The variation of equilibrium constant with temperature is given below : Temperature Equilibrium Constant T1 = 25ºC K1 = 10 T2 = 100ºC K2 = 100 The values of Hº, Gº at T1 and Gº at T2 (in kJ mol–1) res

Question

The variation of equilibrium constant with temperature is given below : Temperature Equilibrium Constant T1 = 25ºC K1 = 10 T2 = 100ºC K2 = 100 The values of Hº, Gº at T1 and Gº at T2 (in kJ mol–1) respectively, are close to [use R = 8.314 JK–1 mol–1]

(A) 0.64, –7.14 and –5.71
(B) 28.4, –5.71 and –14.29
(C) 28.4, –7.14 and –5.71
(D) 0.64, –5.71 and –14.29

Prashant Kotian · Accepted Answer

Correct answer: B. T1 = 323 K T2 = 373 K k1 = 10 k2 = 100       2 k k log = R . H         1 T T | JEE MAIN-2020 | DATE : 06-09-2020 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 4       100 log = . 303 . H       373 298 log 10 = . 303 . H       373 75 H = 373 314 . 303 .    = 28.4 KJ At T1 = 25ºC = 298 K, K1 = 10 G = –2.303 RT1logK1 = –2.303 × 8.314 × 298 × log(10) = –2.303 × 8.314 × 298 × 1 = –5.7 KJ At T2 = 100ºC = 373 K K2 = 100 G = –2.303 RT2logK2 = –2.303 × 8.314 × 373 × log(10)2 = –2.303 × 2 × 8.314 × 373 × 1 = –14283.7 J = –14.29 KJ

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