A catalyst lowers the activation energy for a certain reaction from 75 kJ to `25" kJ mol"^(–1)`. What will be the effect on the rate of reaction at `25^(@)C`, after things being equal.

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constant of a reaction to the activation energy and temperature. The equation is given by:

\[ k = A e^{-\frac{E_a}{RT}} \]

Where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor (frequency factor)
- \( E_a \) = activation energy (in Joules per mole)
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