Question

at 227 degree Celsius, the presence of a catalyst lowers the activation energy of a reaction by 4.606 kcal. the rate of reaction will be increased by

Answer 1

From Arrhenius equation we know that,

K = A e ^(-Ea/RT)

Hence k₁ = A e^(-Ea₁/RT)

k₂ = A e^(-Ea₂/RT)

dividing both the equation we get :

k₁/k₂  = e^(Ea₂-Ea₁/RT)

Given the value of Ea₂-Ea₁ = 4606

R = 2 cal/k mole (constant)

T = 227 degree celcius = 273 + 227 = 500

Hence,

k₁/k₂  = e^(4606/2*500)

= 100

Hence the rate of reaction will increase by 100 times.