The rate constant `k_(1)` and `k_(2)` for two different reactions are `10^(16) e^(-2000//T)` and `10^(15) e^(-1000//T)`, respectively. The temperature at which `k_(1) = k_(2)` is

Options

A. `(2000)/(2.303) K`
B. `2000 K`
C. `(1000)/(2.303)K`
D. `1000 K`

Explanation

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

\[ k_1 = 10^{16} e^{-\frac{2000}{T}} \]
\[ k_2 = 10^{15} e^{-\frac{1000}{T}} \]

### Step 1: Set the rate constants equal to each other
We set \( k_1 \) equal to \( k_2 \):

\[ 10^{16} e^{-\frac{2000}{T}} = 10^{15} e^{-\frac{1000}{T}} \]

### Step 2: Simplify the equation
Dividing both sides by \( 10^{15} \):

\[ 10 e^{-\frac{2000}{T}} = e^{-\frac{1000}{T}} \]

### Step 3: Rearrange the equation
Now we can rearrange the equation:

\[ 10 = e^{-\frac{1000}{T}} \cdot e^{\frac{2000}{T}} \]

This simplifies to:

\[ 10 = e^{\frac{1000}{T}} \]

### Step 4: Take the natural logarithm of both sides
Taking the natural logarithm (ln) of both sides:

\[ \ln(10) = \frac{1000}{T} \]

### Step 5: Solve for T
Now we can solve for \( T \):

\[ T = \frac{1000}{\ln(10)} \]

### Step 6: Substitute the value of \( \ln(10) \)
We know that \( \ln(10) \approx 2.303 \):

\[ T = \frac{1000}{2.303} \approx 434.2 \, \text{K} \]

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

### Final Answer:
\[ T \approx 434.2 \, \text{K} \]

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