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))`

Explanation

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

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

### Step 1: Identify the given values
- \( k_1 = 2.5 \times 10^{-4} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \) at \( T_1 = 327^\circ C = 600 \, K \)
- \( k_2 = 1.0 \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \) at \( T_2 = 527^\circ C = 800 \, K \)
- Activation energy \( E_a = 1.65 \times 10^x \, \text{J/mol} \)
- Gas constant \( R = 8314 \, \text{J/(K mol)} \)

### Step 2: Substitute the values into the Arrhenius equation
Substituting the known values into the equation, we have:

\[ \log \left( \frac{1.0}{2.5 \times 10^{-4}} \right) = \frac{1.65 \times 10^x}{2.303 \times 8314} \left( \frac{1}{600} - \frac{1}{800} \right) \]

### Step 3: Calculate \( \frac{1}{600} - \frac{1}{800} \)
Calculating the difference:

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

### Step 4: Calculate \( \log \left( \frac{1.0}{2.5 \times 10^{-4}} \right) \)
Calculating the logarithm:

\[ \log \left( \frac{1.0}{2.5 \times 10^{-4}} \right) = \log(4000) = \log(4 \times 10^3) = \log(4) + 3 \approx 0.602 + 3 = 3.602 \]

### Step 5: Substitute back into the equation
Now substituting back into the equation:

\[ 3.602 = \frac{1.65 \times 10^x}{2.303 \times 8314} \times \frac{1}{2400} \]

### Step 6: Solve for \( E_a \)
Calculating the denominator:

\[ 2.303 \times 8314 \approx 19116.662 \]

Now rearranging the equation to solve for \( E_a \):

\[ E_a = 3.602 \times 19116.662 \times 2400 \]

Calculating this gives:

\[ E_a \approx 16500 \, \text{J/mol} \]

### Step 7: Relate \( E_a \) to \( 1.65 \times 10^x \)
We know \( E_a = 1.65 \times 10^x \), so we set:

\[ 1.65 \times 10^x = 16500 \]

### Step 8: Solve for \( x \)
Dividing both sides by 1.65:

\[ 10^x = \frac{16500}{1.65} \approx 10000 \]

This simplifies to:

\[ 10^x = 10^4 \]

Thus, \( x = 4 \).

### Final Answer
The numerical value of \( x \) is:

\[ \boxed{4} \]