The energy of activation for a reaction is `100 (kJ) mol^(-1)`. Presence of a catalyst lowers the energy of activation by 75 %. What will be the effect on the rate of reaction at `(20^oC)`, other things being equal.

In presence of a catalyst, the activation energy is lowered by 3 kcal at 27^@C . Hence, the rate of reaction will increase by:

The activation energy of a reaction is zero. The rate constant of this reaction

The activation energy of a reaction is 24.0 kcal mol^(-1) at 27^(@)C and the presence of catalyst changes its activation energy to one-fourth at the same temperature. The approximate ratio of rate in the presence of catalyst to rate in the absence of catalyst will be (use R = 2 cal mol^(-1) K^(-1) ).

A catalyst lowers the activation energy of the forward reaction by 10 kJ "mol"^(-1) ,What effect it has on the activation energy of the backward reaction?

Assuming that energy of activation for most of the reactions is 52 (kJ), what conclusion you draw-about the effect of temperature on the rate of a reaction [Based on Arrhenius equation]

The activation energy of a reaction is zero. Will the rate constant of the reaction depend upon temperature? Give reason.