Arrhenius studies the effect of temperature on the rate of a reaction and postulted that rate constant varies with temperature exponentially as `k=Ae^(E_(a)//RT)` . Thuis method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied.
If the rate of reaction doubles for `10^(@)C` rise of temperature form 290K to 300K, the activation energy of the reaction will be approximately :

Options

A. 40 Kcal `mol^(-1)`
B. 12 Kcal `mol^(-1)`
C. 60 Kcal `mol^(-1)`
D. 70 Kcal `mol^(-1)`

Answer

To solve the problem of finding the activation energy (Ea) of a reaction where the rate doubles for a temperature increase of 10°C (from 290 K to 300 K), we can use the Arrhenius equation and the relationship between rate constants and temperature.

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Set Up the Problem**:
We know that the rate of reaction doubles when the temperature increases from 290 K to 300 K. This means:
\[ k_2 = 2k_1 \]
where \( k_1 \) is the rate constant at 290 K and \( k_2 \) is the rate constant at 300 K.

3. **Apply the Arrhenius Equation**:
For the two temperatures, we can write:
\[ \ln k_1 = \ln A - \frac{E_a}{R \cdot 290} \]
\[ \ln k_2 = \ln A - \frac{E_a}{R \cdot 300} \]

4. **Subtract the Two Equations**:
By subtracting the first equation from the second, we get:
\[ \ln k_2 - \ln k_1 = -\frac{E_a}{R \cdot 300} + \frac{E_a}{R \cdot 290} \]
This simplifies to:
\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{290} - \frac{1}{300}\right) \]

5. **Substituting Known Values**:
Since \( k_2 = 2k_1 \):
\[ \ln 2 = \frac{E_a}{R} \left(\frac{1}{290} - \frac{1}{300}\right) \]
The difference in the fractions can be calculated as:
\[ \frac{1}{290} - \frac{1}{300} = \frac{300 - 290}{290 \cdot 300} = \frac{10}{87000} = \frac{1}{8700} \]

6. **Rearranging for Activation Energy**:
Now substituting this back into the equation:
\[ \ln 2 = \frac{E_a}{R} \cdot \frac{1}{8700} \]
Rearranging gives:
\[ E_a = R \cdot 8700 \cdot \ln 2 \]

7. **Using the Value of R**:
The gas constant \( R \) is approximately \( 2 \, \text{cal/mol K} \) when using kilocalories:
\[ E_a = 2 \cdot 8700 \cdot 0.693 \]

8. **Calculating the Activation Energy**:
\[ E_a = 2 \cdot 8700 \cdot 0.693 \approx 12058.2 \, \text{cal/mol} \]
Converting to kilocalories:
\[ E_a \approx 12.06 \, \text{kcal/mol} \approx 12 \, \text{kcal/mol} \]

### Final Answer:
The activation energy of the reaction is approximately **12 kcal/mol**.