Question

A first-order reaction is carried out at two different temperatures of \(300\,K\) and \(310\,K\), where the rate constants are \(k_1\) and \(k_2\) respectively. The activation energies of the reaction in absence and presence of catalyst are \(E_{a1}\) and \(E_{a2}\) respectively. Which of the following options is correct?

• A. \(k_1 \gt k_2,\; E_{a1}\gt E_{a2}\)
• B. \(k_1 \lt k_2,\; E_{a1}\lt E_{a2}\)
• C. \(k_1 \lt k_2,\; E_{a1}\gt E_{a2}\)
• D. \(k_1 \lt k_2,\; E_{a1}=E_{a2}\)

Hint:

\[ \text{Higher Temperature} \Rightarrow \text{Higher } k \] \[ \text{Catalyst} \Rightarrow \text{Lower } E_a \] A catalyst changes the rate of reaction but does not change the equilibrium constant.

Answer 1

The Correct Option is C

Step 1: Bring in the Arrhenius idea.
The rate constant follows $k = A e^{-E_a/RT}$. So $k$ depends on temperature and on activation energy.

Step 2: Effect of temperature.
Raising the temperature makes the exponential term larger, so $k$ goes up.

Step 3: Compare $k_1$ and $k_2$.
$k_1$ is at $300\,K$ and $k_2$ is at the higher $310\,K$, so $k_2 > k_1$, that is $k_1 < k_2$.

Step 4: Role of a catalyst.
A catalyst gives the reaction a new path with a lower energy barrier, so it lowers the activation energy.

Step 5: Compare activation energies.
$E_{a1}$ is without catalyst and $E_{a2}$ is with catalyst, so $E_{a2}$ is smaller, that is $E_{a1} > E_{a2}$.

Step 6: Combine the two results.
Putting them together gives $k_1 < k_2$ and $E_{a1} > E_{a2}$.
\[ \boxed{k_1 < k_2,\ E_{a1} > E_{a2}} \]