At what temperature rate becomes double than at 300 K ? Given lnk `=10 - (69(KJ))/(RT)`

Options

A. 329
B. 307.7
C. 292.03
D. 323.5

Answer

To solve the problem of finding the temperature at which the rate constant becomes double than at 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given equation**:
We have the equation for the natural logarithm of the rate constant:
\[ \ln k = 10 - \frac{69 \, \text{kJ}}{RT} \]
Here, \( R \) is the universal gas constant.

2. **Convert the activation energy to the correct units**:
The activation energy \( E_a = 69 \, \text{kJ} = 69000 \, \text{J} \).

3. **Set up the Arrhenius equation**:
From the Arrhenius equation, we know:
\[ \ln \frac{k_2}{k_1} = -\frac{E_a}{RT} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Since we are looking for the temperature \( T_2 \) where the rate constant \( k_2 \) is double \( k_1 \), we can set \( k_2 = 2k_1 \).

4. **Substitute values into the equation**:
This gives us:
\[ \ln \frac{2k_1}{k_1} = \ln 2 \]
Therefore:
\[ \ln 2 = -\frac{69000}{R} \left( \frac{1}{T_2} - \frac{1}{300} \right) \]

5. **Substituting the value of \( R \)**:
Using \( R = 8.314 \, \text{J/(mol K)} \):
\[ \ln 2 = -\frac{69000}{8.314} \left( \frac{1}{T_2} - \frac{1}{300} \right) \]

6. **Calculate \( \ln 2 \)**:
The value of \( \ln 2 \) is approximately \( 0.693 \).

7. **Rearranging the equation**:
Rearranging gives:
\[ 0.693 = -\frac{69000}{8.314} \left( \frac{1}{T_2} - \frac{1}{300} \right) \]

8. **Calculate the left side**:
Calculate \( -\frac{69000}{8.314} \):
\[ -\frac{69000}{8.314} \approx -8307.76 \]
Thus:
\[ 0.693 = 8307.76 \left( \frac{1}{T_2} - \frac{1}{300} \right) \]

9. **Solving for \( \frac{1}{T_2} \)**:
Rearranging gives:
\[ \frac{1}{T_2} - \frac{1}{300} = \frac{0.693}{8307.76} \]
Calculate the right side:
\[ \frac{0.693}{8307.76} \approx 0.0000834 \]
Therefore:
\[ \frac{1}{T_2} = 0.0000834 + \frac{1}{300} \]

10. **Finding \( \frac{1}{300} \)**:
Calculate \( \frac{1}{300} \):
\[ \frac{1}{300} \approx 0.003333 \]
Thus:
\[ \frac{1}{T_2} = 0.0000834 + 0.003333 \approx 0.003416 \]

11. **Calculating \( T_2 \)**:
Finally, take the reciprocal to find \( T_2 \):
\[ T_2 \approx \frac{1}{0.003416} \approx 292.03 \, \text{K} \]

### Conclusion:
The temperature at which the rate constant becomes double than at 300 K is approximately \( 292.03 \, \text{K} \).