Two reactions `R_(2) 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. `C_(6)H_(5)CH=underset(overset(|)(CH_(3)))(C)-CH_(2)-CH_(3)`
B. `C_(6)H_(5)CH_(2)-underset(overset(|)(CH_(3)))(C)=CHCH_(3)`
C. `C_(6)H_(5)CH_(2)-underset(overset(|)(OC_(2)H_(5)))overset(underset(|)(CH_(3)))(C)-CH_(2)CH_(3)`
D. `C_(6)H_(5)CH_(2)-underset(overset(|)(CH_(2)CH_(3)))(C)=CH_(2)`

Answer

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

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

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

### Step-by-Step Solution:

1. **Identify the Rate Constants**:
For the two reactions \( R_1 \) and \( R_2 \), we can express their rate constants as follows:
\[ k_1 = A e^{-\frac{E_{a1}}{RT}} \]
\[ k_2 = A e^{-\frac{E_{a2}}{RT}} \]

2. **Given Information**:
We know that the activation energy of \( R_1 \) exceeds that of \( R_2 \) by 10 kJ/mol:
\[ E_{a1} = E_{a2} + 10 \text{ kJ/mol} \]
Converting this to Joules (since \( R \) is in J):
\[ E_{a1} = E_{a2} + 10000 \text{ J/mol} \]

3. **Calculate the Ratio of Rate Constants**:
We want to find \( \ln\left(\frac{k_2}{k_1}\right) \):
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_{a2}}{RT}}}{A e^{-\frac{E_{a1}}{RT}}} \]
The \( A \) terms cancel out:
\[ \frac{k_2}{k_1} = e^{-\frac{E_{a2}}{RT} + \frac{E_{a1}}{RT}} = e^{\frac{E_{a1} - E_{a2}}{RT}} \]

4. **Substituting Activation Energies**:
Substitute \( E_{a1} - E_{a2} = 10000 \text{ J/mol} \):
\[ \frac{k_2}{k_1} = e^{\frac{10000}{RT}} \]

5. **Taking the Natural Logarithm**:
Now, taking the natural logarithm:
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{10000}{RT} \]

6. **Substituting Values**:
Substitute \( R = 8.314 \text{ J/mol K} \) and \( T = 300 \text{ K} \):
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{10000}{8.314 \times 300} \]

7. **Calculating the Result**:
Calculate the denominator:
\[ 8.314 \times 300 = 2494.2 \]
Now calculate:
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{10000}{2494.2} \approx 4.009 \]

### Final Answer:
Thus, the final result is:
\[ \ln\left(\frac{k_2}{k_1}\right) \approx 4.009 \]