For the given reaction:
`H_(2) + I_(2) rarr 2HI`
Given: `{:(T(K),1//T(K^(-1)),log k,),(769,1.3 xx 10^(-3),2.9,),(67,1.5 xx 10^(-3),1.1,):}`
The activation energy will be

Options

A. `41.4 kcal mol^(-1)`
B. `40 kcal mol^(-1)`
C. `-41.4 kcal mol^(-1)`
D. `-40 kcal mol^(-1)`

Explanation

To calculate the activation energy (Ea) for the reaction \( H_2 + I_2 \rightarrow 2HI \) using the provided data, we can use the Arrhenius equation in the logarithmic form. Here’s a step-by-step solution:

### Step 1: Identify the given data
We have the following data from the problem:
- \( T_1 = 667 \, K \) (Temperature 1)
- \( T_2 = 769 \, K \) (Temperature 2)
- \( \log K_1 = 1.1 \) (Rate constant at \( T_1 \))
- \( \log K_2 = 2.9 \) (Rate constant at \( T_2 \))

### Step 2: Calculate the difference in logarithm of rate constants
Using the values of \( \log K_1 \) and \( \log K_2 \):
\[ \log K_2 - \log K_1 = 2.9 - 1.1 = 1.8 \]

### Step 3: Use the Arrhenius equation
The Arrhenius equation in logarithmic form is given by:
\[ \log \frac{K_2}{K_1} = \frac{E_a}{2.303 R} \cdot \frac{T_2 - T_1}{T_1 T_2} \]
Where:
- \( R \) is the universal gas constant. In calories, \( R = 2 \, cal/(K \cdot mol) \).

### Step 4: Calculate \( T_2 - T_1 \)
\[ T_2 - T_1 = 769 - 667 = 102 \, K \]

### Step 5: Calculate \( T_1 T_2 \)
\[ T_1 T_2 = 667 \cdot 769 \]
Calculating this gives:
\[ T_1 T_2 = 513,763 \, K^2 \]

### Step 6: Substitute values into the Arrhenius equation
Now substituting the values into the Arrhenius equation:
\[ 1.8 = \frac{E_a}{2.303 \cdot 2} \cdot \frac{102}{513763} \]

### Step 7: Rearranging to solve for \( E_a \)
Rearranging gives:
\[ E_a = 1.8 \cdot 2.303 \cdot 2 \cdot \frac{513763}{102} \]

### Step 8: Calculate \( E_a \)
Calculating the above expression:
1. Calculate \( 2.303 \cdot 2 = 4.606 \)
2. Then, \( \frac{513763}{102} \approx 5036.4 \)
3. Now, \( E_a = 1.8 \cdot 4.606 \cdot 5036.4 \)

Calculating this gives:
\[ E_a \approx 41400 \, cal/mol = 41.4 \, kcal/mol \]

### Final Answer
Thus, the activation energy \( E_a \) is approximately \( 41.4 \, kcal/mol \).

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