If energy of activation of the reaction is 53.6 ~kJ ~mol ⁻¹ and the temperature changes from 27^ C to 37^ C then the value of ( k_ 37^ C k_ 27^ C ) is

Step-by-step solution

aligned & (c) : Given : T₁=27^ C =27+273=300 ~K & T₂=37^ C =37+273=310 ~K & and E_a=53.6 ~kJ ~mol ⁻¹ & aligned ( k₂ k₁ )= E_a 2.303 R ( T₂-T₁ T₁ T₂ ) ( k_ 310 ~K k_ 300 ~K )= 53.6 10^3 2.303 8.314 ( 310-300 300 310 ) =0.3010 aligned & k_ 310 ~K k_ 300 ~K =2 aligned