The rate constant of a reaction at 300 K is`1.6 xx 10^(-3) sec^(-1)` and at 310K it is `3.2 xx 10^(-3) sec^(-1)` the activation energy of the reaction approximately in kcals is

Options

A. 41609
B. 20-25
C. 30-40
D. 40-50

Explanation

To find the activation energy (Ea) of the reaction using the given rate constants at two different temperatures, we will use the Arrhenius equation in its logarithmic form:

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

### Step 1: Identify the given values
- \( k_1 = 1.6 \times 10^{-3} \, \text{s}^{-1} \) at \( T_1 = 300 \, \text{K} \)
- \( k_2 = 3.2 \times 10^{-3} \, \text{s}^{-1} \) at \( T_2 = 310 \, \text{K} \)
- The gas constant \( R = 1.987 \, \text{cal/(mol K)} \) (which can be approximated to \( 2 \, \text{cal/(mol K)} \))

### Step 2: Calculate the natural logarithm of the rate constants
\[ \ln \left( \frac{k_2}{k_1} \right) = \ln \left( \frac{3.2 \times 10^{-3}}{1.6 \times 10^{-3}} \right) = \ln(2) \]
Using the approximation \( \ln(2) \approx 0.693 \).

### Step 3: Calculate the difference in the reciprocal of the temperatures
\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{310} \]
Calculating this gives:
\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} \approx 0.0001075 \, \text{K}^{-1} \]

### Step 4: Substitute the values into the Arrhenius equation
Now we can substitute the values into the equation:
\[ \ln(2) = -\frac{E_a}{R} \left( \frac{1}{300} - \frac{1}{310} \right) \]
\[ 0.693 = -\frac{E_a}{1.987} \times 0.0001075 \]

### Step 5: Solve for activation energy (Ea)
Rearranging gives:
\[ E_a = -\frac{0.693 \times 1.987}{0.0001075} \]
Calculating this gives:
\[ E_a \approx 12892.5 \, \text{cal/mol} \]

### Step 6: Convert to kilocalories
To convert calories to kilocalories:
\[ E_a \approx \frac{12892.5}{1000} \approx 12.89 \, \text{kcal/mol} \]

### Final Answer
The activation energy of the reaction is approximately \( 12.89 \, \text{kcal/mol} \).

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