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160, , CHEMISTRY, , UNIT 6, , THERMODYNAMICS, , It is the only physical theory of universal content concerning, which I am convinced that, within the framework of the, applicability of its basic concepts, it will never be overthrown., After studying this Unit, you will, be able to, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, •, , explain the terms : system and, surroundings;, discriminate between close,, open and isolated systems;, explain internal energy, work, and heat;, state, first, law, of, thermodynamics and express, it mathematically;, calculate energy changes as, work and heat contributions, in chemical systems;, explain state functions: U, H., correlate ∆U and ∆H;, measure experimentally ∆U, and ∆H;, define standard states for ∆H;, calculate enthalpy changes for, various types of reactions;, state and apply Hess’s law of, constant heat summation;, differentiate between extensive, and intensive properties;, define spontaneous and nonspontaneous processes;, explain entropy as a, thermodynamic state function, and apply it for spontaneity;, explain Gibbs energy change, (∆G); and, establish relationship between, ∆G and spontaneity, ∆G and, equilibrium constant., , Albert Einstein, , Chemical energy stored by molecules can be released as heat, during chemical reactions when a fuel like methane, cooking, gas or coal burns in air. The chemical energy may also be, used to do mechanical work when a fuel burns in an engine, or to provide electrical energy through a galvanic cell like, dry cell. Thus, various forms of energy are interrelated and, under certain conditions, these may be transformed from, one form into another. The study of these energy, transformations forms the subject matter of thermodynamics., The laws of thermodynamics deal with energy changes of, macroscopic systems involving a large number of molecules, rather than microscopic systems containing a few molecules., Thermodynamics is not concerned about how and at what, rate these energy transformations are carried out, but is, based on initial and final states of a system undergoing the, change. Laws of thermodynamics apply only when a system, is in equilibrium or moves from one equilibrium state to, another equilibrium state. Macroscopic properties like, pressure and temperature do not change with time for a, system in equilibrium state. In this unit, we would like to, answer some of the important questions through, thermodynamics, like:, How do we determine the energy changes involved in a, chemical reaction/process? Will it occur or not?, What drives a chemical reaction/process?, To what extent do the chemical reactions proceed?, , 2021-22

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THERMODYNAMICS, , 161, , 6.1 THERMODYNAMIC TERMS, We are interested in chemical reactions and the, energy changes accompanying them. For this, we need to know certain thermodynamic, terms. These are discussed below., 6.1.1 The System and the Surroundings, A system in thermodynamics refers to that, part of universe in which observations are, made and remaining universe constitutes the, surroundings. The surroundings include, everything other than the system. System and, the surroundings together constitute the, universe ., The universe = The system + The surroundings, However, the entire universe other than, the system is not affected by the changes, taking place in the system. Therefore, for, all practical purposes, the surroundings, are that portion of the remaining universe, which can interact with the system., Usually, the region of space in the, neighbourhood of the system constitutes, its surroundings., For example, if we are studying the, reaction between two substances A and B, kept in a beaker, the beaker containing the, reaction mixture is the system and the room, where the beaker is kept is the surroundings, (Fig. 6.1)., , be real or imaginary. The wall that separates, the system from the surroundings is called, boundary. This is designed to allow us to, control and keep track of all movements of, matter and energy in or out of the system., 6.1.2 Types of the System, We, further classify the systems according to, the movements of matter and energy in or out, of the system., 1. Open System, In an open system, there is exchange of energy, and matter between system and surroundings, [Fig. 6.2 (a)]. The presence of reactants in an, open beaker is an example of an open system*., Here the boundary is an imaginary surface, enclosing the beaker and reactants., 2. Closed System, In a closed system, there is no exchange of, matter, but exchange of energy is possible, between system and the surroundings, [Fig. 6.2 (b)]. The presence of reactants in a, closed vessel made of conducting material e.g.,, copper or steel is an example of a closed, system., , Fig. 6.1 System and the surroundings, , Note that the system may be defined by, physical boundaries, like beaker or test tube,, or the system may simply be defined by a set, of Cartesian coordinates specifying a, particular volume in space. It is necessary to, think of the system as separated from the, surroundings by some sort of wall which may, , *, , Fig. 6.2 Open, closed and isolated systems., , We could have chosen only the reactants as system then walls of the beakers will act as boundary., , 2021-22

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162, , CHEMISTRY, , 3. Isolated System, In an isolated system, there is no exchange of, energy or matter between the system and the, surroundings [Fig. 6.2 (c)]. The presence of, reactants in a thermos flask or any other closed, insulated vessel is an example of an isolated, system., 6.1.3 The State of the System, The system must be described in order to make, any useful calculations by specifying, quantitatively each of the properties such as, its pressure (p), volume (V), and temperature, (T ) as well as the composition of the system., We need to describe the system by specifying, it before and after the change. You would recall, from your Physics course that the state of a, system in mechanics is completely specified at, a given instant of time, by the position and, velocity of each mass point of the system. In, thermodynamics, a different and much simpler, concept of the state of a system is introduced., It does not need detailed knowledge of motion, of each particle because, we deal with average, measurable properties of the system. We specify, the state of the system by state functions or, state variables., The state of a thermodynamic system is, described by its measurable or macroscopic, (bulk) properties. We can describe the state of, a gas by quoting its pressure (p), volume (V),, temperature (T ), amount (n) etc. Variables like, p, V, T are called state variables or state, functions because their values depend only, on the state of the system and not on how it is, reached. In order to completely define the state, of a system it is not necessary to define all the, properties of the system; as only a certain, number of properties can be varied, independently. This number depends on the, nature of the system. Once these minimum, number of macroscopic properties are fixed,, others automatically have definite values., The state of the surroundings can never, be completely specified; fortunately it is not, necessary to do so., 6.1.4 The Internal Energy as a State, Function, When we talk about our chemical system, losing or gaining energy, we need to introduce, a quantity which represents the total energy, , of the system. It may be chemical, electrical,, mechanical or any other type of energy you, may think of, the sum of all these is the energy, of the system. In thermodynamics, we call it, the internal energy, U of the system, which may, change, when, • heat passes into or out of the system,, • work is done on or by the system,, • matter enters or leaves the system., These systems are classified accordingly as, you have already studied in section 6.1.2., (a) Work, Let us first examine a change in internal, energy by doing work. We take a system, containing some quantity of water in a thermos, flask or in an insulated beaker. This would not, allow exchange of heat between the system, and surroundings through its boundary and, we call this type of system as adiabatic. The, manner in which the state of such a system, may be changed will be called adiabatic, process. Adiabatic process is a process in, which there is no transfer of heat between the, system and surroundings. Here, the wall, separating the system and the surroundings, is called the adiabatic wall (Fig 6.3)., Let us bring the change in the internal, energy of the system by doing some work on, , Fig. 6.3 An adiabatic system which does not, permit the transfer of heat through its, boundary., , it. Let us call the initial state of the system as, state A and its temperature as TA. Let the, internal energy of the system in state A be, called UA. We can change the state of the system, in two different ways., , 2021-22

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THERMODYNAMICS, , 163, , One way: We do some mechanical work, say, 1 kJ, by rotating a set of small paddles and, thereby churning water. Let the new state be, called B state and its temperature, as TB. It is, found that T B > T A and the change in, temperature, ∆T = T B–TA. Let the internal, energy of the system in state B be UB and the, change in internal energy, ∆U =UB– UA., Second way: We now do an equal amount (i.e.,, 1kJ) electrical work with the help of an, immersion rod and note down the temperature, change. We find that the change in temperature, is same as in the earlier case, say, TB – TA., In fact, the experiments in the above, manner were done by J. P. Joule between, 1840–50 and he was able to show that a given, amount of work done on the system, no matter, how it was done (irrespective of path) produced, the same change of state, as measured by the, change in the temperature of the system., So, it seems appropriate to define a, quantity, the internal energy U, whose value, is characteristic of the state of a system,, whereby the adiabatic work, wad required to, bring about a change of state is equal to the, difference between the value of U in one state, and that in another state, ∆U i.e.,, ∆U = U 2 − U 1 = w ad, Therefore, internal energy, U, of the system, is a state function., By conventions of IUPAC in chemical, thermodynamics. The positive sign expresses, that wad is positive when work is done on the, system and the internal energy of system, increases. Similarly, if the work is done by the, system,wad will be negative because internal, energy of the system decreases., Can you name some other familiar state, functions? Some of other familiar state, functions are V, p, and T. For example, if we, bring a change in temperature of the system, from 25°C to 35°C, the change in temperature, is 35°C–25°C = +10°C, whether we go straight, up to 35°C or we cool the system for a few, degrees, then take the system to the final, temperature. Thus, T is a state function and, the change in temperature is independent of, , *, , the route taken. Volume of water in a pond,, for example, is a state function, because, change in volume of its water is independent, of the route by which water is filled in the, pond, either by rain or by tubewell or by both., (b) Heat, We can also change the internal energy of a, system by transfer of heat from the, surroundings to the system or vice-versa, without expenditure of work. This exchange, of energy, which is a result of temperature, difference is called heat, q. Let us consider, bringing about the same change in temperature, (the same initial and final states as before in, section 6.1.4 (a) by transfer of heat through, thermally conducting walls instead of, adiabatic walls (Fig. 6.4)., , Fig. 6.4, , A system which allows heat transfer, through its boundary., , We take water at temperature, TA in a, container having thermally conducting walls,, say made up of copper and enclose it in a huge, heat reservoir at temperature, TB. The heat, absorbed by the system (water), q can be, measured in terms of temperature difference ,, TB – TA. In this case change in internal energy,, ∆U= q, when no work is done at constant, volume., By conventions of IUPAC in chemical, thermodynamics. The q is positive, when, heat is transferred from the surroundings to, the system and the internal energy of the, system increases and q is negative when, heat is transferred from system to the, surroundings resulting in decrease of the, internal energy of the system.., , Earlier negative sign was assigned when the work is done on the system and positive sign when the work is done by the, system. This is still followed in physics books, although IUPAC has recommended the use of new sign convention., , 2021-22

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164, , CHEMISTRY, , (c) The general case, Let us consider the general case in which a, change of state is brought about both by, doing work and by transfer of heat. We write, change in internal energy for this case as:, ∆U = q + w, , (6.1), , For a given change in state, q and w can, vary depending on how the change is carried, out. However, q +w = ∆U will depend only on, initial and final state. It will be independent of, the way the change is carried out. If there is, no transfer of energy as heat or as work, (isolated system) i.e., if w = 0 and q = 0, then, ∆ U = 0., The equation 6.1 i.e., ∆U = q + w is, mathematical statement of the first law of, thermodynamics, which states that, The energy of an isolated system is, constant., It is commonly stated as the law of, conservation of energy i.e., energy can neither, be created nor be destroyed., Note: There is considerable difference between, the character of the thermodynamic property, energy and that of a mechanical property such, as volume. We can specify an unambiguous, (absolute) value for volume of a system in a, particular state, but not the absolute value of, the internal energy. However, we can measure, only the changes in the internal energy, ∆U of, the system., Problem 6.1, Express the change in internal energy of, a system when, (i) No heat is absorbed by the system, from the surroundings, but work (w), is done on the system. What type of, wall does the system have ?, (ii) No work is done on the system, but, q amount of heat is taken out from, the system and given to the, surroundings. What type of wall does, the system have?, (iii) w amount of work is done by the, system and q amount of heat is, supplied to the system. What type of, system would it be?, , Solution, (i) ∆ U = w ad, wall is adiabatic, (ii) ∆ U = – q, thermally conducting walls, (iii) ∆ U = q – w, closed system., 6.2 APPLICATIONS, Many chemical reactions involve the generation, of gases capable of doing mechanical work or, the generation of heat. It is important for us to, quantify these changes and relate them to the, changes in the internal energy. Let us see how!, 6.2.1 Work, First of all, let us concentrate on the nature of, work a system can do. We will consider only, mechanical work i.e., pressure-volume work., For understanding pressure-volume, work, let us consider a cylinder which, contains one mole of an ideal gas fitted with a, frictionless piston. Total volume of the gas is, Vi and pressure of the gas inside is p. If, external pressure is pex which is greater than, p, piston is moved inward till the pressure, inside becomes equal to pex. Let this change, , Fig. 6.5(a) Work done on an ideal gas in a, cylinder when it is compressed by a, constant external pressure, p ex, (in single step) is equal to the shaded, area., , 2021-22

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THERMODYNAMICS, , 165, , be achieved in a single step and the final, volume be Vf . During this compression,, suppose piston moves a distance, l and is, cross-sectional area of the piston is A, [Fig. 6.5(a)]., then, volume change = l × A = ∆V = (Vf – Vi ), We also know, pressure =, , force, area, , Vf, , w=−, , Therefore, force on the piston = pex . A, If w is the work done on the system by, movement of the piston then, w = force × distance = pex . A .l, = pex . (–∆V) = – pex ∆V = – pex (Vf – Vi ), , If the pressure is not constant but changes, during the process such that it is always, infinitesimally greater than the pressure of the, gas, then, at each stage of compression, the, volume decreases by an infinitesimal amount,, dV. In such a case we can calculate the work, done on the gas by the relation, , (6.2), , The negative sign of this expression is, required to obtain conventional sign for w,, which will be positive. It indicates that in case, of compression work is done on the system., Here (Vf – Vi ) will be negative and negative, multiplied by negative will be positive. Hence, the sign obtained for the work will be positive., If the pressure is not constant at every, stage of compression, but changes in number, of finite steps, work done on the gas will be, summed over all the steps and will be equal, to − ∑ p ∆V [Fig. 6.5 (b)], , Fig. 6.5 (b) pV-plot when pressure is not constant, and changes in finite steps during, compression from initial volume, Vi to, final volume, Vf . Work done on the gas, is represented by the shaded area., , ∫p, , ex, , dV, , ( 6.3), , Vi, , Here, pex at each stage is equal to (pin + dp) in, case of compression [Fig. 6.5(c)]. In an, expansion process under similar conditions,, the external pressure is always less than the, pressure of the system i.e., pex = (pin– dp). In, general case we can write, pex = (pin + dp). Such, processes are called reversible processes., A process or change is said to be, reversible, if a change is brought out in, such a way that the process could, at any, moment, be reversed by an infinitesimal, change. A reversible process proceeds, infinitely slowly by a series of equilibrium, states such that system and the, surroundings are always in near, equilibrium with each other. Processes, , Fig. 6.5 (c) pV-plot when pressure is not constant, and changes in infinite steps, (reversible conditions) during, compression from initial volume, Vi to, final volume, Vf . Work done on the gas, is represented by the shaded area., , 2021-22

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166, , CHEMISTRY, , other than reversible processes are known, as irreversible processes., In chemistry, we face problems that can, be solved if we relate the work term to the, internal pressure of the system. We can, relate work to internal pressure of the system, under reversible conditions by writing, equation 6.3 as follows:, Vf, , w rev = −, , ∫p, , Vf, ex, , dV = −, , Vi, , ∫ (p, , in, , ± dp ) dV, , Vi, , Since dp × dV is very small we can write, , Isothermal and free expansion of an, ideal gas, For isothermal (T = constant) expansion of an, ideal gas into vacuum ; w = 0 since pex = 0., Also, Joule determined experimentally that, q = 0; therefore, ∆U = 0, Equation 6.1, ∆ U = q + w can be, expressed for isothermal irreversible and, reversible changes as follows:, 1., For isothermal irreversible change, q = – w = pex (Vf – Vi ), 2., For isothermal reversible change, , Vf, , Vf, , w rev = − ∫ pin dV, , q = – w = nRT ln V, i, , (6.4), , Vi, , Now, the pressure of the gas (pin which we, can write as p now) can be expressed in terms, of its volume through gas equation. For n mol, of an ideal gas i.e., pV =nRT, , = 2.303 nRT log, 3., , nRT, V, Therefore, at constant temperature (isothermal, process),, ⇒p =, , Vf, , w rev = − ∫ nRT, Vi, , Vf, dV, = −nRT ln, V, Vi, , = – 2.303 nRT log, , Vf, Vi, , (6.5), , Free expansion: Expansion of a gas in, vacuum (pex = 0) is called free expansion. No, work is done during free expansion of an ideal, gas whether the process is reversible or, irreversible (equation 6.2 and 6.3)., Now, we can write equation 6.1 in number, of ways depending on the type of processes., Let us substitute w = – pex∆V (eq. 6.2) in, equation 6.1, and we get, ∆U = q − pex ∆V, , If a process is carried out at constant volume, (∆V = 0), then, ∆U = qV, the subscript V in qV denotes that heat is, supplied at constant volume., , 2021-22, , Vf, Vi, , For adiabatic change, q = 0,, ∆U = wad, Problem 6.2, Two litres of an ideal gas at a pressure of, 10 atm expands isothermally at 25 °C into, a vacuum until its total volume is 10 litres., How much heat is absorbed and how much, work is done in the expansion ?, Solution, We have q = – w = pex (10 – 2) = 0(8) = 0, No work is done; no heat is absorbed., Problem 6.3, Consider the same expansion, but this, time against a constant external pressure, of 1 atm., Solution, We have q = – w = pex (8) = 8 litre-atm, Problem 6.4, Consider the expansion given in problem, 6.2, for 1 mol of an ideal gas conducted, reversibly., Solution, , Vf, Vs 10, = 2.303 × 1 × 0.8206 × 298 × log, , We have q = – w = 2.303 nRT log, , 2

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THERMODYNAMICS, , 167, , = 2.303 x 0.8206 x 298 x log 5, = 2.303 x 0.8206 x 298 x 0.6990, = 393.66 L atm, 6.2.2 Enthalpy, H, (a) A Useful New State Function, We know that the heat absorbed at constant, volume is equal to change in the internal, energy i.e., ∆U = q . But most of chemical, V, reactions are carried out not at constant, volume, but in flasks or test tubes under, constant atmospheric pressure. We need to, define another state function which may be, suitable under these conditions., We may write equation (6.1) as, ∆U = qp – p∆V at constant pressure, where qp, is heat absorbed by the system and –p ∆V, represent expansion work done by the system., Let us represent the initial state by, subscript 1 and final state by 2, We can rewrite the above equation as, U2–U1 = qp – p (V2 – V1), On rearranging, we get, , Remember ∆H = qp, heat absorbed by the, system at constant pressure., ∆H is negative for exothermic reactions, which evolve heat during the reaction and, ∆H is positive for endothermic reactions, which absorb heat from the surroundings., At constant volume (∆V = 0), ∆U = qV,, therefore equation 6.8 becomes, ∆H = ∆U = q, V, , The difference between ∆H and ∆U is not, usually significant for systems consisting of, only solids and / or liquids. Solids and liquids, do not suffer any significant volume changes, upon heating. The difference, however,, becomes significant when gases are involved., Let us consider a reaction involving gases. If, VA is the total volume of the gaseous reactants,, VB is the total volume of the gaseous products,, nA is the number of moles of gaseous reactants, and nB is the number of moles of gaseous, products, all at constant pressure and, temperature, then using the ideal gas law, we, write,, pVA = nART, , qp = (U2 + pV2) – (U1 + pV1), (6.6), Now we can define another thermodynamic, function, the enthalpy H [Greek word, enthalpien, to warm or heat content] as :, H = U + pV, (6.7), so, equation (6.6) becomes, qp= H2 – H1 = ∆H, Although q is a path dependent function,, H is a state function because it depends on U,, p and V, all of which are state functions., Therefore, ∆H is independent of path. Hence,, qp is also independent of path., For finite changes at constant pressure, we, can write equation 6.7 as, ∆H = ∆U + ∆pV, Since p is constant, we can write, ∆H = ∆U + p∆V, (6.8), It is important to note that when heat is, absorbed by the system at constant pressure,, we are actually measuring changes in the, enthalpy., , and, , pVB = nBRT, , Thus, pVB – pVA = nBRT – nART = (nB–nA)RT, or, , p (VB – VA) = (nB – nA) RT, , or, , p ∆V = ∆ngRT, , (6.9), , Here, ∆ng refers to the number of moles of, gaseous products minus the number of moles, of gaseous reactants., Substituting the value of p∆V from, equation 6.9 in equation 6.8, we get, ∆H = ∆U + ∆ngRT, (6.10), The equation 6.10 is useful for calculating, ∆H from ∆U and vice versa., , 2021-22, , Problem 6.5, If water vapour is assumed to be a perfect, gas, molar enthalpy change for, vapourisation of 1 mol of water at 1bar, and 100°C is 41kJ mol–1. Calculate the, internal energy change, when

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168, , CHEMISTRY, , 1 mol of water is vapourised at 1 bar, pressure and 100°C., Solution, (i) The change H2O (l ) → H2O ( g ), , ∆H = ∆U + ∆n g RT, , Fig. 6.6(a) A gas at volume V and temperature T, , or ∆U = ∆H – ∆n g RT , substituting the, values, we get, , ∆U = 41.00 kJ mol −1 − 1, × 8.3 J mol −1K −1 × 373 K, = 41.00 kJ mol-1 – 3.096 kJ mol-1, = 37.904 kJ mol–1, (b) Extensive and Intensive Properties, In thermodynamics, a distinction is made, between extensive properties and intensive, properties. An extensive property is a, property whose value depends on the, quantity or size of matter present in the, system. For example, mass, volume,, internal energy, enthalpy, heat capacity, etc., are extensive properties., Those properties which do not depend, on the quantity or size of matter present are, known as intensive properties. For, example temperature, density, pressure etc., are intensive properties. A molar property,, χm, is the value of an extensive property χ of, the system for 1 mol of the substance. If n, χ, is, is the amount of matter, χ m =, n, independent of the amount of matter. Other, examples are molar volume, Vm and molar, heat capacity, Cm. Let us understand the, distinction between extensive and intensive, properties by considering a gas enclosed in, a container of volume V and at temperature, T [Fig. 6.6(a)]. Let us make a partition such, that volume is halved, each part [Fig. 6.6, (b)] now has one half of the original volume,, V, , but the temperature will still remain the, 2, , same i.e., T. It is clear that volume is an, extensive property and temperature is an, intensive property., , Fig. 6.6 (b) Partition, each part having half the, volume of the gas, , (c) Heat Capacity, In this sub-section, let us see how to measure, heat transferred to a system. This heat appears, as a rise in temperature of the system in case, of heat absorbed by the system., The increase of temperature is proportional, to the heat transferred, , q = coeff × ∆T, The magnitude of the coefficient depends, on the size, composition and nature of the, system. We can also write it as q = C ∆T, The coefficient, C is called the heat capacity., Thus, we can measure the heat supplied, by monitoring the temperature rise, provided, we know the heat capacity., When C is large, a given amount of heat, results in only a small temperature rise. Water, has a large heat capacity i.e., a lot of energy is, needed to raise its temperature., C is directly proportional to amount of, substance. The molar heat capacity of a, , C, substance, Cm =   , is the heat capacity for, n, one mole of the substance and is the quantity, of heat needed to raise the temperature of, one mole by one degree celsius (or one, kelvin). Specific heat, also called specific heat, capacity is the quantity of heat required to, raise the temperature of one unit mass of a, , 2021-22

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THERMODYNAMICS, , 169, , substance by one degree celsius (or one, kelvin). For finding out the heat, q, required, to raise the temperatures of a sample, we, multiply the specific heat of the substance,, c, by the mass m, and temperatures change,, ∆T as, , q = c × m × ∆T = C ∆T, , (6.11), , (d) The Relationship between Cp and CV, for an Ideal Gas, At constant volume, the heat capacity, C is, denoted by CV and at constant pressure, this, is denoted by Cp . Let us find the relationship, between the two., We can write equation for heat, q, , i) at constant volume, qV, ii) at constant pressure, qp, (a) ∆U Measurements, For chemical reactions, heat absorbed at, constant volume, is measured in a bomb, calorimeter (Fig. 6.7). Here, a steel vessel (the, bomb) is immersed in a water bath. The whole, device is called calorimeter. The steel vessel is, immersed in water bath to ensure that no heat, is lost to the surroundings. A combustible, substance is burnt in pure dioxygen supplied, , at constant volume as qV = CV ∆T = ∆U, at constant pressure as qp = C p ∆T = ∆H, The difference between Cp and CV can be, derived for an ideal gas as:, For a mole of an ideal gas, ∆H = ∆U + ∆(pV ), = ∆U + ∆(RT ), = ∆U + R∆T, , ∴ ∆H = ∆U + R ∆T, , (6.12), , On putting the values of ∆H and ∆U,, we have, , C p ∆T = CV ∆T + R∆T, C p = CV + R, Cp – CV = R, , Fig. 6.7 Bomb calorimeter, , (6.13), , 6.3 MEASUREMENT OF ∆ U AND ∆ H:, CALORIMETRY, We can measure energy changes associated, with chemical or physical processes by an, experimental technique called calorimetry. In, calorimetry, the process is carried out in a, vessel called calorimeter, which is immersed, in a known volume of a liquid. Knowing the, heat capacity of the liquid in which calorimeter, is immersed and the heat capacity of, calorimeter, it is possible to determine the heat, evolved in the process by measuring, temperature changes. Measurements are, made under two different conditions:, , in the steel bomb. Heat evolved during the, reaction is transferred to the water around the, bomb and its temperature is monitored. Since, the bomb calorimeter is sealed, its volume does, not change i.e., the energy changes associated, with reactions are measured at constant, volume. Under these conditions, no work is, done as the reaction is carried out at constant, volume in the bomb calorimeter. Even for, reactions involving gases, there is no work, done as ∆V = 0. Temperature change of the, calorimeter produced by the completed, reaction is then converted to qV, by using the, known heat capacity of the calorimeter with, the help of equation 6.11., , 2021-22

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170, , CHEMISTRY, , (b) ∆H Measurements, Measurement of heat change at constant pressure, (generally under atmospheric pressure) can be, done in a calorimeter shown in Fig. 6.8. We know, that ∆Η = qp (at constant p) and, therefore, heat, absorbed or evolved, qp at constant pressure is, also called the heat of reaction or enthalpy of, reaction, ∆rH., In an exothermic reaction, heat is evolved,, and system loses heat to the surroundings., Therefore, qp will be negative and ∆rH will also, be negative. Similarly in an endothermic, reaction, heat is absorbed, qp is positive and, ∆rH will be positive., , of the bomb calorimeter is 20.7kJ/K,, what is the enthalpy change for the above, reaction at 298 K and 1 atm?, Solution, Suppose q is the quantity of heat from the, reaction mixture and CV is the heat capacity, of the calorimeter, then the quantity of heat, absorbed by the calorimeter., q = CV × ∆T, Quantity of heat from the reaction will, have the same magnitude but opposite, sign because the heat lost by the system, (reaction mixture) is equal to the heat, gained by the calorimeter., q = − CV × ∆T = − 20.7 kJ/K × (299 − 298) K, = − 20.7 kJ, , (Here, negative sign indicates the, exothermic nature of the reaction), Thus, ∆U for the combustion of the 1g of, graphite = – 20.7 kJK–1, For combustion of 1 mol of graphite,, , =, , 12.0 g mol −1 × ( −20.7 kJ ), 1g, , = – 2.48 ×102 kJ mol–1 ,, Since ∆ ng = 0,, 2, ∆ H = ∆ U = – 2.48 ×10 kJ mol–1, , Fig. 6.8 Calorimeter for measuring heat changes, at constant pressure (atmospheric, pressure)., , Problem 6.6, 1g of graphite is burnt in a bomb, calorimeter in excess of oxygen at 298 K, and 1 atmospheric pressure according to, the equation, C (graphite) + O2 (g) → CO2 (g), During the reaction, temperature rises, from 298 K to 299 K. If the heat capacity, , 6.4 ENTHALPY CHANGE, ∆ r H OF A, REACTION – REACTION ENTHALPY, In a chemical reaction, reactants are converted, into products and is represented by,, Reactants → Products, The enthalpy change accompanying a, reaction is called the reaction enthalpy. The, enthalpy change of a chemical reaction, is given, by the symbol ∆rH, ∆rH = (sum of enthalpies of products) – (sum, of enthalpies of reactants), , =, , ∑a H, i, , i, , products, , − ∑ bi H reactants (6.14), , Here symbol, , i, , ∑, , (sigma) is used for, , summation and ai and bi are the stoichiometric, , 2021-22

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THERMODYNAMICS, , 171, , coefficients of the products and reactants, respectively in the balanced chemical, equation. For example, for the reaction, CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l), ∆r H = ∑ a i H Pr oducts − ∑ bi H reac tan ts, i, , i, , = [Hm (CO2 ,g) + 2Hm (H2O, l)]– [Hm (CH4 , g), , ethanol at 298 K is pure liquid ethanol at, 1 bar; standard state of solid iron at 500 K is, pure iron at 1 bar. Usually data are taken, at 298 K., Standard conditions are denoted by adding, the superscript 0 to the symbol ∆H, e.g., ∆H0, , + 2Hm (O2, g)], where Hm is the molar enthalpy., , (b) Enthalpy Changes during Phase, Transformations, , Enthalpy change is a very useful quantity., Knowledge of this quantity is required when, one needs to plan the heating or cooling, required to maintain an industrial chemical, reaction at constant temperature. It is also, required to calculate temperature dependence, of equilibrium constant., , Phase transformations also involve energy, changes. Ice, for example, requires heat for, melting. Normally this melting takes place at, constant pressure (atmospheric pressure) and, during phase change, temperature remains, constant (at 273 K)., , (a) Standard Enthalpy of Reactions, , 0, , H2O(s) → H2O(l); ∆fusH = 6.00 kJ moI–1, 0, , Enthalpy of a reaction depends on the, conditions under which a reaction is carried, out. It is, therefore, necessary that we must, specify some standard conditions. The, standard enthalpy of reaction is the, enthalpy change for a reaction when all, the participating substances are in their, standard states., The standard state of a substance at a, specified temperature is its pure form at, 1 bar. For example, the standard state of liquid, , Here ∆fusH is enthalpy of fusion in standard, state. If water freezes, then process is reversed, and equal amount of heat is given off to the, surroundings., The enthalpy change that accompanies, melting of one mole of a solid substance, in standard state is called standard, enthalpy of fusion or molar enthalpy of, fusion, ∆fusH 0., Melting of a solid is endothermic, so all, enthalpies of fusion are positive. Water requires, , Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation, , (Tf and Tb are melting and boiling points, respectively), , 2021-22

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172, , CHEMISTRY, , heat for evaporation. At constant temperature, of its boiling point Tb and at constant pressure:, 0, , H2O(l) → H2O(g); ∆vapH = + 40.79 kJ moI–1 ∆vapH, is the standard enthalpy of vaporisation., , 0, , Amount of heat required to vaporize, one mole of a liquid at constant, temperature and under standard pressure, (1bar) is called its standard enthalpy of, vaporization or molar enthalpy of, vaporization, ∆vapH 0., Sublimation is direct conversion of a solid, into its vapour. Solid CO2 or ‘dry ice’ sublimes, 0, at 195K with ∆ sub H =25.2 kJ mol –1 ;, naphthalene sublimes slowly and for this, ∆sub H0 = 73.0 kJ mol–1 ., Standard enthalpy of sublimation,, 0, ∆subH is the change in enthalpy when one, mole of a solid substance sublimes at a, constant temperature and under standard, pressure (1bar)., The magnitude of the enthalpy change, depends on the strength of the intermolecular, interactions in the substance undergoing the, phase transfomations. For example, the strong, hydrogen bonds between water molecules hold, them tightly in liquid phase. For an organic, liquid, such as acetone, the intermolecular, dipole-dipole interactions are significantly, weaker. Thus, it requires less heat to vaporise, 1 mol of acetone than it does to vaporize 1 mol, of water. Table 6.1 gives values of standard, enthalpy changes of fusion and vaporisation, for some substances., Problem 6.7, A swimmer coming out from a pool is, covered with a film of water weighing, about 18g. How much heat must be, supplied to evaporate this water at, 298 K ? Calculate the internal energy of, vaporisation at 298K., ∆vap H 0 for water, at 298K= 44.01kJ mol–1, , Solution, We can represent the process of, evaporation as, vaporisation, H 2O(1) , → H2 O(g), 1mol, 1mol, , No. of moles in 18 g H2O(l) is, , =, , 18g, = 1mol, 18 g mol −1, , Heat supplied to evaporate18g water at, 0, 298 K, = n × ∆vap H, = (1 mol) × (44.01 kJ mol–1), = 44.01 kJ, (assuming steam behaving as an ideal, gas)., ∆vapU = ∆vap H V − p ∆V = ∆vap H V − ∆n g RT, , ∆vap H V − ∆n g RT = 44.01 kJ, − (1)(8.314 JK −1 mol −1 )(298K )(10 −3 kJ J −1 ), ∆vapU V = 44.01 kJ − 2.48 kJ, = 41.53 kJ, Problem 6.8, Assuming the water vapour to be a perfect, gas, calculate the internal energy change, when 1 mol of water at 100°C and 1 bar, pressure is converted to ice at 0°C. Given, the enthalpy of fusion of ice is 6.00 kJ mol1, heat capacity of water is 4.2 J/g°C, The change take place as follows:, Step - 1, 1 mol H2O (l, 100°C) à 1, mol (l, 0°C) Enthalpy, change ∆H1, Step - 2, , 1 mol H2O (l, 0°C) à 1 mol, H2O( S, 0°C) Enthalpy, change ∆H2, Total enthalpy change will be ∆H = ∆H1 + ∆H2, ∆H1 = - (18 x 4.2 x 100) J mol-1, = - 7560 J mol-1 = - 7.56 k J mol-1, ∆H2 = - 6.00 kJ mol-1, , 2021-22

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THERMODYNAMICS, , 173, , y, Table 6.2 Standard Molar Enthalpies of Formation (∆, ∆f H ) at 298K of a, Few Selected Substances, , Therefore,, ∆H = - 7.56 kJ mol-1 + (-6.00 kJ mol-1), = -13.56 kJ mol-1, There is negligible change in the volume, during the change form liquid to solid, state., Therefore, p∆v = ∆ng RT = 0, ∆H = ∆U = - 13.56kJ mol-1, (c) Standard Enthalpy of Formation, The standard enthalpy change for the, formation of one mole of a compound from, its elements in their most stable states of, , aggregation (also known as reference, states) is called Standard Molar Enthalpy, of Formation. Its symbol is ∆f H 0, where, the subscript ‘ f ’ indicates that one mole of, the compound in question has been formed, in its standard state from its elements in their, most stable states of aggregation. The reference, state of an element is its most stable state of, aggregation at 25°C and 1 bar pressure., For example, the reference state of dihydrogen, is H2 gas and those of dioxygen, carbon and, sulphur are O 2 gas, C graphite and S rhombic, respectively. Some reactions with standard, molar enthalpies of formation are as follows., , 2021-22

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174, , CHEMISTRY, , H2(g) + ½O2 (g) → H2O(1);, , Here, we can make use of standard enthalpy, of formation and calculate the enthalpy, change for the reaction. The following general, equation can be used for the enthalpy change, calculation., , ∆f H0 = –285.8 kJ mol–1, C (graphite, s) + 2H2(g) → Ch4 (g);, ∆f H0 = –74.81 kJ mol–1, 2C (graphite, s)+3H2 (g)+ ½O2(g) → C2H5OH(1);, ∆f H = – 277.7kJ mol, 0, , ∆rH0 = ∑ ai ∆f H0 (products) – ∑ bi ∆f H0 (reactants), i, , –1, , It is important to understand that a, 0, standard molar enthalpy of formation, ∆f H ,, 0, is just a special case of ∆rH , where one mole, of a compound is formed from its constituent, elements, as in the above three equations,, where 1 mol of each, water, methane and, ethanol is formed. In contrast, the enthalpy, change for an exothermic reaction:, CaO(s) + CO2(g) → CaCo3(s);, ∆rH0 = – 178.3kJ mol–1, is not an enthalpy of formation of calcium, carbonate, since calcium carbonate has been, formed from other compounds, and not from, its constituent elements. Also, for the reaction, given below, enthalpy change is not standard, 0, enthalpy of formation, ∆fH for HBr(g)., H2(g) + Br2(l) → 2HBr(g);, ∆r H0 = – 178.3kJ mol–1, Here two moles, instead of one mole of the, product is formed from the elements, i.e., ., ∆r H0 = 2∆f H0, Therefore, by dividing all coefficients in the, balanced equation by 2, expression for, enthalpy of formation of HBr (g) is written as, ½H2(g) + ½Br2(1) → HBr(g);, ∆f H0 = – 36.4 kJ mol–1, Standard enthalpies of formation of some, common substances are given in Table 6.2., By convention, standard enthalpy for, formation, ∆f H 0, of an element in reference, state, i.e., its most stable state of aggregation, is taken as zero., Suppose, you are a chemical engineer and, want to know how much heat is required to, decompose calcium carbonate to lime and, carbon dioxide, with all the substances in their, standard state., CaCO3(s) → CaO(s) + CO2(g); ∆r H0 = ?, , i, , (6.15), where a and b represent the coefficients of the, products and reactants in the balanced, equation. Let us apply the above equation for, decomposition of calcium carbonate. Here,, coefficients ‘a’ and ‘b’ are 1 each. Therefore,, ∆rH0 = ∆f H0 = [CaO(s)]+ ∆f H0 [CO2(g)], – ∆f H0 = [CaCO3(s)], =1 (–635.1 kJ mol–1) + 1(–393.5 kJ mol–1), –1(–1206.9 kJ mol–1), –1, = 178.3 kJ mol, Thus, the decomposition of CaCO3 (s) is an, endothermic process and you have to heat it, for getting the desired products., (d) Thermochemical Equations, A balanced chemical equation together with, the value of its ∆rH is called a thermochemical, equation. We specify the physical state, (alongwith allotropic state) of the substance in, an equation. For example:, C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l);, ∆rH0 = – 1367 kJ mol–1, The above equation describes the, combustion of liquid ethanol at constant, temperature and pressure. The negative sign, of enthalpy change indicates that this is an, exothermic reaction., It would be necessary to remember the, following conventions regarding thermochemical equations., 1. The coefficients in a balanced thermochemical equation refer to the number of, moles (never molecules) of reactants and, products involved in the reaction., 0, , 2. The numerical value of ∆rH refers to the, number of moles of substances specified, by an equation. Standard enthalpy change, 0, ∆rH will have units as kJ mol–1., , 2021-22

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THERMODYNAMICS, , 175, , To illustrate the concept, let us consider, the calculation of heat of reaction for the, following reaction :, , Fe2 O3 (s ) + 3H2 ( g ) → 2Fe ( s) + 3H2 O ( l ) ,, From the Table (6.2) of standard enthalpy of, formation (∆f H 0), we find :, ∆f H 0 (H2O,l) = –285.83 kJ mol–1;, ∆f H 0 (Fe2O3,s) = – 824.2 kJ mol–1;, Also ∆f H 0 (Fe, s) = 0 and, ∆f H 0 (H2, g) = 0 as per convention, Then,, ∆f H 10 = 3(–285.83 kJ mol–1), – 1(– 824.2 kJ mol–1), = (–857.5 + 824.2) kJ mol–1, = –33.3 kJ mol–1, Note that the coefficients used in these, calculations are pure numbers, which are, equal to the respective stoichiometric, 0, coefficients. The unit for ∆ r H, is, –1, kJ mol , which means per mole of reaction., Once we balance the chemical equation in a, particular way, as above, this defines the mole, of reaction. If we had balanced the equation, differently, for example,, , 1, 3, 3, Fe2 O3 (s ) + H2 ( g ) → Fe ( s) + H 2O ( l ), 2, 2, 2, then this amount of reaction would be one, 0, mole of reaction and ∆rH would be, ∆f H, , 0, 2, , 3, = (–285.83 kJ mol–1), 2, , –, , (e) Hess’s Law of Constant Heat, Summation, We know that enthalpy is a state function,, therefore the change in enthalpy is, independent of the path between initial state, (reactants) and final state (products). In other, words, enthalpy change for a reaction is the, same whether it occurs in one step or in a, series of steps. This may be stated as follows, in the form of Hess’s Law., If a reaction takes place in several steps, then its standard reaction enthalpy is the, sum of the standard enthalpies of the, intermediate reactions into which the, overall reaction may be divided at the same, temperature., Let us understand the importance of this, law with the help of an example., Consider the enthalpy change for the, reaction, , 1, O (g) → CO (g); ∆r H 0 = ?, 2 2, Although CO(g) is the major product, some, CO2 gas is always produced in this reaction., Therefore, we cannot measure enthalpy change, for the above reaction directly. However, if we, can find some other reactions involving related, species, it is possible to calculate the enthalpy, change for the above reaction., C (graphite,s) +, , Let us consider the following reactions:, C (graphite,s) + O2 (g) → CO2 (g);, ∆r H 0 = – 393.5 kJ mol–1 (i), , 1, (–824.2 kJ mol–1), 2, , CO (g) +, , = (– 428.7 + 412.1) kJ mol–1, = –16.6 kJ mol–1 = ½ ∆r H 10, It shows that enthalpy is an extensive quantity., 3. When a chemical equation is reversed, the, 0, value of ∆rH is reversed in sign. For, example, N2(g) + 3H2 (g) → 2NH3 (g);, ∆r H 0 = – 91.8 kJ. mol–1, 2NH3(g) → N2(g) + 3H2 (g);, ∆r H 0 = + 91.8 kJ mol–1, , 1, O (g) → CO2 (g), 2 2, ∆r H 0 = – 283.0 kJ mol–1 (ii), , We can combine the above two reactions, in such a way so as to obtain the desired, reaction. To get one mole of CO(g) on the right,, we reverse equation (ii). In this, heat is, absorbed instead of being released, so we, 0, change sign of ∆rH value, CO2 (g) → CO (g) +, , 2021-22, , 1, O (g);, 2 2, , ∆r H 0 = + 283.0 kJ mol–1 (iii)

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176, , CHEMISTRY, , Adding equation (i) and (iii), we get the, desired equation,, , C (graphite, s) +, for which ∆r H, , 0, , 1, O2 ( g ) → CO ( g ) ;, 2, , C6 H12 O6 ( g ) + 6O2 ( g ) → 6CO2 ( g ) + 6H 2 O(1);, ∆C H 0 = – 2802.0 kJ mol–1, , = (– 393.5 + 283.0), , = – 110.5 kJ mol–1, In general, if enthalpy of an overall reaction, A→B along one route is ∆rH and ∆rH1, ∆rH2,, ∆rH3..... representing enthalpies of reactions, leading to same product, B along another, route,then we have, ∆rH = ∆rH1 + ∆rH2 + ∆rH3 ..., , Similarly, combustion of glucose gives out, 2802.0 kJ/mol of heat, for which the overall, equation is :, , Our body also generates energy from food, by the same overall process as combustion,, although the final products are produced after, a series of complex bio-chemical reactions, involving enzymes., , (6.16), , It can be represented as:, A, , ∆rH, , B, ∆rH3, , ∆H1, , C, , ∆rH2, , CO2(g) and H2 O(l) are –393.5 kJ mol–1, and – 285.83 kJ mol–1 respectively., , D, , 6.5 ENTHALPIES FOR DIFFERENT TYPES, OF REACTIONS, It is convenient to give name to enthalpies, specifying the types of reactions., (a), , Standard Enthalpy of Combustion, 0, (symbol : ∆cH ), Combustion reactions are exothermic in, nature. These are important in industry,, rocketry, and other walks of life. Standard, enthalpy of combustion is defined as the, enthalpy change per mole (or per unit amount), of a substance, when it undergoes combustion, and all the reactants and products being in, their standard states at the specified, temperature., Cooking gas in cylinders contains mostly, butane (C4H10). During complete combustion, of one mole of butane, 2658 kJ of heat is, released. We can write the thermochemical, reactions for this as:, , C4 H10 ( g ) +, , Problem 6.9, The combustion of one mole of benzene, takes place at 298 K and 1 atm. After, combustion, CO2(g) and H2O (1) are, produced and 3267.0 kJ of heat is, liberated. Calculate the standard, 0, enthalpy of formation, ∆f H of benzene., Standard enthalpies of formation of, , 13, O2 ( g ) → 4CO2 ( g ) + 5H2 O(1);, 2, ∆C H 0 = – 2658.0 kJ mol–1, , 2021-22, , Solution, The formation reaction of benezene is, given by :, , 6C (graphite ) + 3H2 ( g ) → C6 H6 ( l ) ;, ∆f H 0 = ? ... (i), The enthalpy of combustion of 1 mol of, benzene is :, , C 6 H6 ( l ) +, , 15, O2 → 6CO2 ( g ) + 3H2 O ( l ) ;, 2, ∆C H 0 = – 3267 kJ mol–1... (ii), , The enthalpy of formation of 1 mol of, CO2(g) :, , C ( graphite ) + O2 ( g ) → CO2 ( g ) ;, ∆f H 0 = – 393.5 kJ mol–1... (iii), , The enthalpy of formation of 1 mol of, H2O(l) is :, , H2 ( g ) +, , 1, O 2 ( g ) → H2 O ( l ) ;, 2, ∆C H 0 = – 285.83 kJ mol–1... (iv), , multiplying eqn. (iii) by 6 and eqn. (iv), by 3 we get:

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THERMODYNAMICS, , 177, , 6C ( graphite ) + 6O2 ( g ) → 6CO2 (g ) ;, ∆f H 0 = – 2361 kJ mol–1, , 3 H2 ( g ) +, , 3, O2 ( g ) → 3H 2 O (1) ;, 2, ∆f H 0 = – 857.49 kJ mol–1, , Summing up the above two equations :, 6C ( graphite ) + 3H2 ( g ) +, , 15, O2 ( g ) → 6CO2 ( g ), 2, + 3 H2 O ( l ) ;, , ∆f H 0 = – 3218.49 kJ mol–1... (v), Reversing equation (ii);, , 15, 6CO2 ( g ) + 3H2 O ( l ) → C6 H6 ( l ) +, O2 ;, 2, ∆f H 0 = – 3267.0 kJ mol–1... (vi), Adding equations (v) and (vi), we get, , 6C ( graphite ) + 3H2 (g ) → C6 H6 ( l ) ;, ∆f H 0 = – 48.51 kJ mol–1... (iv), (b) Enthalpy of Atomization, 0, (symbol: ∆aH ), Consider the following example of atomization, of dihydrogen, 0, , H2(g) → 2H(g); ∆aH = 435.0 kJ mol, You can see that H atoms are formed by, breaking H–H bonds in dihydrogen. The, enthalpy change in this process is known as, enthalpy of atomization, ∆ aH 0. It is the, enthalpy change on breaking one mole of, bonds completely to obtain atoms in the gas, phase., –1, , In case of diatomic molecules, like, dihydrogen (given above), the enthalpy of, atomization is also the bond dissociation, enthalpy. The other examples of enthalpy of, atomization can be, 0, , CH4(g) → C(g) + 4H(g); ∆aH = 1665 kJ mol, , –1, , Note that the products are only atoms of C, and H in gaseous phase. Now see the following, reaction:, 0, , Na(s) → Na(g) ; ∆aH = 108.4 kJ mol, , –1, , In this case, the enthalpy of atomization is, same as the enthalpy of sublimation., 0, (c) Bond Enthalpy (symbol: ∆bondH ), Chemical reactions involve the breaking and, making of chemical bonds. Energy is required, to break a bond and energy is released when, a bond is formed. It is possible to relate heat, of reaction to changes in energy associated, with breaking and making of chemical bonds., With reference to the enthalpy changes, associated with chemical bonds, two different, terms are used in thermodynamics., , (i), , Bond dissociation enthalpy, , (ii), , Mean bond enthalpy, , Let us discuss these terms with reference, to diatomic and polyatomic molecules., Diatomic Molecules: Consider the following, process in which the bonds in one mole of, dihydrogen gas (H2) are broken:, 0, –1, H2(g) → 2H(g) ; ∆H–HH = 435.0 kJ mol, The enthalpy change involved in this process, is the bond dissociation enthalpy of H–H bond., The bond dissociation enthalpy is the change, in enthalpy when one mole of covalent bonds, of a gaseous covalent compound is broken to, form products in the gas phase., Note that it is the same as the enthalpy of, atomization of dihydrogen. This is true for all, diatomic molecules. For example:, 0, , Cl2(g) → 2Cl(g) ; ∆Cl–ClH = 242 kJ mol, 0, , –1, , O2(g) → 2O(g) ; ∆O=OH = 428 kJ mol, In the case of polyatomic molecules, bond, dissociation enthalpy is different for different, bonds within the same molecule., Polyatomic Molecules: Let us now consider, a polyatomic molecule like methane, CH4. The, overall thermochemical equation for its, atomization reaction is given below:, –1, , CH4 ( g ) → C(g ) + 4H( g );, ∆a H 0 = 1665 kJ mol–1, In methane, all the four C – H bonds are, identical in bond length and energy. However,, the energies required to break the individual, C – H bonds in each successive step differ :, , 2021-22

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178, , CHEMISTRY, , CH4(g) → CH3(g)+H(g);∆bond H0 = +427 kJ mol–1, CH3(g) → CH2(g)+H(g);∆bond H0 = +439 kJ mol–1, CH2(g) → CH(g)+H(g);∆bond H0 = +452 kJ mol–1, CH(g) → C(g)+H(g);∆bond H0 = +347 kJ mol–1, Therefore,, CH4(g) → C(g)+4H(g);∆a H0 = 1665 kJ mol–1, In such cases we use mean bond enthalpy, of C – H bond., 0, For example in CH4, ∆C–HH is calculated as:, 0, ∆C–HH = ¼ (∆a H0) = ¼ (1665 kJ mol–1), = 416 kJ mol–1, We find that mean C–H bond enthalpy in, methane is 416 kJ/mol. It has been found that, mean C–H bond enthalpies differ slightly from, compound, to, compound,, as, in, CH3CH2Cl,CH3NO2, etc, but it does not differ, in a great deal*. Using Hess’s law, bond, enthalpies can be calculated. Bond enthalpy, values of some single and multiple bonds are, , given in Table 6.3. The reaction enthalpies are, very important quantities as these arise from, the changes that accompany the breaking of, old bonds and formation of the new bonds., We can predict enthalpy of a reaction in gas, phase, if we know different bond enthalpies., The standard enthalpy of reaction, ∆rH0 is, related to bond enthalpies of the reactants and, products in gas phase reactions as:, , ∑ bond enthalpiesreactants, − ∑ bond enthalpies products, , ∆r H 0 =, , (6.17)**, This relationship is particularly more, useful when the required values of ∆f H0 are, not available. The net enthalpy change of a, reaction is the amount of energy required to, break all the bonds in the reactant molecules, minus the amount of energy required to break, all the bonds in the product molecules., Remember that this relationship is, –1, , H, , Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol, , at 298 K, , C, , Br, , 435.8 414, 347, , N, , O, , F, , Si, , P, , 389, 293, 159, , 464, 351, 201, 138, , 569, 439, 272, 184, 155, , 293, 289, 368, 540, 176, , 318, 264, 209, 351, 490, 213, 213, , S, , Cl, , 339, 259, 327, 226, 230, 213, , 431, 330, 201, 205, 255, 360, 331, 251, 243, , 368, 276, 243, 197, 289, 272, 213, 218, 192, , –1, , Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol, N=N, , 418, , C=C, , 611, , N≡N, , 946, , C≡ C, , 837, , C=N, , 615, , C=O, , 741, , C≡N, , 891, , C≡O, , 1070, , I, 297, 238, 201, 213, 213, 209, 180, 151, , H, C, N, O, F, Si, P, S, Cl, Br, I, , at 298 K, , O=O, , 498, , * Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same., 0 ), which is the enthalpy change when one mole of a particular type of, ** If we use enthalpy of bond formation, (∆f H bond, bond is formed from gaseous atom, then ∆f H 0 =, , ∑ ∆f H 0 bonds of products – ∑ ∆f H 0 bonds of reactants, , 2021-22

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THERMODYNAMICS, , 179, , approximate and is valid when all substances, (reactants and products) in the reaction are in, gaseous state., (d) Lattice Enthalpy, The lattice enthalpy of an ionic compound is, the enthalpy change which occurs when one, mole of an ionic compound dissociates into its, ions in gaseous state., , Na + Cl − ( s) → Na + ( g ) + Cl − ( g ) ;, 0, ∆latticeH = +788 kJ mol–1, Since it is impossible to determine lattice, enthalpies directly by experiment, we use an, indirect method where we construct an, enthalpy diagram called a Born-Haber Cycle, (Fig. 6.9)., Let us now calculate the lattice enthalpy, of Na+Cl–(s) by following steps given below :, 1. Na(s) → Na( g ) , sublimation of sodium, 0, metal, ∆subH = 108.4 kJ mol–1, , 2. Na( g ) → Na + ( g ) + e −1 ( g ) , the ionization of, sodium atoms, ionization enthalpy, ∆iH0 = 496 kJ mol–1, 3., , 1, Cl2 ( g ) → Cl( g ) , the dissociation of, 2, chlorine, the reaction enthalpy is half the, bond dissociation enthalpy., 1, 0, –1, ∆ H = 121 kJ mol, 2 bond, , 4. Cl( g ) + e −1 ( g ) → Cl( g ) electron gained by, chlorine atoms. The electron gain enthalpy,, ∆egH 0 = – 348.6 kJ mol –1., You have learnt about ionization enthalpy, and electron gain enthalpy in Unit 3. In, fact, these terms have been taken from, thermodynamics. Earlier terms, ionization, energy and electron affinity were in, practice in place of the above terms (see, the box for justification)., Ionization Energy and Electron Affinity, Ionization energy and electron affinity are, defined at absolute zero. At any other, temperature, heat capacities for the, reactants and the products have to be, taken into account. Enthalpies of reactions, for, +, , M(g) →, M(g) + e, , M (g), –, , →, , +e, , –, , (for ionization), , –, , M (g) (for electron gain), , at temperature, T is, T, 0, , 0, , ∆rH (T ) = ∆rH (0) +, , ∫∆C, r, , 0, P, , dT, , 0, , The value of C p for each species in the, above reaction is 5/2 R (CV = 3/2R), So, ∆rCp0 = + 5/2 R (for ionization), 0, ∆rCp = – 5/2 R (for electron gain), Therefore,, 0, ∆rH (ionization enthalpy), ∆rH, , Fig. 6.9 Enthalpy diagram for lattice enthalpy, of NaCl, , 0, , = E0 (ionization energy) + 5/2 RT, (electron gain enthalpy), = – A( electron affinity) – 5/2 RT, , +, −, +, −, 5. Na ( g ) + Cl ( g ) → Na Cl ( s ), The sequence of steps is shown in Fig. 6.9,, and is known as a Born-Haber cycle. The, , 2021-22

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180, , CHEMISTRY, , importance of the cycle is that, the sum of, the enthalpy changes round a cycle is, zero. A p p l y i n g H e s s ’ s l a w , w e g e t ,, 0, , ∆latticeH = 411.2 + 108.4 + 121 + 496 – 348.6, 0, , ∆latticeH = + 788kJ, for NaCl(s) d Na+(g) + Cl–(g), Internal energy is smaller by 2RT ( because, ∆ng = 2) and is equal to + 783 kJ mol–1., Now we use the value of lattice enthalpy to, calculate enthalpy of solution from the, expression:, 0, , 0, , 0, , ∆solH = ∆latticeH + ∆hydH, For one mole of NaCl(s),, lattice enthalpy = + 788 kJ mol–1, 0, and ∆hydH = – 784 kJ mol –1( from the, literature), 0, ∆sol H = + 788 kJ mol –1 – 784 kJ mol –1, = + 4 kJ mol–1, The dissolution of NaCl(s) is accompanied, by very little heat change., (e) Enthalpy of Solution (symbol : ∆solH0 ), Enthalpy of solution of a substance is the, enthalpy change when one mole of it dissolves, in a specified amount of solvent. The enthalpy, of solution at infinite dilution is the enthalpy, change observed on dissolving the substance, in an infinite amount of solvent when the, interactions between the ions (or solute, molecules) are negligible., When an ionic compound dissolves in a, solvent, the ions leave their ordered positions on, the crystal lattice. These are now more free in, solution. But solvation of these ions (hydration, in case solvent is water) also occurs at the same, time. This is shown diagrammatically, for an, ionic compound, AB (s), , 0, , The enthalpy of solution of AB(s), ∆solH , in, water is, therefore, determined by the selective, 0, values of the lattice enthalpy,∆latticeH and, 0, , enthalpy of hydration of ions, ∆hydH as, 0, , 0, , 0, , ∆sol H = ∆latticeH + ∆hydH, 0, For most of the ionic compounds, ∆sol H is, positive and the dissociation process is, endothermic. Therefore the solubility of most, salts in water increases with rise of, temperature. If the lattice enthalpy is very, high, the dissolution of the compound may not, take place at all. Why do many fluorides tend, to be less soluble than the corresponding, chlorides? Estimates of the magnitudes of, enthalpy changes may be made by using tables, of bond energies (enthalpies) and lattice, energies (enthalpies)., (f) Enthalpy of Dilution, It is known that enthalpy of solution is the, enthalpy change associated with the addition, of a specified amount of solute to the specified, amount of solvent at a constant temperature, and pressure. This argument can be applied, to any solvent with slight modification., Enthalpy change for dissolving one mole of, gaseous hydrogen chloride in 10 mol of water, can be represented by the following equation., For convenience we will use the symbol aq. for, water, HCl(g) + 10 aq. → HCl.10 aq., ∆H = –69.01 kJ / mol, Let us consider the following set of enthalpy, changes:, (S-1) HCl(g) + 25 aq. → HCl.25 aq., ∆H = –72.03 kJ / mol, (S-2) HCl(g) + 40 aq. → HCl.40 aq., ∆H = –72.79 kJ / mol, (S-3) HCl(g) + ∞ aq. → HCl. ∞ aq., ∆H = –74.85 kJ / mol, The values of ∆H show general dependence, of the enthalpy of solution on amount of solvent., As more and more solvent is used, the enthalpy, of solution approaches a limiting value, i.e, the, value in infinitely dilute solution. For, hydrochloric acid this value of ∆H is given, above in equation (S-3)., , 2021-22

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THERMODYNAMICS, , 181, , If we subtract the first equation (equation, S-1) from the second equation (equation S-2), in the above set of equations, we obtain–, HCl.25 aq. + 15 aq. → HCl.40 aq., ∆H = [ –72.79 – (–72.03)] kJ / mol, = – 0.76 kJ / mol, This value (–0.76kJ/mol) of ∆H is enthalpy, of dilution. It is the heat withdrawn from the, surroundings when additional solvent is, added to the solution. The enthalpy of dilution, of a solution is dependent on the original, concentration of the solution and the amount, of solvent added., 6.6 SPONTANEITY, The first law of thermodynamics tells us about, the relationship between the heat absorbed, and the work performed on or by a system. It, puts no restrictions on the direction of heat, flow. However, the flow of heat is unidirectional, from higher temperature to lower, temperature. In fact, all naturally occurring, processes whether chemical or physical will, tend to proceed spontaneously in one, direction only. For example, a gas expanding, to fill the available volume, burning carbon, in dioxygen giving carbon dioxide., But heat will not flow from colder body to, warmer body on its own, the gas in a container, will not spontaneously contract into one, corner or carbon dioxide will not form carbon, and dioxygen spontaneously. These and many, other spontaneously occurring changes show, unidirectional change. We may ask ‘what is the, driving force of spontaneously occurring, changes ? What determines the direction of a, spontaneous change ? In this section, we shall, establish some criterion for these processes, whether these will take place or not., Let us first understand what do we mean, by spontaneous reaction or change ? You may, think by your common observation that, spontaneous reaction is one which occurs, immediately when contact is made between, the reactants. Take the case of combination of, hydrogen and oxygen. These gases may be, mixed at room temperature and left for many, years without observing any perceptible, change. Although the reaction is taking place, , between them, it is at an extremely slow rate., It is still called spontaneous reaction. So, spontaneity means ‘having the potential to, proceed without the assistance of external, agency’. However, it does not tell about the, rate of the reaction or process. Another aspect, of spontaneous reaction or process, as we see, is that these cannot reverse their direction on, their own. We may summarise it as follows:, A spontaneous process is an, irreversible process and may only be, reversed by some external agency., (a) Is Decrease in Enthalpy a Criterion, for Spontaneity ?, If we examine the phenomenon like flow of, water down hill or fall of a stone on to the, ground, we find that there is a net decrease in, potential energy in the direction of change. By, analogy, we may be tempted to state that a, chemical reaction is spontaneous in a given, direction, because decrease in energy has, taken place, as in the case of exothermic, reactions. For example:, , 3, 1, N (g) + H2(g) = NH3(g) ;, 2 2, 2, 0, ∆r H = – 46.1 kJ mol–1, 1, 1, H2(g) + Cl2(g) = HCl (g) ;, 2, 2, ∆r H0 = – 92.32 kJ mol–1, 1, O (g) → H2O(l) ;, 2 2, 0, ∆r H = –285.8 kJ mol–1, The decrease in enthalpy in passing from, reactants to products may be shown for any, exothermic reaction on an enthalpy diagram, as shown in Fig. 6.10(a)., Thus, the postulate that driving force for a, chemical reaction may be due to decrease in, energy sounds ‘reasonable’ as the basis of, evidence so far !, Now let us examine the following reactions:, 1, N (g) + O2(g) → NO2(g);, 2 2, 0, ∆r H = +33.2 kJ mol–1, , H2(g) +, , C(graphite, s) + 2 S(l) → CS2(l);, 0, ∆r H = +128.5 kJ mol–1, , 2021-22

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182, , Fig. 6.10 (a), , CHEMISTRY, , Enthalpy diagram for exothermic, reactions, , These reactions though endothermic, are, spontaneous. The increase in enthalpy may, be represented on an enthalpy diagram as, shown in Fig. 6.10(b)., , Fig. 6.11 Diffusion of two gases, , Fig. 6.10 (b), , Enthalpy diagram for endothermic, reactions, , Therefore, it becomes obvious that while, decrease in enthalpy may be a contributory, factor for spontaneity, but it is not true for all, cases., (b) Entropy and Spontaneity, Then, what drives the spontaneous process in, a given direction ? Let us examine such a case, in which ∆H = 0 i.e., there is no change in, enthalpy, but still the process is spontaneous., Let us consider diffusion of two gases into, each other in a closed container which is, isolated from the surroundings as shown in, Fig. 6.11., The two gases, say, gas A and gas B are, represented by black dots and white dots, , respectively and separated by a movable, partition [Fig. 6.11 (a)]. When the partition is, withdrawn [Fig.6.11( b)], the gases begin to, diffuse into each other and after a period of, time, diffusion will be complete., Let us examine the process. Before, partition, if we were to pick up the gas, molecules from left container, we would be, sure that these will be molecules of gas A and, similarly if we were to pick up the gas, molecules from right container, we would be, sure that these will be molecules of gas B. But,, if we were to pick up molecules from container, when partition is removed, we are not sure, whether the molecules picked are of gas A or, gas B. We say that the system has become less, predictable or more chaotic., We may now formulate another postulate:, in an isolated system, there is always a, tendency for the systems’ energy to become, more disordered or chaotic and this could be, a criterion for spontaneous change !, At this point, we introduce another, thermodynamic function, entropy denoted as, S. The above mentioned disorder is the, manifestation of entropy. To form a mental, , 2021-22

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THERMODYNAMICS, , 183, , picture, one can think of entropy as a measure, of the degree of randomness or disorder in the, system. The greater the disorder in an isolated, system, the higher is the entropy. As far as a, chemical reaction is concerned, this entropy, change can be attributed to rearrangement of, atoms or ions from one pattern in the reactants, to another (in the products). If the structure, of the products is very much disordered than, that of the reactants, there will be a resultant, increase in entropy. The change in entropy, accompanying a chemical reaction may be, estimated qualitatively by a consideration of, the structures of the species taking part in the, reaction. Decrease of regularity in structure, would mean increase in entropy. For a given, substance, the crystalline solid state is the, state of lowest entropy (most ordered), The, gaseous state is state of highest entropy., Now let us try to quantify entropy. One way, to calculate the degree of disorder or chaotic, distribution of energy among molecules would, be through statistical method which is beyond, the scope of this treatment. Other way would, be to relate this process to the heat involved in, a process which would make entropy a, thermodynamic concept. Entropy, like any, other thermodynamic property such as, internal energy U and enthalpy H is a state, function and ∆S is independent of path., Whenever heat is added to the system, it, increases molecular motions causing, increased randomness in the system. Thus, heat (q) has randomising influence on the, system. Can we then equate ∆S with q ? Wait !, Experience suggests us that the distribution, of heat also depends on the temperature at, which heat is added to the system. A system, at higher temperature has greater randomness, in it than one at lower temperature. Thus,, temperature is the measure of average, chaotic motion of particles in the system., Heat added to a system at lower temperature, causes greater randomness than when the, same quantity of heat is added to it at higher, temperature. This suggests that the entropy, change is inversely proportional to the, temperature. ∆S is related with q and T for a, reversible reaction as :, , ∆S =, , qrev, T, , (6.18), , The total entropy change ( ∆Stotal) for the, system and surroundings of a spontaneous, process is given by, , ∆Stotal = ∆Ssystem + ∆Ssurr > 0, , (6.19), , When a system is in equilibrium, the, entropy is maximum, and the change in, entropy, ∆S = 0., We can say that entropy for a spontaneous, process increases till it reaches maximum and, at equilibrium the change in entropy is zero., Since entropy is a state property, we can, calculate the change in entropy of a reversible, process by, ∆Ssys =, , qsys ,rev, T, , We find that both for reversible and, irreversible expansion for an ideal gas, under, isothermal conditions, ∆U = 0, but ∆Stotal i.e.,, , ∆Ssys + ∆Ssurr is not zero for irreversible, process. Thus, ∆U does not discriminate, between reversible and irreversible process,, whereas ∆S does., , 2021-22, , Problem 6.10, Predict in which of the following, entropy, increases/decreases :, (i) A liquid crystallizes into a solid., (ii) Temperature of a crystalline solid is, raised from 0 K to 115 K., , ( iii ), , 2NaHCO3 ( s) → Na 2 CO3 ( s) +, , (iv), , H2 ( g ) → 2 H ( g ), , CO2 (g ) + H2 O (g ), , Solution, (i) After freezing, the molecules attain an, ordered state and therefore, entropy, decreases., (ii) At 0 K, the contituent particles are, static and entropy is minimum. If, temperature is raised to 115 K, these

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184, , CHEMISTRY, , begin to move and oscillate about their, equilibrium positions in the lattice and, system becomes more disordered,, therefore entropy increases., (iii) Reactant, NaHCO3 is a solid and it, has low entropy. Among products, there are one solid and two gases., Therefore, the products represent a, condition of higher entropy., (iv) Here one molecule gives two atoms, i.e., number of particles increases, leading to more disordered state., Two moles of H atoms have higher, entropy than one mole of dihydrogen, molecule., Problem 6.11, For oxidation of iron,, , 4Fe ( s) + 3O 2 ( g ) → 2Fe2 O 3 ( s), entropy change is – 549.4 JK–1mol–1at, 298 K. Inspite of negative entropy change, of this reaction, why is the reaction, spontaneous?, 0, , (∆ r H for, this, reaction, is, –1648 × 103 J mol–1), Solution, One decides the spontaneity of a reaction, by considering, , (, , = 4980.6 JK–1 mol–1, This shows that the above reaction is, spontaneous., (c) Gibbs Energy and Spontaneity, We have seen that for a system, it is the total, entropy change, ∆S total which decides the, spontaneity of the process. But most of the, chemical reactions fall into the category of, either closed systems or open systems., Therefore, for most of the chemical reactions, there are changes in both enthalpy and, entropy. It is clear from the discussion in, previous sections that neither decrease in, enthalpy nor increase in entropy alone can, determine the direction of spontaneous change, for these systems., For this purpose, we define a new, thermodynamic function the Gibbs energy or, Gibbs function, G, as, G = H – TS, (6.20), Gibbs function, G is an extensive property, and a state function., The change in Gibbs energy for the system,, ∆Gsys can be written as, ∆Gsys = ∆H sys − T ∆Ssys − Ssys ∆T, , At constant temperature, ∆T = 0, , ), , ∴ ∆G sys = ∆H sys − T ∆Ssys, , ∆Stotal ∆Ssys + ∆Ssurr . For calculating, ∆S surr, we have to consider the heat, absorbed by the surroundings which is, 0, equal to – ∆rH . At temperature T, entropy, change of the surroundings is, , ( −1648 ×10, =−, , 3, , J mol −1, , ), , 298 K, , –1, , = 5530 JK mol–1, Thus, total entropy change for this, reaction, ∆r Stotal = 5530 JK –1mol –1 +, , ( −549.4 JK, , –1, , mol –1, , ), , Usually the subscript ‘system’ is dropped, and we simply write this equation as, ∆G = ∆H − T ∆S, , (6.21), , Thus, Gibbs energy change = enthalpy, change – temperature × entropy change, and, is referred to as the Gibbs equation, one of the, most important equations in chemistry. Here,, we have considered both terms together for, spontaneity: energy (in terms of ∆H) and, entropy (∆S, a measure of disorder) as, indicated earlier. Dimensionally if we analyse,, we find that ∆G has units of energy because,, both ∆H and the T∆S are energy terms, since, T∆S = (K) (J/K) = J., Now let us consider how ∆G is related to, reaction spontaneity., , 2021-22

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THERMODYNAMICS, , 185, , We know,, ∆Stotal = ∆Ssys + ∆Ssurr, If the system is in thermal equilibrium with, the surrounding, then the temperature of the, surrounding is same as that of the system., Also, increase in enthalpy of the surrounding, is equal to decrease in the enthalpy of the, system., Therefore,, entropy, change, of, surroundings,, , ∆Ssurr =, , ∆H sys, ∆H surr, =−, T, T, ,  ∆H sys , ∆Stotal = ∆Ssys +  −, T , , Rearranging the above equation:, T∆Stotal = T∆Ssys – ∆Hsys, For spontaneous process, ∆Stotal > 0 , so, T∆Ssys – ∆Hsys > Ο, , (, , ), , ⇒ − ∆H sys − T ∆Ssys > 0, Using equation 6.21, the above equation can, be written as, –∆G > O, , ∆G = ∆H − T ∆S < 0, , (6.22), , ∆Hsys is the enthalpy change of a reaction,, T∆Ssys is the energy which is not available to, do useful work. So ∆G is the net energy, available to do useful work and is thus a, measure of the ‘free energy’. For this reason, it, is also known as the free energy of the reaction., ∆G gives a criteria for spontaneity at, constant pressure and temperature., (i) If ∆G is negative (< 0), the process is, spontaneous., (ii) If ∆G is positive (> 0), the process is non, spontaneous., Note : If a reaction has a positive enthalpy, change and positive entropy change, it can be, spontaneous when T∆S is large enough to, outweigh ∆H. This can happen in two ways;, (a) The positive entropy change of the system, , can be ‘small’ in which case T must be large., (b) The positive entropy change of the system, can be ’large’, in which case T may be small., The former is one of the reasons why reactions, are often carried out at high temperature., Table 6.4 summarises the effect of temperature, on spontaneity of reactions., (d) Entropy and Second Law of, Thermodynamics, We know that for an isolated system the, change in energy remains constant. Therefore,, increase in entropy in such systems is the, natural direction of a spontaneous change., This, in fact is the second law of, thermodynamics. Like first law of, thermodynamics, second law can also be, stated in several ways. The second law of, thermodynamics explains why spontaneous, exothermic reactions are so common. In, exothermic reactions heat released by the, reaction increases the disorder of the, surroundings and overall entropy change is, positive which makes the reaction, spontaneous., (e) Absolute Entropy and Third Law of, Thermodynamics, Molecules of a substance may move in a, straight line in any direction, they may spin, like a top and the bonds in the molecules may, stretch and compress. These motions of the, molecule are called translational, rotational, and vibrational motion respectively. When, temperature of the system rises, these motions, become more vigorous and entropy increases., On the other hand when temperature is, lowered, the entropy decreases. The entropy, of any pure crystalline substance, approaches zero as the temperature, approaches absolute zero. This is called, third law of thermodynamics. This is so, because there is perfect order in a crystal at, absolute zero. The statement is confined to, pure crystalline solids because theoretical, arguments and practical evidences have, shown that entropy of solutions and super, cooled liquids is not zero at 0 K. The, importance of the third law lies in the fact that, , 2021-22

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186, , CHEMISTRY, , it permits the calculation of absolute values, of entropy of pure substance from thermal, data alone. For a pure substance, this can be, q, done by summing rev increments from 0 K, T, to 298 K. Standard entropies can be used to, calculate standard entropy changes by a, Hess’s law type of calculation., 6.7 GIBBS ENERGY, EQUILIBRIUM, , CHANGE, , AND, , We have seen how a knowledge of the sign and, magnitude of the free energy change of a, chemical reaction allows:, (i) Prediction of the spontaneity of the, chemical reaction., (ii) Prediction of the useful work that could, be extracted from it., So far we have considered free energy, changes in irreversible reactions. Let us now, examine the free energy changes in reversible, reactions., ‘Reversible’ under strict thermodynamic, sense is a special way of carrying out a, process such that system is at all times in, perfect equilibrium with its surroundings., When applied to a chemical reaction, the, term ‘reversible’ indicates that a given, reaction can proceed in either direction, simultaneously, so that a dynamic, equilibrium is set up. This means that the, reactions in both the directions should, , proceed with a decrease in free energy, which, seems impossible. It is possible only if at, equilibrium the free energy of the system is, minimum. If it is not, the system would, spontaneously change to configuration of, lower free energy., So, the criterion for equilibrium, A + B l C + D ; is, ∆rG = 0, Gibbs energy for a reaction in which all, reactants and products are in standard state,, 0, ∆rG is related to the equilibrium constant of, the reaction as follows:, 0, 0 = ∆rG + RT ln K, 0, or ∆rG = – RT ln K, 0, (6.23), or ∆rG = – 2.303 RT log K, We also know that, (6.24), For strongly endothermic reactions, the, 0, value of ∆rH may be large and positive. In, such a case, value of K will be much smaller, than 1 and the reaction is unlikely to form, much product. In case of exothermic, reactions, ∆rH0 is large and negative, and ∆rG0, is likely to be large and negative too. In such, cases, K will be much larger than 1. We may, expect strongly exothermic reactions to have, a large K, and hence can go to near, 0, 0, completion. ∆rG also depends upon ∆rS , if, the changes in the entropy of reaction is also, taken into account, the value of K or extent, of chemical reaction will also be affected,, , Table 6.4 Effect of Temperature on Spontaneity of Reactions, , ∆rH, , *, , 0, , ∆rS, , 0, , ∆rG, , 0, , Description*, , –, , +, , –, , Reaction spontaneous at all temperatures, , –, , –, , – (at low T ), , Reaction spontaneous at low temperature, , –, , –, , + (at high T ), , Reaction nonspontaneous at high temperature, , +, , +, , + (at low T ), , Reaction nonspontaneous at low temperature, , +, , +, , – (at high T ), , Reaction spontaneous at high temperature, , +, , –, , + (at all T ), , Reaction nonspontaneous at all temperatures, , The term low temperature and high temperature are relative. For a particular reaction, high temperature could even, mean room temperature., , 2021-22

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THERMODYNAMICS, , 187, , 0, , depending upon whether ∆rS is positive or, negative., Using equation (6.24),, 0, (i) It is possible to obtain an estimate of ∆G, V, 0, from the measurement of ∆H and ∆S ,, and then calculate K at any temperature, for economic yields of the products., (ii) If K is measured directly in the, 0, laboratory, value of ∆G at any other, temperature can be calculated., Using equation (6.24),, , ( –13.6 × 10, =, 2.303 (8.314 JK, , 3, , –1, , mol, , –1, , ), , ) (298 K ), , = 2.38, Hence K = antilog 2.38 = 2.4 × 102., Problem 6.14, At 60°C, dinitrogen tetroxide is 50, per cent dissociated. Calculate the, standard free energy change at this, temperature and at one atmosphere., Solution, N2O4(g), , Problem 6.12, Calculate ∆rG 0 for conversion of oxygen, to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp, for this conversion is 2.47 × 10 –29., Solution, We know ∆rG 0 = – 2.303 RT log Kp and, R = 8.314 JK–1 mol–1, 0, Therefore, ∆rG =, – 2.303 (8.314 J K–1 mol–1), × (298 K) (log 2.47 × 10–29), = 163000 J mol–1, = 163 kJ mol–1., Problem 6.13, Find out the value of equilibrium constant, for the following reaction at 298 K., , J mol –1, , 2NO2(g), , If N 2O4 is 50% dissociated, the mole, fraction of both the substances is given, by, , xN, , 2 O4, , pN, , =, , 2 O4, , 1 − 0.5, 2 × 0.5, : x NO2 =, 1 + 0.5, 1 + 0.5, , =, , 0.5, × 1 atm, p NO =, 2, 1.5, 1, × 1 atm., 1.5, , The equilibrium constant Kp is given by, , (p ), , 2, , Kp =, , NO2, , p N 2O 4, , =, , 1.5, (1.5)2 (0.5 ), , = 1.33 atm., Since, , 0, , Standard Gibbs energy change, ∆rG at, the given temperature is –13.6 kJ mol–1., , 0, , ∆rG = –RT ln Kp, 0, , ∆rG = (– 8.314 JK–1 mol–1) × (333 K), , Solution, , × (2.303) × (0.1239), We know, log K =, , = – 763.8 kJ mol–1, , 2021-22

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188, , CHEMISTRY, , SUMMARY, Thermodynamics deals with energy changes in chemical or physical processes and enables us, to study these changes quantitatively and to make useful predictions. For these purposes, we, divide the universe into the system and the surroundings. Chemical or physical processes, lead to evolution or absorption of heat (q), part of which may be converted into work (w). These, quantities are related through the first law of thermodynamics via ∆U = q + w. ∆U, change in, internal energy, depends on initial and final states only and is a state function, whereas q and, w depend on the path and are not the state functions. We follow sign conventions of q and w by, giving the positive sign to these quantities when these are added to the system. We can measure, the transfer of heat from one system to another which causes the change in temperature. The, magnitude of rise in temperature depends on the heat capacity (C) of a substance. Therefore,, heat absorbed or evolved is q = C∆T. Work can be measured by w = –pex∆V, in case of expansion, of gases. Under reversible process, we can put pex = p for infinitesimal changes in the volume, making wrev = – p dV. In this condition, we can use gas equation, pV = nRT., At constant volume, w = 0, then ∆U = qV , heat transfer at constant volume. But in study of, chemical reactions, we usually have constant pressure. We define another state function, enthalpy. Enthalpy change, ∆H = ∆U + ∆ngRT, can be found directly from the heat changes at, constant pressure, ∆H = qp., There are varieties of enthalpy changes. Changes of phase such as melting, vaporization, and sublimation usually occur at constant temperature and can be characterized by enthalpy, changes which are always positive. Enthalpy of formation, combustion and other enthalpy, changes can be calculated using Hess’s law. Enthalpy change for chemical reactions can be, determined by, ∆r H =, , ∑ (a ∆, i, , f, , f, , ), , (, , H products − ∑ bi ∆ f H reactions, i, , ), , and in gaseous state by, 0, , ∆rH =, , Σ bond enthalpies of the reactants – Σ bond enthalpies of the products, , First law of thermodynamics does not guide us about the direction of chemical reactions, i.e., what is the driving force of a chemical reaction. For isolated systems,, ∆U = 0. We define another state function, S, entropy for this purpose. Entropy is a measure of, disorder or randomness. For a spontaneous change, total entropy change is positive. Therefore,, for an isolated system, ∆U = 0, ∆S > 0, so entropy change distinguishes a spontaneous change,, while energy change does not. Entropy changes can be measured by the equation, q rev, q rev, for a reversible process., is independent of path., T, T, Chemical reactions are generally carried at constant pressure, so we define another state, function Gibbs energy, G, which is related to entropy and enthalpy changes of the system by, the equation:, , ∆S =, , ∆rG = ∆rH – T ∆rS, For a spontaneous change, ∆Gsys < 0 and at equilibrium, ∆Gsys = 0., Standard Gibbs energy change is related to equilibrium constant by, 0, , ∆rG = – RT ln K., K can be calculated from this equation, if we know ∆ rG, , 0, , which can be found from, , . Temperature is an important factor in the equation. Many reactions which, are non-spontaneous at low temperature, are made spontaneous at high temperature for systems, having positive entropy of reaction., , 2021-22

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THERMODYNAMICS, , 189, , EXERCISES, 6.1, , Choose the correct answer. A thermodynamic state function is a quantity, (i), used to determine heat changes, (ii), , whose value is independent of path, , (iii) used to determine pressure volume work, (iv), 6.2, , whose value depends on temperature only., , For the process to occur under adiabatic conditions, the correct condition is:, (i), ∆T = 0, ∆p = 0, , (ii), , (iii) q = 0, (iv), 6.3, , w=0, , The enthalpies of all elements in their standard states are:, (i), unity, (ii), , zero, , (iii) < 0, (iv), 6.4, , different for each element, 0, , 0, , ∆U of combustion of methane is – X kJ mol–1. The value of ∆H is, (i), = ∆U 0, (ii), , > ∆U 0, , (iii) < ∆U, (iv), 6.5, , 0, , =0, , The enthalpy of combustion of methane, graphite and dihydrogen at 298 K, are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy, of formation of CH4(g) will be, (i), –74.8 kJ mol–1, (ii), –52.27 kJ mol–1, (iii) +74.8 kJ mol–1, , 6.6, , (iv), , +52.26 kJ mol–1., , A reaction, A + B → C + D + q is found to have a positive entropy change. The, reaction will be, (i), possible at high temperature, (ii), , possible only at low temperature, , (iii) not possible at any temperature, (v), , possible at any temperature, , 6.7, , In a process, 701 J of heat is absorbed by a system and 394 J of work is done, by the system. What is the change in internal energy for the process?, , 6.8, , The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb, calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy, change for the reaction at 298 K., NH2CN(g) +, , 3, O (g) → N2(g) + CO2(g) + H2O(l), 2 2, , 2021-22

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THERMODYNAMICS, , 191, , 0, , 6.20, , The equilibrium constant for a reaction is 10. What will be the value of ∆G ?, R = 8.314 JK–1 mol–1, T = 300 K., , 6.21, , Comment on the thermodynamic stability of NO(g), given, , 1, 1, 0, N (g) +, O (g) → NO(g) ; ∆rH = 90 kJ mol–1, 2 2, 2 2, NO(g) +, 6.22, , 1, 0, O (g) → NO2(g) : ∆rH = –74 kJ mol–1, 2 2, , Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed, 0, under standard conditions. ∆f H = –286 kJ mol–1., , 2021-22