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www.ncrtsolutions.in, , NCERT Solutions for Class 11 Chemistry Chapter 6, Thermodynamics Class 11, Chapter 6 Thermodynamics Exercise Solutions, , Exercise : Solutions of Questions on Page Number : 182, Q1 :, 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, Tetc. depend only on the state of a system and not on the path., Hence, alternative (ii) is correct., , Q2 :, For the process to occur under adiabatic conditions, the correct condition is:, (i) ΔT = 0, (ii) Δp = 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., , Q3 :, The enthalpies of all elements in their standard states are:, (i) unity, (ii) zero, , www.ncrtsolutions.in
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www.ncrtsolutions.in, , Thus, the desired equation is the one that represents the formation of CH4 (g)i.e.,, , Enthalpy of formation of CH4(g) = –74.8 kJ mol–1, Hence, alternative (i) is correct., , Q6 :, A reaction, A + B → C + D + qis 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, ΔGshould be negative., ΔG= ΔH- TΔS, According to the question, for the given reaction,, ΔS= positive, ΔH= negative (since heat is evolved), ⇒ ΔG= negative, Therefore, the reaction is spontaneous at any temperature., Hence, alternative (iv) is correct., , www.ncrtsolutions.in
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www.ncrtsolutions.in, Q7 :, In a process, 701 J of heat is absorbed by a system and 394 J ofwork is done by the system. What is the, change in internal energy for the process?, , Answer :, According to the first law of thermodynamics,, ΔU= q+ W (i), Where,, ΔU= 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, ΔU= 701 J + (-394 J), ΔU= 307 J, Hence, the change in internal energy for the given process is 307 J., , Q8 :, The reaction of cyanamide, NH2CN(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found, to be -742.7 kJ mol-1at 298 K. Calculate enthalpy change for the reaction at 298 K., , Answer :, Enthalpy change for a reaction (ΔH) is given by the expression,, ΔH = ΔU + ΔngRT, Where,, ΔU = change in internal energy, Δng = change in number of moles, For the given reaction,, Δng = ∠‘ng (products) - ∠‘ng (reactants), = (2 - 1.5) moles, Δng = 0.5 moles, , www.ncrtsolutions.in
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www.ncrtsolutions.in, , And,, ΔU = -742.7 kJ mol-1, T = 298 K, R = 8.314 x 10-3 kJ mol-1 K-1, Substituting the values in the expression of ΔH:, ΔH = (-742.7 kJ mol-1) + (0.5 mol) (298 K) (8.314 x 10-3 kJ mol-1 K-1), = -742.7 + 1.2, ΔH = -741.5 kJ mol-1, , Q9 :, Calculate the number of kJ of heat necessary to raise the temperatureof 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. ΔT, Where,, c = molar heat capacity, m = mass of substance, ΔT = change in temperature, Substituting the values in the expression of q:, , q = 1066.7 J, q = 1.07 kJ, , Q10 :, Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. ΔfusH = 6.03 kJ mol1, 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:, , www.ncrtsolutions.in
