system, law of thermodynamic

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Some thermodynamical termsSome common terms, which are frequently used in the discussion of energetics need tobe known.

System'A specified part of the universe which is under investigation is called the system'. Thesystem is separated from the rest of the universe by a definite (real or imaginary)boundary.

TYPES OF SYSTEMS

Systems are of various types depending upon the exchange of mass and energy and theconstituent between the system and the surroundings.

Isolated systemA system, which can neither exchange mass nor energy with surroundings, is called anisolated system. They are also called insulated systems. Hot coffee (in contact with its

vapour) in a closed and insulated thermos is an example of isolated system. Since thevessel is closed, matter can neither enter nor leave the vessel. Moreover, as the vessel iswell insulated, heat can neither leave the system nor enter from the surroundings.

Closed systemA system, which can exchange energy but not mass with the surroundings is called aclosed system. For example, boiling water in a closed steel vessel is an example of aclosed system. The energy can be gained or lost (through the steel walls) but not matter.Similarly, all reactions carried out in a closed container are examples of closed systems.

Open systemA system, which can exchange matter as well as energy with the surroundings is calledan open system. All reactions carried out in open containers are examples of opensystems. Evaporation of water in a beaker or hot coffee in a cup represents an opensystem. Here vapour of water or coffee (matter) can leave the system and escape intoatmosphere. The heat energy required for this purpose is absorbed from thesurroundings. All physical and chemical processes taking place in open in our daily lifeare open systems because they are continuously exchanging matter and energy with thesurroundings.

Homogeneous systemA system is called homogeneous if physical properties and chemical composition areidentical throughout the system. A pure gas or consistent mixture of gases e.g., anoxygen cylinder, or a pure liquid or solid in a container are examples of homogeneous

systems.Heterogeneous systemA system is said to be heterogeneous if it consists of parts separated by definiteboundaries, each of which has different physical and chemical properties. A mixture offertilizer granules (N P K) or ice with water are typical examples of this system.

STATE OF A SYSTEM

The condition of the system when its macroscopic properties have definite measurablevalues, is the state of a system. If any of the macroscopic properties of the system


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changes such as its temperature, pressure etc., the state of the system is also said tochange. To define the state of the system all its macroscopic properties need not bespecified.

State variableA measurable property required to describe the state of a system is called state variable.

For example, temperature (T), pressure (P), volume (V) etc. are the state variables. Forany system, a certain minimum number of variables are sufficient to define its state, asthe other variables become automatically fixed.

For instance, in a system consisting of an ideal gas, the state may be defined by onlythree variables such as temperature (T), pressure (P) and volume (V). The values ofother variables such as amount of gas, density, etc. will be definite and can be easilycalculated.

State functionsA property whose values depend on only the state of the system and not on the path bywhich the change from initial to final state is brought about is called state function. Thechange in the value of the state functions depends only upon the initial and final state

the system.

Some common state functions used in thermodynamics are, pressure, (P), volume (V),temperature (T), internal energy (E or U), enthalpy (H), entropy (S) free energy (G) etc.

State propertiesState variables and functions may posses either extensive or intensive properties.

Extensive properties are those, whose values depend on the quantity of mattercontained in any system or size of the system. Typical properties are mass, volume, area,energy, number of moles, enthalpy, entropy etc.

Intensive properties are those, which depend on the nature of the substance system butis independent of its amount / size in the system. Examples are density, viscosity,

surface tension, temperature, pressure, boiling point etc.State processesIsothermal process is that in which the temperature of the system remains constant.

Adiabatic process is where there is no exchange of heat between the system and thesurrounding.

Isobaric process is when the pressure on the system remains constant during anyoperation.Isochroic process is that in which the volume of the system does not change.

Cyclic process is one when the system returns to its initial state after having undergone aseries of change.Reversible process occurs when a process is carried out slowly so that the system andthe surrounding are always in equilibrium.

Irreversible process is that which is carried out rapidly so that the system does notreturn to its initial state.SurroundingsThe region outside the boundary of the system is termed as surrounding. The rest of theuniverse, which is not the part of the system, is separated by a real and imaginaryboundary. The boundary sets the limits of the system.


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For example, while studying the effects of adding acid to water in a beaker, the acidwater mixture forms the system, the walls of the container is the boundary and the restof the part that is around it is the surrounding.

ENERGY TERMS IN THERMODYNAMICSEvery system has a definite amount of energy. It can exchange energy (lose or gain) with

the surroundings in a variety of ways. In chemical systems the two important modes oftransference of energy between the system and the surroundings are heat and work.

Heat (Q)Energy exchanged between the system and the surroundings when they are at differenttemperatures is commonly known as heat. If a system is at a higher temperature thanthe surroundings, then there is a flow of heat (or energy) from system to surroundings,causing a decline of the systems temperature and an increase in the surroundingtemperature. These processes continue till the fall in temperature of the system and risein temperature of the surroundings, become equal.

Fig: 5.1 - Heat exchange between system and surroundings

Work (W)Work is another mode of transference of energy. Work is said to be done if the point ofapplication of force undergoes displacement in the direction of the force. If the systemloses energy, we say that the work is done by the system. Alternatively, if the systemgains energy, we say that work is done on the system.

For example, if a gas, enclosed in a cylinder with a piston, has a higher pressure than thesurroundings, the piston will move upward until the pressure inside and outside becomeequal. The gas expands against a constant external pressure 'P' and its volume changesby an amount equal toDV. The energy transfer that takes in this case is called work. Atthis step the work is done by the system on the surroundings. This is given as:

Work done by the system = PDV.Alternatively, if the system is at lower pressure, piston will be pushed down until thepressure of the system becomes equal to that of the surroundings. In this case work isdone on the system by the surroundings.

Obviously, if there is no change in volume, i.e., DV = 0, no work is done by the system,i.e. work = 0.


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In addition to these two modes, radiant energy and electrical energy are also modes oftransference of energy between the system and the surroundings.

Units of heat and workThe heat changes are measured in calories (cal), kilo calories (kcal), joules (J) orkilojoules (kJ). These are related as:

1 cal = 4.184J1 kcal = 4.184kJThe S.I. unit of heat is joule or kilojoule

Work is measured in terms of ergs or joules.I Joule = 107 ergs.

1calorie = 4.184 x 107 ergsThe S.I. unit of work is joule.

SIGN CONVENTIONS FOR HEAT AND WORK

The signs of 'w' and 'q' are related to the internal energy change. When 'w' or 'q' ispositive, it means that energy has been supplied to the system as work or as heat. The

internal energy of the system in such a case increases. On the other hand, if 'w' or 'q' isnegative, it means that energy has left the system as work or heat. The internal energy ofthe system decreases. The signs of 'q' and 'w' are:

Heat absorbed by the system= q positive

Heat evolved by the system= q negativeWork done on the system= w positive

Work done by the system= w negativeProblem1. Assuming ideal behaviour, calculate the work done when 1.6 mole of water evaporatesat 373 K against the atmospheric pressure of 760 mm of Hg.

SolutionVolume of 1.6 mole of water at 373 K in gaseous state

Volume of 1 mol = 18 g of liquid water (density = 1 g ml-1)= 18 x 1.6 x 10-3L = 0.0288 L

Now work done (W) = -P(V2 - V1)= -1(48.93 - 0.0288) = - 48.90 atm L

= - 48.90 x 101.325 J = - 4954.8 J.

Exchange of energy between the system and surroundingExothermic reactionsWhen chemical reactions take place, energy is either absorbed or evolved. A reaction issaid to be exothermic, if heat is 'evolved'. The heat energy produced during the reactions


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is indicated by writing + q or more precisely by giving the actual numerical value on theproducts side. In general, exothermic reactions may be represented as:

In the exothermic reactions, the enthalpy of the products will be less than the enthalpyof the reactants so that the enthalpy change is negative as shown below:

DH = Hp - Hr (since Hp r)DH = -ve

Some examples of reactions evolving energy are given below: Coal is burnt simply for the large amount of energy available during its combustion:

When water is added to quick lime (CaO) for preparing whitewash, a considerableamount of heat is produced during the reaction. The heat produced warms up thewater.

When we add dilute hydrochloric acid to a test tube containing granulated zinc,hydrogen gas is evolved. The reaction is accompanied by evolution of heat.

One mole of carbon reacts with one mole of oxygen to form one mole of carbondioxide and 393.5 kJ of heat is evolved at constant temperature and constantpressure. The reaction may be expressed as:

or,

Similarly,

Endothermic reactions

Reactions that are accompanied by absorption of energy from the surroundings arecalled endothermic reactions. Since heat is added to the reactants in these chemicalreactions, it is indicated by either putting +q or by writing the actual numerical value of

heat on the reactants side.

Alternatively, this may be written as,

We know DH = the heat absorbed at constant temperature and constant pressure.Because of the absorption of heat, the enthalpy of the products will be more than the


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enthalpy of the reactants. Consequently, DH will be positive (+ ve) for the endothermicreactions.DH = Hp - Hr since (Hp > Hr)DH = +veFor example: When a small quantity of ammonium chloride (NH4Cl) is dissolved in water in a

test tube, the tube becomes colder than before. During this chemical reaction heat isabsorbed from the surroundings (test tube).

When crystals of sodium thiosulphate (Na2S2O3.5H2O) commonly called hypo, aredissolved in water a cooling effect takes place.

One mole of nitrogen reacts with one mole of oxygen to form two moles of nitricoxide. 180.5kJ of heat is absorbed at constant temperature and the reaction may beexpressed as:

or,

Similarly,

Fig: 5.2 - Enthalpy changes for exothermic and endothermic reactionsSimilarly, if we consider heat change at constant volume and temperature,

DE is -ve for exothermic reactions and DE is +ve for endothermic reactions.Thus it may be concluded that:For exothermic reactions, DH or DE = -veFor endothermic reactions, DH or DE = +veProblem4. Calculate, Q, W, DE and DH for isothermal reversible expansion of one mole of idealgas from initial pressure of 1 bar to 0.1 bar at a constant temperature of 273 K.


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SolutionFor isothermal reversible expansion of ideal gas

q = -w = -(- 5227.16) = 5227.16 JDE = 0DH = 0.OTHER ENERGY CHANGES IN CHEMICAL REACTIONS

Energy changes in chemical reactions can also take place in forms other than heat,such as light, electricity, mechanical energy, etc. Some common examples are: The burning of petrol or diesel in motor engines of cars, trucks, tractors or buses

produces mechanical energy, which is used in running these vehicles.

Chemical reactions taking place in batteries produce electrical energy to runtransistors radios, torches and watches, etc. For example, in a Daniel cell, thechemical reaction between zinc metal and copper sulphate solution is accompaniedby electrical energy.

PhotosynthesisIn this process chlorophyll in green plants converts carbon dioxide and water intoglucose and oxygen, energy is provided by the energy of sunlight,

This reaction is very important for the life of plants

Different types of thermodynamic processThermodynamic ProcessWhenever one or more of the properties of a system change, a change in the state of the systemoccurs.

The path of the succession of states through which the system passes is called thethermodynamicprocess. One example of a thermodynamic process is increasing the temperatureof a fluid whilemaintaining a constant pressure. Another example is increasing the pressure ofa confined gas whilemaintaining a constant temperature. Thermodynamic processes will bediscussed in more detail in laterchapters.Cyclic ProcessWhen a system in a given initial state goes through a number of different changes in state (goingthroughvarious processes) and finally returns to its initial values, the system has undergone acyclic process orcycle. Therefore, at the conclusion of a cycle, all the properties have the samevalue they had at thebeginning. Steam (water) that circulates through a closed cooling loopundergoes a cycle.


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Reversible ProcessA reversible process for a system is defined as a process that, once having taken place, can bereversed,and in so doing leaves no change in either the system or surroundings. In other wordsthe system andsurroundings are returned to their original condition before the process took place.In reality, there are notruly reversible processes; however, for analysis purposes, one usesreversible to make the analysissimpler, and to determine maximum theoretical efficiencies.Therefore, the reversible process is anappropriate starting point on which to base engineeringstudy and calculation.Although the reversibleprocess can be approximated, it can never be matched by real processes.One way to make realprocesses approximate reversible process is to carry out the process in aseries of small or infinitesimalsteps. For example, heat transfer may be considered reversibleif it occurs due to a small temperaturedifference between the system and its surroundings. Forexample, transferring heat across a temperaturedifference of 0.00001 F "appears" to be more reversible than for transferring heat across a temperaturedifference of 100 F. Therefore, bycooling or heating the system in a number of infinitesamally smallsteps, we can approximate areversible process. Although not practical for real processes, this method isbeneficial forthermodynamic studies since the rate at which processes occur is not important.Irreversible ProcessAn irreversible process is a process that cannot return both the system and the surroundings totheiroriginal conditions. That is, the system and the surroundings would not return to their

The first law of thermodynamicsWe have seen that heat is just a form of energy. A system can be given energy either bysupplying heat to it (by placing it in contact with a hotter object) or by doing mechanicalwork on it. Consider an ideal gas in a cylindrical container, fitted with a piston as shownin the figure given below. The piston is fixed in its position and the walls of the cylinderare kept at a temperature higher than that of the gas. The gas molecules strike the walland rebound. The average kinetic energy of a molecule in the wall is greater than theaverage kinetic energy of a gas molecule. Thus, on collision, the gas molecules receiveenergy from the molecules of the wall. This increased kinetic energy is shared by othermolecules of the gas, and in this way, the total internal energy of the gas increases.

Next, consider the same initial situation but now with the walls at the same temperatureas the gas. When the piston is pushed slowly to compress the gas, the gas molecule

collides with the piston coming towards it and the speed of the molecule increases oncollision (assuming elastic collision, v2= v1 + 2u in the figure below). This way theinternal energy of the molecules increases as the piston is pushed in.


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We see that the total internal energy of the gas can increase either due to thetemperature difference between the walls and the gas (heat transfer) or due to themotion of the piston (work done on the gas).

In a general situation, both modes of energy transfer may happen together. As anexample, consider a gas kept in a cylindrical can, fitted with a movable piston. If the canis kept on a heater, the hot bottom of the cylinder supplies heat to the gas. If the pistonis pushed out to some distance, as the piston moves out, the gas does work and the gasloses that amount of energy. Thus, the gas gains energy as heat gets supplied to it and itloses energy as work is done by it.Suppose, in a process, an amount DQ of heat is given to the gas and an amount DW ofwork is done by it, the total energy of the gas must increase by (DQ - DW). As a result,the entire gas, together with its container, may start moving (systematic motion) or theinternal energy (random motion of the molecules) of the gas may increase. If the energydoes not appear as a systematic motion of the gas, then this net energy (DQ - DW) mustgo in the form of its internal energy. If we denote the change in internal energy as DU,we get

Equation (1) is the statement of the first law of thermodynamics. In an ideal monatomicgas, the internal energy of the gas is simply translational kinetic energy of all itsmolecules. In general, the internal energy may get contributions from the vibrationalkinetic energy of molecules, rotational kinetic energy of molecules as well as from thepotential energy corresponding to the molecular forces. Equation (1) represents astatement of conservation of energy and is applicable to any system, howevercomplicated it might be.

Sign convention of heat and workSIGN CONVENTIONS FOR HEAT AND WORK

The signs of 'w' and 'q' are related to the internal energy change. When 'w' or 'q' is

positive, it means that energy has been supplied to the system as work or as heat. Theinternal energy of the system in such a case increases. On the other hand, if 'w' or 'q' isnegative, it means that energy has left the system as work or heat. The internal energy ofthe system decreases. The signs of 'q' and 'w' are:

Heat absorbed by the system= q positive

Heat evolved by the system= q negativeWork done on the system= w positive

Work done by the system= w negative


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Enthalpy, enthalpy change in chemical reactionsThe change in internal energy gives the heat change accompanying a chemical reactionat constant volume. However, most of the chemical reactions carried out in laboratories,are open to normal atmospheric conditions. A chemical reaction in a laboratory mayincur change in volume but the pressure remains constant i.e. atmospheric pressure.

To study the heat changes for reactions at constant pressure and at constanttemperature, a new term called enthalpy has been introduced.The enthalpy of a system may be defined as the sum of the internal energy and theproduct of its pressure and volume. It is denoted by the symbol H and is given by H = E+ PV

where, E = internal energy, P = pressure and V = the volume of the system.Enthalpy is also called heat content.Enthalpy-Change">

ENTHALPY CHANGE

Every substance has a definite value of enthalpy in a particular state. Like internalenergy, the absolute value of enthalpy cannot be measured. However, the change in

enthalpy accompanying a process can be determined as the difference between theenthalpies of the products and the reactants, i.e.,

DH = Hproducts - Hreactants = Hp - Hr

where, 'Hp' is the enthalpy of the products, 'Hr' is the enthalpy of the reactants and DH isthe enthalpy change.If a reaction is carried out at constant temperature and pressure the heat exchanged(evolved or absorbed) by the system with the surroundings (i.e., heat change - DH) isequal to change in enthalpy.

Origin of enthalpy change in a reactionWe know that in chemical reactions, energy is required to break old bonds of thereactants. Energy is also released to form new bonds, which give the end products. The

net energy change (released or absorbed) in a reaction will be equal to energy requiredto break all the bonds in reactants minus the energy released during the formation ofbonds in the products.

If energy required is greater than energy released, the net result will be the absorption ofenergy and the reaction will be endothermic, i.e. DH = + ve. If energy released is greaterthan energy required, the net result is the release of energy and the reaction will beexothermic (DH = - ve).

This can be illustrated by considering a chemical reaction between hydrogen gas andchlorine gas to form hydrochloric acid gas:

The energy required in breaking one mole of bonds in hydrogen and chlorinemolecules are 437 and 244 kJ respectively.


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It is observed that for the formation of one mole of HCl, 433 kJ of energy is released.Therefore, the energy released during the formation of 2 moles of HCl is 2 x 433 =866 kJ.

Thus, Enthalpy change = (437+244) - (2 x 433) = -185 kJ

DH = - 185kJ

Therefore, 185 kJ of energy is released during the formation of 2 moles of gaseous HClfrom one mole each of gaseous hydrogen and gaseous chlorine.

The amount of heat exchanged with the surroundings for a reaction at constant pressure(DH) is different from that exchanged at constant volume (DE) and temperature. Theenergy changes for reactions at constant pressure, includes energy contributions due toexpansion or contraction against atmospheric pressure i.e. the volume of the reactingsystem changes. If the volume increases, the system expands against the atmosphericpressure and energy is required for this expansion. Therefore, a part of energy will beused for the expansion. Thus the amount of heat exchanged at constant pressure (DH)would be less than the amount of heat exchanged at constant volume (DE).Alternatively, if the system contracts at constant pressure, work is done on the system

and the system absorbs some energy from the surroundings. Therefore, the amount ofheat exchanged at constant pressure is greater than that exchanged at constantvolume.DH-and-DE">

Relationship between DH and DEConsider a reaction

at constant pressure 'P'. Let HAbe the enthalpy of the reactants and HB be the enthalpyof products so that change in enthalpy, DH may be

DH = HB - HA

But H = E + PV.

Let EAand VAbe the internal energy and volume of the reactants and EB and VBcorresponding values for the products. Therefore,

HA= EA+ PVAand HB = EB + PVB

DH = (EB + PVB) - (EA+ PVA)

or DH = (EB - EA) + P(VB - VA)

or DH = E+ PDV

where DE is the change in internal energy and DV is the change in volume of the system.

Hess's law of constant heat summationHESS'S LAW OF CONSTANT HEAT SUMMATION

G.H.Hess proposed a law regarding the heats or enthalpies of reaction in 1840 called theHess's law. This law states that 'the heat change in a particular reaction is the samewhether it takes place in one step or several steps'.

For example, a reactant 'A' changes to a product 'B' in one step and the heat changeduring this process is DH. If the reaction is carried out in two steps where 'A' firstchanges to 'C' an intermediate stage and then 'C' changes to 'B' in the following step then


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let the heat change during the formation of 'A' to 'C' beDH1 and that from 'C' to 'B' beDH2. From Hess's law the heat change for the reaction is given as:

DH = DH1 + DH2

Fig: 5.4 - Illustration of Hess's law

This means that the amount of heat evolved or absorbed in a chemical reaction dependsonly upon the energy of initial reactants and the final products. The heat change isindependent of the path or the manner in which the change has taken place.

The formation of carbon dioxide from carbon and oxygen can be illustrated as follows.Carbon can be converted into carbon dioxide in two ways. Firstly solid carbon combineswith sufficient amount of oxygen to form CO2. The same reaction when carried inpresence of lesser amount of oxygen gives carbon monoxide which then gets convertedto CO2 in step two in the presence of oxygen.

DH = DH1 + DH2

Thus, one can conclude that thermochemical equations can be added, subtracted ormultiplied like algebraic equations to obtain the desired equation.APPLICATION OF HESS'S LAW

Hess's law has been useful in determining the heat changes of reactions, which cannotbe measured directly with calorimeter. Some of its applications are:

Applications and problems

Determination of heat of formationCompounds whose heats of formation cannot be measured directly using calorimetricmethods because they cannot be synthesised from their elements easily e.g. methane,carbon monoxide, benzene etc are determined using Hess's Law.

For example, the heat of formation of carbon monoxide can be calculated from the heatof combustion data for carbon and carbon monoxide as shown above.

Determination of heat of transition


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The heats of transition of allotropic modification of compounds such as diamond tographite, rhombic sulphur to monoclinic sulphur, yellow phosphorous to redphosphorous etc. can be determined using Hess's Law.

For example, the heat of transition of diamond to graphite can be calculated from theheat of combustion data for diamond and graphite, which is -395.4 kJ and -393.5 kJ

respectively.The thermochemical equations showing the combustion reaction of diamond andgraphite are:

The conversion that is required is:

This can be obtained by subtracting the second equation from the first one.

Determination of heat of hydrationThe heats of hydration of substances is calculated using Hess's law.

For example the heats of hydration of copper sulphate can be calculated from the heatsof solution of anhydrous and hydrated salts of copper. The heats of solution of CuSO4and CuSO4.5H2O are -66.5 and -11.7 kJ mol-1. The corresponding thermochemicalequations are:

The process of hydration can be expressed as:

According to Hesss law, DH1 = DH + DH2

DH = DH1 - DH2

= -66.5 11.7 = -78.2 kJ/molDetermination of heats of various reactionsHess's law is useful in calculating the enthalpies of many reactions whose directmeasurement is difficult or impossible.

Problems


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19. Calculate the standard heat of formation of carbon disulphide (l). Given that thestandard heats of combustion of carbon (s) sulphur (s) and carbon disulphide (l) are393.3, -293.72 and -1108.76kJ mol-1respectively.

SolutionThe given data can be written in thermochemical equation form as:

The required equation is:Multiplying equation (ii) by 2 and adding to equation (i) we get,

Subtracting equation (iii) from the above equation we have,

20. Calculate lattice energy for the change,

Given that DHsubl. of Li = 160.67 kJ mol-1, DHDissociation of

Cl2 = 244.34 kJ mol-1, DHionisation of Li(g) = 520.07 kJ mol-1,DHE.Aof Cl(g) = - 365.26 kJ mol-1, DHofof LiCl(s) = - 401.66 kJ mol-1.

SolutionConsidering the different changes that occur in the formation of solid lithium chloridebased on the data given the lattice energy of the above can be constituted as:


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or

= - 839.31 kJ mol-1

Heat of neutralization, heat of solution, heat of combustion,heat of vapourization, heat of formation and bond energyThe calorimeter is kept open to the atmosphere in the determination of change in

enthalpy of a reaction. The calorimeter is immersed in an insulated water bath fittedwith stirrer and thermometer. The temperature of the bath is recorded in the beginningand after the end of the reaction and the change in temperature is calculated. Knowingthe heat capacity of water bath and calorimeter and also the change in temperature theheat absorbed or evolved in the reaction is calculated. This gives the enthalpy change(DH) of the reaction.

ENTHALPY OF COMBUSTION

The enthalpy of combustion of a compound is the enthalpy change at normal pressureand at constant temperature accompanying complete combustion of one mole of thecompound. It is denoted by DHc. Combustion here means the burning of the givencompound to the highest oxides of the constituent elements in the presence of excess of

oxygen.For example, the enthalpy of combustion of benzene at 298 K is the enthalpy change ofthe reaction.

Thus, DHcomb C6H6(l) = -3268kJmol-1.


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Enthalpy of combustion is generally obtained experimentally. For cases, where it is notpossible to measure it experimentally, it is estimated from the enthalpies of formation ofthe various compounds involved in the process.Problems11. The heats of combustion of CH4 and C2H6 are -890.3 and -1559.7kJ mol-1

respectively. Which of the two has greater efficiency of fuel per gram?SolutionThe fuel efficiency can be predicted from the amount of heat evolved for every gram offuel consumed.

(i) The combustion of methane is as follows:

DHc = 890.3 kJmol-1

Molar mass of CH4 = 16

(ii) The combustion of ethane is as follows:

DHc = 1559.7kJmol-1

Molar mass of C2H6 = 30

Thus, methane has greater fuel efficiency than ethane.

12. (a) A cylinder of gas supplied by a company is assumed to contain 14 kg of Butane. Ifa normal family requires 20,000 kJ of energy per day for cooking, how long will the

cylinder last?(b) If the air supplied to the burner is insufficient, a portion of gas escapes withoutcombustion. Assuming that 25% of the gas is wasted due to this inefficiency, how longwill the cylinder last? (Heat of combustion of butane = 2658 kJ/mol).

Solution(a) Molecular formula of butane = C4H10

Molecular mass of butane = 4 x 12 + 10 x 1 = 58

Heat of combustion of butane = 2658kJmol-11 mole of 58 g of butane on complete combustion give heat = 2658 kJ

14 x 103 g of butane on complete combustion gives heat =

The family needs 20,000 kJ of heat for cooking per day.641586 kJ of heat will be used for cooking by a family in

The cylinder will last for 82 days


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(b) 25 per cent of the gas is wasted due to inefficiency. This means that only 75% ofbutane is combusted. Therefore,the energy produced by 75% combustion of butane

13. The enthalpy change involved in the oxidation of glucose is - 2880 kJ mol-1. Twentyfive percent of this energy is available for muscular work. If 100 kJ of muscular work isneeded to walk one kilometer, what is the maximum distance that a person will be ableto walk after eating 120 g of glucose.SolutionDHcomb of Glucose (C6H12O6) = - 2880 kJ mol-1

14. Calculate the enthalpy change of combustion of cyclopropane at 298 K. The enthalpyof formation of CO2(g) , H2O(l) and propane(g) are -393.5, -285.8 and 20.42 kJmol-1respectively. The enthalpy ofisomerisation of cyclopropane to propene is -33.0 kJ mol-1.

SolutionThe required DH is

The given equations are:

Multiply equation (i) and (ii) by 3 and add them. Now subtract equation (iii) andsubsequently add equation (iv) from the resulting expression.

DH = 3DH1 + 3DH2 - DH3 - DH4= 3(- 393.5) + 3(- 285.8) - (20.42) + (-33.0) = - 2091.32 kJ.

HEAT OR ENTHALPY OF NEUTRALISATION

The reaction in which an acid and a base react to give a salt and water is calledneutralization reaction. Neutralization reactions are exothermic in nature. The heat


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change when one gram equivalent of an acid is completely neutralised by a base or viceversa in dilute solution, is called heat of neutralization.

Examples of heat of neutralization are:

Neutralization of HCl with NaOHNeutralization of CH3COOH with NaOH

DH =-55.9 kJ

It is important to note that the term gram equivalent is used in the definition of heat ofneutralization. This is because neutralization involves 1 mole of H+ ions and 1 mole ofOH- ions to form 1mole of water and 57.1 kJ of heat is liberated.

Now, one gram equivalent of various acids on complete dissociation liberates one moleof H+ ions. But one mole of the acid may produce more than one mole of H+ ions insolution depending upon its basicity; for example 1mol of H2SO4 gives 2 mol of H+ ionsand 1mol of H3PO4 gives 3 mol of H+ ions on complete dissociation. But 1gramequivalent of both (H2SO4 or H3PO4) produces only 1 mol of H+ions.

Thus, it is more appropriate to use the term gram equivalent in the definition ofenthalpy of neutralization.

The average enthalpy of neutralization of any strong acid by a strong base is found to be- 57.7 kJ (- 13.7 kcal) irrespective of the nature of acid or the base. This suggests that thenet chemical reaction in all neutralization reactions is the same, viz.,This is because strong acids and strong bases are completely ionized in aqueoussolutions. The aqueous solution of one gram equivalent of all strong acids contains thesame number of H+ ions. Similarly, aqueous solution of one gram equivalent of allstrong bases also contains same number of OH-. The neutralization reactions betweenstrong acids and strong bases in aqueous solutions involve simply the combination of H+ions (from an acid) and OH- ions (from a base) to form unionized water molecules.

For example, in the reaction between hydrochloric acid and sodium hydroxide ion. Theneutralization can be represented as:

DH = -57.1kJ

Cancelling common ions:

Neutralization of weak acids and weak basesThe heat of neutralization of a weak acid or a weak base is less than

-57.1 kJ and is also different for different weak acids or bases.

For example for acetic acid the enthalpy of neutralization is -54.9 kJ. This can beexplained as follows:


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Acetic acid (CH3COOH) is a weak acid. Weak acids (or weak bases) are ionised to a smallextent in solutions. So acetic acid is only partially ionised in solution. Since theneutralization involves a reaction between H+(from the acid) and OH- (from the base),hence acetic acid must be fully ionised as per the reaction,

DH = +1.2 kJ

The ionization reaction is endothermic reaction. So, during ionization of acetic acid asmall amount of heat (1.2 kJ) is absorbed. As a result, the enthalpy of neutralization ofacetic acid is 1.2 kJ less than that for a strong acid-strong base pair.Therefore, the aqueous solutions containing one gram equivalent of different weak acidsdo not contain 1 gm equivalent of H+ ions. Similarly, the aqueous solutions containing 1gram equivalent of different weak bases do not contain 1 gram equivalent of OH- ions.The weak acids, weak bases therefore, have to be dissociated to give 1 gram equivalent ofH+ or OH- ions but, neutralisation of weak acid and strong base (or a weak base andstrong acid) not only involves the combination of H+ and OH- ions but also thedissociation of a weak acid (or a weak base). The dissociation process is accompanied bythe absorption of energy. This energy is called the heat of dissociation. Therefore, theoverall liberated energy is less than 57.1 kJ (i.e., 57.1 of dissociation of acid or base).

The neutralization of acetic acid with sodium hydroxide can be explained as follows:

DH = -55.9kJThus, heat of neutralization of acetic acid and sodium hydroxide is

-55.9 kJ, because 1.2 kJ of heat energy is used up in dissociating acetic acid. Similarly,heat of neutralization of ammonium hydroxide and hydrochloric acid is -51.5 kJ as 5.6kJ is the heat of dissociation of NH4OH.Problems15. 100 ml of 1N of an acid and 100 ml of 1N of a base are mixed at 298K. During theexperiment, the temperature arose to 298.0067 K. Calculate the heat of neutralization.

SolutionHeat capacity of solution = Mass of solution x Specific heat capacity

Total mass of solution = 100 + 100 = 200 ml

Heat capacity of solution = 200 x 4.2 = 840 JK-1Heat change in the reaction = Heat capacity x Rise in temperature

= (840 JK-1) (298.0067 - 298)K= 840 x 0.0067 J = 5.63 J

Now, one gram equivalent of acid = 1N HCl in 1000 ml


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100 ml of 1N acid on neutralization gives heat = 5.63 J

= 56.3J

Heat of neutralization = -56.3 J16. Whenever an acid is neutralized by a base, the net reaction is:

Calculate the heat evolved for the following experiments:

(a) 0.60 mol of HNO3 solution is mixed with 0.30 mol of KOH solution.(b) 400 cm3 of 0.2 M H2SO4 is mixed with 600 cm3 of 0.1 M NaOH solution.

SolutionAccording to the reaction,

When 1 mole of H+ ions and 1 mol of OH- ions are neutralized to form 1 mol of water,

57.1 kJ of energy is released.(a) Heat evolved on mixing 0.60 mol of HNO3 with 0.30 mol of KOH solution.

Since HNO3 and KOH are strong acids and bases,O.60 mol of HNO3 0.60 mol of H+ ions

0.30 mol of KOH 0.30 mol of OH- ionsIn this case, out of 0.60 mol of H+ ions (from HNO3) only 0.30 mol will be neutralised(equal to OH- ions present) by the base. 0.3 mol of H+ ions of HNO3 will remainunreacted. The net reaction is:

Now, heat evolved during the formation of 1 mol of H2O = 57.1kJ

Heat evolved in the formation of 0.3 mol of H2O=57.1 x 0.3=17.13 kJ.DHsol)">

Enthalpy of solution (DHsol)When a solute is dissolved in a solvent a solution is formed. During dissolution of asolute in any solvent, a certain amount of heat is either absorbed or evolved. Such heatchanges under constant pressure conditions are known as the enthalpy of solution. 'Thechange in enthalpy when one mole of a solute is dissolved in a specified quantity of asolvent at a given temperature is called enthalpy of solution'.

To avoid the amount of solvent, heat of solution is usually defined for an infinite dilutesolution. Thus, heat of solution at infinite dilution is the heat change when one mole of a

substance is dissolved in such a large quantity of solvent so that further dilution doesnot give any further heat change.

For example, dissolution of sodium chloride

Here 'aq' represents aqueous meaning a large excess of water.


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For substances, which dissolve with the absorption of heat (endothermic), the enthalpyof solution is positive while for the substances which dissolve by liberating heat(exothermic), the enthalpy of solution is negative.

For example, when KCl is dissolved in water, heat is absorbed. Thus, the enthalpy ofsolution of KCl is positive. For a 200 times dilution

(water : KCl = 200 : 1), the enthalpy change during the process,

So, the enthalpy of solution of KCl at a dilution of 200 is 18.6 kJ mol-1. The dissolutionof CaCl2(s) in water is an exothermic process. So, the enthalpy of solution of calciumchloride (CaCl2) is negative. At a dilution of 400, the enthalpy change for the reaction,

So, the enthalpy of solution of CaCl2(s) at a dilution of 400 is

-75.3 kJ mol-1.

DHfus)">

Enthalpy of fusion (DHfus)The enthalpy of fusion of a substance is defined as 'the change in enthalpy when onemole of a solid substance is melted at its melting temperature'. For example, theenthalpy change of the reaction,

is the enthalpy of fusion of ice. Enthalpy of fusion for some common substances aregiven below:

Substance: Ethanol Oxygen Hydrogen sulphide Sodium chloride

(C2H5OH) (O2) (H2S) (NaCl)

Hfus kJmol-1: +4.8 +0.45 +2.0 +29

Melting point: 156 K 55 K 188 K 1074 KFrom the data given above, we see that the enthalpies of fusion for ionic substances arevery high. This is due to strong Coulombic forces between the constituent ions in ionicsolids. The solids such as O2and H2S, which are molecular solids, have low heats offusion because the forces of attraction between their molecules are weak forces. Thus,the heats of fusion of substances give an idea of nature of solid and the magnitude offorces acting between the particles constituting the solids.vap)">

Enthalpy of Vapourisation (vap)'The change in enthalpy when one mole of a liquid is converted into vapours at itsboiling temperature is called enthalpy of vapourisation' (DHvap).

Thus, the enthalpy change of the reaction

is the enthalpy of vapourisation of water.

The enthalpies of vapourisation for certain common liquids are:

Substance He(l) H2(l) O2(l) HCl(l) H2O(l) NaCl(l)

Boiling point 4K 20K 90K 188K 373K 1738 K


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Hvap kJmol -1 0.1 0.9 6.7 16.2 40.6 170

From the data given above, we can say

DHvap (ionic liquids) > DHvap (polar liquids) > DHvap (non-polar liquids)

Thus, the DHvap depends upon the strength of intermolecular forces in any liquid.Problem17. Determine the value of DH and DE for the reversible isothermal evaporation of 90.0g of water at 100C. Assume that water vapours behave as ideal gas and heat ofevaporation of water is 540 cal g -1.

(R = 2 cal mol-1 K-1)

Solution

Heat of evaporation of 1 g of water = 540 cal

Heat of evaporation of 90 g of water = 540 x 90 = 48600 Cal.D H = 48600 Cal

The evaporation of 5 mol of water is represented as

Dn = (5 - 0) = 5

DH = DE + DnRT or DE = DH - DnRT

= 48600 - (5) (2.0) (373) = 44870 cal

DHsub)">

Enthalpy of Sublimation (DHsub)Sublimation is a process in which a solid substance directly changes into its vapours atany temperature below its melting point. Enthalpy of sublimation is defined as follows:

The change in enthalpy when one mole of a solid substance is converted into its vapourswithout melting at a temperature below its melting point is called the enthalpy ofsublimation.

For example, when one mole of solid iodine is converted into its vapours at roomtemperature, heat equal to 62.4 kJ is absorbed. So, the enthalpy of sublimation of iodineis + 62.4 kJ mol-1, i.e.,

Compounds, which sublime on heating are camphor, dry ice, ammonium chloride etc.

The heat of sublimation is related to heat of fusion and heat of vaporization as:

DHsublimation = DHfusion + DHvaporization

Problem18. When 1 g of liquid naphthalene (C10H8) solidifies, 149 J of heat is evolved. Calculatethe heat of fusion of naphthalene.

SolutionThe molecular mass of naphthalene is C10H8, = 10 x 12 + 8 x 1 =128

Heat evolved when 1 g of naphthalene solidifies = 149 J


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Heat evolved when 128 g of naphthalene solidifies

= 149 x 128 = 19072 J

For the fusion reaction,

This reaction is the reverse of the above solidification reaction so that

DHfusion = - DHsolidification

DHfusion = 19072 J or = 19.072 kJ

Enthalpy of hydrationThis is defined as the heat change (evolved or absorbed) when one mole of theanhydrous salt combines with the required number of moles of water to form thespecific hydrated salt

Concept of equivalent massIt is defined as the mass of an element/compound/ion which combines or displaces 1part of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine by mass.It is not alwayspossible to apply this classic definition to determine equivalent weights of chemicalentities. It is so because, we can not conceive of reactions involving chemical entitieswith three named reference of hydrogen, oxygen and chlorine. Generally, we are limitedto determination of equivalent weights of elements and few compounds by using thisdefinition of equivalent weight. A more workable definition is given as :

Equivalent weight,E=Molecular weight

Valence factor

=

MO

x

Clearly, determination of equivalent weight amounts to determining valence factor x.

Here, we shall classify chemical entities and the techniques to determine x.or,

Equivalent mass is equal to the molecular or atomic mass divided by the number ofelectrons involved in the reaction per molecule, atom or ion. For example in thereaction,


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two electrons are needed to produce one molecule of hydrogen gas. So, 2 Faraday ofelectricity is needed to produce one mole of hydrogen gas.

Hence,

For the reaction,per mole of copper atoms, 2 faradays are needed. So, the equivalent mass of copper ishalf of its atomic mass.

The equivalent mass of any species is not simply a property of the species, but dependsupon the reaction in which it participates, i.e., one chemical species can have more thanone value for its equivalent mass depending upon the reaction it participates.The equivalent mass of a substance is the quantity of material deposited or dissolved by1 F (= 96500 C) of electricity.

Equivalent weight of elements, and compoundsEquivalent weight of an elementIn the case of an element, the equivalent weight is defined as :Equivalent weight,E=Atomic weight

Valency

=A

x

Note that atomic weight substitutes molecular weight and valency substitutes valence

factor in the definition. Valencies of hydrogen, calcium and oxygen are 1,2 and 2

respectively. Hence, their equivalent weights are 1/1 =1, 40/2 = 20 and 16/2 = 8

respectively.Equivalent weight of an acid


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The valence factor of an acid is equal to its basicity. The basicity of an acid is equal to

furnishable hydrogen ion (proton) in its aqueous solution. Importantly, basicity is not

same as the number of hydrogen atoms in acid molecule. Consider acetic acid

(CH3COOH). It contains 4 hydrogen atoms in it, but only 1 furnishable hydrogen ion. As

such, basicity of acetic acid is 1. With this background, we define equivalent weight of an

acid as :Equivalent weight,E=Molecular weight of acid

Basicity

Basicity of sulphuric acid is 2. Hence, equivalent weight of sulphuric acid (H2SO4) is

(2X1 + 32 + 4X16)/2 = 98/2 = 49. Similarly, basicity of oxalic acid is 2. Hence,equivalent weight of oxalic acid (H2C2O4 ) is (2X1 + 2X12 + 4X16)/2 = 90/2=45.Phosphorous based acids like phosphoric acid (H3PO4 ), phosphorous acid (H3PO3) and

hypo-phosphorous acid (H3PO2) need special mention here to understand their basicity.

The structures of three acids are shown here. From the structure, it appears that these

compounds may furnish OH ions, but bond strengths between phosphorous and oxygen

(P-O) and phosphorous and hydrogen (P-H) are stronger than between oxygen and

hydrogen (O-H) in OH group. As such, these molecules release hydrogen ions from

OH group and behave as acid. Clearly, basicities of phosphoric acid (H3PO4),

phosphorous acid (H3PO3) and hypo-phosphorous acid (H3PO2) are 3, 2 and 1

respectively.Phosphorous based acids

Figure 1: Furnishable hydrogen ions of acids

Equivalent weight of a base


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The valence factor of a base is equal to its acidity. The acidity of a base is equal to

furnishable hydroxyl ion (OH-) in its aqueous solution. With this background, we define

equivalent weight of a base as :

Equivalent weight,E=Molecular weight of base

Acidity

Acidity of KOH is 1, whereas acidity ofCa(OH)2 is 2. Hence, equivalent weight of KOH is

(39 + 16 + 1)/1 = 56/1 = 56. Similarly, equivalent weight ofCa(OH)2 is {40 +

2X(16+1)}/2 = 74/2=37.Equivalent weight of a compoundThe valence factor of a compound depends on the manner a compound is involved in a

reaction. The compounds of alkali metal salts and alkaline earth metal salts are,

however, constant. These compounds are ionic and they dissociate in ionic components

in aqueous solution. In this case, valence factor is equal to numbers of electronic charge

on either cation or anion.Equivalent weight,E=

Molecular weight of compound

Numbers of electronic charge on cation or anion

The numbers of electronic charge on cation ofNaHCO3 is 1. Hence, equivalent weight of

NaHCO3 is (23 + 1 + 12 + 3X16)/1 = 84.If we look at the defining ratio of equivalent weight of a compound (AB) formed of two

radicals (say A and B), then we can rearrange the ratio as :

Equivalent weight, E=Molecular weight of Radical A

Numbers of electronic charge

+


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Molecular weight of Radical B


Thus,Equivalent weight of AB=Equivalent weight of A+Equivalent weight of BEquivalent weight of an ionThe valence factor of an ion is equal to numbers of electronic charge on the ion.

Therefore, we define equivalent weight of an ion as :Equivalent weight,E=

Molecular weight of ion


The numbers of electronic charge on carbonate ion (CO32 ) is 2. Hence, equivalent

weight of carbonate ion is (12 + 3X16)/1 = 60/2 = 30. Similarly, equivalent weight of

aluminum ion (Al3+) is 27/3 = 9.Equivalent weight of an oxidizing or reducing agentIn a redox reaction, one of the reacting entities is oxidizing agent (OA). The other entity

is reducing agent (RA). The oxidizer is recipient of electrons, whereas reducer is releaser

of electrons. The valence factor for either an oxidizing or reducing agent is equal to the

numbers of electrons transferred from one entity to another.Equivalent weight,E=


Numbers of electrons transferred in redox reaction

Alternatively,Equivalent weight,E=


Change in oxidation number in redox reaction


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Potassium dichromate in acidic medium is a strong oxidizer. It means it gains electrons

during redox reaction. Potassium dichromate in acidic solution results in :K2Cr2O7+14H++6e2K++2Cr3++7H2OEquivalent weight ofK2Cr2O7=294.2

6

=49Study of redox reaction is in itself an exclusive and extensive topic. We shall, therefore,

discuss redox reaction separately

Gram equivalent weightGRAM EQUIVALENT(GEQ)

It is equal to mass in grams numerically equal to equivalent weight. If the mass of achemical entity is g grams, then the given mass contains gram equivalents given by :Gram equivalent (geq)=g

E

This formula is widely used to express grams of substance in terms of gram equivalentand vice-versa.

Relation between moles and gram equivalents (geq)Gram equivalents is given by :geq=g

E

Substituting for equivalent weight, we have :


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geq=g

E

=xg

MO

Moles is given by :n=g

MO

Combining expressions, we have :geq=xngram equivalent=valence factorXmolesGRAM EQUIVALENT CONCEPT

Consider the example of formation of water :

2H2+O22H2OHere, 2 moles of hydrogen combines with 1 mole of oxygen to form 2 moles of watermolecule. In terms of mass, 4 gm of hydrogen combines with 32 gm of oxygen to form36 gm of water molecule. The relevant proportions involved with this equation are :Coefficients of balanced equation = 2:1:2Molecules/moles = 2:1:2Mass = 4:32:36 = 1:8:9On the other hand, equivalent weights of hydrogen, oxygen and water are 1, 8 and 9.Clearly, proportion of mass in which chemical entities participate is exactly theproportion of equivalent weights! Note that mole concept depends on the coefficient ofbalanced chemical equation. On the other hand, the equivalent weight concept isindependent of coefficient of balanced chemical equation. If we know that hydrogen andoxygen combines to form water molecule, then we can say straightway that entities arein the proportion of equivalent weights - without any reference to coefficients in thebalanced chemical equation.


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equivalent weight of hydrogenequivalent weight of oxygenequivalent weight of waterNote that there is no mention of coefficients of balanced chemical equation in theequivalent weight relation. We should, however, understand that two techniques ofanalyzing chemical reactions are essentially equivalent. We need to consider coefficientsinvolved in balanced chemical equation for applying mole concept. On the other hand,coefficients are not considered when using equivalent weight concept, but we need toknow the corresponding valence factor of each entity. It is important to realize thatabove relation is not a relation connected by "equal to (=)" sign. Rather, they areconnected by equivalent sign (). As such, we still need to apply unitary method toanalyze the relation.Gram equivalent concept is a step head in this context. The gram equivalents ofparticipating entities are same. For the case of formation of water, the proportion ofmass of hydrogen, oxygen and water is 1 gm: 8 gm : 9 gm. Now, we know that the gramequivalents of entities are obtained by dividing mass by equivalent weight. Hence, gramequivalents of three entities are 1/1 = 1, 8/8=1 and 9/9 = 1. Thus, gram equivalents of

participating entities are same. If gram equivalents of hydrogen is 2, then gramequivalents of oxygen and water will also be 2. As such :x gm equivalents of hydrogen=x gm equivalents of oxygen=x gm equivalents of waterSignificant aspect of this relation is that it is connected with equal to (=) sign and assuch relieves us from applying unitary method altogether

Relation between equivalent weight, valency and atomic

weightConsider an element X of atomic weight A and valency n.Suppose the element X combines with hydrogen to give the compound hydride, XHn.X + nH ?XHn1 atom of X combines with n atoms of hydrogen.1 gram-atom of X combines with n gram-atoms of hydrogen.Atomic weight of X = AAtomic weight of hydrogen = 1.008n 1.008g of hydrogen combines with = A gram of XA1.008 grams of hydrogen combines with = ------------- 1.008

n 1.008

A= ------n

By definition, this gives the equivalent weight of the element X.


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Atomic weightEquivalent weight = -------------------Valency

In the case of univalent elements, the atomic weight itself gives the equivalent weight.

Elements having varying valency have different equivalent weight.Equivalent weight of copper in cuprous oxide Cu2O is 63.5. (here, valency of Cu is one).Equivalent weight of copper in cupricoxide,63.5CuO is ------ = 31.75 (here, valency of Cu is 2).

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