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

148 times

Catalysts provide the surface area for the reaction, and when the surface area for the reaction is increased, more molecules come together and the rate of reaction increases. The catalyst itself does not get consumed in the reaction and is left at the end of the reaction as it is.

Any reaction proceeds by consuming some amount of activation energy. Catalysts provide a new reaction pathway, in which activation energy is lowered compared to the reaction where the catalyst is not used. A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller energy barrier. The relation of activation energy and rate constant is given by the Arrhenius equation,

\(k = Ae^{\frac{−E_a}{RT}}\)

Where k is the rate constant,

E_a  is the activation energy,

R is the universal gas constant=2cal

T is the temperature in Kelvin= 27^∘C +273 = 300K

When compared with the rate of reaction, with and without the use of a catalyst, the expression will be written as,

\(\frac{k_1}{k_2} = \frac{e^{\frac{-E_{a_1}}{RT}}}{e^{\frac{-E_{a_2}}{RT}}}\)

Where \(E_{a_1}\)  = activation energy without the use of catalyst

\(E_{a_2}\) = activation energy with the use of catalyst

\(k_1\) = rate of reaction without the use of catalyst

\(k_2\)= rate of reaction with the use of catalyst

\(\frac{k_1}{k_2} = e^{\frac{E_{a_1} − E_{a_2}}{RT}}\)

Given in the question value of the difference between the two activation energies is 3kcal, putting the values in the equation, we get,

\(\frac{k_1}{k_2} = e^{\frac{3000}{2 × 300}}\)

⇒ \(\frac{k_1}{k_2} = e^{5}\)

⇒ \(k_2 = 148k_1\)