Rate constant for a reaction `H_(2) + I_(2) rarr 2HI` is `49`, then rate constant for reaction `2HI rarr H_(2) + I_(2)` is

Options

A. `7`
B. `1//49`
C. `49`
D. `21`

Explanation

To find the rate constant for the reverse reaction \(2HI \rightarrow H_2 + I_2\), we can use the relationship between the rate constants of forward and reverse reactions.

### Step-by-Step Solution:

1. **Identify the Forward Reaction and Its Rate Constant**:
The forward reaction is given as:
\[ H_2 + I_2 \rightarrow 2HI \]
The rate constant for this reaction is provided as \(k_f = 49\).

2. **Understand the Reverse Reaction**:
The reverse reaction is:
\[ 2HI \rightarrow H_2 + I_2 \]
We need to find the rate constant for this reaction, denoted as \(k_r\).

3. **Use the Relationship Between Forward and Reverse Rate Constants**:
The relationship between the rate constants of the forward and reverse reactions is given by:
\[ k_f \cdot k_r = K \]
where \(K\) is the equilibrium constant for the reaction. For a simple case where the reaction is at equilibrium, we can state that:
\[ k_r = \frac{1}{k_f} \]

4. **Calculate the Rate Constant for the Reverse Reaction**:
Substituting the known value of \(k_f\):
\[ k_r = \frac{1}{49} \]

5. **Final Answer**:
Therefore, the rate constant for the reverse reaction \(2HI \rightarrow H_2 + I_2\) is:
\[ k_r = \frac{1}{49} \]

### Summary:
The rate constant for the reaction \(2HI \rightarrow H_2 + I_2\) is \(\frac{1}{49}\).