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

wher is the rest of the question