A reactant `(A)` forms two products
`A overset (k_(1))rarr B`, Activation energy `E_(a1)`
`A overset (k_(2))rarr C`, Activation energy `E_(a2)`
If `E_(a_(2)) = 2E_(a_(1))` then `k_(1)` and `k_(2)` are related as

Options

A. `k_(1)=2k_(2) e^(E_(a_(2))//RT)`
B. `k_(1)=k_(2) e^(E_(a_(1))//RT)`
C. `k_(2) =k_(1) e^(E_(a_(2))//RT)`
D. `k_(1)=Ak_(2)e^(E_(a_(1))//RT)`

Explanation

To solve the problem of how the rate constants \( k_1 \) and \( k_2 \) are related given the activation energies, we can follow these steps:

### Step 1: Write the Arrhenius Equation
The Arrhenius equation relates the rate constant \( k \) to the activation energy \( E_a \):
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant (8.314 J/(K·mol)),
- \( T \) is the temperature in Kelvin.

### Step 2: Write the Equations for \( k_1 \) and \( k_2 \)
For the two reactions, we can write:
\[ k_1 = A_1 e^{-\frac{E_{a1}}{RT}} \]
\[ k_2 = A_2 e^{-\frac{E_{a2}}{RT}} \]

### Step 3: Substitute the Given Relation
We are given that \( E_{a2} = 2E_{a1} \). Thus, we can substitute this into the equation for \( k_2 \):
\[ k_2 = A_2 e^{-\frac{2E_{a1}}{RT}} \]

### Step 4: Find the Ratio \( \frac{k_2}{k_1} \)
Now, we can find the ratio \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = \frac{A_2 e^{-\frac{2E_{a1}}{RT}}}{A_1 e^{-\frac{E_{a1}}{RT}}} \]
This simplifies to:
\[ \frac{k_2}{k_1} = \frac{A_2}{A_1} e^{-\frac{2E_{a1}}{RT} + \frac{E_{a1}}{RT}} = \frac{A_2}{A_1} e^{-\frac{E_{a1}}{RT}} \]

### Step 5: Rearranging the Equation
From the above equation, we can express \( k_2 \) in terms of \( k_1 \):
\[ k_2 = k_1 \cdot \frac{A_2}{A_1} e^{-\frac{E_{a1}}{RT}} \]

### Step 6: Conclusion
Thus, the relationship between \( k_1 \) and \( k_2 \) can be summarized as:
\[ k_2 = k_1 \cdot \frac{A_2}{A_1} e^{-\frac{E_{a1}}{RT}} \]