The rate constants `k_(1)` and `k_(2)` of two reactions are in the ratio `2:1`. The corresponding energies of acativation of the two reaction will be related by

Options

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

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constant of a reaction to its activation energy. The Arrhenius 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 equation for both reactions:**
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. **Use the given ratio of rate constants:**
We are given that:
\[ \frac{k_1}{k_2} = \frac{2}{1} \]
This implies:
\[ k_1 = 2k_2 \]

3. **Substitute the Arrhenius equations into the ratio:**
\[ \frac{A_1 e^{-\frac{E_1}{RT}}}{A_2 e^{-\frac{E_2}{RT}}} = 2 \]

4. **Rearranging the equation:**
\[ \frac{A_1}{A_2} e^{-\frac{E_1 - E_2}{RT}} = 2 \]

5. **Assuming equal pre-exponential factors (A1 = A2):**
If we assume that the pre-exponential factors are approximately equal (which is a common assumption), we can simplify the equation to:
\[ e^{-\frac{E_1 - E_2}{RT}} = 2 \]

6. **Taking the natural logarithm of both sides:**
\[ -\frac{E_1 - E_2}{RT} = \ln(2) \]
Rearranging gives:
\[ E_1 - E_2 = -RT \ln(2) \]

7. **Interpreting the result:**
Since \( \ln(2) \) is a positive value, this means:
\[ E_1 < E_2 \]
Therefore, we conclude that the activation energy for the second reaction (E2) is greater than that for the first reaction (E1).

### Conclusion:
The correct relationship between the activation energies is:
\[ E_2 > E_1 \]