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
Text Solution
Verified by Experts
The correct Answer is:
A
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.
Given : AhArrB+C, DeltaH=-"10 kcals" , the energy of activation of backward reaction is "15 kcal mol"^(-1) . If the energy of activation of forward reaction in the presence of a catalyst is "3 kcal mol"^(-1) the catalyst will increase the rate of reaction at 300 K by the number of times equal to
Consider the reaction A to 2B + C, Delta H = -15 kcal . The energy of activation of backward reaction is 20 kcal mol^(-1) . In presence of catalyst, the energy of activation of forward reaction is 3 kcal mol^(-1) .At 400 K the catalystcauses the rate of the forward reaction to increase by the number of times equal to -
