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CHAPTER 6, THERMODYNAMICS”, Brief Summary of the chapter:, 1., , Thermodynamics: Science which deals with study of different forms of, energy and quantitative relationship., , 2., , System & Surroundings: The part of universe for study is called system and, remaining portion is surroundings., , 3., , State of system & state function: State of system is described in terms of T,, P, V etc. The property which depends only on state of system not upon path, is called state function eg. P, V, T, E, H, S etc., , 4., , Extensive & Intensive Properties: Properties which depends on quantity of, matter called extensive prop. eg. mass, volume, heat capacity, enthalpy,, entropy etc. The properties which do not depends on matter present, depends upon nature of substance called Intensive properties. eg. T,P,, density, refractive index, viscosity, bp, pH, mole fraction etc., , 5., , Internal energy: The total energy with a system., i.e. U = Ee + En + Ec + Ep + Ek + -----U = U2 – U1 or UP – UR & U is state function and extensive properly. If, U1 > U2 energy is released., , 6., , Heat (q): It I a form of energy which is exchanged between system and, surrounding due to difference of temperature. Unit is Joule (J) or Calorie (1, Calorie = 4.18, , 7., , J)., , First Law of Thermodynamics: It is law of conservation energy. Energy can, neither be created not destroyed, it may be converted from one from into, another., Mathematically U = q + w, w = –p. V (work of expansion), U = q – p. V or q = U + p. V, q,w are not state function., 82

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15., , Bond enthalpy: It is amount of energy released when gaseous atoms, combines to form one mole of bonds between them or heat absorbed when, one mole of bonds between them are broken to give free gaseous atoms., Further, , 16., , rH, , = B.E. (Reactants) - B.E. (Products), , Spontaneous & Non Spontaneous Processes: A process which can take, place by itself is called spontaneous process. A process which can neither, take place by itself or by initiation is called non Spontaneous., , 17., , Driving forces for spontaneous process: (i) Tendency for minimum energy, state. (ii) Tendency for maximum randomness., , 18., , Entropy (S): It is measure of randomness or disorder of system., i.e. Gas>Liquid>Solid., Entropy change ( S) =, , 19., , q ( rev.), T, , J.K 1 .mol, , 1, , Spontaneity in term of ( S), S(total) = S(universe) = S(system) + S(surrounding), If S(total) is +ve, the process is spontaneous., If S(total) is –ve, the process is non spontaneous., , 20., , Second Law of thermodynamics: In any spontaneous process, the entropy of, the universe always increases. A spontaneous process cannot be reversed., , 21., , Gibb’s free energy (G): defined as G = H – T.S & G = H – T. S (Gibb’s, Helmholts equation) it is equal useful work i.e. - G = W(useful) = W(max.), If G = ve, process is spontaneous., , 22., , Effects of T on spontaneity of a process: G = H – T. S., (i) For endothermic process may be non spontaneous at law temp., , 84

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(ii) For exothermic process may be non spontaneous at high temp. and, spontaneous at law temp., 23., , Calculation of ( rG0), rG, , 24., , 0, , =, , fG, , 0, , (p) -, , fG, , 0, , (r), , Relationship between ( rG0) & equilibrium constant (k), G = G0 + RTlnQ & G0 = –2.303RT logk., , 25., , Calculation of entropy change:, 0, rS, , =, , S0 (p) - S0 (r), , ONE MARK QUESTIONS:, 1., , State First Law of thermodynamics., , 2., , What is a thermodynamic state function?, , 3., , Give enthalpy (H) of all elements in their standard state., , 4., , From thermodynamic point to which system the animals and plants belong?, , 5., , Predict the sign of S for the following reactions., CaCO3(s) + CO2(g), , heat, , CaO(s) + CO2(g), , 6., , For the reaction 2Cl(g), , Cl2(g), What will be the sign of H and S?, , 7., , State Hess’s Law for constant heat summation?, , 8., , What is Gibb’s Helmhaltz equation?, , 9., , Define extensive properties., , 10., , Give relationship between H, U for a reaction in gaseous state., , 85

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ANSWERS FOR ONE MARK QUESTIONS, 1., , Energy can neither be created nor destroyed. The energy of an isolated, system is constant. U = q + w., , 2., , A function whose value is independent of path. eg. P, V, E, H, , 3., , In standard state enthalpies of all elements is zero., , 4., , Open system., , 5., , S is positive (entropy increases), , 6., , H: (–ve) b/c energy is released in bond formation and, S: (–ve) b/c atoms combines to form molecules., , 7., , The change of enthalpy of reaction remains same, whether the reaction is, carried out in one step or several steps., , 8., 9., , G = H – T. S, Properties which depends upon amount of substance called extensive, properties. Volume, enthalpy, entropy., , 10., , H = U + ng. RT., TWO MARKS QUESTIONS:-, , Q.1, , In a process, 701J heat is absorbed and 394J work is done by system. What is, change in Internal energy for process?, , Q.2, , Given: N2(g) + 3H2(g), , 2NH3(g),, , 0, rH, , = –92.4KJ.mol–1. What is the standard, , enthalpy of formation of NH3(g)., Q.3, , Calculate entropy change in surroundings when 1.0 mol of H2O(l) is formed, under standard conditions? Given H0 = –286KJmol–1., , Q.4, , Give relationship between entropy change and heat absorbed/evolved in a, reversible reaction at temperature T., , Q.5, , What is spontaneous change? Give one example., 86

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Q.6, , A real crystal has more entropy than an Ideal Crystal. Why?, , Q.7, , Under what condition, the heat evolved/absorbed in a reaction is equal to its, free energy change?, , Q.8, , Q.9, , Predict the entropy change in(i), , A liquid crystallizes into solid, , (ii), , Temperature of a crystallize solid raised from OK to 115K, , What is bond energy? Why is it called enthalpy of atomization?, , Q.10 Calculate entropy change for the following process., H2O(l), is 6.0 KJ mol-1 at 00C., , H2O(s), , ANSWER FOR TWO MARKS QUESTIONS:, 1., 2., 3., , q = 701J, w = 394J, so U = q + w = 701 – 394 = 307J., , 5., , 92.4, 46.2KJ .mol 1, 2, , q(rev.) = – H0 = –286 KJmol-1 = 286000Jmol-1, S, , 4., , NH3(g) = –, , fH, , q ( rev.), , S=, , T, , 286000 J.mol, 298 K, , 1, , 959 J.K 1mol, , 1, , q ( rev), T, , A process which can take place of its own or initiate under some condition., eg. Common salt dissolve in water of its own., , 6., , A real crystal has some disorder due to presence of defects in their, structural arrangement, and Ideal crystal does not have any disorder., , 7., , In G = H – T. S, when reaction is carried out at OK or S = 0, then G, = H., 87

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8., , (i) Entropy decreases b/c molecules attain an ordered state., (ii) entropy increase b/c from OK to 115K particles begin to move., , 9., , It is the amount of energy required to dissociate one mole of bonds present, b/w atoms in gas phase. As molecules dissociates into atoms in gas phase so, bond energy of diatomic molecules is called enthalpy of atomization., , 10., , H2O(s), , H2O(l) at 00C,, , fusH, , = 6KJ mol-1, , = 6000J mol-1, Tf = 00C = 0 + 273 = 273K, Do, , fuss =, , fus, , T, , H, , 6000 J.mol, 273 K, , 1, , 21 .98 J.K 1mol, , 1, , THREE MARKS QUESTIONS:, Q.1, , For oxidation of iron, 4Fe(s) + 3O2(g), , 2Fe2O3(s), , S is –549.4J.K-1 mol-1, at 298K. Inspite of –ve entropy change of this reaction,, Why the reaction is spontaneous? ( rH0 = –1648x103 J.mol-1), Q.2, , Using the bond energy of Hr = 435 KJ mol-1, Br2 = 192 KJ mol-1, HBr = 368, KJmol-1. Calculate enthalpy change for the reaction H2(g) + Br2(g), , Q.3, , Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) and –110, –393, 81, and 9.7 KJ mol-1 respectively. Find value, , Q.4, , 2HBr(g), , For the reaction at 298K, 2A+B, , rH, , for the reaction N2O4(g) + 3CO(g), , C, H = 400 KJ mol-1, S = 0.2 KJ mol-1 K-, , 1, , . At what temperature will the reaction become spontaneous, considering H,, S be constant at the temp., , Q.5, , The equilibrium constant for a reaction is 10. What will be the value of G0? R, = 8.314J.K-1mol-1 T = 300K., , Q.6, , What do you understand by state function? Neither q nor w is a state function, but q + w is a state function? Explain., , Q.7, , Justify the following statements:, 88

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= [81 + 1179] – [9.7 – 330] = –777.7KJ, Ans.4 H = 400 KJ mol-1, S = 0.2 KJK-1 mol-1., G = H – T. S, O = 400 – 0.2 x T ( G = 0 at equilibrium), 400, 0.2, , T=, , 2000K, so reaction will be spontaneous above 2000K., , 0, , = –2.303 RT logK, = –2.303 x 8.314 x 300 x log10, = –19.147 x 300 x 1 = –5744.1J, 0, -1, rG = –5.7441KJ.mol, Ans.6 The property whose value depends upon state of system and is independent of, path. q + w = U, which is a state function as value of U does not depends upon, path., Ans.7 (a) It is false, exothermic reaction is not always spontaneous. If S = +ve and, T. S> H. The process will be non spontaneous even it. It is endothermic., (b) The entropy of vapour is more than that of liquid, so entropy increases during, vaporization., Ans.5, , rG, , Ans.8, , G=, , H – T., H, S, , T=, , S, at equilibrium, , 108 .4x10 3 J.mol, 190 J.K 1mol 1, , G = 0,, , H = T., , S, , 1, , 570 .526 K, , So the reaction will be spontaneous above 570.52K, as above this temperature, G will be –ve., Ans.9, , rH, , =, , f, , H ( CO 2 ) 2, , f, , H CH 4 ( g ), , –890.5KJ = –393.5KJ + 2v – 286 –, =, , H ( O2 ), Hf(CH4) – O, , Hf(CH4) = –75.0 KJ.mol-1., , = Hf (CH ), 4, , Ans.10, , f, , 75.0KJ.mol 1 ., , H = 177 KJ mol-1,, G=, , H – T., , S = 285 JK-1 mol-1, , S = 177KJ –, , 298x 285, KJ, 1000, , = 177 KJ – 84.93 KJ = 92.07 KJ.mol-1., Since, , G is +ve, so the reaction is non spontaneous., 90

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FIVE MARKS QUESTIONS:Q.1, , What is entropy? Why is the entropy of a substance taken as zero at 0K?, Calculate the rG for the reaction?, N2(g) + 3H2(g), 2NH3(g) at 298K, The value of equilibrium constant (K) is 6.6x105, R = 8.314JK-1mol-1., , Ans: It is measure of randomness or disorder of system., Because at O K there is complete order in the system., G0 = –2.303 RT logK = –2.303 x 8.314 x 298 x log6.6 x 105, = -5705.8[log6.6 + log 105], = –5705.8[0.8195 + 5.0] = –5705.8 + 5.8195J, = –33204.903J, G0 = –33.205 KJ mol-1., Q.2, , (i) What are extensive property and intensive properties?, (ii)Calculate the value of equilibrium constant (K) at 400K for, 2 NOCl(g) →2NO(g) + Cl2(g)., H0 = 77.2KJ.mol-1,, , S0 = 122J.K-1mol-1 at 400K, R = 8.314 J.K-1mol-1., , Ans. (i) An extensive property is a property whose value depends on the, quantity or size of matter present in the system Those properties which do not, depend on the quantity or size of matter present are known as intensive, properties, (ii), , H0 =, , H0 – T., , S0 = 77.2KJ -, , 400x122, KJ .mol 1, 1000, , = 77.2 – 48.8 = 28.4 KJ mol-1, and, , G0 = –2.303 RT logK., , 28400 = –2.303 x 8.314 x 400 log K., logK =, , 7.1, 2.303x8.314, , 3.7081 4.2919K, , K – antilog ( 4.2919 ) 1.95 x10 4 Ans., 91

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Q.3, , Define standard enthalpy of formation. Calculate the enthalpy of formation of, benzene from data, 15, O 2( g ), 2, , C 6 H 6(l), , C(s) + O2(g), , 6CO 2( g ), , 3H 2 O ( l) ,, , c, , f, , H0, , CO2(g), , 1, 2, , H2(g)+ O2(g), , H2O(l), , f, , H0, , H0, , 3266.0KJ, 393 .1KJ, , 286 .0KJ, , Ans. Ans. The standard enthalpy change for the formation of one mole of a, compound from its elements in their most stable states of aggregation (also, known as reference states) is called Standard Molar Enthalpy of Formation., c, , H0, , 6, , f, , H (0co22 ), , 3, , f, , H 0 ( H 2O ), , f, , H (0C6 H 6 ), , 15, H 0 f (O2 ), 2, , = –3266KJ = 6 x – 393.1 + 3x – 286 – f H0(C H ) - 0, 6, , 6, , = -3218 kJ/mol, HOTS QUESTIONS, 1. Why standard entropy of an elementary substance is not zero whereas standard, enthalpy of formation is taken as zero?, Ans. A substance has a perfectly ordered arrangement only at absolute zero., Hence , entropy is zero only at absolute zero. Enthalpy of formation is the heat, change involved in the formation of one mole of the substance from its, elements. An element formed from it means no heat change., 2. The equilibrium constant for a reaction is one or more if ∆G° for it is less than, zero. Explain, Ans. ―∆G° = RT ln K, thus if ∆G° is less than zero. i.e., it is negative, then ln, K will be positive and hence K will be greater than one., 3. Many thermodynamically feasible reactions do not occur under ordinary, conditions. Why?, Ans. Under ordinary conditions, the average energy of the reactants may be, less than threshold energy. They require some activation energy to initiate the, reaction., , 92