b with the ordinate (the y-axis) is

ln A.

Example 8.10

The rate constants for the decomposition of acetaldehyde

were measured at five different temperatures. The data are shown in

the table. Plot ln k versus 1/T, and determine the activation energy (in

kJ/mol) for the reaction. This reaction has been experimentally shown

to be ―3/2‖ order in CH3CHO, so k has the units of 1/M1 /2 . s.


250

Strategy

Consider the Arrhenius equation written as a linear equation

A plot of ln k versus 1/T (y versus x) will produce a straight line with a

slope equal to –Ea/R. Thus, the activation energy can be determined

from the slope of the plot.

Solution

First, we convert the data to the following table:


251

Figure 8.4: Plot of ln k versus 1/T. The slope of the line is equal to –Ea/R


252

A plot of these data yields the graph in Figure 8.4. The slope of the line

is calculated from two pairs of coordinates:

From the linear form of Equation

An equation relating the rate constants k1 and k2 at temperatures T1 and

T2 can be used to calculate the activation energy or to find the rate

constant at another temperature if the activation energy is known. To

derive such an equation we start with equation below:

Subtracting ln k2 from ln k1 gives


253

The rate constant of a first-order reaction is 4.68 ×10-2 s-1 at 298 K. What

is the rate constant at 375 K if the activation energy for the reaction is

33.1 kJ/mol?

Strategy

A modified form of the Arrhenius equation relates two rate constants

at two different temperatures [see above]. Make sure the units of R and

Ea are consistent.

Solution

The data are

Substituting in Equation

gives


254

We convert Ea to units of J/mol to match the units of R. Solving the

equation gives

Reaction Mechanisms

As we mentioned earlier, an overall balanced chemical equation does

not tell us much about how a reaction actually takes place. In many

cases, it merely represents the sum of several elementary steps, or

elementary reactions, a series of simple reactions that represent the progress of

the overall reaction at the molecular level. The term for the sequence of

elementary steps that leads to product formation is reaction mechanism.

The reaction mechanism is comparable to the route of travel followed

during a trip; the overall chemical equation specifies only the origin

and destination.

As an example of a reaction mechanism, let us consider the

reaction between nitric oxide and oxygen:


255

2NO(g) + O2(g) → 2NO2(g)

We know that the products are not formed directly from the collision

of two NO molecules with an O2 molecule because N2O2 is detected

during the course of the reaction. Let us assume that the reaction

actually takes place via two elementary steps as follows:

In the first elementary step, two NO molecules collide to form a N2O2

molecule. This event is followed by the reaction between N2O2 and O2

to give two molecules of NO2. The net chemical equation, which

represents the overall change, is given by the sum of the elementary

steps:


256

Species such as N2O2 are called intermediates because they appear in the

mechanism of the reaction (that is, the elementary steps) but not in the overall

balanced equation. Keep in mind that an intermediate is always formed

in an early elementary step and consumed in a later elementary step.

The molecularity of a reaction is the number of molecules reacting

in an elementary step. These molecules may be of the same or different

types. Each of the elementary steps just discussed is called a

bimolecular reaction, an elementary step that involves two molecules. An

example of a unimolecular reaction, an elementary step in which only one

reacting molecule participates, is the conversion of cyclopropane to

propene is an example. Very few termolecular reactions, reactions that

involve the participation of three molecules in one elementary step, are

known, because the simultaneous encounter of three molecules is a far

less likely event than a bimolecular collision.


257

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Science

The behaviour of gases-Igori wallace