The rate constant `(K_(1))` of one reaction is found to be double that of the rate constant of `(K_(2))` another reaction. Then the relationship - ' between the corresponding activation energies of two reactions `(E_(1) "and" E_(2))` can be represented.

Options

A. `E_(1) gt E_(2)`
B. `E_(1) lt E_(2)`
C. `E_(1) = E_(2)`
D. `E_(1) = 4E_(2)`

Explanation

To find the relationship between the activation energies \(E_1\) and \(E_2\) of two reactions with rate constants \(K_1\) and \(K_2\), where \(K_1 = 2K_2\), we can use the Arrhenius equation, which relates the rate constant to the activation energy.

### Step-by-Step Solution:

1. **Write the Arrhenius Equation for Both Reactions:**
The Arrhenius equation is given by:
\[ K = A e^{-\frac{E}{RT}} \]
For the first reaction:
\[ K_1 = A_1 e^{-\frac{E_1}{RT}} \]
For the second reaction:
\[ K_2 = A_2 e^{-\frac{E_2}{RT}} \]

2. **Substitute the Given Relationship:**
According to the problem, \(K_1 = 2K_2\). Thus, we can write:
\[ A_1 e^{-\frac{E_1}{RT}} = 2 \left(A_2 e^{-\frac{E_2}{RT}}\right) \]

3. **Rearranging the Equation:**
We can rearrange this equation to isolate the exponential terms:
\[ \frac{A_1}{A_2} = 2 e^{-\frac{E_1 - E_2}{RT}} \]

4. **Taking Natural Logarithm:**
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{A_1}{A_2}\right) = \ln(2) - \frac{E_1 - E_2}{RT} \]

5. **Rearranging for Activation Energies:**
Rearranging this equation, we can express the difference in activation energies:
\[ \frac{E_1 - E_2}{RT} = \ln(2) - \ln\left(\frac{A_1}{A_2}\right) \]
This implies:
\[ E_1 - E_2 = RT \left(\ln(2) - \ln\left(\frac{A_1}{A_2}\right)\right) \]

6. **Conclusion on Activation Energies:**
Since \(RT \ln(2)\) is a positive term, we can conclude that:
\[ E_2 > E_1 \]
This means the activation energy for the second reaction is greater than that of the first reaction.

### Final Relationship:
\[ E_2 > E_1 \]