Question

(i) and (ii) are two chemical reactions carried out at TK.
\[ (i) \, X \rightarrow Y + W \, \text{with rate constant} \, k_1 \] \[ (ii) \, X \rightarrow Z + W \, \text{with rate constant} \, k_2 \]
The activation energy for reaction (ii) is 3 times that of reaction (i). What is the expression for \( k_1 \)?

• A. \( k_1 = -2k_2 e^{\frac{2E_a}{RT}} \)
• B. \( k_1 = k_2 e^{\frac{E_a}{RT}} \)
• C. \( k_1 = 2k_2 e^{\frac{E_a}{RT}} \)
• D. \( k_1 = \frac{k_2}{2} e^{\frac{E_a}{RT}} \)

Hint:

The Arrhenius equation relates the rate constant to the activation energy. When the activation energy for one reaction is related to another, the rate constants for the reactions can be compared using the ratio of exponentials.

Answer 1

The Correct Option is B

Step 1: Understand the relationship between rate constants and activation energy.
The rate constant (\( k \)) for a chemical reaction is related to the activation energy (\( E_a \)) by the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
Where:
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin.

Step 2: Use the relationship between the activation energies for reactions (i) and (ii).
The problem states that the activation energy for reaction (ii) is 3 times that of reaction (i):
\[ E_a (\text{ii}) = 3E_a (\text{i}) \]
From the Arrhenius equation, we can express the rate constants for the two reactions:
\[ k_1 = A e^{-\frac{E_a (\text{i})}{RT}}, \quad k_2 = A e^{-\frac{E_a (\text{ii})}{RT}} = A e^{-\frac{3E_a (\text{i})}{RT}} \]
Dividing \( k_2 \) by \( k_1 \) gives: \[ \frac{k_2}{k_1} = \frac{e^{-\frac{3E_a (\text{i})}{RT}}}{e^{-\frac{E_a (\text{i})}{RT}}} = e^{\frac{2E_a (\text{i})}{RT}} \]
Therefore:
\[ k_1 = k_2 e^{\frac{E_a (\text{i})}{RT}} \]

Step 3: Conclusion.
Thus, the expression for \( k_1 \) is: \[ k_1 = k_2 e^{\frac{E_a}{RT}} \]
This corresponds to option (B).