For the two gaseous reactions, following data are given `A rarr B, k_(1) =10^(10) e^(-20,000//T), C rarr D, k_(2) =10^(12) e^(-24,606//T)` the temperature at which `k_(1)` becomes equal to `k_(2)` is :

Options

A. 400 K
B. 1000 K
C. 800 K
D. 1500 K

Explanation

To find the temperature at which the rate constants \( k_1 \) and \( k_2 \) are equal, we will start by setting the two equations for the rate constants equal to each other:

1. **Write the equations for the rate constants:**
\[ k_1 = 10^{10} e^{-\frac{20000}{T}} \]
\[ k_2 = 10^{12} e^{-\frac{24606}{T}} \]

2. **Set \( k_1 \) equal to \( k_2 \):**
\[ 10^{10} e^{-\frac{20000}{T}} = 10^{12} e^{-\frac{24606}{T}} \]

3. **Divide both sides by \( 10^{10} \):**
\[ e^{-\frac{20000}{T}} = 10^{2} e^{-\frac{24606}{T}} \]

4. **Rearrange the equation:**
\[ e^{-\frac{20000}{T}} = 100 e^{-\frac{24606}{T}} \]

5. **Take the natural logarithm of both sides:**
\[ -\frac{20000}{T} = \ln(100) - \frac{24606}{T} \]

6. **Simplify the equation:**
\[ -\frac{20000}{T} + \frac{24606}{T} = \ln(100) \]
\[ \frac{4606}{T} = \ln(100) \]

7. **Calculate \( \ln(100) \):**
\[ \ln(100) = 2 \ln(10) = 2 \times 2.303 = 4.606 \]

8. **Substitute \( \ln(100) \) back into the equation:**
\[ \frac{4606}{T} = 4.606 \]

9. **Solve for \( T \):**
\[ T = \frac{4606}{4.606} \approx 1000 \text{ K} \]

Thus, the temperature at which \( k_1 \) becomes equal to \( k_2 \) is approximately **1000 K**.