It is generalized that a `10^(@)C` increases in temperature casues the rate of reaction to double. Applied to a reaction at `295 K`, what is the value of `E_(a)`?

Options

A. `120 "kcal mol"^(-1)`
B. `1200 "kcal mol"^(-1)`
C. `1.2 "kcal mol"^(-1)`
D. `12 "kcal mol"^(-1)`

Explanation

To find the activation energy \( E_a \) for a reaction where a \( 10^\circ C \) increase in temperature causes the rate of reaction to double, we can use the Arrhenius equation and the relationship between rate constants and temperature.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that a \( 10^\circ C \) increase in temperature doubles the rate of reaction. This means that if we denote the rate constants at two temperatures \( k_1 \) (at \( 295 \, K \)) and \( k_2 \) (at \( 305 \, K \)), we have:
\[ \frac{k_2}{k_1} = 2 \]

2. **Using the Arrhenius Equation**:
- The Arrhenius equation relates the rate constants to the activation energy \( E_a \):
\[ k = A e^{-\frac{E_a}{RT}} \]
- Taking the ratio of the rate constants at two temperatures gives:
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{RT_2}}}{A e^{-\frac{E_a}{RT_1}}} = e^{-\frac{E_a}{RT_2} + \frac{E_a}{RT_1}} = e^{\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \]

3. **Taking the Logarithm**:
- Taking the natural logarithm of both sides, we get:
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
- Since \( \frac{k_2}{k_1} = 2 \), we have:
\[ \ln(2) = \frac{E_a}{R} \left(\frac{1}{295} - \frac{1}{305}\right) \]

4. **Calculating the Temperature Difference**:
- Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \):
\[ \frac{1}{295} - \frac{1}{305} = \frac{305 - 295}{295 \times 305} = \frac{10}{295 \times 305} \]

5. **Substituting Values**:
- Now substituting this back into the equation:
\[ \ln(2) = \frac{E_a}{R} \left(\frac{10}{295 \times 305}\right) \]
- The universal gas constant \( R \) in kilocalories is \( 2 \times 10^{-3} \, \text{kcal/mol K} \).

6. **Rearranging for \( E_a \)**:
- Rearranging gives:
\[ E_a = \frac{\ln(2) \cdot R \cdot 295 \cdot 305}{10} \]

7. **Calculating \( E_a \)**:
- Plugging in the values:
\[ E_a = \frac{\ln(2) \cdot (2 \times 10^{-3}) \cdot 295 \cdot 305}{10} \]
- Calculate \( \ln(2) \approx 0.693 \):
\[ E_a \approx \frac{0.693 \cdot (2 \times 10^{-3}) \cdot 295 \cdot 305}{10} \]
- Performing the multiplication and division will yield \( E_a \approx 12.4 \, \text{kcal/mol} \).

### Final Answer:
The activation energy \( E_a \) is approximately \( 12 \, \text{kcal/mol} \).