The activation energy of a reaction is 5 kcal/mol. The increase in the rate constant when its temperature is raised from 300 to 305 K is approximately

Options

A. 0.146
B. 0.5
C. 1
D. 25.7%.

Answer

A

Explanation

From Arrhenius equation,
`log(k_2)/(k_1)=(E_a)/(2.303R)((T_2-T_1)/(T_1T_2))=(500)/(2.303xx2)((5)/(305xx300))=0.0593`
Antilog (0.0593)=1.146
Hence `(k_2)/(k_1)=1.146` or `(k_2)/(k_1)-1=1.146-1,(k_2-k_1)/(k_1)=0.146` or `k_2-k_1=0.146k_1` i.e., `k_2` increases =14.6%