Title DPP-3 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 3 1 (d) ∆n = 0 ∴ ∆H = ∆U 2 (a) H2 + 1 2 O2 →H2O(g); ∆H = ― 57.0 kcal.....(i) H2 + 1 2 O2 →H2O(l); ∆H = ― 68.3 kcal......(ii) By eq. (i) and (ii), H2O(l)→H2O(g); ∆H = + 11.3 kcal 3 (b) ∆H = ∆U + ∆nRT ∴ ∆U = 176 ― 1 × 8.314 × 1240 × 10―3 = 165.6 kJ 4 (a) TV γ―1 = constant T Tfinal = ( V2 V1 ) γ―1 T Tfinal = ( 2 1 ) (5 3―1) = 2 (2 3) T Tfinal = T 2 (2 3) 6 (b) ∆n = ― 2 ∴ ∆H = ∆U + ∆nRT = ― 1415 + ( ―2) × 0.0083 × 300 = ― 1420 kJ 7 (c) Experimental determination of heats of reaction by bomb calorimeter represents its value at constant volume, i.e., ∆U. 8 (b) Graphite possesses sp 2 -hybridisation and has flat layer structure whereas diamond possesses sp 3 -hybridisation and has rigid tetrahedral nature. Topic :- THERMODYNAMICS Solutions
9 (c) nefficiency = T2 ― T1 T2 or 0.25 = T ― 400 T ∴ T = 533.3 K 10 (a) Lower is energy level of a system, more is its stability. 11 (b) ∆H = ∆U + ∆nRT Since, ∆n = ― 2 Thus, ∆H < ∆U 12 (b) K + 1 2 O2 + 1 2 H2 →KOH; ∆H = ? Find ∆H by Eq. [(i) + (ii)] – (iii). 14 (c) The fact for a quantity referred as state function. 15 (c) Bond formation is always exothermic. 16 (d) N2 + 1 2 O2 ⟶N2O; ∆H = 28 kJ 1 2 N2 + 1 2 O2 →NO; ∆H = 90 kJ By eq. [4 × (ii)] ― [2 × (i)], 2N2O + O2 →4NO; ∆H = 304 kJ 17 (b) Calorific value = Heat of combustion per g of fuel, i.e., for C2H4, it is ―1411 28 , the lowest value. 18 (b) ∆G° = ∆H° ― T∆S° = ―54.07 ― 298 × 10 × 10―3 = ―57.05 kJ Also, ∆G° = 2.303 RTlog10K log10K = ―57.05 × 103 2.303 × 8.314 × 298 19 (b) Hess’s law is based upon law of conservation of energy i.e., first law of thermodynamics. 20 (d) ∆Sf = ∆Hf T = 6 × 103 273 = 21.98 J
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. D A B A B B C B C A Q. 11 12 13 14 15 16 17 18 19 20 A. B B C C C D B B B D