ENERGY CHANGES IN PHYSICAL AND CHEMICAL PROCESSES

Energy is needed before any work can be done. Many forms of energy exist, for example, chemical energy, electrical energy, heat energy, kinetic energy and potential energy. All the forms of energy are interconvertible.

Many physical and chemical changes are normally accompanied by heat changes. When substances are heated, they may undergo physical or chemical changes depending on the nature of the substance. In year one, it was shown that if certain substances are sufficiently cooled or heated at atmospheric pressure, they may exist in any of the three states of matter. The following experiment investigates what happens to the heat supplied to the system (apparatus and chemicals used) when solid water (ice) is melting or when liquid water is boiling.

Experiment 2.1: What happens to the heat supplied to water at its melting point and boiling point?

Half-fill a beaker with crushed ice and arrange the apparatus as in figure 2.1.

Record the steady temperature of the ice in the beaker before heating. Then gently heat the ice while stirring. Record the temperature of the ice every half a minute until it starts to melt. Continue heating gently and recording the temperature of the melting ice every half a minute while continuously stirring.

Now heat the melted ice (liquid water) more strongly while stirring continuously and record the temperature every half a minute until it starts to boil. Continue heating and recording the temperature of the boiling water for another two minutes. Record the results as in table 2.1.

Table 2.1Time (minutes) 0 ½ 1 1 ½ 2 2 ½ 3 3 ½ 4 4 ¼ 5 5 ½ 6Temperature of ice/water (0C)

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Answer the following questions 1. Use the experimental values from table 2.1, draw a graph of temperature (on the y-axis) versus time

(on the x-axis).2. Why does the temperature of the system remain constant when:

(a) The ice is melting,(b) The water is boiling?

3. What is the name of the heat energy required during (a) Melting, (b) Boiling?

Discussion

In the solid state bonds hold the water molecules together in specific patterns. However, because of the internal energy within the substances, the molecules in the solid state always vibrate in their fixed positions.

When the solid is heated, its temperature rises as the molecules absorb the heat energy. This heat energy absorbed is converted into vibrational energy and the molecules vibrate more vigorously until they break away from the solid structure. At this point the bonds holding the particles together in the solid state are broken and the solid starts to melt. The temperature remains constant as heating continues until all the ice has melted. The heat supplied during melting is absorbed by the water molecules and transformed into kinetic energy. The molecules now have a higher freedom of movement in the liquid state because of the absence of fixed three dimensional bonds. The amount of heat needed to convert a given amount of a solid substance into a liquid at its melting point is known as latent heat of fusion. Molar heat of fusion is the amount of heat energy required to convert one mole of a solid substance into a liquid at its melting temperature. The same amount of energy is liberated in the reverse process when one mole of this liquid substance turns into a solid at constant temperature.

When the liquid is heated, the molecules absorb heat energy and its temperature rises until boiling commences. The heat absorbed by the molecules is converted into kinetic energy which makes the molecules move faster as the heating continues until they acquire enough kinetic energy to break away from one another and move from the liquid state into the gaseous state. When boiling commences and heating continues, the temperature of the boiling water remains constant until all the liquid has changed into vapour. The boiling temperature remains constant because the heat energy absorbed by the molecules enables them to change from the liquid state into the gaseous state. Heat absorbed by a substance on changing from the liquid state into the gaseous state at constant temperature is known as latent heat of vaporization. When molar quantities of substances are used, then the heat absorbed by the particles as the substance changes from the liquid state into the gaseous state at constant temperature is called molar heat of vaporization.

Molar heats of fusion and molar heats of vaporization can be used to estimate the strength of bonds holding the particles together in solids and in liquids. If the forces holding the particles together in solids and in liquids. If the forces holding the particles together in the solid structure are strong, then the melting point of the substance is high, but when these bonds are weak, the melting temperature is low. This explains why the melting point of water (0C) is higher than that of ethanol (-1170C); and why the boiling point of water (1000C) is higher than that of ethanol (780C). The intermolecular forces holding water molecules together are stronger than those holding the ethanol molecules together. Latent heats of fusion and vaporization of water are thus higher than those for ethanol.

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Experiment 2.2: What happens when sodium hydroxide pellets and potassium nitrate crystals dissolve in water at room temperature?

Half-fill a small plastic beaker (or boiling tube) with water and note its temperature. To the water in the plastic beaker add two spatula-fuls of sodium hydroxide pellets. Using the thermometer, stir the mixture vigorously but carefully and record any temperature changes in the resulting solution. Rinse the thermometer with water.

Using another clean plastic beaker, repeat the experiment using potassium nitrate crystals instead of sodium hydroxide pellets.

Caution: Do not allow the sodium hydroxide pellets to be in contact with your skin because they are caustic and can burn the skin.

Answer the following question1. What observations were made when sodium hydroxide pellets and potassium nitrate crystals dissolved

in water? Explain.

Discussion

Experiment 2.2 shows that dissolving certain substances in water can be accompanied by heat loss or heat gain. For example, sodium hydroxide pellets dissolve in water with considerable release of heat energy while potassium nitrate crystals dissolve in water with absorption of heat. The temperature of the resulting sodium hydroxide solution is higher than the temperature of water at room temperature, while the temperature of the resulting potassium nitrate solution is lower than the temperature of water.

When sodium hydroxide pellets dissolve in water, energy is needed to separate the sodium and hydroxide ions. Energy is released when these ions are attracted by the water molecules and become hydrated in solution. Since there is more energy released during the formation of hydrated ions than the energy absorbed during the formation of ions from the solid, there is a net release of energy as the pellets dissolve and the temperature of the solution rises.

When solid potassium nitrate dissolves in water, the energy absorbed during the formation of potassium ions and nitrate ions is greater than the energy released when these ions become hydrated in solution. There is a net absorption of heat energy and the temperature of the resulting solution drops.

If a change is accompanied by a rise in temperature, energy is liberated to the surroundings; the reaction is exothermic. However, if the change is accompanied by a fall in temperature, energy is absorbed from the surroundings and the reaction is endothermic. The heat changes are normally referred to as enthalpy changes, denoted by the symbol ∆H. Figure 2.2 shows the relative energies of reactants and products in exothermic and endothermic reactions respectively. If H1 is the heat content of the reactants and H2 the heat content of the products, then ∆H = H2 – H1.

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For an exothermic reaction, H1 > H2 and ∆H will have a negative value to show that the heat evolved in the reaction is lost to the surroundings. For an endothermic reaction H1 < H2 and ∆H has a positive value showing that the products absorbs heat from the surroundings.

RELATIONSHIP BETWEEN HEAT OF SOLUTION, HYDRATION ENERGY AND LATTICE ENERGY

When one mole of a crystalline solid dissolves in water to give an infinitely dilute solution, (i.e. a solution which shows no change in its properties when more water is added) then the heat liberated or absorbed is known as the molar heat of solution. For example, when sodium chloride crystals dissolve in water, the change is represented as: -

NaCl(s) + Water → Na+(aq) + Cl-(aq).

This process involves first the interaction of water molecules with the ions in the lattice to such an extent that the lattice is broken and the ions begin to scatter in all directions in the water. These free ions then become hydrated (i.e. surrounded by water molecules). In breaking the crystalline lattice an amount of energy equivalent to the lattice energy is absorbed by the salt. However, during the formation of the hydrated ions, energy is released as the polar water molecules attract the salt ions.

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These changes are illustrated in figure 2.3.

Sodium chloride in the cold crystalline state is at a lower energy level than its component gaseous ions. The difference between those two energy levels is the lattice energy. Lattice energy has a positive sign in this case because energy is required to break the lattice. If the reverse process takes place, lattice energy has a negative sign. The three energies are related as shown by the following equation:

∆Hsoln = ∆Hlatt + ∆Hhyd

= 776 + (-771)= +5KJ mol-1

The overall enthalpy change of the solution, known as the enthalpy of solution ∆Hsoln depends on the magnitude of the two processes shown in the diagram and is the sum of the two enthalpy changes. In the

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case of sodium chloride, the endothermic process is slightly greater than the exothermic one, and the dissolving of sodium chloride in water is an endothermic process.

When the lattice energy and the hydration energy for a given salt are of the same magnitude, the salt will dissolve in water without any overall enthalpy change. When ions are hydrated, the amount of energy released depends on the size of the ion and the charge on the ion.

QUANTITATIVE DETERMINATION OF ENTHALPY CHANGE IN SOME CHEMICAL PROCESSES

In order to compare enthalpy changes for different processes, the temperature and pressure of the reactants and products should be the same before and after the reaction. The standard conditions of temperature and pressure adopted for reactions are 250C (298K) and one atmosphere pressure respectively. The enthalpy changes measured for molar quantities of substances under these conditions are called standard enthalpies of reaction, ∆HØ

298. The symbol ∆HØ298 implies that all the substances

involved in the reaction are in their normal physical states at 250C and one atmosphere pressure.

In this section some quantitative enthalpy changes during some chemical processes are investigated.

Experiment 2.3: What is the molar enthalpy change for the dissolution of concentrated sulphuric acid?Wrap a clean 250cm3 glass or plastic beaker with a newspaper leaf. Measure 50cm3 of tap water into the beaker and note the steady temperature of the water. Hold the beaker in a tilted position and very carefully pour 2cm3 of concentrated sulphuric acid into the beaker as shown in figure 2.4.

Caution: Concentrated sulphuric acid should ALWAYS be added to water and never nice versa.

Stir the mixture carefully but vigorously using the thermometer and record the highest temperature of the solution. Record the results as in table 2.2.

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Table 2.2 Concentrated sulphuric acid

Ammonium nitrate solid

Potassium nitrate solid

Sodium hydroxide pellets

Temperature of water (0C)Highest temperature of solution (0C)Change in temperature (0C)

The experiment can be repeated using 0.8g of ammonium nitrate, 1.01g of potassium nitrate or 0.4g of sodium hydroxide pellets instead of concentrated sulphuric acid but using 50cm3 of water in each case.

Answer the following questions 1. State whether the changes in the experiment above are exothermic or endothermic.2. Calculate the enthalpy change in each experiment done using the following information:

i. The density of concentrated sulphuric acid is 1.84gcm -3.ii. The density of each solution is 1gcm-3.iii. There is no change in volume when any of the solids dissolve in water.iv. The specific heat capacity of the solutions is 4.2kJ Kg-1K-1.

Discussion When a solution dissolves in water, the enthalpy change can be calculated using the formula:Mass of solution (m) x specific heat capacity of solution (C) x temperature change (∆T) = mC ∆T.

Suppose on dissolving 2cm3 of concentrated sulphurc acid in 98cm3 of water there is a temperature rise of 50C.

If it is assumed that the density of the solution is 1gcm-3 and that the final volume of the solution is (98 + 2)cm3 i.e. 100cm3;

Enthalpy change = mC ∆TWhere: m = 100cm3 x 1gcm-3 = 100g = 0.1kg C = 4.2 kJ kg-1 K-1

∆T = 5KEnthalpy change is:0.1kg x 4.2kJ kg-1K-1 x 5KOr 0.1 x 5 x 4.2 kJBut the mass of acid used = 1.84 x 2 = 3.68g and formula mass of sulphuric acid = 98g 3.68g of acid liberate 4.2 x 5 kJ

10398g of acid liberate 4.2 x 5 x 98kJ

10 3.68= 56kJ

This is the molar enthalpy change when concentrated sulphuric acid is dissolved in water as described in the experiment. The same procedure can be used to calculate the molar enthalpy change when ammonium nitrate, sodium hydroxide pellets or potassium nitrate solid is dissolved in a given amount of

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water. The dissolutions of concentrated sulphuric acid and sodium hydroxide pellets are exothermic while those of ammonium nitrate and potassium nitrate are endothermic.

Experiment 2.4: What is the heat of neutralization of hydrochloric acid solution by sodium hydroxide solution?Wrap a clean dry 250cm3 glass or plastic beaker with a newspaper leaf. Transfer exactly 50cm3 of 2.0M hydrochloric acid solution into the beaker. Note the steady temperature T1 of the acid solution. Usng another clean dry measuring cylinder, measure exactly 50cm3 of 2.0M sodium hydroxide solution and note its steady temperature T2. Carefully stir the contents of the beaker with the thermometer while the sodium hydroxide solution to the acid. Note the highest temperature T4 attained by the resulting solution.

The apparatus is set up as shown in figure 2.5.

Record the results as shown below.Temperature of the acid, T1 = 0CTemperature of the hydroxide, T2 = 0CAverage temperature of the two solutions, T1 + T2 = T3 = 0C

2The highest temperature of the mixture, T4 = 0CTemperature change, T = (T4 – T3) = 0C

Repeat the experiment using some of the following pairs:(a) 2M hydrochloric acid and 2M potassium hydroxide,(b) 1M sulphuric acid and 2M sodium hydroxide,(c) 2M nitric acid and 2M potassium hydroxide,(d) 2M ethanoic acid and 2M sodium hydroxide,(e) 2M hydrochloric acid and 2M ammonium hydroxide

Answer the following questions 1. For each experiment

(a) Write an ionic equation for each pair,(b) Calculate the heats of neutralization for each pair of reagents. Comment on the values obtained.

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2. If a little concentrated sulphuric acid accidentally spilled on someone’s skin, the place should be washed immediately with a large quantity of running water rather than neutralize the acid with an alkali. Explain.

Discussion Strong acids and strong alkalis are dissociated completely into their ions in water. During neutralization, hydrogen ions react with hydroxide ions to form water molecules with liberation of energy. The reactions can be represented by the following equations:

(a) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)(b) HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)(c) H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)(d) HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)(e) CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)(f) HCl(aq) + NH4OH(aq) → NH4Cl(aq) + H2O(l)

The fundamental reaction considered in acid-base neutralization process is shown by the equation:H+(aq) + OH-(aq) → H2O(l)

The other ions from the acid and the base remain in solution. A sample calculation on enthalpy of neutralization is shown below:Temperature of hydrochloric acid solution, T1 = 22.750CTemperature of sodium hydroxide solution, T2 = 22.800CAverage temperature of the acid and Alkali T1 + T2 = T3 = 22.780C2

2The highest temperature of the acid and alkali mixture, T4 = 36.400C.Temperature change, ∆T = T4 – T3 = 36.400C – 22.780C = 13.620CThe specific heat capacity of the solution = 4.2kJ kg -1 K-1 If in the experiment, 50cm3 of 2M hydrochloric acid are neutralized by -50cm3 of sodium hydroxide, then the final volume of the mixture = (50 + 50)cm3 = 100cm3.

If the density of the resulting solution is taken to be 1gcm -3 then heat evolved = mass of solution x its specific heat capacity x the temperature change = 100 x 4.2 x 13.62kJ

1000But 50cm3 of 2M hydrochloric acid contains 2 x 50 moles

1000= 0.1 moles of H+ ions.

Similarly 50cm3 of 2M sodium hydroxide contains 2 x 50 moles 1000

= 0.1 moles of H- ions

This implies that the two solutions neutralize each other completely. The heat liberated when one mole of each reagent is used = 5.72 x 1 kJ = 57.2kJ

0.1This neutralization reaction can then be written incorporating enthalpy change as shown:H+(aq) + OH-1(aq) → H2O(l), ∆ H = -57.2kJ.mol-11 mole 1 mole 1 mole (neutr)

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Most strong acid – strong base reactions liberate about 57.2kJ when they form one mole of water in the neutralization process. When one of the reactants used is a weak acid or weak base the molar enthalpy of neutralization may be less than is the case when strong acids are used with strong bases. For example, when one mole of hydrochloric acid (strong acid) is neutralized by one mole of ammonium hydroxide (weak base), only 51.4kJ of heat energy is evolved for every mole of water formed.

HCL(aq) + NH4OH(aq) → NH4Cl(aq) + H2O(l),∆HØ (neutr) = -51.4kJ mol-1

Similarly if ethanoic acid (weak acid) is neutralized by sodium hydroxide (strong base) only 55.2kJ of heat is produced for every mole of water formed.CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l),∆H (neutr) = - 55.2kJmol-1

The weak acids, such as ethanoic acid or weak bases such as ammonium hydroxide, are only partly ionized in solution. Some energy is therefore used up to ionize them completely before the neutralization is complete.Enthalpy change obtained when a weak acid and a weak base react can be very low. For example, hydrogen cyanide reacts with ammonium hydroxide to produce only 5.4kJ.

HCN(aq) + NH4OH(aq) → NH4CN(aq) + H2O(l), ∆HØ (neutr) = -5.4kJ mol-1

Experiment 2.5: What is the molar enthalpy change in the reaction between copper (II) ions and zinc?

Wrap a plastic cup or a glass beaker with a newspaper leaf and transfer into it 25.0cm3 of 0.2M copper sulphate solution. Note the steady temperature of the solution. Carefully transfer 0.5g of zinc powder into the plastic cup and stir carefully with the thermometer. See Fig. 2.6. Record the highest temperature attained by the solution.

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Sample of results:Mass of zinc powder weighed = gTemperature of copper sulphate solution T1 = 0CHighest temperature of the mixture T2 = 0CChange in temperature, ∆T = T2 – T1 = 0C

The experiment may be done using iron powder in place of zinc.

Answer the following questions.1. Is the reaction endothermic or exothermic?2. What other observations were made? Explain.3. Why is it necessary to use excess of zinc powder (or iron filings) in this reaction?4. Calculate the enthalpy change for this reaction given that:

(a) The density of the solution formed is 1 gcm-3,(b) The volume of the solution remains unchanged after the reaction,(c) The specific heat capacity of the solution is 4.2kJ kg-1 K-1,

(Cu = 63.5, S = 32, O = 16, Zn = 65, Fe = 56)

Discussion From the equation of the reaction Zn(s) + CuSO4(aq) Cu(s) + ZnSO4(aq) it can be seen that zinc displaces copper ions from its solution as shown by the following ionic equation:

Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq).

The hydrated copper (II) ions in copper sulphate solution give the solution its blue colour. During the reaction, these copper (II) ions in solution are replaced by the colourless zinc ions, and so the blue colour of the solution fades as brown deposits of copper metal are formed in the plastic cup.

The following results were obtained from an experiment.

Initial temperature of copper sulphate solution, T1 = 230CHighest temperature of the mixture T2 = 330CChange in temperature, ∆T = T2 – T1 = 100C

Volume of copper sulphate solution used = 25.0cm3

Mass of zinc powder taken = 0.6gSpecific heat capacity of the solution = 4.2kJ kg-1K-1

Density of the solution = 1gcm-3

In this reaction = 25 kg x 10K x 4.2kJ kg-1k-1 = 1.050kJ1000

No. of moles of copper ions displaced = 25 x 0.2 1000

= 5 x 10-3 moles

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When 5 x 10-3 moles of copper ions are displaced in solution by zinc atoms, 1.050kJ of heat energy is released.Displacement of one mole of copper (II) ions would give

1.050 x 1 kJ = 210kJ5 x 10-3

The complete equation for this reaction would be:Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq), ∆H = 210kJ mol-1.

The theoretical value for the heat liberated in this reaction is 216kJ mol-1. The experimental value obtained in the above experiment is low because the heat lost to the surroundings and by the apparatus is not accounted for in the calculation. The reaction between copper sulphate solution and zinc metal is an oxidation – reduction reaction where the blue copper (II) ions are reduced to brown copper metal while the zinc atoms are oxidized to colourless zinc ions. The zinc atoms displace the copper (II) ions from the copper sulphate solution. This type of reaction is called displacement reaction.

Experiment 2.6: What is the molar enthalpy of combustion of ethanol?

Pour exactly 100cm3 of distilled water into a tin. Nearly half-fill a small bottle with ethanol and close the bottle with a rubber bung fitted with a clean wick. Weigh the bottle and its contents. Arrange the apparatus as shown in figure 2.7.

Light the wick. Stir the water carefully with the thermometer and after the temperature of the water has risen by about 100C, extinguish the flame; note the temperature of the water and reweigh the bottle.

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Record the results.Initial mass of bottle with its contents, W1 = gFinal mass of bottle with its contents, W2 = gMass of ethanol burned, W1 – W2 = gFinal temperature of water, T2 = 0CInitial temperature of water, T1 = 0CRise in temperature of water, ∆T = T2 – T1 = 0CMass of water used = gSpecific heat capacity of water = 4.2Jg-1k-1

Answer the following questions 1. Using your results, calculate the molar heat of combustion of ethanol.2. The theoretical enthalpy of combustion of ethanol is 1368kJ mol -1. Why does the value calculated from

experimental results differ from this?3. Write the equation for the complete combustion of ethanol, showing the enthalpy change for the

reaction.

DiscussionIn a combustion experiment the following results were obtained:Initial mass of bottle + its contents = 28.25gFinal mass of bottle + its contents = 28.02gMass of ethanol burnt = 0.23gFinal temperature of water = 43.00CInitial temperature of water = 28.00CRise in temperature of water = 150CMass of water heated = 100g

Specific heat capacity of water = 4.2kJ kg-1K-1

= x x

= kJ

= 6.3kJFormula mass of ethanol, CH3, CH2OH = 46gBut 0.23g of ethanol liberated 6.3kJ of heat 46g of ethanol would liberate 6.3 x 46 = 1260 kJ mol-1

0.23

The heat of combustion ∆Hc for ethanol would be given as ∆HcØ (CH3CH2OH) = -1260 kJ mol-1 and the

equation for the combustion reaction is:CH3CH2OH(l) + 302(g) → 2CO2(g) + 3H2O(l), ∆Hc = - 1260kJ mol-1

The determination of exact heats of combustion of substances requires the knowledge of the heat capacities of the materials used in constructing the apparatus. Complete combustion of the substance must be ensured. The apparatus is calibrated so that its heat capacity is used in the calculation. Many

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Heat evolved in this reaction

Mass of water

Specific heat capacity

Temperature change

100 x 4.2 x 151000

experimentally determined heats of combustion are usually lower than the expected theoretical values because of heat loss to the surroundings. The theoretical heat of combustion for ethanol is about 1370 kJ mol-1.

The heat produced in a combustion process can be transformed into various forms of energy which can be used to perform useful work, for example in internal combustion engines, rocket propulsion etc.

FUELS

A fuel is a substance that releases energy when burned. The energy produced in such a combustion process is used either directly as heat, or is converted into other forms e.g. electrical energy. Fuels can be solid, such as coal, charcoal, and wood; they can be liquids such as petrol, kerosene and fuel oil; or they can be gases like natural gas which consists mainly methane.

Heats of combustion are especially important because they provide one way of comparing the heat energy produced when equal quantities of different fuels are burned. Generally not all heat energy liberated by a combustion process is utilized. For example, when coal is burned in a power station, only about 50% of the heat energy liberated is converted into electricity.

Heating values of fuelsHeating value has the units kJg-1 (kilojoules per gram). It is obtained by dividing the molar enthalpy of combustion by the formula mass of the fuel. For example, ethanol has an enthalpy of combustion of -1360kJ mol-1. On dividing this by the formula mass of ethanol, 46, we get the heating value of ethanol as 30kJg-1. The heating value of a fuel is therefore the amount of heat energy given out when a unit mass, or a unit volume in case the fuel is a gas, is completely burned in oxygen. The heating values for some other fuels are given in table 2.3.

Table 2.3: The heating values for some fuels.Fuels Heating value in kJ/gSolid fuels:Charcoal 33Coal 29Wood 17

Liquid fuels Ethanol 30Fuel oil 45Paraffin 48

Gaseous fuels Methane (natural gas) 55Propane 50Butane 48

How to choose a fuel The choice of a fuel depends on the purpose it is to be used for. The following points have to be considered in each case.1. Heating value2. Ease and rate of combustion 3. Availability

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4. Ease of transportation 5. Ease of storage 6. Environmental effects

For example, if a fuel is to be used for domestic heating, then a slow burning, cheap, easily transportable and non-polluting fuel is preferred. Wood and charcoal are usually chosen because they are cheap and relatively easily available.

Suppose we want a fuel that will propel a rocket at say 50,000 kilometres an hour away from the earth’s space, then the fuel must be able to produce a huge amount of heat energy in a very short time. The fuel burns to produce hot gases which are directed towards the back of the rocket where they expand rapidly and produce a huge thrust on the rocket while moving away from it. An example of such a fuel is methylhydrazine, CH3NHNH2.

Environmental effects of fuelsFossils fuels contain sulphur compounds and on burning they produce sulphur dioxide, which is a toxic gas. Sulphur dioxide can interfere with many forms of biological systems on land and in water. When it dissolves in rain water, it becomes one of the components of acid rain which can corrode iron sheets and destroy buildings. Petrol, one of the products obtained from the fractional distillation of crude oil, is a hydrocarbon that has had lead compounds added to it. If it is burnt in a limited amount of air such as in the internal combustion engine, it produces some unburned hydrocarbons, carbon monoxide, lead compounds as well as oxides of nitrogen. These gaseous products are released into the atmosphere as dangerous pollutants. Carbon monoxide, lead compounds, oxides of nitrogen and incompletely burnt hydrocarbons are especially harmful to both plant and animal life.

ENERGY LEVEL DIAGRAMS AND HESS’S LAW

The standard heats of formation for many compounds for example, carbon dioxide and magnesium oxide, can be measured directly using a calorimeter. However, there are many compounds, for example, carbon tetrachloride (tetrachloromethane) and methane for which the standard heats of formation, ∆H f

Ø cannot be measured directly in a calorimeter. The standard heats of formation of these compounds can only be obtained indirectly. One such method involves measuring the heat of combustion of the compound and of its constituent elements.

Another method involves knowledge of intermediate steps followed in order to obtain the products. These values are then linked in an energy level diagram or energy cycle diagram with the heat of formation of the compound. Figures 2.8 (a) and (b) show the energy level diagram and the energy cycle diagram respectively for the formation of a carbon dioxide with carbon monoxide as an intermediate product during the formation of carbon dioxide from its elements in their standard states.

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Both the energy level and the energy cycle diagrams for the formation of carbon dioxide show that there are two ways of oxidizing graphite to carbon dioxide. It can be done by either burning graphite in excess oxygen, or by first burning the element in a limited amount of oxygen to form carbon monoxide, and then burning the carbon monoxide gas formed in more oxygen to convert it to carbon dioxide. According to Hess’s law the overall enthalpy change for the conversion of graphite to carbon dioxide does not depend on the route taken, and so we can connect the energy changes from the energy level diagram as follows: -

∆H1 = ∆H2 + ∆H3 But ∆H1 = ∆HØ

c (C(graphite)) = -393 kJ mol-1and ∆H3 = ∆HØ

c (CO(g)) = -283 kJ mol-1hence ∆H2 = ∆HØ

f (CO(g)) = ∆H1 – ∆H3 = -393 – (-283) kJ mol-1= -393 + 283 Kj mol-1= -110 kJ mol-1

The foregoing argument is the basis of Hess’s law of “Constant Heat Summation” which states that when substances A and B react to give product(s), then the energy change that accompanies the conversion of the reactants into the product(s) does not depend on the route followed.

Hess’s law is basically an application of the more fundamental law of conversion of energy which states that matter and energy remain constant during a chemical reaction. Hess’s law therefore provides an invaluable basis on which the standard heats of formation can be calculated. It is also used to calculate the energy changes in various types of reactions because the energy cycle diagrams (the thermochemical cycle) gives the connection of the various enthalpy changes for a particular reaction.

The following example illustrates the use of Hess’s law to calculate the heat of formation of butane.

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Butane is a compound containing carbon and hydrogen atoms. A butane molecule cannot be obtained by a direct reaction of elemental carbon and hydrogen. If butane, carbon and hydrogen are each completely burned in oxygen, then one mole of each substance liberates 2877 kJ, 393 kJ and 286 kJ of energy respectively.

(a) Write the equation for the formation of butane from its elements.(b) Write the equations for the combustion of:-

i. Butaneii. Carboniii. Hydrogen

Showing the heat changes.(c) Draw an energy level diagram to show the connection of all the reactions listen in (a) and (b) above.(d) Draw an energy cycle diagram and use it to calculate the heat of formation of butane from its elements.

Solution (a) 4C(s) + 5H2(g) → C4H10(g)(b) (i) C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(l), ∆HØ

c = -28.77 kJ mol-1(ii) C(s) + O2(g) → CO2(g), ∆HØ

c = -393 kJ mol-1(iii) H2(g) + ½ O2(g) → H2O (l), ∆HØ

c = -286 kJ mol-1

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From the energy level diagram,∆H1 = ∆H2 + ∆H3

where ∆H1 = enthalpy of combustion of 4 moles of graphite and 5 moles of hydrogen.∆H2 = enthalpy of formation of butane.∆H3 = enthalpy of combustion of one mole of butane.

(d) The changes can be connected in our energy cycle as shown below:

Hence ∆H2 = ∆H1 – ∆H3 = 4(-393) + 5(-286) – (-2877) kJ mol-1= -3002 + 2877 kJ mol-1= -125 kJ mol-1

The heat of formation of butane is -125kJ mol-1. Similar calculations can be used to calculate the enthalpy changes for other compounds that cannot be formed in a calorimeter.

HEAT EXCHANGERS

In industries the heat generated during certain process is normally removed using heat exchangers. This ensures that the temperature at which the process is taking place is maintained within a range that favours best yields.

Heat exchangers are constructed using good conductors of heat. The hot and cold liquids or gases conveyed through the exchanger are separated by the walls of the exchanger. Usually the hot liquid or gas is conveyed in a small pipe enclosed in a larger one where the cold liquid or gas is conveyed in the opposite direction. This way, the heat in some industrial processes is recycled.

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Summary 1. The energy changes in chemical reactions are due to bond breaking in the reactants and new bond

formation in the products.2. The amount of heat liberated or absorbed during a chemical reaction is the difference between the

energy content of the reactants and the products.3. In an exothermic reaction, heat is produced along with the products and is lost to the surroundings as

the system cools to attain room temperature. ∆H for such a reaction is given a negative value.In an endothermic reaction, the products form with a reduction in temperature and the products gain heat fro the surroundings in order to attain room temperature. ∆H for such a reaction is given to a positive value.

4. The standard conditions under which the enthalpy changes are normally measured are 250C (298K) and one atmosphere pressure.

5. The standard heat of reaction, ∆HØ, is the amount of heat absorbed or evolved, when molar quantities

of reactants in the equation react to give products under standard conditions of temperature and pressure.

6. The standard heat of formation, ∆HØf of a substance is the amount of heat absorbed or liberated

when one mole of the substance is formed from its elements under standard conditions.7. The standard heat of combustion ∆HØ

c is the enthalpy change when one mole of the substance is completely burned in oxygen.

8. The heat of formation of a compound gives a measure of stability of that compound. The more negative ∆HØ

f , value is, the more stable is the compound formed.9. The greater the latent heats of fusion and vaporization, the stronger are the bonds holding the particles

together in solids and liquids.10. In nature, systems tend to change from a state of higher energy to one of lower energy. However,

many endothermic reactions occur readily because systems also tend to move from a more orderly state to a less orderly one during chemical reactions.

Further Work 1. In an experiment to compare the heats of combustion of methanol, CH3OH, and ethanol, CH3CH2OH, it

was established that the heat evolved during combustion of methanol is about 970 kJ mol -1 while that for ethanol is about 1250 kJ mol-1.(a) Write equations, including the heat changes, to show the complete combustion of methanol and

ethanol.(b) Suggest a reason why one mole of ethanol liberates more heat energy on complete combustion

than one mole of methanol.2. When 8.0g of sulphur is completely burned in oxygen in a calorimeter, the heat evolved raises the

temperature of 500cm3 of water by 350C. Calculate the heat of combustion of sulphur. (The specific heat capacity of water is 4.2 kJ kg-1K-1, S = 32)

3. Sulphur dioxide reacts completely with oxygen to form sulphur trioxide, liberating 185 kilojoules of energy.(a) Write the equation for this reaction, showing the heat change.(b) What is the heat of combustion when one mole of sulphur dioxide is converted to sulphur trioxide in

excess oxygen?(c) What would be the heat evolved if only 0.2 moles of sulphur trioxide were formed?

4. Some of the fuels commonly used at home are charcoal, wood, kerosene, natural gas, biogas, because they burn in oxygen to give out heat energy. Comment on the suitability of these substances as fuels.

5. Why are coal and oil described as fossils fuels?6. Why do many athletes take glucose after a race?7. Assuming that an oil reserve has been discovered in Kenya;

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i. What would be the advantage of extracting the newly discovered oil?ii. How would this affect the use of other fuels commonly used in Kenyan rural and urban homes?

8. (a) Define the standard heat of formation of a substance.(b) Ethanol, CH3CH2OH, cannot be prepared directly from its elements and so its standard heat of formation must be obtained indirectly.

i. Write an equation for the formation of ethanol from its elements in their normal physical states at standard conditions of temperature and pressure.

ii. Draw an energy cycle diagram linking the heat of formation of ethanol with its heat of combustion and the heats of combustion of its constituent elements.

iii. If the following heats of combustion are given;∆Hc (graphite) = -393 kJ mol-1∆Hc (H2(g)) = -286 kJ mol-1∆Hc (ethanol) = -393 kJ mol-1

Calculate the heat of formation of ethanol ∆HØf (C2H5OH(l)

9. One mole of butane (C4H10) burns completely in oxygen and liberates 2877 kJ.(a) Write the equation for the combustion of butane.(b) Draw an energy level diagram for the reactions concerned. Draw an energy level cycle for these

changes.(c) If the following heats of combustion are given

∆HØc (graphite) = -393 kJ mol-1

∆HØc (H2(g)) = -286 kJ mol-1

∆HØc (C4H10) = -2877 kJ mol-1

Calculate the heat of formation of butane,C4H10 from its elements in their normal states at standard conditions of temperature and pressure.

10. Use the table of enthalpies of hydration and lattice energies in the appendix to answer the questions that follow:(a) Why are the hydration energies of both anions and cations negative?(b) Comment on the trend of hydration energies of:

(i) The singly charged positive ions,(ii) The singly charged negative ions,(iii) The positive ions carrying more than one charge.

(c) Calculate the hydration energy for magnesium bromide and aluminium chloride.(d) Sodium, magnesium and aluminium belong to the same period while fluorine, chlorine, bromine

and iodine belong to the same group in the periodic table. Explain the trend in hydration energy values of the cations and anions formed from these atoms.

(e) Calculate enthalpies of solution of:(i) Potassium iodide(ii) Lithium chloride

(f) Calculate the lattice enthalpy of:(i) Aluminium chloride(ii) Sodium chloride (iii) Calcium chloride

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