On icreasing temperature from `200 K` to `220 K` rate of reaction A increases by `3` times and rate of reaction B increases by `9` times then correct relationship between activation energy of A and B is

Options

A. `E_(A) = 3 E_(B)`
B. `3 E_(A) = E_(B)`
C. `E_(B) = 2E_(A)`
D. `E_(A) = 2E_(B)`

Explanation

To find the relationship between the activation energies of reactions A and B based on the given information, we can use the Arrhenius equation. The equation relates the rate constants of reactions at two different temperatures to their activation energies.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The rate of reaction A increases by 3 times when the temperature is increased from 200 K to 220 K.
- The rate of reaction B increases by 9 times under the same temperature change.
- We need to find the relationship between the activation energies \( E_A \) and \( E_B \) of reactions A and B.

2. **Using the Arrhenius Equation**:
The Arrhenius equation in logarithmic form is given by:
\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
where:
- \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \),
- \( E \) is the activation energy,
- \( R \) is the universal gas constant.

3. **For Reaction A**:
- Given that the rate increases by 3 times, we have:
\[ \frac{k_2}{k_1} = 3 \implies \ln(3) = -\frac{E_A}{R} \left(\frac{1}{220} - \frac{1}{200}\right) \]
- Calculate \( \frac{1}{220} - \frac{1}{200} \):
\[ \frac{1}{220} - \frac{1}{200} = \frac{200 - 220}{220 \times 200} = \frac{-20}{44000} = -\frac{1}{2200} \]
- Substitute this back into the equation:
\[ \ln(3) = \frac{E_A}{R} \cdot \frac{1}{2200} \]
- Rearranging gives:
\[ E_A = R \cdot 2200 \cdot \ln(3) \quad \text{(Equation 1)} \]

4. **For Reaction B**:
- Given that the rate increases by 9 times, we have:
\[ \frac{k_2}{k_1} = 9 \implies \ln(9) = -\frac{E_B}{R} \left(\frac{1}{220} - \frac{1}{200}\right) \]
- Since \( \ln(9) = 2 \ln(3) \), we can write:
\[ 2 \ln(3) = \frac{E_B}{R} \cdot \frac{1}{2200} \]
- Rearranging gives:
\[ E_B = R \cdot 2200 \cdot 2 \ln(3) \quad \text{(Equation 2)} \]

5. **Comparing Activation Energies**:
- From Equation 1 and Equation 2:
\[ E_B = 2 \cdot E_A \]
- This indicates that the activation energy for reaction B is twice that of reaction A.

### Final Relationship:
\[ E_B = 2 E_A \]