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]

Step-by-step solution

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 = ln