The plot of logk f versus 1 T for a reversible reaction A ( g ) ⇌ P ( g ) is shown. Pre-exponential factors for the forward and backward reactions are 10 15   s - 1 and 10 11   s - 1 , respectively. If the value of logK for the reaction at 500   K is 6 , the value of logk b at 250   K is K = equilibrium constant of the reaction, k f = rate constant of forward reaction, K b = rate constant of backward reaction]
A. A
From the question A f = 10 15 , A b = 10 11 , f =  Forward reaction  logK  at  500 K = 6 b =  Backward reaction  logk f  at  500 K = 9  (from graph)  logk b  at  500 K : logK = log k f k b     since  ⇒ K = k f k b 6 = logk f - logk b 6 = 9 - logk b log ⁡ k b = 3  at  500 K log ⁡ k 2 k 1 = − E a R 1 T 2 − 1 T 1 k b = A b e − E ab RT ln ⁡ k b = ln ⁡ A b e − E ab RT lnk b = lnA b - E ab RT 2 . 303 logk b = 2 . 303 logA b - E ab 500 R E a 500 R = 2 . 303 logA b - logk b E a 500 R = 2 . 303 log 10 11 - 3 E a 500 R = 2 . 303 ( 11 - 3 ) = 2 . 303 × 8 E a = 2 . 303 × 8 × 500 R ln k 2 k 1 = - E a R 1   T 2 - 1   T 1 ln k 250   K k 500   K = - E a R 1 250 - 1 500 ln k 250 K k 500 K = - 2 . 303 × 8 × 500 R R 1 500 2 . 303 logk 250 K - logk 500   K = - 2 . 303 × 8 logk 250 K - 3 = - 8 logk 250 K = - 5 logk 250 K = 5
Related: Chemistry — Chemical Kinetics · All PYQ Banks