Title DPP-1 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 1 1 (c) ∆H = nCp∆T The process is isothermal therefore, ∆G = 0 ∴ ∆H = 0 2 (b) The system returns to its original state, i.e., cyclic process. 3 (b) ∆G = ∆H ― T∆S; at equilibrium, ∆G = 0, ∴ ∆H = T∆S or ∆H = 273 × (60.01 ― 38.20) = 5954.13 J mol―1 4 (a) EN F ~ EN Cl = 0.2028 ∆ and ∆ = [eF─Cl ― (eF─F × eCl─Cl) 1/2] ∴ ENF ~ ENCl = 0.2028 [eF─Cl ― (eF─F × eCl─Cl) 1/2] 1/2 Or 1 = 0.2028 [eF─Cl ― (38 × 58) 1/2] 1/2 ∴ eF─Cl = 71.26 kcal mol ―1 . 6 (b) 0.2 mole will neutralize 0.2 mole of HNO3 heat evolved = 51 × 0.2 = 11.4 kJ 7 (b) Kirchhoff’s equation is : ∆H2 ― ∆H1 = ∆Cp (T2 ― T1) 8 (d) ∆n depends on stoichiometry of reaction. 9 (a) eA―A = a Also, 1 2 A2 + 1 2 B2 →AB; eA―B = a ∆H = ― 100 kJ mol ―1 eB―B = 0.5a ∴ ∆H = ― [eA―B ] + 1 2 [eA―A + eB―B ] = a + 1 2 [a + 0.5a] Topic :- THERMODYNAMICS Solutions
― 100 = ―0.25 a ∴ a = 400 kJ mol ―1 10 (d) The properties of the system whose value is independent of the amount of substance present in the system are called intensive properties e.g., viscosity, surface tension, temperature, pressure etc. 11 (d) When a real gas is forced through a porous plug into a region of low pressure, it is found that due to expansion, the gas on the side of low pressure gets cooled 12 (b) The room got heated because heat is lost to surroundings. 13 (b) Tb = ∆H ∆S = 30 × 103 75 = 400 K 14 (d) Heat of combustion is always exothermic; Few combustion reactions such as F2 to F2O, N2 to N2O and NO are endothermic but these reactions do not give heat of combustion because the substance should be completely oxidized. In F2O, F2 is reduced and N2O and NO are not completely oxidized state of N2. However, three reactions are exceptions but these do not represent heat of combustion. These are, N2 + O2 ⟶N2O; ∆H = + ve N2 + O2 ⟶NO; ∆H = + ve and F2 +(1/2)O2 ⟶F2O; ∆H = +ve 15 (b) For an isothermal process ∆T = 0 and ∆E = 0 and q ≠ 0. 16 (b) Given: (i)H2 + 1 2 O2→H2O; ∆H = ― 241 kJ (ii)C6H10 + 17 2 O2→6CO2 +5H2O; ∆H = ― 3800 kJ (iii)C6H12 +9O2→6CO2 +6H2O; ∆H = ― 3920 kJ for the reaction C6H10 + H2→C6H12 [It is infact Eq.(i)+Eq.(ii) ― Eq.(iii)] Thus, ∆H = ―241 ― 3800 ― ( ―3920) = ―121 kJ 17 (d) In isothermal reversible process, ideal gas has constant volume and so, ∆E = 0 and ∆H = ∆E = 0 18 (a) ∆H = ― 2 × eH─Cl + eH─H + eCl─Cl ∴ n182 = ― 2 × a + 430 + 242 ∴ a = 245 kJ mol ―1 19 (d) ∆H = ∆U + ∆nRT ∆n = + 1/2
Thus, ∆H > ∆U 20 (c) Cylinder contains 11.2 kg or 193.10 mole butane. ( ∵ molecular mass of butane =58) ∵ Energy released by 1 mole of butane = ―2658 ∴ Energy released by 193.10 mole of butane = ―2658 × 193.10 = 5.13 × 105 kJ ∴ 5.13 × 105 20000 = 25.66 or 26 days