At `400K` , the energy of activation of a reaction is decreased by `0.8Kcal` in the presence of catalyst. Hence, the rate will be a. Increased by `2.73` times b. Increased by `1.18` times c. Decreased by `2.72` times d. Increased by `6.26` times

At `400K` , the energy of activation of a reaction is decreased by `0.

1.	At `400K` , the energy of activation of a reaction is decreased by `0.8Kcal` in the presence of catalyst. Hence, the rate will be a. Increased by `2.73` times b. Increased by `1.18` times c. Decreased by `2.72` times d. Increased by `6.26` times
Answer» We know `(K_(2))/(K_(1))=` Antilog `[(DeltaE)/(2.303RT)]` Given `DeltaE=0.8Kcal=3.344xx10^(3)J` `R=8.314JK^(-1)mol^(-1)` `T=400K` Substituting all the values in Eq. (i), we get `(K_(2))/(K_(1))=` Antilog `[(3.344xx10^(3))/(8.314xx2.303xx400)]=2.73` `:. K_(2)=2.73K_(1)` Therefore, the rate will increase by `2.73` times. We know `(K_(2))/(K_(1))=` Antilog `[(DeltaE)/(2.303RT)]` Given `DeltaE=0.8Kcal=3.344xx10^(3)J` `R=8.314JK^(-1)mol^(-1)` `T=400K` Substituting all the values in Eq. (i), we get `(K_(2))/(K_(1))=` Antilog `[(3.344xx10^(3))/(8.314xx2.303xx400)]=2.73` `:. K_(2)=2.73K_(1)` Therefore, the rate will increase by `2.73` times.