For a gaseous reaction, following data is given:
`ArarrB, 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`

Explanation

To find the temperature at which the rate constants \( k_1 \) and \( k_2 \) are equal for the given gaseous reactions, we can follow these steps:

### Step-by-Step Solution:

1. **Write down the expressions for the rate constants:**
\[ k_1 = 10^{15} e^{-\frac{2000}{T}} \]
\[ k_2 = 10^{14} e^{-\frac{1000}{T}} \]

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 \( 10^{14} \):**
\[ 10^{1} e^{-\frac{2000}{T}} = e^{-\frac{1000}{T}} \]
This simplifies to:
\[ 10 = e^{-\frac{1000}{T} + \frac{2000}{T}} \]
or
\[ 10 = e^{\frac{1000}{T}} \]

4. **Take the natural logarithm of both sides:**
\[ \ln(10) = \frac{1000}{T} \]

5. **Rearrange to solve for \( T \):**
\[ T = \frac{1000}{\ln(10)} \]

6. **Calculate \( \ln(10) \):**
Using the approximate value \( \ln(10) \approx 2.303 \):
\[ T = \frac{1000}{2.303} \approx 434.2 \text{ K} \]

### Final Answer:
The temperature at which \( k_1 = k_2 \) is approximately **434.2 K**.

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