If the half lives of the first order reaction at 350 ~K and 300 ~K are 2 and 20 seconds respectively, the activation energy of the reaction in kJ mol ⁻¹ is

Step-by-step solution

Given: t_ 1 / 2 ( l )=20 ~s aligned & t_ 1 / 2 (2)=2 ~s [ t_ 1 / 2 1 k ] & T₁=300 ~K , T₂=350 ~K & k₂ k₁ =- E_a 2.303 R [ 1 T₂ - 1 T₁ ] & 0.693 2 0.693 20 =- E_a 2.303 8.314 [ 1 350 - 1 300 ] & 20 2 =- E_a 2.303 8.314 [ -50 350 300 ] & E_a= 2.303 8.314 350 300 50 & E_a=40.209 ~kJ / mol & aligned