Two reactions `R_(1) and R_(2)` have identical pre - exponential factors. Activations enery of `R_(1)` exceeds that of `R_(2)` by 10 kJ `mol_(-1)` . If `k_(1) and k_(2)` are rate constants for rate constants for reactions `R_(1) and R_(2)`
respectively at 300k , then In `(k_(2)/k_(1))`is equal to `(R=8.314 J mol^(-1)K^(-1))`

Options

A. 4
B. 8
C. 12
D. 6

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constant \( k \) of a reaction to its activation energy \( E_a \) and the temperature \( T \):

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

where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J mol\(^{-1}\) K\(^{-1}\)),
- \( T \) is the temperature in Kelvin.

### Step 1: Write the Arrhenius equations for both reactions

For reaction \( R_1 \):
\[ k_1 = A e^{-\frac{E_{a1}}{RT}} \]

For reaction \( R_2 \):
\[ k_2 = A e^{-\frac{E_{a2}}{RT}} \]

### Step 2: Relate the activation energies

We know that the activation energy of \( R_1 \) exceeds that of \( R_2 \) by 10 kJ/mol:
\[ E_{a1} = E_{a2} + 10 \text{ kJ/mol} \]

### Step 3: Substitute the activation energies into the equations

Let \( E_{a2} = E_a \). Then:
\[ E_{a1} = E_a + 10 \text{ kJ/mol} \]

Now substituting into the equations:
\[ k_1 = A e^{-\frac{E_a + 10 \times 1000}{RT}} \]
\[ k_2 = A e^{-\frac{E_a}{RT}} \]

### Step 4: Find the ratio of the rate constants

To find the ratio \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{RT}}}{A e^{-\frac{E_a + 10 \times 1000}{RT}}} \]

The \( A \) cancels out:
\[ \frac{k_2}{k_1} = e^{-\frac{E_a}{RT}} \cdot e^{\frac{E_a + 10 \times 1000}{RT}} = e^{\frac{10 \times 1000}{RT}} \]

### Step 5: Substitute the values

Now, substituting \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \) and \( T = 300 \, \text{K} \):
\[ \frac{k_2}{k_1} = e^{\frac{10 \times 1000}{8.314 \times 300}} \]

Calculating the exponent:
\[ \frac{10 \times 1000}{8.314 \times 300} = \frac{10000}{2494.2} \approx 4.01 \]

Thus,
\[ \frac{k_2}{k_1} \approx e^{4.01} \]

### Step 6: Take the natural logarithm

Taking the natural logarithm of both sides:
\[ \ln\left(\frac{k_2}{k_1}\right) \approx 4.01 \]

### Final Answer

Thus, we can conclude that:
\[ \ln\left(\frac{k_2}{k_1}\right) \approx 4 \]