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. Reaction 2 is slow compared to reaction 1
B. Reaction 2 is more exothermic than reaction 1
C. Reaction 2 is more endothermic than reaction 1
D. Reaction 2 is fast compared to reaction 1

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constants of reactions to their activation energies. 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 1: Write the Arrhenius equation for both reactions

For the reaction \( A \overset{k_1}{\rightarrow} B \):
\[ k_1 = A e^{-\frac{E_{a1}}{RT}} \quad \text{(1)} \]

For the reaction \( A \overset{k_2}{\rightarrow} C \):
\[ k_2 = A e^{-\frac{E_{a2}}{RT}} \quad \text{(2)} \]

### Step 2: Divide the two equations

Now, we will divide equation (1) by equation (2):

\[ \frac{k_1}{k_2} = \frac{A e^{-\frac{E_{a1}}{RT}}}{A e^{-\frac{E_{a2}}{RT}}} \]

The \( A \) terms cancel out:

\[ \frac{k_1}{k_2} = e^{-\frac{E_{a1}}{RT}} \cdot e^{\frac{E_{a2}}{RT}} \]

### Step 3: Simplify the exponent

Using the property of exponents that states \( e^a \cdot e^b = e^{a+b} \):

\[ \frac{k_1}{k_2} = e^{-\frac{E_{a1}}{RT} + \frac{E_{a2}}{RT}} = e^{\frac{E_{a2} - E_{a1}}{RT}} \]

### Step 4: Substitute the relationship between activation energies

Given that \( E_{a2} = 2E_{a1} \), we substitute this into the equation:

\[ \frac{k_1}{k_2} = e^{\frac{2E_{a1} - E_{a1}}{RT}} = e^{\frac{E_{a1}}{RT}} \]

### Step 5: Rearranging the equation

Now, we can express \( k_1 \) in terms of \( k_2 \):

\[ k_1 = k_2 e^{\frac{E_{a1}}{RT}} \]

### Conclusion

Thus, the relationship between \( k_1 \) and \( k_2 \) is:

\[ k_1 = k_2 e^{\frac{E_{a1}}{RT}} \]