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154 CHEMISTRY UNIT 6 THERMODYNAMICS d he...

154 CHEMISTRY UNIT 6 THERMODYNAMICS d he It is the only physical theory of universal content concerning pu T which I am convinced that, within the framework of the is applicability of its basic concepts, it will never be overthrown. be able to re ER After studying this Unit, you will Albert Einstein bl explain the terms : system and surroundings; discriminate between close, open and isolated systems; be C explain internal energy, work Chemical energy stored by molecules can be released as heat and heat; during chemical reactions when a fuel like methane, cooking state first law of gas or coal burns in air. The chemical energy may also be thermodynamics and express o N it mathematically; used to do mechanical work when a fuel burns in an engine calculate energy changes as or to provide electrical energy through a galvanic cell like work and heat contributions dry cell. Thus, various forms of energy are interrelated and in chemical systems; under certain conditions, these may be transformed from © explain state functions: U, H. one form into another. The study of these energy correlate ∆U and ∆H; transformations forms the subject matter of thermodynamics. measure experimentally ∆U The laws of thermodynamics deal with energy changes of and ∆H; macroscopic systems involving a large number of molecules define standard states for ∆H; rather than microscopic systems containing a few molecules. calculate enthalpy changes for Thermodynamics is not concerned about how and at what various types of reactions; rate these energy transformations are carried out, but is state and apply Hess’s law of based on initial and final states of a system undergoing the constant heat summation; change. Laws of thermodynamics apply only when a system differentiate between extensive tt and intensive properties; is in equilibrium or moves from one equilibrium state to define spontaneous and non- another equilibrium state. Macroscopic properties like spontaneous processes; pressure and temperature do not change with time for a explain entropy as a system in equilibrium state. In this unit, we would like to no thermodynamic state function answer some of the important questions through and apply it for spontaneity; thermodynamics, like: explain Gibbs energy change (∆G); How do we determine the energy changes involved in a establish relationship between chemical reaction/process? Will it occur or not? ∆G and spontaneity, ∆G and What drives a chemical reaction/process? equilibrium constant. To what extent do the chemical reactions proceed? THERMODYNAMICS 155 6.1 THERMODYNAMIC TERMS be real or imaginary. The wall that separates the system from the surroundings is called We are interested in chemical reactions and the boundary. This is designed to allow us to energy changes accompanying them. For this control and keep track of all movements of we need to know certain thermodynamic matter and energy in or out of the system. terms. These are discussed below. 6.1.2 Types of the System 6.1.1 The System and the Surroundings We, further classify the systems according to A system in thermodynamics refers to that d the movements of matter and energy in or out part of universe in which observations are of the system. made and remaining universe constitutes the 1. Open System he surroundings. The surroundings include everything other than the system. System and In an open system, there is exchange of energy the surroundings together constitute the and matter between system and surroundings universe. [Fig. 6.2 (a)]. The presence of reactants in an pu T open beaker is an example of an open system*. is The universe = The system + The surroundings However, the entire universe other than Here the boundary is an imaginary surface re ER the system is not affected by the changes enclosing the beaker and reactants. 2. Closed System taking place in the system. Therefore, for bl all practical purposes, the surroundings In a closed system, there is no exchange of are that portion of the remaining universe matter, but exchange of energy is possible which can interact with the system. between system and the surroundings Usually, the region of space in the [Fig. 6.2 (b)]. The presence of reactants in a be C neighbourhood of the system constitutes closed vessel made of conducting material e.g., its surroundings. copper or steel is an example of a closed For example, if we are studying the system. o N 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). tt Fig. 6.1 System and the surroundings Note that the system may be defined by physical boundaries, like beaker or test tube, no 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. 156 CHEMISTRY 3. Isolated System a quantity which represents the total energy In an isolated system, there is no exchange of of the system. It may be chemical, electrical, energy or matter between the system and the mechanical or any other type of energy you surroundings [Fig. 6.2 (c)]. The presence of may think of, the sum of all these is the energy reactants in a thermos flask or any other closed of the system. In thermodynamics, we call it insulated vessel is an example of an isolated the internal energy, U of the system, which may system. change, when 6.1.3 The State of the System heat passes into or out of the system, d The system must be described in order to make work is done on or by the system, any useful calculations by specifying matter enters or leaves the system. he quantitatively each of the properties such as These systems are classified accordingly as its pressure (p), volume (V), and temperature you have already studied in section 6.1.2. (T ) as well as the composition of the system. We need to describe the system by specifying (a) Work pu T it before and after the change. You would recall is from your Physics course that the state of a Let us first examine a change in internal energy by doing work. We take a system re ER system in mechanics is completely specified at a given instant of time, by the position and containing some quantity of water in a bl velocity of each mass point of the system. In thermos flask or in an insulated beaker. This thermodynamics, a different and much simpler would not allow exchange of heat between the concept of the state of a system is introduced. system and surroundings through its It does not need detailed knowledge of motion boundary and we call this type of system as be C of each particle because, we deal with average adiabatic. The manner in which the state of measurable properties of the system. We specify such a system may be changed will be called the state of the system by state functions or adiabatic process. Adiabatic process is a o N state variables. process in which there is no transfer of heat The state of a thermodynamic system is between the system and surroundings. Here, described by its measurable or macroscopic the wall separating the system and the (bulk) properties. We can describe the state of surroundings is called the adiabatic wall (Fig 6.3). © 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 tt nature of the system. Once these minimum number of macroscopic properties are fixed, others automatically have definite values. no The state of the surroundings can never Fig. 6.3 An adiabatic system which does not be completely specified; fortunately it is not permit the transfer of heat through its necessary to do so. boundary. 6.1.4 The Internal Energy as a State Let us bring the change in the internal Function energy of the system by doing some work on When we talk about our chemical system it. Let us call the initial state of the system as losing or gaining energy, we need to introduce state A and its temperature as T A. Let the THERMODYNAMICS 157 internal energy of the system in state A be the change in temperature is independent of called UA. We can change the state of the system the route taken. Volume of water in a pond, in two different ways. for example, is a state function, because One way: We do some mechanical work, say change in volume of its water is independent 1 kJ, by rotating a set of small paddles and of the route by which water is filled in the thereby churning water. Let the new state be pond, either by rain or by tubewell or by both, called B state and its temperature, as T B. It is (b) Heat found that T B > T A and the change in We can also change the internal energy of a d temperature, ∆T = T B–TA. Let the internal system by transfer of heat from the energy of the system in state B be UB and the surroundings to the system or vice-versa he change in internal energy, ∆U =U B – UA. without expenditure of work. This exchange Second way: We now do an equal amount of energy, which is a result of temperature (i.e., 1kJ) electrical work with the help of an difference is called heat, q. Let us consider bringing about the same change in temperature pu T immersion rod and note down the temperature (the same initial and final states as before in is change. We find that the change in temperature is same as in the earlier case, say, section 6.1.4 (a) by transfer of heat through TB – TA. re ER thermally conducting walls instead of adiabatic walls (Fig. 6.4). bl 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 be C how it was done (irrespective of path) produced the same change of state, as measured by the change in the temperature of the system. o N 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 Fig. 6.4 A system which allows heat transfer difference between the value of U in one state through its boundary. and that in another state, ∆U i.e., We take water at temperature, TA in a ∆U = U 2 − U1 = w ad container having thermally conducting walls, Therefore, internal energy, U, of the system say made up of copper and enclose it in a huge is a state function. heat reservoir at temperature, TB. The heat The positive sign expresses that wad is absorbed by the system (water), q can be positive when work is done on the system. measured in terms of temperature difference , TB – TA. In this case change in internal energy, tt Similarly, if the work is done by the system,wad will be negative. ∆U= q, when no work is done at constant volume. Can you name some other familiar state functions? Some of other familiar state The q is positive, when heat is transferred no functions are V, p, and T. For example, if we from the surroundings to the system and q is bring a change in temperature of the system negative when heat is transferred from from 25°C to 35°C, the change in temperature system to the surroundings. is 35°C–25°C = +10°C, whether we go straight up to 35°C or we cool the system for a few (c) The general case degrees, then take the system to the final Let us consider the general case in which a temperature. Thus, T is a state function and change of state is brought about both by 158 CHEMISTRY doing work and by transfer of heat. We write Solution change in internal energy for this case as: (i) ∆ U = w , wall is adiabatic ad ∆U = q + w (6.1) (ii) ∆ U = – q, thermally conducting walls For a given change in state, q and w can (iii) ∆ U = q – w, closed system. vary depending on how the change is carried out. However, q +w = ∆U will depend only on 6.2 APPLICATIONS initial and final state. It will be independent of Many chemical reactions involve the generation d the way the change is carried out. If there is of gases capable of doing mechanical work or no transfer of energy as heat or as work the generation of heat. It is important for us to he (isolated system) i.e., if w = 0 and q = 0, then quantify these changes and relate them to the ∆ U = 0. changes in the internal energy. Let us see how! The equation 6.1 i.e., ∆U = q + w is 6.2.1 Work pu T mathematical statement of the first law of First of all, let us concentrate on the nature of is thermodynamics, which states that work a system can do. We will consider only mechanical work i.e., pressure-volume work. constant. re ER The energy of an isolated system is For understanding pressure-volume bl It is commonly stated as the law of work, let us consider a cylinder which conservation of energy i.e., energy can neither contains one mole of an ideal gas fitted with a be created nor be destroyed. frictionless piston. Total volume of the gas is Vi and pressure of the gas inside is p. If be C Note: There is considerable difference between external pressure is p ex which is greater than the character of the thermodynamic property p, piston is moved inward till the pressure energy and that of a mechanical property such inside becomes equal to p ex. Let this change o N 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 tt 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 no surroundings. What type of wall does the system have? (iii) w amount of work is done by the Fig. 6.5(a) Work done on an ideal gas in a system and q amount of heat is cylinder when it is compressed by a supplied to the system. What type of constant external pressure, p ex system would it be? (in single step) is equal to the shaded area. THERMODYNAMICS 159 be achieved in a single step and the final If the pressure is not constant but changes volume be V f. During this compression, during the process such that it is always suppose piston moves a distance, l and is infinitesimally greater than the pressure of the cross-sectional area of the piston is A gas, then, at each stage of compression, the [Fig. 6.5(a)]. volume decreases by an infinitesimal amount, then, volume change = l × A = ∆V = (Vf – Vi ) dV. In such a case we can calculate the work done on the gas by the relation force We also know, pressure = Vf d area w= − ∫p ex dV ( 6.3) Therefore, force on the piston = pex. A Vi he If w is the work done on the system by Here, p ex at each stage is equal to (p in + dp) in movement of the piston then case of compression [Fig. 6.5(c)]. In an w = force × distance = pex. A.l expansion process under similar conditions, the external pressure is always less than the pu T = pex. (–∆V) = – pex ∆V = – pex (V f – V i ) (6.2) pressure of the system i.e., pex = (pin– dp). In is The negative sign of this expression is general case we can write, pex = (pin + dp). Such required to obtain conventional sign for w, re ER which will be positive. It indicates that in case processes are called reversible processes. A process or change is said to be bl of compression work is done on the system. reversible, if a change is brought out in Here (V f – V i ) will be negative and negative such a way that the process could, at any multiplied by negative will be positive. Hence moment, be reversed by an infinitesimal the sign obtained for the work will be positive. change. A reversible process proceeds be C If the pressure is not constant at every infinitely slowly by a series of equilibrium stage of compression, but changes in number states such that system and the of finite steps, work done on the gas will be surroundings are always in near o N summed over all the steps and will be equal equilibrium with each other. Processes to − ∑ p ∆V [Fig. 6.5 (b)] tt © no Fig. 6.5 (c) pV-plot when pressure is not constant Fig. 6.5 (b) pV-plot when pressure is not constant and changes in infinite steps and changes in finite steps during (reversible conditions) during compression from initial volume, Vi to compression from initial volume, Vi to final volume, Vf. Work done on the gas final volume, Vf. Work done on the gas is represented by the shaded area. is represented by the shaded area. 160 CHEMISTRY other than reversible processes are known Isothermal and free expansion of an as irreversible processes. ideal gas In chemistry, we face problems that can For isothermal (T = constant) expansion of an be solved if we relate the work term to the ideal gas into vacuum ; w = 0 since p ex = 0. internal pressure of the system. We can Also, Joule determined experimentally that relate work to internal pressure of the system q = 0; therefore, ∆U = 0 under reversible conditions by writing Equation 6.1, ∆ U = q + w can be equation 6.3 as follows: expressed for isothermal irreversible and d Vf Vf reversible changes as follows: wrev = − ∫p dV =− ∫ (p ± dp )dV 1. For isothermal irreversible change he ex in Vi Vi q = – w = pex (Vf – Vi ) Since dp × dV is very small we can write 2. For isothermal reversible change pu T Vf Vf w rev = − ∫ p in dV is (6.4) q = – w = nRT ln V i Vi re ER Now, the pressure of the gas (pin which we = 2.303 nRT log Vf bl Vi can write as p now) can be expressed in terms of its volume through gas equation. For n mol 3. For adiabatic change, q = 0, of an ideal gas i.e., pV =nRT ∆U = wad nRT be C ⇒p = Problem 6.2 V Therefore, at constant temperature (isothermal Two litres of an ideal gas at a pressure of o N process), 10 atm expands isothermally into a Vf vacuum until its total volume is 10 litres. dV Vf How much heat is absorbed and how w rev = − ∫ nRT = −nRT ln Vi V Vi much work is done in the expansion ? © Solution Vf = – 2.303 nRT log (6.5) We have q = – w = p ex (10 – 2) = 0(8) = 0 Vi No work is done; no heat is absorbed. Free expansion: Expansion of a gas in vacuum (pex = 0) is called free expansion. No Problem 6.3 work is done during free expansion of an ideal Consider the same expansion, but this gas whether the process is reversible or time against a constant external pressure irreversible (equation 6.2 and 6.3). of 1 atm. Now, we can write equation 6.1 in number Solution tt of ways depending on the type of processes. We have q = – w = pex (8) = 8 litre-atm Let us substitute w = – p ex∆V (eq. 6.2) in Problem 6.4 equation 6.1, and we get Consider the same expansion, to a final no ∆U = q − p ex ∆V volume of 10 litres conducted reversibly. Solution If a process is carried out at constant volume (∆V = 0), then 10 We have q = – w = 2.303 × 10 log ∆U = qV 2 the subscript V in qV denotes that heat is = 16.1 litre-atm supplied at constant volume. THERMODYNAMICS 161 6.2.2 Enthalpy, H ∆H is positive for endothermic reactions (a) A useful new state function which absorb heat from the surroundings. We know that the heat absorbed at constant At constant volume (∆V = 0), ∆U = qV, volume is equal to change in the internal therefore equation 6.8 becomes energy i.e., ∆U = qV. But most of chemical ∆H = ∆U = q V reactions are carried out not at constant volume, but in flasks or test tubes under The difference between ∆H and ∆U is not constant atmospheric pressure. We need to usually significant for systems consisting of d define another state function which may be only solids and / or liquids. Solids and liquids suitable under these conditions. do not suffer any significant volume changes upon heating. The difference, however, he We may write equation (6.1) as becomes significant when gases are involved. ∆U = q p − p ∆V at constant pressure, where Let us consider a reaction involving gases. If qp is heat absorbed by the system and –p∆V VA is the total volume of the gaseous reactants, pu T represent expansion work done by the system. VB is the total volume of the gaseous products, is Let us represent the initial state by nA is the number of moles of gaseous reactants subscript 1 and final state by 2 and nB is the number of moles of gaseous re ER We can rewrite the above equation as products, all at constant pressure and bl temperature, then using the ideal gas law, we U2–U 1 = qp – p (V2 – V1) write, On rearranging, we get pVA = n ART qp = (U 2 + pV2) – (U 1 + pV1) (6.6) be C Now we can define another thermodynamic and pVB = n BRT function, the enthalpy H [Greek word Thus, pVB – pVA = nB RT – nART = (n B–nA)RT enthalpien, to warm or heat content] as : o N H = U + pV (6.7) or p (VB – VA) = (nB – nA) RT so, equation (6.6) becomes or p ∆V = ∆n gRT (6.9) qp= H 2 – H1 = ∆H Here, ∆ng refers to the number of moles of © Although q is a path dependent function, gaseous products minus the number of moles H is a state function because it depends on U, of gaseous reactants. p and V, all of which are state functions. Substituting the value of p∆V from Therefore, ∆H is independent of path. Hence, equation 6.9 in equation 6.8, we get qp is also independent of path. For finite changes at constant pressure, we ∆H = ∆U + ∆n gRT (6.10) can write equation 6.7 as The equation 6.10 is useful for calculating ∆H = ∆U + ∆pV ∆H from ∆U and vice versa. Since p is constant, we can write tt Problem 6.5 ∆H = ∆U + p∆V (6.8) If water vapour is assumed to be a perfect It is important to note that when heat is gas, molar enthalpy change for absorbed by the system at constant pressure, no vapourisation of 1 mol of water at 1bar we are actually measuring changes in the and 100°C is 41kJ mol–1. Calculate the enthalpy. internal energy change, when Remember ∆H = qp , heat absorbed by the (i) 1 mol of water is vaporised at 1 bar system at constant pressure. pressure and 100°C. ∆H is negative for exothermic reactions (ii) 1 mol of water is converted into ice. which evolve heat during the reaction and 162 CHEMISTRY Solution halved, each part [Fig. 6.6 (b)] now has one V half of the original volume, , but the (i) The change H2 O (l ) → H 2O ( g ) 2 temperature will still remain the same i.e., T. ∆H = ∆U + ∆ n g RT It is clear that volume is an extensive property and temperature is an intensive property. or ∆U = ∆H – ∆n g R T , substituting the values, we get d ∆U = 41.00 kJ mol −1 − 1 × 8.3 J mol −1 K −1 × 373 K he = 41.00 kJ mol −1 − 3.096 kJ mol −1 = 37.904 kJ mol–1 Fig. 6.6(a) A gas at volume V and temperature T (ii) The change H2O ( l ) → H2O ( s) pu T is There is negligible change in volume, re ER So, we can put p ∆V = ∆n g R T ≈ 0 in this bl case, ∆H ≅ ∆U so, ∆U = 41.00 kJ mol − 1 Fig. 6.6 (b) Partition, each part having half the volume of the gas be C (b) Extensive and Intensive Properties (c) Heat Capacity In thermodynamics, a distinction is made In this sub-section, let us see how to measure o N between extensive properties and intensive heat transferred to a system. This heat appears properties. An extensive property is a as a rise in temperature of the system in case property whose value depends on the quantity of heat absorbed by the system. or size of matter present in the system. For The increase of temperature is proportional © example, mass, volume, internal energy, to the heat transferred enthalpy, heat capacity, etc. are extensive properties. q = coeff × ∆T Those properties which do not depend on The magnitude of the coefficient depends the quantity or size of matter present are on the size, composition and nature of the known as intensive properties. For example system. We can also write it as q = C ∆T temperature, density, pressure etc. are The coefficient, C is called the heat capacity. intensive properties. A molar property, χm, is Thus, we can measure the heat supplied the value of an extensive property χ of the by monitoring the temperature rise, provided tt system for 1 mol of the substance. If n is the we know the heat capacity. χ When C is large, a given amount of heat amount of matter, χm = is independent of n results in only a small temperature rise. Water no the amount of matter. Other examples are has a large heat capacity i.e., a lot of energy is molar volume, Vm and molar heat capacity, Cm. needed to raise its temperature. Let us understand the distinction between C is directly proportional to amount of extensive and intensive properties by substance. The molar heat capacity of a considering a gas enclosed in a container of volume V and at temperature T [Fig. 6.6(a)]. C  substance, C m =   , is the heat capacity for Let us make a partition such that volume is n  THERMODYNAMICS 163 one mole of the substance and is the quantity heat capacity of the liquid in which calorimeter of heat needed to raise the temperature of one is immersed and the heat capacity of mole by one degree celsius (or one kelvin). calorimeter, it is possible to determine the heat Specific heat, also called specific heat capacity evolved in the process by measuring is the quantity of heat required to raise the temperature changes. Measurements are temperature of one unit mass of a substance made under two different conditions: by one degree celsius (or one kelvin). For i) at constant volume, qV finding out the heat, q, required to raise the ii) at constant pressure, qp d temperatures of a sample, we multiply the specific heat of the substance, c, by the mass (a) ∆U measurements m, and temperatures change, ∆T as For chemical reactions, heat absorbed at he constant volume, is measured in a bomb q = c × m × ∆T = C ∆T (6.11) calorimeter (Fig. 6.7). Here, a steel vessel (the (d) The relationship between Cp and CV for bomb) is immersed in a water bath. The whole pu T an ideal gas device is called calorimeter. The steel vessel is is At constant volume, the heat capacity, C is immersed in water bath to ensure that no heat denoted by CV and at constant pressure, this is lost to the surroundings. A combustible re ER is denoted by C p. Let us find the relationship substance is burnt in pure dioxygen supplied bl between the two. in the steel bomb. Heat evolved during the We can write equation for heat, q reaction is transferred to the water around the bomb and its temperature is monitored. Since at constant volume as qV = CV ∆T = ∆U the bomb calorimeter is sealed, its volume does be C not change i.e., the energy changes associated at constant pressure as qp = C p ∆T = ∆H with reactions are measured at constant The difference between Cp and CV can be volume. Under these conditions, no work is o N 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 C p − CV = R tt (6.13) 6.3 MEASUREMENT OF ∆ U AND ∆ H: no 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 Fig. 6.7 Bomb calorimeter 164 CHEMISTRY done as the reaction is carried out at constant Problem 6.6 volume in the bomb calorimeter. Even for reactions involving gases, there is no work 1g of graphite is burnt in a bomb done as ∆V = 0. Temperature change of the calorimeter in excess of oxygen at 298 K calorimeter produced by the completed and 1 atmospheric pressure according to reaction is then converted to qV , by using the the equation known heat capacity of the calorimeter with C (graphite) + O2 (g) → CO2 (g) the help of equation 6.11. During the reaction, temperature rises d (b) ∆ H measurements from 298 K to 299 K. If the heat capacity Measurement of heat change at constant of the bomb calorimeter is 20.7kJ/K, he pressure (generally under atmospheric what is the enthalpy change for the above pressure) can be done in a calorimeter shown reaction at 298 K and 1 atm? in Fig. 6.8. We know that ∆H = q p (at Solution constant p) and, therefore, heat absorbed or Suppose q is the quantity of heat from pu T evolved, qp at constant pressure is also called the reaction mixture and CV is the heat is the heat of reaction or enthalpy of reaction, ∆ rH. capacity of the calorimeter, then the re ER In an exothermic reaction, heat is evolved, quantity of heat absorbed by the calorimeter. bl and system loses heat to the surroundings. Therefore, qp will be negative and ∆r H will also q = C V × ∆T be negative. Similarly in an endothermic Quantity of heat from the reaction will reaction, heat is absorbed, qp is positive and have the same magnitude but opposite ∆ rH will be positive. sign because the heat lost by the system be C (reaction mixture) is equal to the heat gained by the calorimeter. o N q = − C V × ∆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 ∆ n g = 0, ∆ H = ∆ U = – 2.48 ×10 kJ mol–1 2 tt 6.4 ENTHALPY CHANGE, ∆ r H OF A REACTION – REACTION ENTHALPY no In a chemical reaction, reactants are converted into products and is represented by, Reactants → Products The enthalpy change accompanying a Fig. 6.8 Calorimeter for measuring heat changes reaction is called the reaction enthalpy. The at constant pressure (atmospheric enthalpy change of a chemical reaction, is pressure). given by the symbol ∆rH THERMODYNAMICS 165 ∆ rH = (sum of enthalpies of products) – (sum melting. Normally this melting takes place at of enthalpies of reactants) constant pressure (atmospheric pressure) and during phase change, temperature remains = ∑ a i H products − ∑ bi Hreactants (6.14) constant (at 273 K). i i H2O(s) → H2O(l ); ∆ fus H V = 6.00 kJ mol −1 (Here symbol ∑ (sigma) is used for summation and ai and bi are the stoichiometric Here ∆fus H 0 is enthalpy of fusion in standard coefficients of the products and reactants state. If water freezes, then process is reversed d respectively in the balanced chemical and equal amount of heat is given off to the equation. For example, for the reaction surroundings. he CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) The enthalpy change that accompanies melting of one mole of a solid substance ∆r H = ∑ a i H products − ∑ bi Hreactants in standard state is called standard pu T i i enthalpy of fusion or molar enthalpy of 0 fusion, ∆fusH. is = [H m (CO2 ,g) + 2Hm (H2O, l)]– [H m (CH4 , g) + 2Hm (O2, g)] Melting of a solid is endothermic, so all re ER where Hm is the molar enthalpy. enthalpies of fusion are positive. Water requires bl Enthalpy change is a very useful quantity. heat for evaporation. At constant temperature Knowledge of this quantity is required when of its boiling point Tb and at constant pressure: one needs to plan the heating or cooling H2O(l ) → H2 O(g); ∆vap H V = + 40.79 kJ mol − 1 required to maintain an industrial chemical be C reaction at constant temperature. It is also ∆vapH 0 is the standard enthalpy of vaporization. required to calculate temperature dependence Amount of heat required to vaporize of equilibrium constant. one mole of a liquid at constant o N (a) Standard enthalpy of reactions temperature and under standard pressure Enthalpy of a reaction depends on the (1bar) is called its standard enthalpy of conditions under which a reaction is carried vaporization or molar enthalpy of out. It is, therefore, necessary that we must vaporization, ∆ vapH 0. © specify some standard conditions. The Sublimation is direct conversion of a solid standard enthalpy of reaction is the enthalpy change for a reaction when all into its vapour. Solid CO2 or ‘dry ice’ sublimes the participating substances are in their at 195K with ∆ s u bH 0=25.2 kJ mol –1 ; standard states. naphthalene sublimes slowly and for this The standard state of a substance at a ∆sub H V = 73.0 kJ mol −1. specified temperature is its pure form at Standard enthalpy of sublimation, 1 bar. For example, the standard state of liquid ∆subH 0 is the change in enthalpy when one ethanol at 298 K is pure liquid ethanol at mole of a solid substance sublimes at a tt 1 bar; standard state of solid iron at 500 K is constant temperature and under standard pure iron at 1 bar. Usually data are taken at pressure (1bar). 298 K. no The magnitude of the enthalpy change Standard conditions are denoted by adding depends on the strength of the intermolecular the superscript V to the symbol ∆H, e.g., ∆H V interactions in the substance undergoing the (b) Enthalpy changes during phase phase transfomations. For example, the strong transformations hydrogen bonds between water molecules hold Phase transformations also involve energy them tightly in liquid phase. For an organic changes. Ice, for example, requires heat for liquid, such as acetone, the intermolecular 166 CHEMISTRY Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation d he pu T is re ER (T f and Tb are melting and boiling points, respectively) bl dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise ∆ vapH V − ∆ng RT = 40.66 kJ mol −1 1 mol of acetone than it does to vaporize 1 mol − (1)(8.314 JK −1mol −1 )(373K )(10 −3 kJ J −1 ) of water. Table 6.1 gives values of standard be C enthalpy changes of fusion and vaporisation ∆vapU V = 40.66 kJ mol −1 − 3.10 kJ mol −1 for some substances. = 37.56 kJ mol −1 o N Problem 6.7 (c) Standard enthalpy of formation A swimmer coming out from a pool is The standard enthalpy change for the covered with a film of water weighing about 18g. How much heat must be formation of one mole of a compound from © supplied to evaporate this water at its elements in their most stable states of 298 K ? Calculate the internal energy of aggregation (also known as reference vaporisation at 100°C. states) is called Standard Molar Enthalpy of Formation. Its symbol is ∆ f H 0, where ∆vapHV for water the subscript ‘ f ’ indicates that one mole of the compound in question has been formed in at 373K = 40.66 kJ mol–1 its standard state from its elements in their Solution most stable states of aggregation. The reference We can r epresent the process of state of an element is its most stable state of evaporation as tt aggregation at 25°C and 1 bar pressure. For example, the reference state of dihydrogen 18 gH2O(l) → vaporisation 18 g H2O(g) is H2 gas and those of dioxygen, carbon and No. of moles in 18 g H2O(l) is sulphur are O 2 gas, Cgraphite and Srhombic no respectively. Some reactions with standard 18g = = 1 mol molar enthalpies of formation are given below. 18 g mol − 1 H2 (g) + ½O2 (g) → H2 O(1); ∆vap U = ∆vap H V − p ∆V = ∆ vap H V − ∆ n gRT ∆ f H V = −285.8 kJ mol −1 (assuming steam behaving as an ideal gas). C (graphite, s) +2H2 (g) → CH4 (g); THERMODYNAMICS 167 ∆ f H y) at 298K of a Table 6.2 Standard Molar Enthalpies of Formation (∆ Few Selected Substances d he pu T is re ER bl be C o N © ∆ f H V = −74.81 kJ mol−1 is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been 2C( graphite,s) + 3H2 (g ) + ½O2 (g) → C2 H5OH(1); formed from other compounds, and not from ∆ f H V = −277.7kJ mol −1 its constituent elements. Also, for the reaction tt given below, enthalpy change is not standard 0 It is important to understand that a enthalpy of formation, ∆ fH for HBr(g). standard molar enthalpy of formation, ∆f H 0, is just a special case of ∆r H 0, where one mole H2 (g) + Br2 (l ) → 2HBr(g); no of a compound is formed from its constituent ∆r H V = −72.8 kJ mol −1 elements, as in the above three equations, Here two moles, instead of one mole of the where 1 mol of each, water, methane and product is formed from the elements, i.e., ethanol is formed. In contrast, the enthalpy change for an exothermic reaction: ∆ r H V = 2∆ f H V. CaO(s) + CO2 (g) → CaCO3 (s); Therefore, by dividing all coefficients in the −1 ∆r H V = −178.3kJ mol balanced equation by 2, expression for 168 CHEMISTRY enthalpy of formation of HBr (g) is written as (alongwith allotropic state) of the substance in an equation. For example: ½H 2 (g) + ½Br2 (1) → HBr(g ); ∆ f H V = − 36.4 kJ m ol− 1 C2H 5 OH(l ) + 3O 2 (g ) → 2CO 2 (g ) + 3H 2 O(l ) : Standard enthalpies of formation of some ∆ r H V = −1367 kJ mol −1 common substances are given in Table 6.2. The abov e equation describes the By convention, standard enthalpy for combustion of liquid ethanol at constant formation, ∆f H 0, of an element in reference d temperature and pressure. The negative sign state, i.e., its most stable state of aggregation of enthalpy change indicates that this is an is taken as zero. exothermic reaction. he Suppose, you are a chemical engineer and It would be necessary to remember the want to know how much heat is required to following conventions regarding thermo- decompose calcium carbonate to lime and chemical equations. pu T carbon dioxide, with all the substances in their 1. The coefficients in a balanced thermo- is standard state. chemical equation refer to the number of re ER CaCO3 (s) → CaO(s) + CO2 (g); ∆r H V = ? moles (never molecules) of reactants and bl Here, we can make use of standard enthalpy products involved in the reaction. 0 of formation and calculate the enthalpy 2. The numerical value of ∆r H refers to the change for the reaction. The following general number of moles of substances specified equation can be used for the enthalpy change by an equation. Standard enthalpy change be C 0 calculation. ∆ rH will have units as kJ mol–1. ∆ r H V = ∑ ai ∆ f H V (products) − ∑ bi ∆ f H V ( reactants) To illustrate the concept, let us consider i i the calculation of heat of reaction for the o N (6.15) following reaction : where a and b represent the coefficients of the products and reactants in the balanced Fe2O3 ( s) + 3H2 ( g ) → 2Fe ( s ) + 3H2O ( l ) , equation. Let us apply the above equation for From the Table (6.2) of standard enthalpy of © decomposition of calcium carbonate. Here, formation (∆ f H 0), we find : coefficients ‘a’ and ‘b’ are 1 each. Therefore, ∆ f H V ( H2O, l ) = –285.83 kJ mol–1; ∆r H V = ∆ f H V [CaO(s)] + ∆ f H V [CO2 (g)] ∆ f H V ( Fe 2O 3 ,s ) = – 824.2 kJ mol–1; − ∆ f H V [CaCO3 (s)] A lso ∆ f H V(Fe, s) = 0 and ∆ f H V (H 2 ,g ) = 0 a s per convention =1( − 635.1 kJ mol− 1 ) + 1( − 393.5 kJ mol − 1) tt − 1( −1206.9 kJ mol− 1 ) Then, V ∆r H 1 = 3(–285.83 kJ mol–1) = 178.3 kJ mol–1 Thus, the decomposition of CaCO3 (s) is an – 1(– 824.2 kJ mol–1) no endothermic process and you have to heat it = (–857.5 + 824.2) kJ mol –1 for getting the desired products. = –33.3 kJ mol–1 (d) Thermochemical equations Note that the coefficients used in these A balanced chemical equation together with calculations are pure numbers, which are the value of its ∆ rH is called a thermochemical equal to the respective stoichiometric equation. We specify the physical state coefficients. The unit for ∆ r H 0 is THERMODYNAMICS 169 kJ mol–1, which means per mole of reaction. Consider the enthalpy change for the Once we balance the chemical equation in a reaction particular way, as above, this defines the mole 1 of reaction. If we had balanced the equation C ( graphite,s) + O2 ( g ) → CO ( g ) ; ∆r H V = ? 2 differently, for example, 1 3 3 Although CO(g) is the major product, some Fe2O3 ( s) + H2 ( g ) → Fe ( s ) + H2O ( l ) CO2 gas is always produced in this reaction. 2 2 2 Therefore, we cannot measure enthalpy change d then this amount of reaction would be one 0 for the above reaction directly. However, if we mole of reaction and ∆ rH would be can find some other reactions involving related he 3 species, it is possible to calculate the enthalpy V ∆ r H2 = 2 (− 285.83 kJ mol −1 ) change for the above reaction. Let us consider the following reactions: 1 − ( − 824.2 kJ mol− 1 ) pu T C ( graphite,s) + O2 ( g ) → CO2 ( g ) ; is 2 (i) = (– 428.7 + 412.1) kJ mol –1 ∆r H V = − 393.5 kJ mol − 1 re ER V bl = –16.6 kJ mol–1 = ½ ∆r H 1 1 CO ( g ) + O2 ( g ) → CO2 ( g ) ; It shows that enthalpy is an extensive quantity. 2 (ii) 3. When a chemical equation is reversed, the ∆r H V = −283.0 kJ mol −1 value of ∆ r H 0 is reversed in sign. For be C example We can combine the above two reactions in such a way so as to obtain the desired N 2(g) + 3H2 ( g ) → 2NH3 (g); reaction. To get one mole of CO(g) on the right, o N ∆r H V = −91.8 kJ mol − 1 we reverse equation (ii). In this, heat is absorbed instead of being released, so we 2NH3 (g) → N 2(g) + 3H2 (g); change sign of ∆rH 0 value ∆r H V = + 91.8 kJ mol −1 © 1 CO2 ( g) → CO ( g ) + O2 ( g ) ; (e) Hess’s Law of Constant Heat 2 Summation ∆r H V = + 283.0 kJ mol − 1 (iii) We know that enthalpy is a state function, therefore the change in enthalpy is Adding equation (i) and (iii), we get the independent of the path between initial state desired equation, (reactants) and final state (products). In other 1 words, enthalpy change for a reaction is the C ( graphite,s) + O2 ( g ) → CO ( g ) ; same whether it occurs in one step or in a 2 tt series of steps. This may be stated as follows in the form of Hess’s Law. for which ∆r H V = ( −393.5 + 283.0) If a reaction takes place in several steps = – 110.5 kJ mol–1 no then its standard reaction enthalpy is the In general, if enthalpy of an overall reaction sum of the standard enthalpies of the A→B along one route is ∆ rH and ∆r H1, ∆rH 2, intermediate reactions into which the ∆rH 3..... representing enthalpies of reactions overall reaction may be divided at the same leading to same product, B along another temperature. route,then we have Let us understand the importance of this law with the help of an example. ∆rH = ∆ rH 1 + ∆ rH 2 + ∆r H3... (6.16) 170 CHEMISTRY It can be represented as: combustion, CO2(g) and H2O (1) are ∆rH produced and 3267.0 kJ of heat is A B liberated. Calculate the standard enthalpy of formation, ∆ f H 0 of benzene. Standard ∆rH1 ∆rH3 enthalpies of formation of CO 2(g) and ∆rH2 H 2O(l) are –393.5 kJ mol–1 and – 285.83 C D kJ mol –1 respectively. d 6.5 ENTHALPIES FOR DIFFERENT TYPES Solution OF REACTIONS The formation reaction of benezene is he It is convenient to give name to enthalpies given by : specifying the types of reactions. (a) Standard enthalpy of combustion 6C ( graphite ) + 3H2 ( g ) → C6 H6 ( l ) ; (symbol : ∆ cH 0 ) ∆ f H V = ?... ( i ) pu T is Combustion reactions are exothermic in The enthalpy of combustion of 1 mol of nature. These are important in industry, benzene is : re ER rocketry, and other walks of life. Standard bl 15 enthalpy of combustion is defined as the C6H 6 ( l ) + O2 → 6CO2 ( g ) + 3H2 O ( l ) ; enthalpy change per mole (or per unit amount) 2 of a substance, when it undergoes combustion ∆c H V = −3267 kJ mol -1... ( ii ) and all the reactants and products being in The enthalpy of formation of 1 mol of be C their standard states at the specified temperature. CO 2(g) : Cooking gas in cylinders contains mostly C ( graphite ) + O2 ( g ) → CO2 ( g ) ; o N butane (C4H10). During complete combustion ∆ f H V = −393.5 kJ mol-1... ( iii ) of one mole of butane, 2658 kJ of heat is released. We can write the thermochemical The enthalpy of formation of 1 mol of reactions for this as: H 2O(l) is : © 13 1 C4 H10 (g) + O2 (g) → 4CO2 (g) + 5H2 O(1); H2 ( g ) + O 2 ( g ) → H 2O ( l ) ; 2 2 ∆c H V = −2658.0 kJ mol −1 ∆ f H V = − 285.83 kJ mol -1... ( iv ) Similarly, combustion of glucose gives out multiplying eqn. (iii) by 6 and eqn. (iv) 2802.0 kJ/mol of heat, for which the overall by 3 we get: equation is : 6C ( graphite ) + 6O 2 ( g ) → 6CO 2 ( g) ; C6H12O6 (g) + 6O2 (g) → 6CO2 (g) + 6H2O(1); ∆ f H V = −2361 kJ mol -1 tt ∆c H V = −2802.0 kJ mol −1 3 Our body also generates energy from food 3H 2 ( g ) + O 2 ( g) → 3H 2O ( l ) ; by the same overall process as combustion, 2 no although the final products are produced after ∆ f H V = − 857.49 kJ mol –1 a series of complex bio-chemical reactions involving enzymes. Summing up the above two equations : 15 6C ( graphite ) + 3H2 ( g ) + O2 ( g ) → 6CO2 ( g ) Problem 6.8 2 The combustion of one mole of benzene + 3H2 O ( l) ; takes place at 298 K and 1 atm. After THERMODYNAMICS 171 (i) Bond dissociation enthalpy ∆ f H V = −3218.49 kJ mol -1... (v ) (ii) Mean bond enthalpy Reversing equation (ii); Let us discuss these terms with reference 15 to diatomic and polyatomic molecules. 6CO2 ( g ) + 3H 2O ( l ) → C 6H6 ( l ) + O2 ; Diatomic Molecules: Consider the following 2 ∆ f H V = 3267.0 kJ mol- 1... ( vi ) process in which the bonds in one mole of dihydrogen gas (H2) are broken: d Adding equations (v) and (vi), we get H2(g) → 2H(g) ; ∆H–HH 0 = 435.0 kJ mol–1 6C ( graphite ) + 3H2 ( g ) → C6 H6 ( l ) ; The enthalpy change involved in this process he is the bond dissociation enthalpy of H–H bond. ∆ f H V = 48.51 kJ mol -1 The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds (b) Enthalpy of atomization of a gaseous covalent compound is broken to pu T (symbol: ∆aH 0 ) form products in the gas phase. is Consider the following example of atomization Note that it is the same as the enthalpy of re ER of dihydrogen atomization of dihydrogen. This is true for all diatomic molecules. For example: H2(g) → 2H(g); ∆aH 0 = 435.0 kJ mol–1 bl You can see that H atoms are formed by Cl2(g) → 2Cl(g) ; ∆Cl–ClH 0 = 242 kJ mol–1 breaking H–H bonds in dihydrogen. The 0 –1 enthalpy change in this process is known as O2(g) → 2O(g) ; ∆O=OH = 428 kJ mol enthalpy of atomization, ∆aH 0. It is the In the case of polyatomic molecules, bond be C enthalpy change on breaking one mole of dissociation enthalpy is different for different bonds completely to obtain atoms in the gas bonds within the same molecule. phase. Polyatomic Molecules: Let us now consider o N In case of diatomic molecules, like a polyatomic molecule like methane, CH4. The dihydrogen (given above), the enthalpy of overall thermochemical equation for its atomization is also the bond dissociation atomization reaction is given below: enthalpy. The other examples of enthalpy of © CH4 (g) → C(g) + 4H(g); atomization can be CH4(g) → C(g) + 4H(g); ∆ aH 0 = 1665 kJ mol–1 ∆a H V = 1665 kJ mol −1 Note that the products are only atoms of C In methane, all the four C – H bonds are and H in gaseous phase. Now see the following identical in bond length and energy. However, reaction: the energies required to break the individual Na(s) → Na(g) ; ∆aH 0 = 108.4 kJ mol–1 C – H bonds in each successive step differ : In this case, the enthalpy of atomization is CH4(g) → CH3(g) + H(g); ∆bond HV = +427 kJ mol−1 same as the enthalpy of sublimation. tt CH3(g) → CH2(g) + H(g); ∆bond H V = +439kJ mol− 1 (c) Bond Enthalpy (symbol: ∆ bondH 0) Chemical reactions involve the breaking and CH2 (g) → CH(g) + H(g); ∆bo nd H V = +452 kJ mol −1 making of chemical bonds. Energy is required no to break a bond and energy is released when CH(g) → C(g) + H(g);∆bond H V = +347 kJ mol −1 a bond is formed. It is possible to relate heat Therefore, of reaction to changes in energy associated with breaking and making of chemical bonds. CH4(g) → C(g) + 4H(g); ∆a H V = 1665 kJ mol −1 With reference to the enthalpy changes associated with chemical bonds, two different In such cases we use mean bond enthalpy terms are used in thermodynamics. of C – H bond. 172 CHEMISTRY For example in CH4, ∆C–HH 0 is calculated as: products in gas phase reactions as: ∆C− H H V = ¼( ∆a H V ) = ¼ (1665 kJ mol −1 ) ∆r H V = ∑ bond enthalpiesreact ants = 416 kJ mol–1 − ∑ bond enthalpies product s We find that mean C–H bond enthalpy in (6.17)** methane is 416 kJ/mol. It has been found that This relationship is particularly more mean C–H bond enthalpies differ slightly from useful when the required values of ∆ f H0 are compound to compound, as in not available. The net enthalpy change of a d CH 3CH 2 Cl, CH 3 NO 2 , etc, but it does not differ reaction is the amount of energy required to in a great deal*. Using Hess’s law, bond break all the bonds in the reactant molecules he enthalpies can be calculated. Bond enthalpy minus the amount of energy required to break values of some single and multiple bonds are all the bonds in the product molecules. given in Table 6.3. The reaction enthalpies are Remember that this relationship is approximate and is valid when all substances pu T very important quantities as these arise from is the changes that accompany the breaking of (reactants and products) in the reaction are in old bonds and formation of the new bonds. We gaseous state. re ER can predict enthalpy of a reaction in gas phase, (d) Enthalpy of Solution (symbol : ∆solH 0 ) bl if we know different bond enthalpies. The standard enthalpy of reaction, ∆r H0 is related Enthalpy of solution of a substance is the to bond enthalpies of the reactants and enthalpy change when one mole of it dissolves –1 be C Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol at 298 K H C N O F Si P S Cl Br I o N 435.8 414 389 464 569 293 318 339 431 368 297 H 347 293 351 439 289 264 259 330 276 238 C 159 201 272 - 209 - 201 243 - N 138 184 368 351 - 205 - 201 O © 155 540 490 327 255 197 - F 176 213 226 360 289 213 Si 213 230 331 272 213 P 213 251 213 - S 243 218 209 Cl 192 180 Br 151 I –1 Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol at 298 K tt N=N 418 C=C 611 O=O 498 N ≡N 946 C ≡ C 837 C=N 615 C=O 741 no C ≡N 891 C ≡O 1070 * Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same. ** If we use enthalpy of bond formation, (∆f H bond 0 ), which is the enthalpy change when one mole of a particular type of bond is formed from gaseous atom, then ∆ H = ∑ ∆ H − ∑∆ H V V V r f bonds of products f bonds of reactants THERMODYNAMICS 173 in a specified amount of solvent. The enthalpy Na +Cl− ( s ) → Na + (g) + Cl − ( g ) ; of solution at infinite dilution is the enthalpy change observed on dissolving the substance ∆lat tice H V = +788 kJ mol −1 in an infinite amount of solvent when the Since it is impossible to determine lattice interactions between the ions (or solute enthalpies directly by experiment, we use an molecules) are negligible. indirect method where we construct an When an ionic compound dissolves in a enthalpy diagram called a Born-Haber Cycle solvent, the ions leave their ordered positions (Fig. 6.9). d on the crystal lattice. These are now more free in Let us now calculate the lattice enthalpy solution. But solvation of these ions (hydration of Na+ Cl–(s) by following steps given below : he in case solvent is water) also occurs at the same time. This is shown diagrammatically, for an 1. Na(s) → Na(g) , sublimation of sodium ionic compound, AB (s) metal, ∆sub H V = 108.4 kJ mol −1 pu T 2. Na(g) → Na +(g) + e −1 (g) , the ionization of is sodium atoms, ionization enthalpy re ER ∆ iH 0 = 496 kJ mol–1 bl 1 3. Cl 2 (g) → Cl(g) , the dissociation of 2 chlorine, the reaction enthalpy is half the be C The enthalpy of solution of AB(s), ∆ solH 0, in o N water is, therefore, determined by the selective values of the lattice enthalpy,∆latticeH 0 and enthalpy of hydration of ions, ∆hyd H0 as © ∆sol HV = ∆lattice HV + ∆hyd H V For most of the ionic compounds, ∆sol H0 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 tt 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 no energies (enthalpies). 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 Fig. 6.9 Enthalpy diagram for lattice enthalpy ions in gaseous state. of NaCl 174 CHEMISTRY bond dissociation enthalpy. Internal energy is smaller by 2RT ( because ∆n g = 2) and is equal to + 783 kJ mol–1. 1 ∆bon d H = 121kJ mol−1. V Now we use the value of lattice enthalpy to 2 calculate enthalpy of solution from the 4. Cl(g) + e−1(g) → Cl(g) electron gained by expression: chlorine atoms. The electron gain enthalpy, ∆sol H V = ∆la ttic eH V + ∆hyd H V ∆ egH 0 = –348.6 kJ mol–1. You have learnt about ionization enthalpy For one mole of NaCl(s), d and electron gain enthalpy in Unit 3. In lattice enthalpy = + 788 kJ mol–1 0 fact, these terms have been taken from and ∆ hydH = – 784 kJ mol –1 ( from the he thermodynamics. Earlier terms, ionization literature) energy and electron affinity were in practice ∆sol H 0 = + 788 kJ mol –1 – 784 kJ mol –1 in place of the above terms (see the box for = + 4 kJ mol–1 justification). The dissolution of NaCl(s) is accompanied pu T is by very little heat change. Ionization Energy and Electron Affinity 6.6 SPONTANEITY re ER Ionization energy and electron affinity are The first law of thermodynamics tells us about bl defined at absolute zero. At any other temperature, heat capacities for the the relationship between the heat absorbed reactants and the products have to be and the work performed on or by a system. It taken into account. Enthalpies of reactions puts no restrictions on the direction of heat for flow. However, the flow of heat is unidirectional be C + – M(g) → M (g) + e (for ionization) from higher temperature to lower temperature. – – M(g) + e → M (g) (for electron gain) In fact, all naturally occurring processes at temperature, T is whether chemical or physical will tend to o N T proceed spontaneously in one direction only. For example, a gas expanding to fill the ∫∆C V 0 ∆r H (T ) = ∆rH (0) + 0 r p dT 0 available volume, burning carbon in dioxygen The value of Cp for each species in the giving carbon dioxide. © above reaction is 5/2 R (CV = 3/2R) But heat will not flow from colder body to 0 So, ∆rCp = + 5/2 R (for ionization) warmer body on its own, the gas in a container 0 ∆rCp = – 5/2 R (for electron gain) will not spontaneously contract into one corner Therefore, or carbon dioxide will not form carbon and 0 ∆r H (ionization enthalpy) dioxygen spontaneously. These and many = E0 (ionization energy) + 5/2 RT other spontaneously occurring changes show 0 ∆r H (electron gain enthalpy) unidirectional change. We may ask ‘what is the = – A( electr on affinity) – 5/2 RT driving force of spontaneously occurring changes ? What determines the direction of a 5. Na + (g ) + Cl − (g) → Na + Cl − (s) tt spontaneous change ? In this section, we shall The sequence of steps is shown in Fig. 6.9, establish some criterion for these processes and is known as a Born-Haber cycle. The whether these will take place or not. importance of the cycle is that, the sum of Let us first understand what do we mean no the enthalpy changes round a cycle is zero. by spontaneous reaction or change ? You may Applying Hess’s law, we get, think by your common observation that spontaneous reaction is one which occurs ∆lattice H V = 411.2 +108.4 + 121 + 496 − 348.6 immediately when contact is made between the ∆lattic e H V = +788 kJ reactants. Take the case of combination of hydrogen and oxygen. These gases may be for NaCl(s) → Na + (g) + Cl −(g) mixed at room temperature and left for many THERMODYNAMICS 175 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 d of spontaneous reaction or process, as we see is that these cannot reverse their direction on their own. We may summarise it as follows: he A spontaneous process is an irreversible process and may only be reversed by some external agency. Fig. 6.10 (a) Enthalpy diagram for exothermic pu T reactions (a) Is decrease in enthalpy a criterion for is spontaneity ? C(graphite, s) + 2 S(l) → CS2(l); re ER If we examine the phenomenon like flow of 0 ∆r H = +128.5 kJ mol–1 bl water down hill or fall of a stone on to the These reactions though endothermic, are ground, we find that there is a net decrease in spontaneous. The increase in enthalpy may be potential energy in the direction of change. By represented on an enthalpy diagram as shown analogy, we may be tempted to state that a in Fig. 6.10(b). be C chemical reaction is spontaneous in a given direction, because decrease in energy has taken place, as in the case of exothermic reactions. For example: o N 1 3 N (g) + H2(g) = NH3(g) ; 2 2 2 ∆r H0 = – 46.1 kJ mol–1 © 1 1 H2(g) + Cl2(g) = HCl (g) ; 2 2 ∆r H 0 = – 92.32 kJ mol–1 1 H2(g) + O (g) → H 2O(l) ; 2 2 Fig. 6.10 (b) Enthalpy diagram for endothermic ∆r H0 = –285.8 kJ mol–1 reactions The decrease in enthalpy in passing from reactants to products may be shown for any Therefore, it becomes obvious that while tt exothermic reaction on an enthalpy diagram decrease in enthalpy may be a contributory as shown in Fig. 6.10(a). factor for spontaneity, but it is not true for all Thus, the postulate that driving force for a cases. no chemical reaction may be due to decrease in (b) Entropy and spontaneity energy sounds ‘reasonable’ as the basis of Then, what drives the spontaneous process in evidence so far ! a given direction ? Let us examine such a case Now let us examine the following reactions: in which ∆H = 0 i.e., there is no change in 1 enthalpy, but still the process is spontaneous. N (g) + O2(g) → NO2(g); Let us consider diffusion of two gases into 2 2 ∆r H 0 = +33.2 kJ mol–1 each other in a closed container which is 176 CHEMISTRY isolated from the surroundings as shown in At this point, we introduce another Fig. 6.11. thermodynamic function, entropy denoted as S. The above mentioned disorder is the manifestation of entropy. To form a mental 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 d chemical reaction is concerned, this entropy change can be attributed to rearrangement of atoms or ions from one pattern in the reactants he 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 pu T increase in entropy. The change in entropy is accompanying a chemical reaction may be estimated qualitatively by a consideration of re ER the structures of the species taking part in the bl 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 be C gaseous state is state of highest entropy. Fig. 6.11 Diffusion of two gases Now let us try to quantify entropy. One way to calculate the degree of disorder or chaotic o N The two gases, say, gas A and gas B are distribution of energy among molecules would represented by black dots and white dots be through statistical method which is beyond respectively and separated by a movable the scope of this treatment. Other way would partition [Fig. 6.11 (a)]. When the partition is be to relate this process to the heat involved in withdrawn [Fig.6.11( b)], the gases begin to © a process which would make entropy a diffuse into each other and after a period of thermodynamic concept. Entropy, like any time, diffusion will be complete. other thermodynamic property such as Let us examine the process. Before internal energy U and enthalpy H is a state partition, if we were to pick up the gas function and ∆S is independent of path. molecules from left container, we would be Whenever heat is added to the system, it sure that these will be molecules of gas A and increases molecular motions causing similarly if we were to pick up the gas increased randomness in the system. Thus molecules from right container, we would be heat (q) has randomising influence on the sure that these will be molecules of gas B. But, system. Can we then equate ∆S with q ? Wait ! tt if we were to pick up molecules from container Experience suggests us that the distribution when partition is removed, we are not sure of heat also depends on the temperature at whether the molecules picked are of gas A or which heat is added to the system. A system no gas B. We say that the system has become less at higher temperature has greater randomness predictable or more chaotic. in it than one at lower temperature. Thus, We may now formulate another postulate: temperature is the measure of average in an isolated system, there is always a chaotic motion of particles in the system. tendency for the systems’ energy to become Heat added to a system at lower temperature more disordered or chaotic and this could be causes greater randomness than when the a criterion for spontaneous change ! same quantity of heat is added to it at higher THERMODYNAMICS 177 temperature. This suggests that the entropy (ii) At 0 K, the contituent particles are change is inversely proportional to the static and entropy is minimum. If temperature. ∆S is related with q and T for a temperature is raised to 115 K, these reversible reaction as : begin to move and oscillate about qrev their equilibrium positions in the ∆S = (6.18) lattice and system becomes more T The total entropy change ( ∆Stotal) for the disordered, therefore entropy system and surroundings of a spontaneous increases. d process is given by (iii) Reactant, NaHCO3 is a solid and it has low entropy. Among products ∆Stot al = ∆S syst em + ∆S surr > 0 he (6.19) there are one solid and two gases. When a system is in equilibrium, the Therefore, the products represent a entropy is maximum, and the change in condition of higher entropy. entropy, ∆S = 0. (iv) Here one molecule gives two atoms pu T is We can say that entropy for a spontaneous i.e., number of particles increases process increases till it reaches maximum and leading to more disordered state. re ER at equilibrium the change in entropy is zero. Two moles of H atoms have higher bl Since entropy is a state property, we can entropy than one mole of dihydrogen calculate the change in entropy of a reversible molecule. process by Problem 6.10 q sys ,rev For oxidation of iron, ∆Ssys = be C T 4Fe ( s ) + 3O2 ( g ) → 2Fe2O3 ( s ) We find that both for reversible and entropy change is – 549.4 JK–1mol–1at o N irreversible expansion for an ideal gas, under 298 K. Inspite of negative entropy change isothermal conditions, ∆U = 0, but ∆Stotal i.e., of this reaction, why is the reaction ∆Ssys + ∆Ssurr is not zero for irreversible spontaneous? process. Thus, ∆U does not discriminate (∆ r H 0for this reaction is © between reversible and irreversible process, –1648 × 103 J mol –1) whereas ∆S does. Solution One decides the spontaneity of a reaction Problem 6.9 by considering Predict in which of the following, entropy increases/decreases : ∆St otal ( ∆S sys + ∆S surr ). For calculating (i) A liquid crystallizes into a solid.