For the reaction of `H_(2)` with `I_(2)`, the constant is `2.5 xx 10^(-4) dm^(3) mol^(-1)s^(-1)` at `327^(@)C` and `1.0 dm^(3) mol^(-1) s^(-1)` at `527^(@)C`. The activation energy for the reaction is `1.65 xx 10x J//"mole"`. The numerical value of x is __________. `(R = 8314JK^(-1) mol^(-1))`

Options

A. 150
B. 72
C. 166
D. 59

Explanation

To solve the problem, we will use the Arrhenius equation in the form of the logarithmic relationship between the rate constants and temperatures. The equation we will use is:

\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 1: Convert temperatures from Celsius to Kelvin
Given:
- \( T_1 = 327^\circ C = 327 + 273 = 600 \, K \)
- \( T_2 = 527^\circ C = 527 + 273 = 800 \, K \)

### Step 2: Identify the rate constants
Given:
- \( k_1 = 2.5 \times 10^{-4} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \)
- \( k_2 = 1.0 \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \)

### Step 3: Substitute values into the logarithmic equation
Now, we can substitute the values into the equation:

\[ \log \frac{1.0}{2.5 \times 10^{-4}} = \frac{E_a}{2.303 \times 8314} \left( \frac{1}{600} - \frac{1}{800} \right) \]

### Step 4: Calculate the left side
Calculating the left side:

\[ \log \frac{1.0}{2.5 \times 10^{-4}} = \log(4000) \approx 3.602 \]

### Step 5: Calculate the right side
Now, calculate \( \left( \frac{1}{600} - \frac{1}{800} \right) \):

\[ \frac{1}{600} - \frac{1}{800} = \frac{4 - 3}{2400} = \frac{1}{2400} \]

Now substitute this back into the equation:

\[ 3.602 = \frac{E_a}{2.303 \times 8314} \cdot \frac{1}{2400} \]

### Step 6: Rearranging to find \( E_a \)
Rearranging gives:

\[ E_a = 3.602 \times 2.303 \times 8314 \times 2400 \]

### Step 7: Calculate \( E_a \)
Calculating the right side:

1. Calculate \( 2.303 \times 8314 \approx 19115.622 \)
2. Now calculate \( 3.602 \times 19115.622 \approx 68866.68 \)
3. Finally, multiply by \( 2400 \):

\[ E_a \approx 68866.68 \times 2400 \approx 165000000 \, \text{J/mol} = 165 \times 10^3 \, \text{J/mol} \]

### Step 8: Express \( E_a \) in the form \( 1.65 \times 10^x \)
We can express \( 165000 \) as \( 1.65 \times 10^5 \, \text{J/mol} \).

### Conclusion
Thus, the numerical value of \( x \) is:

\[ \boxed{5} \]