For a gaseous reaction , following date is given :
`A rarr B , k_1=10^(15)e^(-2000//T)`
`C rarrD , K _2 = 10^(14) e^(-1000//T)`
The temperature at which `k_1 = k_2` is

Options

A. 1000 K
B. 2000 K
C. 868.82 K
D. 434.2 K

Answer

To find the temperature at which \( k_1 = k_2 \), we start with the given equations for the rate constants \( k_1 \) and \( k_2 \):

1. **Given equations:**
\[ k_1 = 10^{15} e^{-\frac{2000}{T}} \]
\[ k_2 = 10^{14} e^{-\frac{1000}{T}} \]

2. **Set \( k_1 \) equal to \( k_2 \):**
\[ 10^{15} e^{-\frac{2000}{T}} = 10^{14} e^{-\frac{1000}{T}} \]

3. **Divide both sides by \( 10^{14} \):**
\[ 10 e^{-\frac{2000}{T}} = e^{-\frac{1000}{T}} \]

4. **Rearranging gives:**
\[ e^{-\frac{2000}{T}} = \frac{1}{10} e^{-\frac{1000}{T}} \]

5. **Taking the natural logarithm of both sides:**
\[ -\frac{2000}{T} = \ln\left(\frac{1}{10}\right) - \frac{1000}{T} \]

6. **Using the property of logarithms, \( \ln\left(\frac{1}{10}\right) = -\ln(10) \):**
\[ -\frac{2000}{T} = -\ln(10) - \frac{1000}{T} \]

7. **Multiply through by -1:**
\[ \frac{2000}{T} = \ln(10) + \frac{1000}{T} \]

8. **Rearranging gives:**
\[ \frac{2000}{T} - \frac{1000}{T} = \ln(10) \]
\[ \frac{1000}{T} = \ln(10) \]

9. **Solving for \( T \):**
\[ T = \frac{1000}{\ln(10)} \]

10. **Using the value of \( \ln(10) \approx 2.303 \):**
\[ T \approx \frac{1000}{2.303} \approx 434.2 \, \text{K} \]

Thus, the temperature at which \( k_1 = k_2 \) is approximately **434.2 K**.