The activation energy and enthalpy of chemisorption of oxygen on a metal surface is `37.3kJ mol^(-1)` and `-72.1kJ mol^(-1)` At a certain pressure, the rate constant for chemisorption is `1.2xx10^(-3)` at `318 K` . What will be the value of the rate constant at `308K` ?

Options

A. `7.6 xx 10^(-4)s^(-1)`
B. `1.6xx10^(-6)s^(-1)`
C. `7.6xx10^(-2)s^(-1)`
D. `1.6xx10^(-5)s^(-1)`

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constants at two different temperatures to the activation energy. The formula we will use is:

\[ \log \left( \frac{K_2}{K_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Where:
- \( K_1 \) is the rate constant at temperature \( T_1 \)
- \( K_2 \) is the rate constant at temperature \( T_2 \)
- \( E_a \) is the activation energy
- \( R \) is the universal gas constant
- \( T_1 \) and \( T_2 \) are the absolute temperatures in Kelvin

### Step-by-step Solution:

1. **Identify the given values:**
- Activation energy, \( E_a = 37.3 \, \text{kJ/mol} = 37300 \, \text{J/mol} \)
- Rate constant at \( T_1 = 318 \, \text{K} \), \( K_1 = 1.2 \times 10^{-3} \)
- Temperature \( T_2 = 308 \, \text{K} \)
- Universal gas constant, \( R = 8.314 \, \text{J/(mol K)} \)

2. **Substitute the values into the equation:**
\[ \log \left( \frac{K_2}{1.2 \times 10^{-3}} \right) = \frac{37300}{2.303 \times 8.314} \left( \frac{1}{318} - \frac{1}{308} \right) \]

3. **Calculate the right-hand side:**
- First, calculate \( \frac{1}{318} - \frac{1}{308} \):
\[ \frac{1}{318} \approx 0.003145 \quad \text{and} \quad \frac{1}{308} \approx 0.003247 \]
\[ \frac{1}{318} - \frac{1}{308} \approx 0.003145 - 0.003247 = -0.000102 \]

- Now calculate \( \frac{37300}{2.303 \times 8.314} \):
\[ 2.303 \times 8.314 \approx 19.188 \]
\[ \frac{37300}{19.188} \approx 1945.25 \]

- Now multiply by \( -0.000102 \):
\[ 1945.25 \times -0.000102 \approx -0.198 \]

4. **Combine the values:**
\[ \log K_2 - \log (1.2 \times 10^{-3}) = -0.198 \]
- Calculate \( \log (1.2 \times 10^{-3}) \):
\[ \log (1.2) \approx 0.079 \quad \text{and} \quad \log (10^{-3}) = -3 \]
\[ \log (1.2 \times 10^{-3}) \approx 0.079 - 3 = -2.921 \]

5. **Substitute back to find \( \log K_2 \):**
\[ \log K_2 = -2.921 - 0.198 = -3.119 \]

6. **Convert from logarithmic form to find \( K_2 \):**
\[ K_2 = 10^{-3.119} \approx 7.6 \times 10^{-4} \]

### Final Answer:
The value of the rate constant at \( 308 \, \text{K} \) is approximately \( K_2 = 7.6 \times 10^{-4} \, \text{s}^{-1} \).

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