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. 1000K
B. `(2000)/(2.303)K`
C. 2000K
D. `(1000)/(23.03)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 \( k_1 \) equal to \( k_2 \)

We set the two expressions equal to each other:

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

### Step 2: Divide both sides by \( 10^{15} \)

Dividing both sides by \( 10^{15} \):

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

This simplifies to:

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

### Step 3: Rearrange the equation

Next, we can rearrange the equation:

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

This can be rewritten as:

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

### Step 4: Take the natural logarithm of both sides

Taking the natural logarithm (ln) of both sides gives:

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

### Step 5: Solve for \( T \)

Now, we can solve for \( T \):

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

### Step 6: Calculate \( \ln(10) \)

Using the approximation \( \ln(10) \approx 2.303 \):

\[ T \approx \frac{1000}{2.303} \]

### Step 7: Final calculation

Calculating this gives:

\[ T \approx 434.3 \, \text{K} \]

Thus, the temperature at which \( k_1 = k_2 \) is approximately \( 434.3 \, \text{K} \).

---