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

Answer

To find the rate constant for the reverse reaction \(2HI \rightarrow H_2 + I_2\), we can follow these steps:

### Step 1: Understand the given reaction and its rate constant
The forward reaction is:
\[ H_2 + I_2 \rightarrow 2HI \]
The rate constant for this reaction is given as \(K_f = 49\).

### Step 2: Identify the reverse reaction
The reverse reaction is:
\[ 2HI \rightarrow H_2 + I_2 \]

### Step 3: Relate the rate constants of forward and reverse reactions
For a reaction and its reverse, the relationship between the rate constants is given by:
\[ K_r = \frac{1}{K_f} \]
where \(K_r\) is the rate constant for the reverse reaction and \(K_f\) is the rate constant for the forward reaction.

### Step 4: Calculate the rate constant for the reverse reaction
Substituting the given value of \(K_f\):
\[ K_r = \frac{1}{49} \]

### Step 5: Final answer
Thus, the rate constant for the reaction \(2HI \rightarrow H_2 + I_2\) is:
\[ K_r = \frac{1}{49} \]

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

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