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 find the numerical value of \( x \) in the activation energy expression \( 1.65 \times 10^x \, \text{J/mol} \), we can use the Arrhenius equation, which relates the rate constant \( k \) to the temperature \( T \) and the activation energy \( E_a \):

\[ k = A e^{-\frac{E_a}{RT}} \]

Where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8314 J/(K·mol)),
- \( T \) is the absolute temperature in Kelvin.

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

### Step 2: Write the Arrhenius equation for both temperatures
We can write the Arrhenius equation for both temperatures:
\[ k_1 = A e^{-\frac{E_a}{RT_1}} \quad \text{(for } T_1\text{)} \]
\[ k_2 = A e^{-\frac{E_a}{RT_2}} \quad \text{(for } T_2\text{)} \]

### Step 3: Take the ratio of the two equations
Dividing the two equations gives:
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{RT_2}}}{A e^{-\frac{E_a}{RT_1}}} \]
This simplifies to:
\[ \frac{k_2}{k_1} = e^{-\frac{E_a}{RT_2} + \frac{E_a}{RT_1}} \]

### Step 4: Substitute the known values
Given:
- \( k_1 = 2.5 \times 10^{-4} \, \text{dm}^3 \text{mol}^{-1} \text{s}^{-1} \)
- \( k_2 = 1.0 \, \text{dm}^3 \text{mol}^{-1} \text{s}^{-1} \)

Substituting these values into the equation:
\[ \frac{1.0}{2.5 \times 10^{-4}} = e^{-\frac{E_a}{R \cdot 800} + \frac{E_a}{R \cdot 600}} \]
Calculating the left side:
\[ \frac{1.0}{2.5 \times 10^{-4}} = 4000 \]

### Step 5: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \ln(4000) = -\frac{E_a}{R \cdot 800} + \frac{E_a}{R \cdot 600} \]

### Step 6: Simplify the equation
Let’s express \( E_a \) in terms of \( R \):
\[ \ln(4000) = E_a \left( \frac{1}{R \cdot 600} - \frac{1}{R \cdot 800} \right) \]
\[ \ln(4000) = E_a \left( \frac{800 - 600}{R \cdot 600 \cdot 800} \right) \]
\[ \ln(4000) = E_a \left( \frac{200}{R \cdot 600 \cdot 800} \right) \]

### Step 7: Solve for \( E_a \)
Now, substituting \( R = 8314 \, \text{J/(K·mol)} \):
\[ E_a = \frac{\ln(4000) \cdot R \cdot 600 \cdot 800}{200} \]

### Step 8: Calculate \( \ln(4000) \)
Calculating \( \ln(4000) \):
\[ \ln(4000) \approx 8.294 \]

### Step 9: Substitute and calculate \( E_a \)
\[ E_a = \frac{8.294 \cdot 8314 \cdot 600 \cdot 800}{200} \]
Calculating this gives:
\[ E_a \approx 1.65 \times 10^x \]

### Step 10: Find \( x \)
After performing the calculations, we find that \( x \) is approximately \( 3 \).

Thus, the numerical value of \( x \) is:
\[ \boxed{3} \]