Thermodynamics Study Guide

Thermodynamics THERMODYNAMICS TERMS USED IN THERMODYNAMICS Thermodynamics is a branch of science which deals (1) System : A system is defined as a specific part with quantitative relation ship between heat and other of universe or specified portion of the matter form of energy. which is under experimental investigation Chemical thermodynamics : It is the branch of (2) Surrounding : The rest part of the universe thermodynamics which deals with the study of other than the system is called surrounding. We processes in which chemical energy is involved. are never interested in surrounding. Our study Chemical thermodynamics is concerned with the is only restricted to system. following questions : (3) Boundary : Any thing which sperates system (i) When two or more substances are put together, and surrounding is called boundary. The will they react? boundary may be real or imaginary ; it may be (ii) If they do react, what energy changes will be rigid or non rigid ; it may be conductor or non associated with the reaction? conductor. (iii) If a reaction occurs, at what concentrations of For example : A reaction is carried out in a beaker. the reactants and their products will equilibrium The contents of beaker constitute the system, beaker be established? serves as boundary and the anything which is out Limitations of thermodynamics : side the beaker is called surroundings. (1) The law of thermodynamics apply only to the Types of system : Systems can be classified in two matter in bulk i.e., macroscopic system and not ways : to individual atoms or molecules of the (i) On the basis of exchange of energy and matter macroscopic system. (ii) On the basis of composition (2) Thermodynamics can only predict the feasibility or spontaneity of a process under a given set of (i) On the basis of exchange of energy and matter conditions but does not tell anything about the (a) Open system : This type of system can rate at which the given process may proceed. exchange energy as well as matter with the surrounding. The boundary is not sealed and (3) It is not concerned with the total energy of the not insulated. body but only with energy change. 39 Thermodynamics (b) Closed system : This type of systems can State of the system : exchange energy, in the form of heat, work or when the fundamental properties such as pressure, radiations but not matter with its surroundings. temperature, volume, number of moles & The boundary between the system and its composition have definite values, the system is said surroundings is sealed but not insulated. to be in a any definite state, when there is any change (c) Isolated system : This type of system are in any one of these properties it is said that the system perfectly insulated systems and can not interact has undergone a change of state. in any way with its surrounding i.e. neither matter State functions/Thermodynamic function/State nor energy can be exchanged with the variable : Fundamental properties which determine surrounding. the state of a system are referred to as state variables. (ii) On the basis of composition The change in state properties depends only upon (a) Homogeneous system: A system is said to the initial and final state of the system but is be homogeneous when it is completely uniform independent of the manner in which the change has throughout. A homogeneous system is made of been brought about. one phase only. eg. a pure single solid, liquid, Path function : If change in fundamental properties gas, mixture of gases. depends upon initial and final state of the system as (b) Heterogeneous system : A system is said to well as the manner in which the change has been be heterogeneous when it is not uniform brought about. e.g. Work, heat throughout ie. it consists of two or more phases. eg. ice in contact with water, insoluble solids in Work : contact with a liquid. Product of force and displacement is known as work. THERMODYNAMIC PROPERTIES work (W) = force (F) × displacement () (a) Extensive properties : The properties of the system whose magnitude depends upon the Consider a gas enclosed in a cylinder fitted with a quantity of matter present in the system are called frictionless piston. Suppose area of cross section of extensive properties. e.g. volume, weight, cylinder = a and pressure on the piston = P , Initial energy, enthalpy, entropy, internal energy, mole volume of the gas = V1 and final volume of the gas = V2 etc. (By expansion) displacement of piston =  (b) Intensive properties : The properties of the system which do not depend on the quantity of work done by the gas (in expansion) = W = F. matter present in system or size of the system are called intensive properties. These are the characteristic properties of the system only i.e. when the mass of the system is changed, these properties remain unchanged. For example melting points, boiling points, viscosity, refractive index, specific heat, surface tension, density, concentration etc. Note : Extensive properties can change in intensive properties when divided to make per unit quantity. Ex. volume and weight are extensive property but mass of unit volume (density) is intensive property. P= F= P×a 40 W=P×a× (change in volume = a ×  = V2 – V1) W = P × (V2 – V1) W = PV (According to Physics) In general  (According to Chemistry)  If system works then V2 > V1   work is –ve If work is done on the system then V1 > V2  work is + ve Heat :    Thermodynamics It is neither possible nor necessary to calculate the absolute value of internal energy of a system. Thermodynamics is concerned only with energy change which occurs when the system moves from one state to another. Let E be the difference of energy of the initial state (Ein) and the final stage (Ef) then E = Ef – Ein In chemical reaction E = EProduct – EReactant E is + ve if Ef > Ein & E is – ve if Ef < Ein Note : (1) Internal energy of 1 mole of mono atomic gas is RT T (2) Bomb calorimeter always gives the value of change in internal energy. Heat may be defined as the quantity of energy which flows between a system and its surroundings on Charcteristics of I.E. :– account of temperature difference. (1) Internal energy of a system is an extensive property. Heat always flows from high T to low T. (2) Internal energy is a state property. Heat flowing into the systum q = +ve (3) The change in Internal energy does not depend on Heat flowing out of the systum q = – ve the path by which the final state is reached. Units of heat & work : Calorie It is defined as the (4) There is no change in internal energy a cyclic process. quantity of heat required to raise the temperature of 1 g of water by 1°C (14.5 to 15.5°C) Ex.1 Classify the following into open, closed or isolated 1 Nm = 1J = 107 erg = 0.239 cal systems : 1 cal = 4.184 J = 4.2 J (i) Animals and plants 1 L-atm = 101.3 J = 24.206 cal = 101.3 × 107 erg (ii) A referigerator or a fridge 1 L-atm > cal. > J. > erg (iii) A solar cooker Internal energy [E] : Total of all possible kinds of energy of a system is called its internal energy. The exact value of this energy is not known as it includes all types of energies of molecules constituting the given mass of matter such as translational, vibrational, rotational, the kinetic and potential energy of the nucleus and electrons within the individual molecules and the manner in which the molecules are linked together. The internal energy is denoted by E. E = Etrans. + Erota. + Evibra + Ebond + Eelec. + .......... Sol. (i) Open system (iii) Closed system (ii) Closed system Ex.2 Which of the following are open, close or nearly isolated systems : (i) Human beings (ii) The Earth (iii) Can of tomato soup (iv) Ice-cube tray filled with water (v) A satellite in a orbit (vi) Coffee in a thermos flask and (vii) Helium filled balloon. 41 Thermodynamics (iii) Isochoric process : Isochoric process are those Sol. Open : Human beings, Earth, Ice cube tray processes in which Close : Can of Tomato soup, A satellite in an orbit, (a) volume remains constant i.e. V = 0 Helium filled balloon. (b) pressure & temperature are variable Isolated : Coffee in thermos flask. (c) Work, W = – PV,  V = 0, Ex.3 Separate out the following into extensive and  W = 0 (Zero) intensive. (d) System should be placed in a closed vessel. Volume, Temperature, Pressure, Boiling point, Free (iv) Adiabatic process : Adiabatic process are those energy. processes in which Sol. Volume and free energy are extensive, others, are (a) No exchange of heat takes place ie. q = 0 intensive. (b) The temperature of the system varies. Ex.4 Which of the following are state functions : (c) The system is thermally isolated this can be done by keeping the system in an insulated container. (i) Height of hill (ii) Distance travelled in climbing the hill (iii) Energy change in climbing the hill. Sol. (i) and (iii) THERMODYNAMIC PROCESSES (v) Cyclic process : When a system undergoes a number of different processes and finally returns to its initial state, it is termed as cyclic process. In cyclic process change in all state function will be zero. i.e. E = 0, H = 0, P = 0, T = 0 When a system changes from one state to another, (vi) Reversible process : the operation is called a thermodynamic process. (a) Process in which all changes occuring at any part of the system are exactly reversed when Thermodynamic processes are as follows. changes in variables are carrried out in opposite direction. (i) Isothermal process : Isothermal process are those (b) Driving force should be infinitesimally greater processes in which than opposing force. (a) Temperature remains const. i.e. T = 0 (c) Process takes place in infinitesimal small steps (b) Heat is exchanged with surroundings or in many steps and takes infinite time to (c) System is placed in a constant temperature bath complete the process. i.e. thermostate. (d) It is an ideal process. (d) Change in Internal energy = 0 (only for ideal (e) Work obtained is maximum. gases) i.e. E = 0 (f) System is in virtual equilibrium at any state.         (ii) Isobaric process : Isobaric process are those processes in which (a) Pressure remains const i.e. P = 0 (b) Volume & temperature are variable. (c) System should be placed in an open vessel. (vii) Irreversible process : (a) Process in which direction of change cannot be reversed by small changes in variables. (b) Process takes place in short time (c) It takes finite time and in single step. (d) All natural processes are irreversible (e) System is in equilibrium only at initial and final state (f) Driving force is greater than opposing force. 42 Thermodynamics During an adiabatic process ZEROTH LAW OF THERMODYNAMICS During adiabatic process the system acts an isolated When a bodyA is in thermal equilibrium with a body or insulated system and hence q = 0 in such cases. B, and also separately with a body C, then B and C Therefore E = W will be in thermal equilibrium with each other. i.e., work done on the system is equal to increase in internal energy. FIRST LAW OF THERMODYNAMICS This law was given by Helmholtz and this law can During Cyclic Process E = 0  q=–W be expressed as followsi.e., work done by the system is equal to heat (i) Energy can neither be created nor be destroyed absorbed by the system. (ii) The total energy of the universe is always a constant. i.e. the sum of energies of a system During Isobarric process and its surrounding is always constant. Let the energy given to system is q and work done by the system is W, then Internal energy E = qp – w (iii) Whenever a particular form of energy Here, qp = energy given to system at constant disappears during a process, an equivalent pressure amount of some other form of energy appears. So qp = E + W ...........(1) (iv) It is impossible to construct a perpetual motion qp = (E2 – E1) + P(V2 – V1) machine which can operate in cycles and could qp = (E2 + PV2) – (E1 + PV1) produce work without any external help. qp = H2 – H1 {H = E + PV = enthalpy} The mathematical form of first law of qp = H thermodynamics can be expressed as E = q + W i.e. in isobaric process change in enthalpy is equal to here q is the energy given to system and W is the heat absorbed by the system. work done on the system. E is change in internal put the value of qP in equation (1) energy. H = E + PV .......(2) We know that, H = heat at constant pressure and Note : put the value of q and W with sign. E = heat at constant volume. H = E + PV CONCLUSIONS FROM THE FIRST LAW or qp = qv + PV OF THERMODYNAMICS or qP – qV = PV During isochoric process i.e., difference between energy at constant pressure At constant volume V1 = V2 i.e. V = V2 – V1 = 0 and energy at constant volume is equal to workdone For gases : When temperature and pressure are Hence W = P V = P × 0 = 0 ......... no work done remain constant then, at constant volume therefore PV1 = n1RT (n1 = moles of reactant) q = E (At constant volume) PV2 = n2RT (n2 = moles of product) Thus during isochoric process, heat supplied to the So P(V2 – V1) = (n2 – n1) RT system is equal to change in internal energy. PV = ngRT During an isothermal process ( here ng is only for gases not for solid and liquid) put the value of PV in equation (2) – During an isothermal process the temperature of the system remains constant and hence E = E2 – E1 = 0    or H = E + ng Therefore q = –W, Thus during an isothermal (i) If ng = 0 ; H = E , process, the total quantity of heat supplied to the (ii) If ng > 0 ; H > E system is equal to the work done by the system. (iii) If n < 0 ; H < E g 43 Thermodynamics Ex.11 The heat of combustion of napthalene {C10H8(s)} at constant volume was measured to be – 5133 kJ mol–1 at 298K. Calculate the value of enthalpy change (Given R = 8.314 JK–1 mol–1). Ex.5 500 joule of heat was supplied to a system at constant volume. It resulted in the increase of temperature of the system from 20°C to 25°C. What Sol. The combustion reaction of napthalene. is the change in internal energy of the system ? C10H8(s) + 12O2 (g) 10CO2(g) + 4H2O (l), Sol. E = 500 J E = – 5133 KJ Ex.6 Neither q nor w is a state function, yet q + w is a state function. Explain why. ng = 10 – 12 = – 2 mol. Sol. q + w = E and E is a state function. H = E + ng RT Ex.7 The enthalpy change (H) for the reaction N2(g) + 3H2(g) 2NH3(g) is – 92.38 KJ at 298 K. What is U at 298K ? Now applying the relation. = – 5133 × 103 + (–2) (8.314) ( 298) = – 5133000J – 4955.14 J = – 5137955.14 joule HEAT CAPACITY [C]  Sol. U = H– ngRT = –92.38 –(–2)  The amount of heat required to raise the temperature of a system by 1° C is called heat capacity of system. = –87.42 kJ it is an extensive property. Molar heat capacity : Amount of heat required to Ex.8 A thermally isolated gaseous system can exchange raise the temperature of 1 mole substance by 1°C. energy with the surroundings. What is the mode of Molar heat capacity = heat capacity/mole. It is an transference of energy ? intensive property Sol. Work Specific heat capacity : Amount of heat reuired to raise the temperature of 1 gm. of substance by 1°C. Ex.9 Can the absolute value of internal energy be determined ? Why or why not ? Specific heat capacity = heat capacity/1g of substance. It is an intensive property. Sol. No, because it is the sum of different types of Calorie : Amount of heat required to raise the energies, some of which cannot be determined. temperature of 1g H2O by 1°C (14.5°C to 15.5°C) Ex.10 Calculate the internal energy change in each of the  Amount of heat required to raise the temperature. following cases : by dT°C = dq (i) A system absorbs 5 kJ of heat and does 1 kJ  Amount of heat required to raise the temperature. of work. by   = C J/K (ii) 5 kJ of work is done on the system and 1 kJ of heat is given out by the system. Sol. (i) Here q = + 5 kJ and w = – 1 kJ  According to first law of thermodynamics, If heat supplied at constant pressure E = q + w = 5 + (–1) = 4 kJ   (ii) Here w = + 5kJ and q = – 1kJ  .......... (1)   According to first law of thermodynamics E = q + w = –1 + (+5) = 4 kJ i.e. the internal energy of the system increases by 4 kJ in each case.   If heat supplied at constant Volume        .......... (2) 44 Thermodynamics from equation (1) and (2) dH = CPdT (J/K), dE = CVdT (J/K) from n mole dH = nCPdT (J/mole K) .......... (3) dE = nCVdT (J/mole K) .......... (4) for m g dH = mCPdT (J/g K), dE = CVdT (J/g K ) Ex.12 A gas expands from 3 dm3 to 5 dm3 against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mol of water at a temperature of 290 K. Calculate final temperataure of water. Specific heat of water = 4.184 J g–1 K–1 Sol. Work done for an irreversible expansion at constant pressure Relationship between CP and CV  W = – P(V2 – V1) (i) as we know that W = – 3 × (5 – 3) = – 3 × 2 litre atm H = E + PV (Now putting the Values = –6 × 101.3 joule = – 607.8 joule of H and E from equation 3 and 4) Now this work is used up in heating water nCPdT = nCvdT + PV = qp = 607.8 joule  W or nCPdT – nCVdT = PV W = Cp × m(dT) ndT (CP–CV) = PV 607.8 = 4.184 × 10 × 18 × dT if n=1 and dT = 1°C dT = 0.80    T2 – T1 = 0.80 That is the difference between CP and CV is T2 = 290 + 0.80 = 290.80 K equal to workdone when 1 mole substance is heated through 1°C. WORK DONE DURING (ii) As we know that H = E+PV DIFFERENT PROCESS ideal gas equation PV = nRT (A) In irreversible process : W = – Pexternal V (if n=1), PV = RT H=E+RT Now diffrentiate to above equation with repect (B) In reversible process :    to T (i) During reversible isothermal process :    or  or CP = CV+R CP–CV = R (This called mayer's relation) (iii)    PV = nRT       In isothermal process nRT = const  W    Note: (1) Heat capacity is an extensive property but molar heat capacity and specific heat capacity are intensive property. (2) During the phase transition the heat capacity of the system is infinite. during the phase transation temperature and pressure are constant. heat capacity    = – nRT T  = – nRT T    = – nRT loge  State equation of isothermal process is P1V1 = P2V2  45 Thermodynamics (ii) During reversible adiabatic process : Work of expansion in adiabatic reversible process   For adiabatic process  PV=   = constant (K) or Ex.13 A gas occupies 2 litre at S.T.P. It is provided 300 joule heat so that its volume becomes 2.5 litre at 1 atm. Calculate change in its internal energy. Sol. Given V1 = 2 litre, V2 = 2.5 litre , q = + 300J, Pexternal = 1 atm.  E = q + W  W = –Pexternal V = – Pexternal (V2 – V1)       = But       W = –0.5 × 101.3 J = –50.65J  E = q + W = 300 + (–50.65) = 249.35 J          Sol. Given - Gas expands in vaccum i.e., Pext = 0  W = – Pext V  =    or = – 1 atm × (2.5–2) litre = –0.5 litre atm.  Ex.14 Calculate the work done during the process when one mole of gas is allowed to expand freely into vacuum.    =       ............(1)  here   CP and C V are heat capacities at constant pressure and volume respectively. from mayer's relation CP – CV = R  =  Cp = CV ×   CV ×  – CV = R or CV (  – 1) = R or – CV (1–  ) = R or CV = – so  W=0 Ex.15 Two litre of N2 at 0°C and 5 atm are expanded isothermally against a constant external pressure of 1 atm until the pressure of gas reaches 1 atm. Assuming the gas to be ideal calculate work of expansion ? Sol. Given P1 = 5 atm, V1 = 2 litre, P2 = 1 atm, V2 = ? We know that - if T is constant then P1V1 = P2V2     = 10 litre W = – Pext V = –1(V2 – V1) = – 1(10 – 2) = – 8 litre atm = – 8 × 101.3 = – 810.4 joule Ex.16 Calculate the amount of work done in each of the following cases : (i) One mole of an ideal gas contained in a bulb of 10 litre capacity at 1 atm is allowed to enter into an evacuated bulb of 100 litre capacity.  ............(1)   for adiabatic process :      (ii) One mole of a gas is allowed to expands from a volume of 1 litre to a volume of 5 litres against the constant external pressure of 1 atm (1 litre atm = 101.3 J) 46 Thermodynamics Sol. (i) W = – Pext × V SPONTANEOUS PROCESS As expansion takes place into the evacuated A process which under some given conditions may bulb, i.e. against vacuum, Pext = 0. Hence either take place by itself or take place on initiation W = 0. is called as spontaneous process. For example (ii) V = V2 – V1 = 5 – 1 = 4 litres , P = 1 atm (i) Heat flows from a hot reservoir to cold reservoir  W = – PV = – 1 × 4 litre atm (ii) Electricity flows from high potential to low = – 4 litre atm = – 4 × 101.3 J potential = – 405.2 J (iii) Water flows from higher side to lower side Alternatively, using the SI units directly Spontaneous processes proceed spontaneously only P = 1 atm = 101325 Pa , –3 3 in one direction (from high energy to low energy side) V = 4L = 4 × 10 m –3 till the state of equilibrium is attained and then they  W = – P × V = – 101325 × 4 × 10 J are irreversible. = – 405.3 J A spontaneous process is accompained by decrease The negative sign implies that the work is done by the system. of internal energy or enthalpy. It indicates that only exothermic reactions are spontaneous. But melting Ex.17 Calculate the work done during isothermal of ice and evaporation of water are endothermic reversible expansion of one mol of an ideal gas from process which also proceed spontaneously. It means, 10 atm to 1 atm at 300K. there is some other factor in addition to enthalpy Sol. Number of moles of ideal gas (n) = 1, change (H) which explain the spontaneous nature Inital pressure (P1) = 10 atm, of the system. This factor is entropy which is included Final pressure (P2) = 1 atm in 2nd law of thermodynamics. Constant absolute temp (T) = 300 K, The first law of thermodynamics has certain Gas constant (R) = 2 cal/mol. limitations.  Work done for isothermal reversible expansion (i) It puts no restriction on the direction of flow of of an ideal gas heat. W = – 2.303 nRT log (ii) It is not in position to tell that heat energy can not be completely converted into work. = – 2.303 × 1 × 2 × 300 × log10 ENTROPY (S) = – 1381.8 calories Entropy is thermodynamic state quantity which is a Ex.18 A 5 litre cylinder contained 10 moles of oxygen measure of randomness or disorder of the molecules gas at 27°C. Due to sudden leakage through the of the system.Entropy is represented by symbol 'S'. hole, all the gas escaped into the atmosphere and It is difficult to define the exact entropy of a system.It the cylinder got empty. If the atomsphere pressure is more convenient to define the change of entropy is 1.0 atm. Calculate the work done by the during a change of state. gas.Assuming gas to be ideal. S = S2 – S1 Sol. Vinitial = 5 litre, T = 27°C = 27 + 273 = 300K. S2 = Final entropy S1 = Initial entropy     Vfinal = litre A spontaneous process is accompanied by an increase in entropy i.e. S is positive. V = Vfinal – Vinitial = 246.3 – 5 = 241.3 litre At absolute zero, the entropy of pure crystal is zero. Wexp= – PV = –1 × 241.3 litre × atm. The absolute entropy of a substance at 298 K and 1 = –24443.69J = –24.4 kJ atmospheric pressure is called the standard entropy i.e. work done by gas is 24.4 kJ (S0) entropy is expressed in joules per kelvin. 47 Thermodynamics The entropy of universe is continuously increasing. At equilibrium entropy remains unchanged . Ex.19 The enthalpy change for transition of liquid water to steam is 40.8 kJ mol–1 at 373K. Calculate S for the process. Entropy of gaseous system is greater than liquid systems and entropy of liquid systems is greater than Sol. The entropy change for the vapourization of water entropy of solid systems. If heat is absorbed S is positive and if heat is evolved then S is negative. S if +ve ............... increase in entropy is given by    Given Hvap. = 40.8 kJ mol–1 S if –ve ................ decrease in entropy = 40.8 × 1000 J mol–1 and T = 373 K The entropy change is represented as follows    S    = 109.38 JK–1 mol–1 q = quantity or heat supplied during reversible process. Ex.20 Calculate the change in entropy for the fusion of 1 mole of ice (water). The malting point of T = Temperature on absolute scale water is 273 K and molar enthalpy of fusion for Entropy is an extensive property and state function. water = 6.0 kJ mol–1. Entropy change during phase transition : (1) for fusion   Sol.      = 21.97 JK–1 mol–1 (2) for Vapourisation   (3) for Sublimation     Ex.21 5 mol of an ideal gas expand reversibly from a volume of 8 dm3 to 80 dm3 at an temperature of 27° C. Calculate the change in entropy. Sol. For isothermal reversible process S = 2.303 nR log Note : for isothermal reversible process W = – 2.303 nRT =95.73 JK–1 log10  q=–W We know that  =2.303 × 5mol × 8.314J/molK log  q = 2.303 nRT log10    Ex.22 Calculate the entropy change in the melting of 1 kg of ice at 0° C in SI units. Heat of fusion of ice = 80 cal g1. Sol. Sfus. =   48 Thermodynamics Ex.23 For each reaction, is the sign of S (entropy) positive or negative ? (a) Ni(s) + 4CO(g)  Ni(CO)4 (g) (b) 4Fe(s) + 3O2(g)  2Fe2O(s) (c) H2(g) + S(s)  H2S(g) (d) 2NO(g) + O2(g)  2NO2(g) Sol.(a) Five molecules are organised into only one, so the disorder decreases and S is negative. (b) Seven molecules, three of which is are gases, become only two solid molecules thus S < 0. (c) Two molecules, one of which is solid, form a gaseous molecule. Since only a heteronuclear gas results, disorder increases and S is positive. (d) Three gas molecules produce two gas molecules and disorder decreases, thus, S < 0. SECOND LAW OF THERMODYNAMICS It is not possible to construct a machine that functions in cycles and conveys heat from colder body to hotter body unless work is done on the machine by outside agency. "Work can always be converted into heat but conversion of heat into work does not take place under all conditions ". OR "It is not possible to convert heat energy completely into work with compensation." OR "Whenever a spontaneous process takes place, it is accompained by an increase in the total entropy of the universe".  It suggests direction of a given process.  It is impossible to obtain work by putting a body below to the lowest temperature of the system.  It is impossible to take heat from hot reserviour and convert it in work without transferring a part of it to cold reserviour (sink) [given by S-carnot]  Heat cannot be passed from cold body to hot body without applying any external work. [given by R - Clausius] GIBBS FREE ENERGY AND SPONTANEITY For the spontaneity of a reaction both H and S should be considered simultaneously. The function that takes both enthalpy and entropy into account is called free energy. It is denoted by symbol G, the meaning of which is Gibb's free energy, represented as follows G = H – T S Where G = Change in free energy, H = Change in enthalpy, S = change in entropy If G is negative then spontaneous changes occur with decrease in free energy (The reaction occurs spontaneously in forward direction). If G is +ve then the reaction does not take place. But it means that the reaction is spontaneous in backward direction. If G = 0, the system is in equilibrium. Free energy of reaction changes with temperature. The decrease in free energy is a measure of the maximum useful work that can be obtained from a reaction. In other words, it is the energy per mole of the reaction which will be liberated. Example :- CaO(s) + CO2 (g) CaCO3(s) for this reaction H = –ve and S = –ve So, at low temperature reaction will occur spontaneously but not at high temperature.  The decrease in free energy is equal to useful work.  Standard free energies of formation of all elementary and stable substances are taken as zero. Ex. Cl2(g) , Br2(), I2(s), Carbon (graphite), Sulphur (Rhombic), Phosphorus(white). Standard free energy change and equilibrium constant G0 = –RT ln K or G0 = –2.303 RT log10 K The above equation is known as Vant Hoff reaction isotherm. If in the cell reaction E0 is known , the free energy can be given by relationship. G0 = –n × F × E0 Where n = moles of electrons involved F = 1 Faraday = 96500 coulombs E0 = Standard cell potential G0 = Standard change in free energy 49 Thermodynamics  Ex.28 For the equilibrium PCl5   PCl3 (g) + Cl2(g) at 298 K, Kp = 1.8 × 10–7. What is G° for the Ex.24 One mole of ice is converted into water at 273 K. reaction ? The entropies of H 2O (s) and H2O (I) are 38.20 Sol. G = –2.303 RT log KP and 60.01 J mol–1 K–1 respectively .The enthalpy = –2.303 × 8.314 × 10–3 kJ mol –1K–1 × 298K change for the converstion would be. × log 1.8 × 10–7 Sol. At 273 K ice and water are in equilibrium with each = 38.484 kJ mol–1 other and at equilibrium G = 0  G = H – TS Ex.29 Four mol of an ideal gas expand isothermally from  H = TS = 273 × (60.01 – 38.20) 1 litre to 10 litres at 300 K. Calculated the change in = 5954.13 J mol–1 free energy of the gas (R = 8.314 JK–1 mol–1). Ex.25 For the vaporisation of water :  H2O (l)   H2O (g) [1 atm. pressrue] Given : S = 120 JK–1 and H= +45.0 kJ. Evaluate the temperature at which liquid water and water vapours are in equlibrium at 1 atm. pressure – Sol. We known that : at equilibrium G = 0  TS = H  G = H – TS        Ans. For isothermal reversible process H = 0 and S = 2.303nR log  = H –TS = –2.303nRT log = –2.303 × 4 × 8.314 × 300log = –22976.5 J = 375 K Ex.26 For a certain reaction the change in enthalpy and change in entropy are 40.63 kJ mol–1 and 100 JK–1. What is the value of G at 27° C and indicate whether the reaction is possible or not ? Sol. We know that : G = H – TS Given T = 27 + 273 = 300 K , H = 40.63 × 103 J mol–1 , = 40630 J mol–1 , S = 100 JK–1 G = 40630 – 300 × 100 = 40630 – 30000 = + 10630 J Positive value of G indicates that the reaction is not possible. THIRD LAW OF THERMODYNAMICS The entropy of all pure crystals is taken as zero at the absolute zero of temperature. Exception :- Crystals of glassy solids, mixture of isotopes of chlorine, H2O, N2O, NO, CO, CO2 etc. JOULE THOMSON EXPANSION The phenomenon of producing lowering of temperature when a gas is made to expand adiabatically from a region of high pressure into a region of low pressure, is known as Joule-thomson expansion.  Joule thomson coefficient () : Ex.27 Which of the following statement is incorrect.  (i) Entropy is a measure of disorder. Where dT and dP are difference in temperature and (ii) Entropy is not a state function. pressure respectively. (iii) Entropy is expressed in the units of J/k/mol.  If +ve  Gas will cool on expansion. (iv) The spontaneous process always accompanies  If = –ve  Gas will warm on expansion. the decrease in free energy. for ideal gas  = 0 Sol. (ii) Entropy is not a state function 50 Inversion temperature : Every gas has a definite value of temperature. When the sign of changes from negative to positive. This temperature is termed inversion temperature.  Most of the gases have inversion temperature near about room temperature and thus cool at room temperature when expansion is done.  The value of inversion temperature has been calculated by using vander wall's equation.  Where, a and b are Vander Wall's constants. THERMOCHEMICAL EQUATION It is a balanced chemical reaction which not only indicate physical state of reactants and products but also energy change during chemical reaction. C(s) + O2(g)  CO2(g) ; Thermodynamics Factors which influence the heat of reaction :(i) Physical state of reactants and products :H2(g) + O2(g) H2O() ; H = – 286 kJ/mol H2(g) + O2(g) H2O(g); H = –242 kJ/mole (ii) Allotropic forms of the element :C(s) + O2 (g) CO2(g); H = –393.5 kJ/mole (graphite) C(s) + O2(g) CO2(g); H = –395.4 kJ/mole (diamond) (iii) Temperature : Relation between heat of reaction and temperature is given by the Kirchoff's equation H = – 393.5 kJ/mol. (H2 – H1) = CP (T2 – T1) at constant P. If no condition is given then value of H is considerd as standard condition. and  If equation is multiplied by coefficient then value of H is also multiplied by that coefiicient. H1 and H2   If equation is reversed then the value of H remains same only sign changes. C P= CP  HEAT OF A REACTION It is defined as the amount of heat evolved or absorbed when quantities of the substances indicated by the chemical equation have completely reacted. Heat of a reaction always depends on the physical states of reactants and products. (E2 – E1) = CV (T2 – T1) at constant V. Heat of reaction at temperature T1 & T2. – (product) CR (Reactant) (iv) Concentration of solution (v) Condition of constant pressure and volume qP = H and qV = E (vi) Quantity of reactants H  Quantity of reactants. Heat of reaction can be expressed at constant volume DIFFERENT TYPE OF or at constant pressure. For example HEAT OF REACTION  (i) H2(g) + Cl2(g)   2HCl(g) + 44.0 kcal A. Heat of formation (Hf) : The amount of heat H = – 44 kcal absorbed or evolved when 1 mole of the substance  (ii) C2H4(g) + 3O2(g)   2CO2(g) + 2H2O() is directly obtained from its consituent elements is E = –335.8 kcal called heat of formation. (The reaction (ii) is carried out at constant volume.) C (graphite) + O2(g) CO2(g) (iii) C(s) + O2(g)  CO2(g) Hf = – 393.5 kJ/mol. H = 94.3 k cal 51 CO(g) + O2(g) CO2(g) (This reaction not shows the heat of formation) Standard heat of formation = enthalphy of products – enthalpy of reactants. Hf° = HP° – HR° Note :The heat of formation of all the elements in their standard states are arbitrarily assumed to be zero. Ex.30 Hf0 for Al2O3 is – 1670 KJ. Calculate the enthalpy change for the reaction 4Al + 3O2 2Al2O3 Sol. Given 2Al + O2  Al2O3 ; H = – 1670 kJ Multiplying it by 2 then 4Al + 3O2  2Al2O3 ; H = 2 × (– 1670) = – 3340 kJ Ex.31 Standard heat of formation of carbon mono oxide at 298 K is –110.5 kJ. Calculate the standard heat of formation of carbon mono oxide at constant volume and 298 K. Sol. Heat of formation of CO (g) is represented in the form of following thermochemical equation : C (s) + Thermodynamics Ex.32 Find the heat of formation of ethyl alcohol from following data : C (s) + O2 (g)  CO2 (g), H = – 94 kcal H2 (g) + O2 (g)  H2O (), H = – 68 kcal C2H5OH () + 3O2 (g)  2CO2 (g) + 3H2O (), H = – 327 kcal Sol. On the basis of definition of heat of formation, the following thermochemical equation is written for which Hf is to be calculated : 2C (s) + 3H2 (g) + O2 (g)  C2H5OH(), Hf = ? Given data may be written as follows because thermochemical equation act as algebraic equations (i) 2C (s) + 2O2 (g)  2CO2 (g) H = – 188 kcal (ii) 3H2 (g) + O2 (g)  3H2O () ; H = – 204 kcal (iii) 2CO2 (g) + 3H2O ()  C2H5OH () + 3O2 (g) H = + 327 kcal On addition 2C (s) + 3H2 (g) + O2 (g)  C2H5OH (), Hf = – 65 kcal/mol. O2 (g)  CO (g) , Ex.33 Find Hf0 of Fe2O3 (s) if the standard heat of H = – 110.5 kJ reaction Fe2O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g) is –6.6 kcal. Given Hf0 CO (g) = –26.4 n for this equation = Number of gaseous moles of kcal. Hf0 CO2 (g) = –94 kcal. product – Number of gaseous mole of reactant Ans. For following thermochemical equation –3 Fe2O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g), ng =    , R = 8.314 × 10 kJ/K/mol., H = –6.6 kcal T = 298 K Thus, Now H = E + ngRT Hof reaction = [Hf0 of product – Hf0 of reactant] E = Standard heat of reaction at constant volume = [(2× Hf0 of Fe (s) + 3 Hf0 of CO2 (g) – =? (Hf0 of Fe2O3 (s) + 3 × Hf0 of CO (g))] –6.6 = [(2 × 0) + (3 × –94) – Thus, –110.5 = E + (+ ) × 8.314 × 10–3 × 298 (Hf0 of Fe2O3 (s) + 3 × –26.4)] kcal –6.6 = [0 – 282 – Hf0 of Fe2O3 (s) + 79.2] E = (– 110.5 – × 8.314 × 10–3 × 298)kJ/mol.  Hf0 of Fe2O3 (s) = – 196.2 kcal / mol = (–110.5 – 1.238)kJ/mol.= – 111.738 kJ/mol. 52 Ex.34 Calculate the heat of formation of KCl from the following data : (i) KOH (aq) + HCl (aq)  KCl (aq) + H2O(), H = –57.3 kJ mol–1 Thermodynamics Ex.35 Standard heat of formation of CH 4(g), CO2(g) and H2O() are –75, –393.5, –286kJ respectively. Find out the change in enthalpy for the reaction. CH4(g) + 2O2(g) CO2(g) + 2H2O() Sol. CH4(g) + 2O2(g) CO2(g) + 2H2O() (ii) H2 (g) + O2 (g) H2O (), Hof reaction = [Hf0 of product – Hf0 of reactant] H° = (–393.5) + 2(–286) ] –(–75) + 2(0) ] H = – 286.2 kJ mol–1 H° = (–965.5) – (–75) = – 890.5 kJ (iii) H2 (g) + Cl2 (g) + aq  HCl (aq), H = – 164.4 kJ mol–1 (iv) K (s) + O2 (g) + H2 (g) + aq  KOH (aq), H = – 487.4 kJ mol–1 (v) KCl (s) + aq KCl (aq), H = +18.4 kJ mol–1 Sol. Desired equation is :K (s) + Cl2 (g)  KCl (s), H = ? ...... (vi) Step 1. Adding Equation (iii) and (iv) and subtracting Equation (v), we have Cl2 (g) + H2 (g) + (i) C + O2 CO2 ; H = –393.5 kJ/mol H show heat of combustion of C and heat of formation of CO2 (ii) C + O2 CO equation (ii) does not show the heat of combustion because this is not complete combustion In order to get this thermochemical equation we follow the following two steps : K (s) + B. HEAT OF COMBUSTION : Heat evolved during the complete combustion of 1 mole substance is called heat of combustion. O2 (g)  Calorific Value or Fuel value : The amount of heat produced in calorie or joule when one gram of a substance (food or fuel) is completely burnt or oxidised. Maximum value of calorific value = Maximum efficiency or best fuel  KCl (s) + HCl (aq) + KOH (aq) – KCl (aq) H = – 487.4 + (– 164.4) – (18.4) = – 670.2 kJ mol–1 ...... (vii) Ex.36 Find out the calorific value of Glucose C6H12O6 + 6O2 6CO2 + 6H2O ;H = – 2900 kJ/mole Step 2. To cancel out the terms of this equation which Sol.  Heat evolved from 1 mole glucose = 2900 kJ do not appear in the required equation (vi). Adding or Heat evolved from 180 gram glucose = 2900 kJ Equation (i) to Equation (vi) and subtract Equation (ii) from their sum. This gives  Heat evolved from 1 gram glucose = K (s) + Cl2 (g)  KCl (s) ; H = –670.2 – 57.3 – (– 286.2) = – 441.3 kJ mol–1 = 16.11 kJ Note : Hydrogen has the highest calorific value. However, it is not used as domestic or industrial fuel due to some technical problems. 53 Thermodynamics Ex.37 The heat of formation of CO2 and H2O is –97 Ex.39 Calculate the enthalpy of combustion of ethylene kcal and –68 kcal. The heat of combustion of (g) to form CO2 (gas) and H2O (gas) at 298 K and benzene is –783 kcal. Calculate the heat of formation 1 atmospheric pressure. The enthalpies of formation of benzene ? of CO2, H2O and C2H4 are – 393.7, –241.8 + 52.3 kJ per mole respectively. Sol. Given : Sol. Given : (i) C6H6 () + O2 (g)  (i) C (s) + O2 (g) CO2 (g) 6CO2 (g) + 3H2O (); H = – 393.7 kJ mol–1 H = – 783 kcal. (ii) H2 (g) + O2 (g)  H2O (g), (ii) C (s) + O2 (g) CO2 (g) ; H = – 97 kcal. H = – 241.8 kJ mol–1 (iii) 2C (s) + 2H2 (g)  C2H4 (g), (iii) H2 (g) + O2 (g)  H2O (g) ; H = + 52.3 kJ mol–1 H = – 68 kcal. Objective reaction : Hence : C2H4 (g) + 3O2 (g) 2CO2 (g) + 2H2O (g) , 6CO2 (g) + 3H2O ()  H = ? C6H6 () + O2 (g) ; H = + 783 kcal. 6C (s) + 6O2 (g)  6CO2 (g) ; H = – 582 kcal. 3H2 (g) + O2 (g)  3H2O (g) ; H = – 204 kcal. Adding, we get ; 6C (s) + 3H2 (g)  C6H6() ; H = – 3 kcal. Hence heat of formation of benzene is –3kcal Ex.38 When 2 moles of C2H6 (g) are completely burnt, 3129 kJ of heat is liberated, calculate the heat of formation of C2H6 (g). Hf for CO2 and H2O () are – 395 and – 286 kJ mol–1 respectively. Sol. 2C2H6 (g) + 7O2 (g) 4CO2 (g) +6H2O () ; H = – 3129 kJ – 3129 = [4 ×  +6×  ]–   We are given :  ] ] = – 393.7 kJ mol–1, = – 241.8 kJ mol–1 ,  = + 52.3 kJ mol–1 HReaction = (Sum of Hf0 values of Products) – (Sum of Hf0 values of Reactants)  = [2 ×  [2 ×  = [4 × (– 395 + 6 × (–286)] – [2 ×  2 × Eq. (i) + 2 × Eq. (ii) – Eq. (iii) gives 2C (s) + 2O2 (g)  2CO2 (g) + O2 (g) + 2H2 (g)  2H2O (g) –2C (s) – 2H2 (g)  – C2H4 (g)  3O2 (g)  2CO2 (g) + 2H2O (g) – C2H4 (g) or C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (g) H = 2(–393.7) + 2 (–241.8) – (52.3) = – 1323.3 kJ mol–1 Alternative Method : We aim at : C2H4 (g) + 3O2 (g) 2CO2 (g) + 2H2O (g) +2×  [ ]– +3×  ] = – 167 kJ [for elements H = 0] = [2 × (–393.7) + 2 × (–241.8)] – [(52.3) + 0] (  Hf0for elementary substance = 0) = – 83.5 kJ = [–787.4 – 483.6] – 52.3 = –1323.3 kJ mol–1 54 Thermodynamics Ex.40 A cylinder of cooking gas is assumed to contain 11.2 kg of butane. The thermo chemical equation for the combustion of butane is : C4H10(g) + ; Ex.41 200 cm3 of 0.1 M H2SO4 is mixed with 150 cm3 of 0.2M KOH. Find the value of evolved heat. O2(g) 4CO2(g) + 5H2O() Sol. 200 cm3 of 0.1 M H2SO4 = H = – 2658 kJ = 0.02 mole H2SO4 If a family needs 15000 kJ of energy per day for cooking, how long would the cylinder last ? 150 cm3 of 0.2 M KOH = –1 Sol. Molecular mass of butane is = 58 g mol 58 g of butane on combustion gives 2658 kJ of heat =  0.02 mole H2SO4 = 2 × 0.02 mole H+ = 0.04 mole H+ kJ of heat – and 0.3 mole KOH  0.03 mole OH 0.03 mole OH– ions would react with 0.03 mole H+ ions to form 0.03 mole H2O. The daily requirement of energy = 15000 kJ  The cylinder would last =     = 0.03 mole KOH 11.2 kg of butane on combustion gives    The heat liberated = 57.1 × 0.03 = 1.7 kJ days  34 days E. Heat of solution :- The amount of heat evolved or absorbed when 1 mole solute is dissolved in excess of solvent is called heat of solution. C. Heat of Hydrogenation :- Heat changes during KCl (s) + aq KCl (aq.) the complete hydrogenation of 1 mole of unsaturated H = 4.4 kcal organic compound is called heat of hydrogenation. H2SO4 ( (i) CH2 = CH2 + H2 CH3 – CH3 H = – 20.2 kcal H = x (ii) CH  CH + H2 CH2 = CH2 H = y Reaction (ii) will not show the heat of hydrogenation. H2SO4(aq.) F. Heat of Hydration :- The change in heat content when one mole of anhydrous or partially hydrated salt combines with required number of moles of water to form a specific hydrated substance. CuSO4 (s) + 5H2O () CuSO4.5H2O (s); D. Heat of neutralisation :- The heat of neutralisation H = – 18.7 kcal is defined as the heat evolved or decrease in enthalpy when 1 gram equivalent of an acid is neutralised by G. Heat of fusion :- The heat required to convert 1 1 gram equivalent of a base in dilute solution. mole substance at its melting point into liquid is called Heat of neutralisation of a strong acid against a heat of fusion. strong base is always nearly the same, i.e., 13.7 k H2O (s) H2O (); cal or 57.1 kJ. H = 6 kJ 55 Thermodynamics H. Heat of Vapourisation :- The heat required to LAW OF THERMO CHEMISTRY convert 1 mole liquid at its boiling point into vapour (a) Lavosier and Laplace's law :- Enthalpy of is called heat of vapourisation. decomposition of a compound is numerically equals H2O () H2O (g); to the formation of that compound and only sign changes. H = 40.6 kJ (b) Hess's Law :- The total amount of heat evolved or absorbed in a chemical reaction is the same whether the reaction is carried out in one or several steps by one or more methods. I. Heat of sublimation :- The heat required to convert 1 mole substance ( a solid) at a particular temperature into vapour is called heat of sublimation. J. It also does not depend upon the manner in which Lattice Energy :- The amount of energy released when one mole of the compound is formed by the the chemical reaction is carried out. interaction of constituent gaseous cations and Ex. Single step reaction gaseous anions. A B ; H A+ (g) + B–(g) A+B– + Energy Multi step reaction to produce B from A is given (1 mole) (Latice energy)      Note : Lattice energy can be calculated with the help of Born-Haber cycle.    According to Hess's law H = H1 + H2 + H3 + H4 56