Question

Which among the following is correct when energy of activation, $E_a$, of the catalyzed reaction decreases at constant temperature and for same concentration?

• A. $\dfrac{E_a}{RT}$ decreases
• B. $K$ decreases
• C. $e^{-E_a/RT}$ decreases
• D. $-\dfrac{E_a}{RT}$ decreases

Hint:

Catalyst lowers activation energy, increasing the rate constant without changing equilibrium constant.

Answer 1

The Correct Option is A

Step 1: Arrhenius equation.
The Arrhenius equation is given by:
\[ k = A e^{-E_a/RT} \]
Step 2: Effect of catalyst on activation energy.
A catalyst lowers the activation energy $E_a$ without affecting temperature or concentration.

Step 3: Analyze each term.
When $E_a$ decreases at constant $T$, the term $\dfrac{E_a}{RT}$ decreases directly.
As a result, the exponential term $e^{-E_a/RT}$ increases and the rate constant $k$ increases.

Step 4: Option analysis.
(A) $\dfrac{E_a}{RT}$ decreases — Correct
(B) $K$ decreases — Incorrect, rate constant increases
(C) $e^{-E_a/RT}$ decreases — Incorrect, it increases
(D) $-\dfrac{E_a}{RT}$ decreases — Incorrect, its value becomes less negative

Step 5: Conclusion.
Thus, the correct statement is (A) $\dfrac{E_a{RT}$ decreases}.