Question

A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given : log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^-1 mol^-1 )

Answer 1

k₂ = 0.693 / 20,

k₁ = 0.693/40 log

Ea = 27663.8 J/mol

or 27.66 kJ/ mol