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Chemistry, XI, Question 6.1:, Choose the correct answer. A thermodynamic state function is a quantity, (i) used to determine heat changes, (ii) whose value is independent of path (iii), used to determine pressure volume work, (iv) whose value depends on temperature only., Answer, A thermodynamic state function is a quantity whose value is independent of a path., Functions like p, V, T etc. depend only on the state of a system and not on the path., Hence, alternative (ii) is correct., Question 6.2:, For the process to occur under adiabatic conditions, the correct condition is:, (i) ∆T = 0, (ii) ∆p = 0, (iii) q = 0, (iv) w = 0, Answer, A system is said to be under adiabatic conditions if there is no exchange of heat between, the system and its surroundings. Hence, under adiabatic conditions, q = 0., Therefore, alternative (iii) is correct., Question 6.3:, The enthalpies of all elements in their standard states are:, (i) unity, (ii) zero, (iii) < 0, (iv) different for each element, Answer, The enthalpy of all elements in their standard state is zero., Therefore, alternative (ii) is correct., , 1

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Enthalpy of formation of CH4(g) = –74.8 kJ mol–1 Hence,, alternative (i) is correct., , Question 6.6:, A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction, will be, (i) possible at high temperature, (ii) possible only at low temperature, (iii) not possible at any temperature, (iv) possible at any temperature, Answer, For a reaction to be spontaneous, ∆G should be negative., ∆G = ∆H – T∆S, According to the question, for the given reaction,, ∆S = positive, ∆H = negative (since heat is evolved), ⇒ ∆G = negative, Therefore, the reaction is spontaneous at any temperature., Hence, alternative (iv) is correct., Question 6.7:, In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the, system. What is the change in internal energy for the process?, Answer, According to the first law of thermodynamics,, ∆U = q + W (i), , 3

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Where,, ∆U = change in internal energy for a process q, = heat, W = work Given,, q = + 701 J (Since heat is absorbed), W = –394 J (Since work is done by the system), Substituting the values in expression (i), we get, ∆U = 701 J + (–394 J), ∆U = 307 J, Hence, the change in internal energy for the given process is 307 J., Question 6.8:, The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter,, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the, reaction at 298 K., , Answer, Enthalpy change for a reaction (∆H) is given by the expression,, ∆H = ∆U + ∆ngRT, Where,, ∆U = change in internal energy, ∆ng = change in number of moles, For the given reaction,, ∆ng = ∑ng (products) – ∑ng (reactants), = (2 – 2.5) moles, ∆ng = –0.5 moles, And,, ∆U = –742.7 kJ mol–1, T = 298 K, R = 8.314 × 10–3 kJ mol–1 K–1, 4

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Substituting the values in the expression of ∆H:, ∆H = (–742.7 kJ mol–1) + (–0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1), = –742.7 – 1.2, ∆H = –743.9 kJ mol–1, , Question 6.9:, Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of, aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol –1 K–1., Answer, From the expression of heat (q),, q = m. c. ∆T Where,, c = molar heat capacity m, = mass of substance, ∆T = change in temperature, Substituting the values in the expression of q:, , q = 1066.7 J q, = 1.07 kJ, Question 6.10:, Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –, 10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C., Cp[H2O(l)] = 75.3 J mol–1 K–1, Cp[H2O(s)] = 36.8 J mol–1 K–1, Answer, Total enthalpy change involved in the transformation is the sum of the following changes:, (a), , Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol, , of water at 0°C., (b), , Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of, , ice at 0°C., 5

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For the given reaction,, N2O4(g) + 3CO(g), , N2O(g) + 3CO2(g), , Substituting the values of ∆fH for N2O, CO2, N2O4, and CO from the question, we get:, , Hence, the value of ∆rH for the reaction is, , ., , Question 6.13:, Given, ; ∆rHθ = –92.4 kJ mol–1, What is the standard enthalpy of formation of NH3 gas?, Answer, Standard enthalpy of formation of a compound is the change in enthalpy that takes place, during the formation of 1 mole of a substance in its standard form from its constituent, elements in their standard state., Re-writing the given equation for 1 mole of NH3(g),, , Standard enthalpy of formation of NH, , 3(g), , = ½ ∆rH, , θ, , = ½ (–92.4 kJ mol–1), = –46.2 kJ mol–1, , Question 6.14:, Calculate the standard enthalpy of formation of CH3OH(l) from the following data:, , 7

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Answer, From the expression,, ∆G = ∆H – T∆S, Assuming the reaction at equilibrium, ∆T for the reaction would be:, , (∆G = 0 at equilibrium), , T = 2000 K, For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction to, be spontaneous, T should be greater than 2000 K., Question 6.18:, For the reaction,, 2Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ?, Answer, ∆H and ∆S are negative, The given reaction represents the formation of chlorine molecule from chlorine atoms., Here, bond formation is taking place. Therefore, energy is being released. Hence, ∆H is, negative., Also, two moles of atoms have more randomness than one mole of a molecule. Since, spontaneity is decreased, ∆S is negative for the given reaction., Question 6.19:, For the reaction, 2A(g) + B(g) → 2D(g), ∆Uθ = –10.5 kJ and ∆Sθ= –44.1 JK–1., Calculate ∆Gθ for the reaction, and predict whether the reaction may occur spontaneously., Answer, 10

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N2(g) +, , NO(g) +, , O2(g) → NO(g) ; ∆rHθ = 90 kJ mol–1, , O2(g) → NO2(g) : ∆rHθ= –74 kJ mol–1, , Answer, The positive value of ∆rH indicates that heat is absorbed during the formation of NO (g)., This means that NO(g) has higher energy than the reactants (N2 and O2). Hence, NO(g) is, unstable., The negative value of ∆rH indicates that heat is evolved during the formation of NO 2(g) from, NO(g) and O2(g). The product, NO2(g) is stabilized with minimum energy., Hence, unstable NO(g) changes to unstable NO2(g)., Question 6.22:, Calculate the entropy change in surroundings when 1.00 mol of H 2O(l) is formed under, standard conditions. ∆fHθ = –286 kJ mol–1., Answer, It is given that 286 kJ mol–1 of heat is evolved on the formation of 1 mol of H2O(l). Thus,, an equal amount of heat will be absorbed by the surroundings. qsurr = +286 kJ mol–1, , Entropy change (∆Ssurr) for the surroundings =, , ∆Ssurr, , = 959.73 J mol–1 K–1, , 12