The rate constant of the elementary reaction CH3OCH3 (g) CH4 (g) + CH2O (g) is k = 8.33*10-6 s-1 at 427°C, and the reaction has an activation energy of 245 kJ mol-1. (a) Compute the rate constant of the reaction at a temperature of 545°C. s-1 (b) At a temperature of 427°C, 8.32*104 s is required for half of the CH3OCH3 originally present to be consumed. How long will it take to consume half of the reactant if an identical experiment is performed at 545°C?

(a) The rate constant is 3.61*10^-3 s^-1

(b) 7.12*10^4 s

Explanation:

(a) Log (K2/K1) = Ea/2.303R * [1/T1 - 1/T2]

K1 = 8.33*10^-6 s^-1

Ea = 245 kJ = 245,000 J

R = 8.314 J/mol. K

T1 = 427°C = 427+273 = 700 K

T2 = 545°C = 546+273 = 818 K

Log (K2/8.33*10^-6) = 245,000/2.303 * [1/700 - 1/818]

Log (K2/8.33*10^-6) = 2.637

K2/8.33*10^-6 = 10^2.637

K2 = 8.33*10^-6 * 433.51 = 3.61*10^-3 s^-1

(b) The relationship between temperature and the time required for reactants to be consumed is inverse

t2 = T1t1/T2

T1 = 427 °C = 700 K

t1 = 8.32*10^4 s

T2 = 545 °C = 818 K

t2 = 700*8.32*10^4/818 = 7.12*10^4 s