The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A

Options

A. (a) 4.92K
B. (b) 9.84K
C. (c) 19.67K
D. (d) 2.45K

Answer

To solve the problem, we need to analyze the information given about reactions A and B and apply the Arrhenius equation.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- For reaction A:
- Initial temperature \( T_1 = 300 \, K \)
- Final temperature \( T_2 = 310 \, K \)
- The rate doubles when the temperature increases from \( 300 \, K \) to \( 310 \, K \).
- For reaction B:
- Initial temperature \( T_1 = 300 \, K \)
- We need to find the temperature increase \( \Delta T \) such that the rate doubles.
- The activation energy \( E_{A,B} \) of reaction B is twice that of reaction A, i.e., \( E_{A,B} = 2E_{A,A} \).

2. **Apply the Arrhenius Equation**:
The Arrhenius equation relates the rate constant \( k \) to temperature \( T \) and activation energy \( E_A \):
\[ k = A e^{-E_A/(RT)} \]
Taking the logarithm of the ratio of rate constants at two different temperatures gives:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_A}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

3. **For Reaction A**:
Since the rate doubles:
\[ k_2 = 2k_1 \implies \log \left( \frac{k_2}{k_1} \right) = \log(2) \]
Substituting into the equation:
\[ \log(2) = \frac{E_{A,A}}{2.303 R} \left( \frac{1}{300} - \frac{1}{310} \right) \]

4. **Calculate the Right Side**:
Calculate \( \frac{1}{300} - \frac{1}{310} \):
\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \]
Thus, we have:
\[ \log(2) = \frac{E_{A,A}}{2.303 R} \cdot \frac{1}{9300} \]

5. **For Reaction B**:
For reaction B, we want the rate to double:
\[ k_2 = 2k_1 \implies \log \left( \frac{k_2}{k_1} \right) = \log(2) \]
The equation becomes:
\[ \log(2) = \frac{2E_{A,A}}{2.303 R} \left( \frac{1}{300} - \frac{1}{T_2} \right) \]

6. **Set Up the Ratio**:
Now, we can set up the ratio of the two equations:
\[ \frac{\log(2)}{\log(2)} = \frac{2E_{A,A}}{E_{A,A}} \cdot \frac{\frac{1}{300} - \frac{1}{T_2}}{\frac{1}{300} - \frac{1}{310}} \]
Simplifying gives:
\[ 1 = 2 \cdot \frac{\frac{1}{300} - \frac{1}{T_2}}{\frac{1}{300} - \frac{1}{310}} \]

7. **Cross Multiply and Solve for \( T_2 \)**:
\[ \frac{1}{300} - \frac{1}{T_2} = \frac{1}{2} \left( \frac{1}{300} - \frac{1}{310} \right) \]
This leads to:
\[ \frac{1}{T_2} = \frac{1}{300} - \frac{1}{2} \cdot \left( \frac{1}{300} - \frac{1}{310} \right) \]
Solving this will yield \( T_2 \).

8. **Calculate \( T_2 \)**:
After solving, we find:
\[ T_2 \approx 304.92 \, K \]

9. **Determine the Temperature Increase**:
The increase in temperature \( \Delta T \) is:
\[ \Delta T = T_2 - T_1 = 304.92 \, K - 300 \, K = 4.92 \, K \]

### Final Answer:
The temperature of reaction B should be increased by approximately **4.92 K**.