If reaction `A` and `B` are given with same temperature and same concentration but rate of `A` is double than `B`. Pre exponential factor is same for both the reaction then difference in activation energy `E_(A) - E_(B)` is ?

Options

A. `-RTln2`
B. `RT ln2`
C. 2RT
D. `(RT)/(2)`

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate of a reaction to its activation energy and temperature. The equation is given by:

\[ k = A e^{-\frac{E_a}{RT}} \]

where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Write the Arrhenius equations for both reactions:**
For reaction A:
\[ r_A = A e^{-\frac{E_A}{RT}} \quad \text{(Equation 1)} \]
For reaction B:
\[ r_B = A e^{-\frac{E_B}{RT}} \quad \text{(Equation 2)} \]

2. **Use the information given:**
We know that the rate of reaction A is double that of reaction B:
\[ r_A = 2 r_B \]

3. **Substitute the rates into the equation:**
Substitute \( r_A \) and \( r_B \) from Equations 1 and 2:
\[ 2 r_B = A e^{-\frac{E_A}{RT}} \quad \text{and} \quad r_B = A e^{-\frac{E_B}{RT}} \]

4. **Divide the equations:**
Dividing the equation for \( r_A \) by the equation for \( r_B \):
\[ \frac{r_A}{r_B} = \frac{A e^{-\frac{E_A}{RT}}}{A e^{-\frac{E_B}{RT}}} \]
Since \( A \) is the same for both reactions, it cancels out:
\[ 2 = e^{-\frac{E_A}{RT}} \cdot e^{\frac{E_B}{RT}} = e^{-\frac{E_A - E_B}{RT}} \]

5. **Take the natural logarithm of both sides:**
\[ \ln(2) = -\frac{E_A - E_B}{RT} \]

6. **Rearrange to find the difference in activation energies:**
\[ E_A - E_B = -RT \ln(2) \]

### Final Result:
The difference in activation energy \( E_A - E_B \) is given by:
\[ E_A - E_B = -RT \ln(2) \]