The rate constant `(K')` of one reaction is double of the rate constant (K") of another reaction. Then the relationship between the corresponding activation energies of the two reactions `(E_(a)^(') "and" E_(a)^(''))` will be

Options

A. `E_(a)^(') gt E_(a)^(")`
B. `E_(a)^(') = E_(a)^(")`
C. `E_(a)^(') lt E_(a)^(")`
D. `E_(a)^(') lt 4E_(a)^(")`

Answer

To solve the problem, we need to establish the relationship between the activation energies of two reactions based on their rate constants using the Arrhenius equation.

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
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.

2. **Relate the Rate Constants**:
We are given that the rate constant \( k' \) of one reaction is double that of another reaction \( k'' \):
\[ k' = 2k'' \]

3. **Express the Rate Constants Using the Arrhenius Equation**:
For the first reaction:
\[ k' = A' e^{-\frac{E_a'}{RT}} \]
For the second reaction:
\[ k'' = A'' e^{-\frac{E_a''}{RT}} \]

4. **Substituting the Relationship**:
Substituting \( k'' \) into the equation for \( k' \):
\[ A' e^{-\frac{E_a'}{RT}} = 2 (A'' e^{-\frac{E_a''}{RT}}) \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{A'}{A''} = 2 e^{-\frac{E_a' - E_a''}{RT}} \]

6. **Analyzing the Exponential Term**:
Since \( e^{-\frac{E_a' - E_a''}{RT}} \) is always positive, for the left side to equal \( 2 \) (which is greater than \( 1 \)), the term \( -\frac{E_a' - E_a''}{RT} \) must be negative. This implies:
\[ E_a' - E_a'' < 0 \quad \Rightarrow \quad E_a' < E_a'' \]

7. **Conclusion**:
Therefore, the relationship between the activation energies is:
\[ E_a' < E_a'' \]

### Final Answer:
The activation energy corresponding to the rate constant \( k' \) is less than that corresponding to \( k'' \):
\[ E_a' < E_a'' \]