CHEM ch-5.pdf

Unit 5 Thermodynamics It is the only physical theory of universal content concerning...

Unit 5 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 Albert Einstein able to explain the terms : system and surroundings; discriminate between close, open and isolated systems; Chemical energy stored by molecules can be released as explain internal energy, work and heat; heat during chemical reactions when a fuel like methane, state first law of thermodynamics cooking gas or coal burns in air. The chemical energy may and express it mathematically; also be used to do mechanical work when a fuel burns calculate energy changes as in an engine or to provide electrical energy through a work and heat contributions in galvanic cell like dry cell. Thus, various forms of energy chemical systems; are interrelated and under certain conditions, these may explain state functions: U, H. be transformed from one form into another. The study correlate ∆U and ∆H; of these energy transformations forms the subject matter measure experimentally ∆U and ∆H; of thermodynamics. The laws of thermodynamics deal define standard states for ∆H; with energy changes of macroscopic systems involving calculate enthalpy changes for a large number of molecules rather than microscopic various types of reactions; systems containing a few molecules. Thermodynamics is state and apply Hess’s law of not concerned about how and at what rate these energy constant heat summation; transformations are carried out, but is based on initial and differentiate between extensive final states of a system undergoing the change. Laws of and intensive properties; thermodynamics apply only when a system is in equilibrium define spontaneous and non- spontaneous processes; or moves from one equilibrium state to another equilibrium explain entropy as a state. Macroscopic properties like pressure and temperature thermodynamic state function do not change with time for a system in equilibrium state. and apply it for spontaneity; In this unit, we would like to answer some of the important explain Gibbs energy change (∆G); questions through thermodynamics, like: and How do we determine the energy changes involved in a establish relationship between ∆G and spontaneity, ∆G and chemical reaction/process? Will it occur or not? equilibrium constant. What drives a chemical reaction/process? To what extent do the chemical reactions proceed? 2024-25 Unit 5.indd 136 9/12/2022 11:53:33 AM  THERMODYNAMICS 137 5.1 Thermodynamic terms the system from the surroundings is called We are interested in chemical reactions and boundary. This is designed to allow us to the energy changes accompanying them. For control and keep track of all movements of this we need to know certain thermodynamic matter and energy in or out of the system. terms. These are discussed below. 5.1.2 Types of the System 5.1.1 The System and the Surroundings We, further classify the systems according A system in thermodynamics refers to that to the movements of matter and energy in or part of universe in which observations are out of the system. made and remaining universe constitutes the surroundings. The surroundings include 1. Open System everything other than the system. System In an open system, there is exchange of energy and the surroundings together constitute the and matter between system and surroundings universe. [Fig. 5.2 (a)]. The presence of reactants in an The universe = The system + The surroundings open beaker is an example of an open system*. Here the boundary is an imaginary surface However, the entire universe other than enclosing the beaker and reactants. the system is not affected by the changes taking place in the system. Therefore, for all 2. Closed System practical purposes, the surroundings are that In a closed system, there is no exchange of portion of the remaining universe which can matter, but exchange of energy is possible interact with the system. Usually, the region between system and the surroundings of space in the neighbourhood of the system [Fig. 5.2 (b)]. The presence of reactants in a constitutes its surroundings. closed vessel made of conducting material For example, if we are studying the e.g., copper or steel is an example of a closed reaction between two substances A and B system. 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. 5.1). Fig. 5.1 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 be real or imaginary. The wall that separates Fig. 5.2 Open, closed and isolated systems. * We could have chosen only the reactants as system then walls of the beakers will act as boundary. 2024-25 Unit 5.indd 137 9/12/2022 11:53:34 AM  138 chemIstry 3. Isolated System a quantity which represents the total energy In an isolated system, there is no exchange of the system. It may be chemical, electrical, of energy or matter between the system and mechanical or any other type of energy you the surroundings [Fig. 5.2 (c)]. The presence may think of, the sum of all these is the energy of reactants in a thermos flask or any other of the system. In thermodynamics, we call it closed insulated vessel is an example of an the internal energy, U of the system, which isolated system. may change, when heat passes into or out of the system, 5.1.3 The State of the System work is done on or by the system, The system must be described in order to matter enters or leaves the system. make any useful calculations by specifying quantitatively each of the properties such as These systems are classified accordingly its pressure (p), volume (V), and temperature as you have already studied in section 5.1.2. (T ) as well as the composition of the system. (a) Work We need to describe the system by specifying it before and after the change. You would Let us first examine a change in internal energy recall from your Physics course that the by doing work. We take a system containing state of a system in mechanics is completely some quantity of water in a thermos flask specified at a given instant of time, by the or in an insulated beaker. This would not position and velocity of each mass point of allow exchange of heat between the system the system. In thermodynamics, a different and surroundings through its boundary and and much simpler concept of the state of a we call this type of system as adiabatic. The system is introduced. It does not need detailed manner in which the state of such a system knowledge of motion of each particle because, may be changed will be called adiabatic we deal with average measurable properties of process. Adiabatic process is a process in the system. We specify the state of the system which there is no transfer of heat between by state functions or state variables. the system and surroundings. Here, the wall The state of a thermodynamic system is separating the system and the surroundings described by its measurable or macroscopic is called the adiabatic wall (Fig. 5.3). (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, Fig. 5.3 An adiabatic system which does not others automatically have definite values. permit the transfer of heat through its The state of the surroundings can never boundary. be completely specified; fortunately it is not Let us bring the change in the internal necessary to do so. energy of the system by doing some work on it. Let us call the initial state of the system 5.1.4 The Internal Energy as a State as state A and its temperature as TA. Let Function the internal energy of the system in state A When we talk about our chemical system be called UA. We can change the state of the losing or gaining energy, we need to introduce system in two different ways. 2024-25 Unit 5.indd 138 9/12/2022 11:53:36 AM  THERMODYNAMICS 139 One way: We do some mechanical work, say the route taken. Volume of water in a pond, for 1 kJ, by rotating a set of small paddles and example, is a state function, because change thereby churning water. Let the new state in volume of its water is independent of the be called B state and its temperature, as route by which water is filled in the pond, TB. It is found that TB > TA and the change either by rain or by tubewell or by both. in temperature, ∆T = TB–TA. Let the internal (b) Heat energy of the system in state B be UB and the change in internal energy, ∆U =UB– UA. We can also change the internal energy of a system by transfer of heat from the Second way: We now do an equal amount surroundings to the system or vice-versa (i.e., 1kJ) electrical work with the help of an immersion rod and note down the temperature without expenditure of work. This exchange change. We find that the change in temperature of energy, which is a result of temperature is same as in the earlier case, say, TB – TA. difference is called heat, q. Let us consider bringing about the same change in temperature In fact, the experiments in the above (the same initial and final states as before manner were done by J. P. Joule between in section 5.1.4 (a) by transfer of heat 1840–50 and he was able to show that a through thermally conducting walls instead given amount of work done on the system, of adiabatic walls (Fig. 5.4). 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 =U2 –U1= wad Fig. 5.4 A system which allows heat transfer Therefore, internal energy, U, of the through its boundary. system is a state function. By conventions of IUPAC in chemical We take water at temperature, TA in a thermodynamics. The positive sign expresses container having thermally conducting walls, that wad is positive when work is done on the say made up of copper and enclose it in a system and the internal energy of system huge heat reservoir at temperature, TB. The increases. Similarly, if the work is done by the heat absorbed by the system (water), q can be system, wad will be negative because internal measured in terms of temperature difference, energy of the system decreases. TB – TA. In this case change in internal energy, ∆U = q, when no work is done at constant Can you name some other familiar state volume. functions? Some of other familiar state functions are V, p, and T. For example, if we By conventions of IUPAC in chemical bring a change in temperature of the system thermodynamics. The q is positive, when heat from 25°C to 35°C, the change in temperature is transferred from the surroundings to the is 35°C–25°C = +10°C, whether we go straight system and the internal energy of the system up to 35°C or we cool the system for a few increases and q is negative when heat is degrees, then take the system to the final transferred from system to the surroundings temperature. Thus, T is a state function and resulting in decrease of the internal energy of the change in temperature is independent 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. 2024-25 Unit 5.indd 139 9/12/2022 11:53:37 AM  140 chemIstry (c) The general case Solution Let us consider the general case in which a change of state is brought about both by (i) ∆ U = w ad, wall is adiabatic doing work and by transfer of heat. We write (ii) ∆ U = – q, thermally conducting change in internal energy for this case as: walls ∆U = q + w (5.1) (iii) ∆ U = q – w, closed system. For a given change in state, q and w can 5.2 Applications vary depending on how the change is carried Many chemical reactions involve the generation out. However, q +w = ∆U will depend only on of gases capable of doing mechanical work or initial and final state. It will be independent the generation of heat. It is important for us of the way the change is carried out. If there to quantify these changes and relate them is no transfer of energy as heat or as work to the changes in the internal energy. Let us (isolated system) i.e., if w = 0 and q = 0, then see how! ∆ U = 0. The equation 5.1 i.e., ∆U = q + w is 5.2.1 Work mathematical statement of the first law of First of all, let us concentrate on the nature of thermodynamics, which states that work a system can do. We will consider only The energy of an isolated system is mechanical work i.e., pressure-volume work. constant. For understanding pressure-volume It is commonly stated as the law of conservation work, let us consider a cylinder which of energy i.e., energy can neither be created contains one mole of an ideal gas fitted with nor be destroyed. a frictionless piston. Total volume of the gas is Vi and pressure of the gas inside is p. If Note: There is considerable difference between external pressure is pex 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 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 5.1 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 Fig. 5.5 (a) Work done on an ideal gas in a cylinder when it is compressed by system and q amount of heat is a constant external pressure, p ex supplied to the system. What type (in single step) is equal to the shaded of system would it be? area. 2024-25 Unit 5.indd 140 9/12/2022 11:53:37 AM  THERMODYNAMICS 141 inside becomes equal to pex. Let this change If the pressure is not constant but be achieved in a single step and the final changes during the process such that it volume be V f. During this compression, is always infinitesimally greater than the suppose piston moves a distance, l and pressure of the gas, then, at each stage of is cross-sectional area of the piston is A compression, the volume decreases by an [Fig. 5.5(a)]. infinitesimal amount, dV. In such a case we then, volume change = l × A = ∆V = (Vf – Vi ) can calculate the work done on the gas by the We also know, pressure = relation Vf Therefore, force on the piston = pex. A If w is the work done on the system by w p Vi ex dV (5.3) movement of the piston then w = force × distance = pex. A.l Here, pex at each stage is equal to (pin + dp) in case of compression [Fig. 5.5(c)]. In an = pex. (–∆V) = – pex ∆V = – pex (Vf – Vi ) (5.2) expansion process under similar conditions, The negative sign of this expression is the external pressure is always less than the required to obtain conventional sign for w, pressure of the system i.e., pex = (pin– dp). In which will be positive. It indicates that in case general case we can write, pex = (pin + dp). Such of compression work is done on the system. processes are called reversible processes. Here (Vf – Vi ) will be negative and negative A process or change is said to be multiplied by negative will be positive. Hence reversible, if a change is brought out in such a the sign obtained for the work will be positive. way that the process could, at any moment, If the pressure is not constant at every be reversed by an infinitesimal change. stage of compression, but changes in number A reversible process proceeds infinitely of finite steps, work done on the gas will be slowly by a series of equilibrium states summed over all the steps and will be equal such that system and the surroundings are to – Σ р ∆V [Fig. 5.5 (b)] always in near equilibrium with each other. Fig. 5.5 (c) pV-plot when pressure is not constant Fig. 5.5 (b) pV-plot when pressure is not constant and changes in infinite steps (reversible and changes in finite steps during conditions) during compression from compression from initial volume, Vi to initial volume, Vi to final volume, Vf. final volume, Vf. Work done on the gas Work done on the gas is represented is represented by the shaded area. by the shaded area. 2024-25 Unit 5.indd 141 9/12/2022 11:53:37 AM  142 chemIstry Processes other than reversible processes Isothermal and free expansion of an are known as irreversible processes. ideal gas In chemistry, we face problems that can For isothermal (T = constant) expansion of be solved if we relate the work term to the an ideal gas into vacuum; w = 0 since pex = 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 5.3 as follows: Equation 5.1, can be Vf Vf expressed for isothermal irreversible and reversible changes as follows: w rev    Vi pex dV    (p Vi in  dp ) dV 1. For isothermal irreversible change q = – w = pex (Vf – Vi ) Since dp × dV is very small we can write 2. For isothermal reversible change Vf w rev    pin dV (5.4) q = – w = nRT ln Vi Now, the pressure of the gas (pin which we Vf can write as p now) can be expressed in terms = 2.303 nRT log V i of its volume through gas equation. For n mol For adiabatic change, q = 0, of an ideal gas i.e., pV =nRT ∆U = wad nRT p  Problem 5.2 V Two litres of an ideal gas at a pressure of 10 Therefore, at constant temperature atm expands isothermally at 25 °C into a (isothermal process), vacuum until its total volume is 10 litres. Vf How much heat is absorbed and how dV Vf w rev    nRT  nRT ln much work is done in the expansion ? V Vi Vi Solution Vf We have q = – w = pex (10 – 2) = 0(8) = 0 = – 2.303 nRT log (5.5) No work is done; no heat is absorbed. Vi Problem 5.3 Free expansion: Expansion of a gas in Consider the same expansion, but vacuum (pex = 0) is called free expansion. this time against a constant external No work is done during free expansion of an pressure of 1 atm. ideal gas whether the process is reversible or irreversible (equation 5.2 and 5.3). Solution Now, we can write equation 5.1 in number We have q = – w = pex (8) = 8 litre-atm of ways depending on the type of processes. Problem 5.4 Let us substitute w = – pex∆V (eq. 5.2) in Consider the expansion given in problem equation 5.1, and we get 5.2, for 1 mol of an ideal gas conducted U  q  pex V reversibly. If a process is carried out at constant volume Solution (∆V = 0), then Vf We have q = – w = 2.303 nRT log ∆U = qV Vs the subscript V in qV denotes that heat is = 2.303 × 1 × 0.8206 × 298 × log 10 supplied at constant volume. 2 2024-25 Unit 5.indd 142 9/12/2022 11:53:38 AM  THERMODYNAMICS 143 Remember ∆H = qp, heat absorbed by the = 2.303 x 0.8206 x 298 x log 5 system at constant pressure. = 2.303 x 0.8206 x 298 x 0.6990 = 393.66 L atm ∆H is negative for exothermic reactions which evolve heat during the reaction and ∆H is positive for endothermic reactions 5.2.2 Enthalpy, H which absorb heat from the surroundings. (a) A Useful New State Function At constant volume (∆V = 0), ∆U = qV, We know that the heat absorbed at constant therefore equation 5.8 becomes volume is equal to change in the internal ∆H = ∆U = qV energy i.e., ∆U = qV. But most of chemical reactions are carried out not at constant The difference between ∆H and ∆U is not volume, but in flasks or test tubes under usually significant for systems consisting constant atmospheric pressure. We need to of only solids and / or liquids. Solids and define another state function which may be liquids do not suffer any significant volume suitable under these conditions. changes upon heating. The difference, however, becomes significant when gases are We may write equation (5.1) as involved. Let us consider a reaction involving ∆U = qp – p∆V at constant pressure, where gases. If VA is the total volume of the gaseous qp is heat absorbed by the system and –p ∆V reactants, VB is the total volume of the gaseous represent expansion work done by the system. products, nA is the number of moles of gaseous Let us represent the initial state by reactants and nB is the number of moles of subscript 1 and final state by 2 gaseous products, all at constant pressure We can rewrite the above equation as and temperature, then using the ideal gas U2–U1 = qp – p (V2 – V1) law, we write, On rearranging, we get pVA = nART qp = (U2 + pV2) – (U1 + pV1) (5.6) and pVB = nBRT Now we can define another thermodynamic function, the enthalpy H [Greek word Thus, pVB – pVA = nBRT – nART = (nB–nA)RT enthalpien, to warm or heat content] as : or p (VB – VA) = (nB – nA) RT H = U + pV (5.7) or p ∆V = ∆ngRT (5.9) so, equation (5.6) becomes qp= H2 – H1 = ∆H Here, ∆ng refers to the number of moles of gaseous products minus the number of moles Although q is a path dependent function, of gaseous reactants. H is a state function because it depends on U, p and V, all of which are state functions. Substituting the value of p∆V from Therefore, ∆H is independent of path. Hence, equation 5.9 in equation 5.8, we get qp is also independent of path. ∆H = ∆U + ∆ngRT (5.10) For finite changes at constant pressure, The equation 5.10 is useful for calculating we can write equation 5.7 as ∆H from ∆U and vice versa. ∆H = ∆U + ∆pV Since p is constant, we can write Problem 5.5 ∆H = ∆U + p∆V (5.8) If water vapour is assumed to be a perfect gas, molar enthalpy change for It is important to note that when heat is vapourisation of 1 mol of water at 1bar absorbed by the system at constant pressure, and 100°C is 41kJ mol–1. Calculate the we are actually measuring changes in the internal energy change, when enthalpy. 2024-25 Unit 5.indd 143 9/12/2022 11:53:38 AM  144 chemIstry 1 mol of water is vapourised at 1 bar pressure and 100°C. Solution (i) The change H2O (l ) → H2O (g) ∆H = ∆U + ∆ngRT Fig. 5.6(a) A gas at volume V and temperature T or ∆U = ∆H – ∆ngRT, substituting the values, we get ∆U = 41.00 kJ mol–1 – 1 × 8.3 J mol–1 K–1 × 373 K = 41.00 kJ mol-1 – 3.096 kJ mol-1 = 37.904 kJ mol–1 Fig. 5.6 (b) Partition, each part having half the volume of the gas (b) Extensive and Intensive Properties (c) Heat Capacity In thermodynamics, a distinction is made In this sub-section, let us see how to measure between extensive properties and intensive heat transferred to a system. This heat properties. An extensive property is a appears as a rise in temperature of the system property whose value depends on the quantity in case 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 q  coeff  T properties. The magnitude of the coefficient depends Those properties which do not depend on the size, composition and nature of the on the quantity or size of matter present system. We can also write it as q = C ∆T are known as intensive properties. For example temperature, density, pressure etc. The coefficient, C is called the heat are intensive properties. A molar property, capacity. χm, is the value of an extensive property χ of Thus, we can measure the heat supplied the system for 1 mol of the substance. If n is by monitoring the temperature rise, provided  we know the heat capacity. the amount of matter,  m  n is independent When C is large, a given amount of heat of the amount of matter. Other examples are results in only a small temperature rise. Water molar volume, Vm and molar heat capacity, has a large heat capacity i.e., a lot of energy C m. Let us understand the distinction is needed to raise its temperature. between extensive and intensive properties by C is directly proportional to amount of considering a gas enclosed in a container of substance. The molar heat capacity of a volume V and at temperature T [Fig. 5.6(a)]. Let us make a partition such that volume substance, Cm =  C  , is the heat capacity is halved, each part [Fig. 5.6 (b)] now has n for one mole of the substance and is one half of the original volume, V , but the 2 the quantity of heat needed to raise the temperature will still remain the same i.e., T. temperature of one mole by one degree It is clear that volume is an extensive property celsius (or one kelvin). Specific heat, also and temperature is an intensive property. called specific heat capacity is the quantity 2024-25 Unit 5.indd 144 9/12/2022 11:53:39 AM  THERMODYNAMICS 145 of heat required to raise the temperature of i) at constant volume, qV one unit mass of a substance by one degree ii) at constant pressure, qp celsius (or one kelvin). For finding out the heat, q, required to raise the temperatures (a) ∆U Measurements of a sample, we multiply the specific heat For chemical reactions, heat absorbed at of the substance, c, by the mass m, and constant volume, is measured in a bomb temperatures change, ∆T as calorimeter (Fig. 5.7). Here, a steel vessel (the q  c  m  T  C T (5.11) bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is (d) The Relationship between Cp and CV immersed in water bath to ensure that no heat for an Ideal Gas is lost to the surroundings. A combustible 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 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 (5.12) On putting the values of ∆H and ∆U, we have C p T  CV T  RT C p  CV  R Fig. 5.7 Bomb calorimeter Cp – CV = R (5.13) substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the 5.3 Measurement of ∆U and ∆H: reaction is transferred to the water around the Calorimetry bomb and its temperature is monitored. Since We can measure energy changes associated the bomb calorimeter is sealed, its volume with chemical or physical processes by an does not change i.e., the energy changes experimental technique called calorimetry. associated with reactions are measured at In calorimetry, the process is carried out in a constant volume. Under these conditions, no vessel called calorimeter, which is immersed work is done as the reaction is carried out in a known volume of a liquid. Knowing at constant volume in the bomb calorimeter. the heat capacity of the liquid in which Even for reactions involving gases, there is no calorimeter is immersed and the heat capacity work done as ∆V = 0. Temperature change of of calorimeter, it is possible to determine the the calorimeter produced by the completed heat evolved in the process by measuring reaction is then converted to qV, by using the temperature changes. Measurements are known heat capacity of the calorimeter with made under two different conditions: the help of equation 5.11. 2024-25 Unit 5.indd 145 9/12/2022 11:53:39 AM  146 chemIstry (b) ∆H Measurements of the bomb calorimeter is 20.7kJ/K, Measurement of heat change at constant what is the enthalpy change for the above pressure (generally under atmospheric pressure) reaction at 298 K and 1 atm? can be done in a calorimeter shown in Fig. 5.8. We know that ∆Η = qp (at constant p) and, Solution therefore, heat absorbed or evolved, qp at Suppose q is the quantity of heat from constant pressure is also called the heat of the reaction mixture and C V is the reaction or enthalpy of reaction, ∆rH. heat capacity of the calorimeter, then the quantity of heat absorbed by the In an exothermic reaction, heat is evolved, calorimeter. and system loses heat to the surroundings. Therefore, qp will be negative and ∆rH will q = CV × ∆T also 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 (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, ∆ H = ∆ U = – 2.48 ×102 kJ mol–1 5.4 Enthalpy change, ∆ r H of a reaction – Reaction Enthalpy In a chemical reaction, reactants are converted Fig. 5.8 Calorimeter for measuring heat changes into products and is represented by, at constant pressure (atmospheric Reactants → Products pressure). The enthalpy change accompanying a reaction is called the reaction enthalpy. The Problem 5.6 enthalpy change of a chemical reaction, is 1g of graphite is burnt in a bomb given by the symbol ∆rH calorimeter in excess of oxygen at 298 K ∆rH = (sum of enthalpies of products) – (sum and 1 atmospheric pressure according of enthalpies of reactants) to the equation C (graphite) + O (g) → CO (g)  a H i products   bi H reactants (5.14) 2 2 i i During the reaction, temperature rises from 298 K to 299 K. If the heat capacity Here symbol ∑ (sigma) is used for summation and ai and bi are the stoichiometric 2024-25 Unit 5.indd 146 9/12/2022 11:53:40 AM  THERMODYNAMICS 147 coefficients of the products and reactants ethanol at 298 K is pure liquid ethanol at respectively in the balanced chemical 1 bar; standard state of solid iron at 500 K equation. For example, for the reaction is pure iron at 1 bar. Usually data are taken CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) at 298 K. =∆r H ∑ a i H Pr oducts − ∑ bi H reac tan ts Standard conditions are denoted by i i adding the superscript  to the symbol ∆H, = [Hm (CO2, g) + 2Hm (H2O, l)]– [Hm (CH4, g) e.g., ∆H + 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 Phase transformations also involve energy one needs to plan the heating or cooling changes. Ice, for example, requires heat for required to maintain an industrial chemical melting. Normally this melting takes place at reaction at constant temperature. It is also constant pressure (atmospheric pressure) and required to calculate temperature dependence during phase change, temperature remains of equilibrium constant. constant (at 273 K). (a) Standard Enthalpy of Reactions H2O(s) → H2O(l); ∆fusH  = 6.00 kJ moI–1 Enthalpy of a reaction depends on the Here ∆fusH is enthalpy of fusion in standard conditions under which a reaction is carried state. If water freezes, then process is reversed out. It is, therefore, necessary that we and equal amount of heat is given off to the must specify some standard conditions. surroundings. The standard enthalpy of reaction is the The enthalpy change that accompanies enthalpy change for a reaction when all melting of one mole of a solid substance in the participating substances are in their standard state is called standard enthalpy standard states. of fusion or molar enthalpy of fusion, The standard state of a substance at a ∆fusH . specified temperature is its pure form at Melting of a solid is endothermic, so 1 bar. For example, the standard state of liquid all enthalpies of fusion are positive. Water Table 5.1 Standard Enthalpy Changes of Fusion and Vaporisation (Tf and Tb are melting and boiling points, respectively) 2024-25 Unit 5.indd 147 9/12/2022 11:53:40 AM  148 chemIstry requires heat for evaporation. At constant temperature of its boiling point Tb and at Solution constant pressure: We can represent the process of H2O(l) → H2O(g); ∆vapH = + 40.79 kJ moI–1 evaporation as ∆vapH is the standard enthalpy of vaporisation. H2 O(1)  vaporisation → H2 O(g) 1mol 1mol Amount of heat required to vaporize one mole of a liquid at constant temperature No. of moles in 18 g H2O(l) is and under standard pressure (1bar) is called 18g = =1 mol its standard enthalpy of vaporization or 18g mol –1 molar enthalpy of vaporization, ∆vapH . Heat supplied to evaporate18g water at Sublimation is direct conversion of a 298 K = n × ∆vap H  solid into its vapour. Solid CO2 or ‘dry ice’ = (1 mol) × (44.01 kJ mol–1) sublimes at 195K with ∆subH=25.2 kJ mol–1; = 44.01 kJ naphthalene sublimes slowly and for this (assuming steam behaving as an ideal ∆sub H = 73.0 kJ mol–1. gas). Standard enthalpy of sublimation, ∆subH is the change in enthalpy when one ∆vapU = ∆vapH – p∆V = ∆vapH – ∆ngRT mole of a solid substance sublimes at a constant temperature and under standard ∆vapHV – ∆ngRT = 44.01 kJ pressure (1bar). –(1)(8.314 JK–1mol–1)(298K)(10–3kJ J–1) The magnitude of the enthalpy change ∆vapUV = 44.01 kJ – 2.48kJ depends on the strength of the intermolecular = 41.53 kJ interactions in the substance undergoing the phase transfomations. For example, Problem 5.8 the strong hydrogen bonds between water molecules hold them tightly in liquid phase. Assuming the water vapour to be a perfect gas, calculate the internal energy change For an organic liquid, such as acetone, the when 1 mol of water at 100°C and 1 bar intermolecular dipole-dipole interactions are pressure is converted to ice at 0°C. Given significantly weaker. Thus, it requires less the enthalpy of fusion of ice is 6.00 kJ heat to vaporise 1 mol of acetone than it does mol-1 heat capacity of water is 4.2 J/g°C to vaporize 1 mol of water. Table 5.1 gives The change take place as follows: values of standard enthalpy changes of fusion Step - 1 1 mol H2O (l, 100°C)  1 and vaporisation for some substances. mol (l, 0°C) Enthalpy change ∆H1 Problem 5.7 Step - 2 1 mol H2O (l, 0°C)  1 mol A swimmer coming out from a pool is H2O( S, 0°C) Enthalpy covered with a film of water weighing change ∆H2 about 18g. How much heat must be Total enthalpy change will be - supplied to evaporate this water at 298 K ? Calculate the internal energy ∆H = ∆H1 + ∆H2 of vaporisation at 298K. ∆H1 = - (18 x 4.2 x 100) J mol-1 ∆vap H  for water = - 7560 J mol-1 = - 7.56 k J mol-1 at 298K= 44.01kJ mol–1 ∆H2 = - 6.00 kJ mol-1 2024-25 Unit 5.indd 148 9/12/2022 11:53:40 AM  THERMODYNAMICS 149 Table 5.2 Standard Molar Enthalpies of Formation (∆f H ) at 298K of a Few Selected Substances of aggregation (also known as reference Therefore, states) is called Standard Molar Enthalpy ∆H = - 7.56 kJ mol-1 + (-6.00 kJ mol-1) of Formation. Its symbol is ∆f H , where the    = -13.56 kJ mol-1 subscript ‘ f ’ indicates that one mole of the There is negligible change in the volume compound in question has been formed in its during the change form liquid to solid state. standard state from its elements in their most stable states of aggregation. The reference Therefore, p∆v = ∆ng RT = 0 state of an element is its most stable state ∆H = ∆U = - 13.56kJ mol-1 of aggregation at 25°C and 1 bar pressure. For example, the reference state of dihydrogen (c) Standard Enthalpy of Formation is H 2 gas and those of dioxygen, carbon The standard enthalpy change for the and sulphur are O2 gas, Cgraphite and Srhombic formation of one mole of a compound from respectively. Some reactions with standard its elements in their most stable states molar enthalpies of formation are as follows. 2024-25 Unit 5.indd 149 9/12/2022 11:53:40 AM  150 chemIstry H2(g) + ½O2 (g) → H2O(1); Here, we can make use of standard enthalpy of formation and calculate the enthalpy ∆f H = –285.8 kJ mol–1 change for the reaction. The following general C (graphite, s) + 2H2(g) → Ch4 (g); equation can be used for the enthalpy change ∆f H = –74.81 kJ mol–1 calculation. 2C (graphite, s)+3H2 (g)+ ½O2(g) → C2H5OH(1); ∆rH = ∑ ai ∆f H (products) – ∑ bi ∆f H (reactants) i ∆f H  = – 277.7kJ mol–1 i (5.15) It is important to understand that a where a and b represent the coefficients of standard molar enthalpy of formation, ∆f H , the products and reactants in the balanced is just a special case of ∆rH , where one mole equation. Let us apply the above equation for of a compound is formed from its constituent decomposition of calcium carbonate. Here, elements, as in the above three equations, coefficients ‘a’ and ‘b’ are 1 each. Therefore, where 1 mol of each, water, methane and ∆rH = ∆f H  = [CaO(s)]+ ∆f H  [CO2(g)] ethanol is formed. In contrast, the enthalpy – ∆f H  = [CaCO3(s)] change for an exothermic reaction: =1 (–635.1 kJ mol–1) + 1(–393.5 kJ mol–1) CaO(s) + CO2(g) → CaCo3(s); –1(–1206.9 kJ mol–1) ∆rH  = – 178.3kJ mol–1 = 178.3 kJ mol –1 is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been Thus, the decomposition of CaCO3 (s) is formed from other compounds, and not from an endothermic process and you have to heat its constituent elements. Also, for the reaction it for getting the desired products. given below, enthalpy change is not standard (d) Thermochemical Equations enthalpy of formation, ∆fH  for HBr(g). A balanced chemical equation together with H2(g) + Br2(l) → 2HBr(g); the value of its ∆rH is called a thermochemical ∆r H  = – 178.3kJ mol–1 equation. We specify the physical state Here two moles, instead of one mole of the (alongwith allotropic state) of the substance product is formed from the elements, i.e., in an equation. For example: ∆r H  = 2∆f H C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l); Therefore, by dividing all coefficients in ∆rH = – 1367 kJ mol–1 the balanced equation by 2, expression for The above equation describes the enthalpy of formation of HBr (g) is written as combustion of liquid ethanol at constant ½H2(g) + ½Br2(1) → HBr(g); temperature and pressure. The negative sign ∆f H = – 36.4 kJ mol–1 of enthalpy change indicates that this is an exothermic reaction. Standard enthalpies of formation of some common substances are given in Table 5.2. It would be necessary to remember the By convention, standard enthalpy for following conventions regarding thermo- formation, ∆f H , of an element in reference chemical equations. state, i.e., its most stable state of aggregation 1. The coefficients in a balanced thermo- is taken as zero. chemical equation refer to the number of moles (never molecules) of reactants and Suppose, you are a chemical engineer and want to know how much heat is required to products involved in the reaction. decompose calcium carbonate to lime and 2. The numerical value of ∆rH  refers to the carbon dioxide, with all the substances in number of moles of substances specified their standard state. by an equation. Standard enthalpy change CaCO3(s) → CaO(s) + CO2(g); ∆r H = ? ∆rH will have units as kJ mol–1. 2024-25 Unit 5.indd 150 9/12/2022 11:53:40 AM  THERMODYNAMICS 151 To illustrate the concept, let us consider (e) Hess’s Law of Constant Heat the calculation of heat of reaction for the Summation following reaction : We know that enthalpy is a state function, Fe2O3  s   3H2  g  2Fe  s   3H2O  l  , therefore the change in enthalpy is independent of the path between initial state (reactants) From the Table (5.2) of standard enthalpy of and final state (products). In other words, formation (∆f H ), we find : enthalpy change for a reaction is the same ∆f H (H2O,l) = –285.83 kJ mol–1; whether it occurs in one step or in a series ∆f H (Fe2O3,s) = – 824.2 kJ mol–1; of steps. This may be stated as follows in the form of Hess’s Law. Also ∆f H (Fe, s) = 0 and If a reaction takes place in several ∆f H (H2, g) = 0 as per convention steps then its standard reaction enthalpy Then, is the sum of the standard enthalpies of ∆f H1 = 3(–285.83 kJ mol–1) the intermediate reactions into which the – 1(– 824.2 kJ mol–1) overall reaction may be divided at the same temperature. = (–857.5 + 824.2) kJ mol–1 Let us understand the importance of this = –33.3 kJ mol–1 law with the help of an example. Note that the coefficients used in these Consider the enthalpy change for the calculations are pure numbers, which reaction are equal to the respective stoichiometric c o e f f i c i e n t s. T h e u n i t f o r ∆ rH  i s C (graphite,s) + O2 (g) → CO (g); ∆r H  = ? kJ mol–1, which means per mole of reaction. Although CO(g) is the major product, some Once we balance the chemical equation in a CO2 gas is always produced in this reaction. particular way, as above, this defines the mole Therefore, we cannot measure enthalpy of reaction. If we had balanced the equation change for the above reaction directly. differently, for example, However, if we can find some other reactions 1 3 3 involving related species, it is possible to Fe2O3  s   H2  g  Fe  s   H2O  l  2 2 2 calculate the enthalpy change for the above reaction. then this amount of reaction would be one mole of reaction and ∆rH  would be Let us consider the following reactions: 3 C (graphite,s) + O2 (g) → CO2 (g); ∆f H 2 = (–285.83 kJ mol–1) 2 ∆r H  = – 393.5 kJ mol–1 (i) 1 – (–824.2 kJ mol–1) 1 2 CO (g) + O (g) → CO2 (g) = (– 428.7 + 412.1) kJ mol–1 2 2 ∆r H  = – 283.0 kJ mol–1 (ii) = –16.6 kJ mol–1 = ½ ∆r H 1 We can combine the above two reactions It shows that enthalpy is an extensive in such a way so as to obtain the desired quantity. reaction. To get one mole of CO(g) on the 3. When a chemical equation is reversed, right, we reverse equation (ii). In this, heat the value of ∆rH  is reversed in sign. For is absorbed instead of being released, so we example change sign of ∆rH  value N2(g) + 3H2 (g) → 2NH3 (g); CO2 (g) → CO (g) + O2 (g); ∆r H  = – 91.8 kJ. mol–1 ∆r H  = + 283.0 kJ mol–1 (iii) 2NH3(g) → N2(g) + 3H2 (g); ∆r H  = + 91.8 kJ mol–1 2024-25 Unit 5.indd 151 9/12/2022 11:53:41 AM  152 chemIstry Adding equation (i) and (iii), we get the Similarly, combustion of glucose gives out desired equation, 2802.0 kJ/mol of heat, for which the overall 1 equation is : C  graphite, s   O2  g  CO  g  ; 2 C6 H12O6 ( g )  6O2 ( g ) 6CO2 ( g )  6H2O(1); for which ∆r H  = (– 393.5 + 283.0) ∆C H  = – 2802.0 kJ mol–1 = – 110.5 kJ mol–1 Our body also generates energy from food by the same overall process as combustion, In general, if enthalpy of an overall although the final products are produced after reaction A→B along one route is ∆rH and a series of complex bio-chemical reactions ∆rH1, ∆rH2, ∆rH3..... representing enthalpies involving enzymes. of reactions leading to same product, B along another route, then we have Problem 5.9 ∆rH = ∆rH1 + ∆rH2 + ∆rH3... (5.16) The combustion of one mole of benzene It can be represented as: takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O (1) are ∆rH produced and 3267.0 kJ of heat is A B liberated. Calculate the standard ∆H1 ∆rH3 enthalpy of formation, ∆f H  of benzene. Standard enthalpies of formation of C D CO2(g) and H2O(l) are –393.5 kJ mol–1 ∆rH2 and – 285.83 kJ mol–1 respectively. 5.5 Enthalpies for different types Solution of reactions The formation reaction of benezene is It is convenient to give name to enthalpies given by : specifying the types of reactions. 6C  graphite   3H2  g  C6 H6  l  ; (a) Standard Enthalpy of Combustion ∆f H  = ?... (i) (symbol : ∆cH ) The enthalpy of combustion of 1 mol Combustion reactions are exothermic in of benzene is : nature. These are important in industry, 15 rocketry, and other walks of life. Standard C6 H6  l   O2 6CO2  g   3H2O  l  ; enthalpy of combustion is defined as the 2 enthalpy change per mole (or per unit amount) ∆C H  = – 3267 kJ mol–1... (ii) of a substance, when it undergoes combustion The enthalpy of formation of 1 mol of and all the reactants and products being CO2(g) : in their standard states at the specified C  graphite   O2  g  CO2  g  ; temperature. ∆f H  = – 393.5 kJ mol–1... (iii) Cooking gas in cylinders contains mostly butane (C4H10). During complete combustion The enthalpy of formation of 1 mol of of one mole of butane, 2658 kJ of heat is H2O(l) is : released. We can write the thermochemical 1 H2  g   O2  g  H2 O  l  ; reactions for this as: 2 13 ∆C H  = – 285.83 kJ mol–1... (iv) C4 H10 ( g )  O2 ( g ) 4CO2 ( g )  5H2O(1); 2 multiplying eqn. (iii) by 6 and eqn. (iv) ∆C H  = – 2658.0 kJ mol–1 by 3 we get: 2024-25 Unit 5.indd 152 9/12/2022 11:53:41 AM  THERMODYNAMICS 153 In this case, the enthalpy of atomization is 6C  graphite   6O2  g  6CO2  g  ; same as the enthalpy of sublimation. ∆f H  = – 2361 kJ mol–1 (c) Bond Enthalpy (symbol: ∆bondH ) 3 Chemical reactions involve the breaking and 3H 2  g   O2  g  3H2O 1 ; making of chemical bonds. Energy is required 2 to break a bond and energy is released when a ∆f H  = – 857.49 kJ mol–1 bond is formed. It is possible to relate heat of Summing up the above two equations : reaction to changes in energy associated with 15 breaking and making of chemical bonds. With 6C  graphite   3H2  g   O2  g  6CO2  g  2 reference to the enthalpy changes associated  3H 2 O  l  ; with chemical bonds, two different terms are used in thermodynamics. ∆f H  = – 3218.49 kJ mol–1... (v) (i) Bond dissociation enthalpy Reversing equation (ii); (ii) Mean bond enthalpy 15 6CO2  g   3H2O  l  C6 H6  l   O2 ; Let us discuss these terms with reference 2 to diatomic and polyatomic molecules. ∆f H  = – 3267.0 kJ mol–1... (vi) Diatomic Molecules: Consider the following Adding equations (v) and (vi), we get process in which the bonds in one mole of 6C  graphite   3H2  g  C6 H6  l  ; dihydrogen gas (H2) are broken: H2(g) → 2H(g); ∆H–HH  = 435.0 kJ mol–1 ∆f H  = – 48.51 kJ mol–1... (iv) The enthalpy change involved in this process is the bond dissociation enthalpy of H–H bond. The bond dissociation enthalpy is the (b) Enthalpy of Atomization change in enthalpy when one mole of covalent (symbol: ∆aH ) bonds of a gaseous covalent compound is Consider the following example of atomization broken to form products in the gas phase. of dihydrogen Note that it is the same as the enthalpy of H2(g) → 2H(g); ∆aH  = 435.0 kJ mol–1 atomization of dihydrogen. This is true for all You can see that H atoms are formed by diatomic molecules. For example: breaking H–H bonds in dihydrogen. The Cl2(g) → 2Cl(g); ∆Cl–ClH  = 242 kJ mol–1 enthalpy change in this process is known as enthalpy of atomization, ∆aH . It is the O2(g) → 2O(g); ∆O=OH = 428 kJ mol–1 enthalpy change on breaking one mole of In the case of polyatomic molecules, bond bonds completely to obtain atoms in the gas dissociation enthalpy is different for different phase. bonds within the same molecule. In case of diatomic molecules, like Polyatomic Molecules: Let us now consider dihydrogen (given above), the enthalpy of a polyatomic molecule like methane, CH4. atomization is also the bond dissociation The overall thermochemical equation for its enthalpy. The other examples of enthalpy of atomization reaction is given below: atomization can be CH4 (g) → C(g) + 4H(g); CH4(g) → C(g) + 4H(g); ∆aH = 1665 kJ mol  –1 ∆a H  = 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 = 108.4 kJ mol–1 C – H bonds in each successive step differ : 2024-25 Unit 5.indd 153 9/12/2022 11:53:42 AM  154 chemIstry CH4(g) → CH3(g)+H(g);∆bond H = +427 kJ mol–1 given in Table 5.3. The reaction enthalpies are CH3(g) → CH2(g)+H(g);∆bond H = +439 kJ mol–1 very important quantities as these arise from the changes that accompany the breaking of CH2(g) → CH(g)+H(g);∆bond H = +452 kJ mol–1 old bonds and formation of the new bonds. CH(g) → C(g)+H(g);∆bond H = +347 kJ mol–1 We can predict enthalpy of a reaction in gas Therefore, phase, if we know different bond enthalpies. The standard enthalpy of reaction, ∆rH is CH4(g) → C(g)+4H(g);∆a H = 1665 kJ mol–1 related to bond enthalpies of the reactants In such cases we use mean bond enthalpy and products in gas phase reactions as: of C – H bond. For example in CH4, ∆C–HH  is calculated as: ∆r H   bond enthalpiesreactants ∆C–HH = ¼ (∆a H) = ¼ (1665 kJ mol–1)   bond enthalpies products (5.17)** = 416 kJ mol–1 This relationship is particularly more We find that mean C–H bond enthalpy useful when the required values of ∆f H are in methane is 416 kJ/mol. It has been not available. The net enthalpy change of a found that mean C–H bond enthalpies differ reaction is the amount of energy required slightly from compound to compound, as to break all the bonds in the reactant in CH3CH2Cl, CH3NO2, etc., but it does not molecules minus the amount of energy differ in a great deal*. Using Hess’s law, bond required to break all the bonds in the product enthalpies can be calculated. Bond enthalpy molecules. Remember that this relationship is values of some single and multiple bonds are approximate and is valid when all substances Table 5.3(a) Some Mean Single Bond Enthalpies in kJ mol–1 at 298 K H C N O F Si P S Cl Br I 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 CI 192 180 Br 151 I Table 5.3(b) Some Mean Multiple Bond Enthalpies in kJ mol–1 at 298 K N=N 418 C = C 611 O = O 498 N≡N 946 C ≡ C 837 C=N 615 C = O 741 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), which is the enthalpy change when one mole of a particular type of bond is formed from gaseous atom, then ∆f H  = ∑ ∆f Hbonds of products – ∑ ∆f H bonds of reactants 2024-25 Unit 5.indd 154 9/12/2022 11:53:42 AM  THERMODYNAMICS 155 (reactants and products) in the reaction are  1 2. Na( g ) Na ( g )  e ( g ) , the ionization of in gaseous state. sodium atoms, ionization enthalpy (d) Lattice Enthalpy ∆iH = 496 kJ mol–1 The lattice enthalpy of an ionic compound is 3. 1 Cl 2 ( g ) → Cl( g ), t h e d i s s o c i a t i