1. Nature of the given reaction

The reaction is free-radical iodination of methane:

$CH_4 + I_2 \xrightleftharpoons[h\nu]{} CH_3I + HI$

Unlike chlorination and bromination, iodination of alkanes is reversible. This is because the propagation steps are not sufficiently favorable thermodynamically, and the formed $HI$ can drive the reverse reaction.

2. How can reversibility be removed?

To make the reaction irreversible, we must remove one of the products, especially $HI$, from the reaction mixture.

If $HI$ is continuously oxidized to $I_2$, the backward reaction is suppressed and the equilibrium shifts to the right by Le Chatelier’s principle.

3. Check the given reagents

Option A: $HOCl$

$HOCl$ is an oxidizing agent, but it is not the standard reagent used here to selectively remove $HI$ and drive alkane iodination forward.

Option B: dilute $HNO_2$

This is not used for this purpose.

Option C: Liquid $NH_3$

Liquid ammonia does not help make this radical iodination irreversible.

Option D: Concentrated $HIO_3$

$Iodic\ acid$ is a strong oxidizing agent and reacts with $HI$:

$HIO_3 + 5HI \rightarrow 3I_2 + 3H_2O$

Thus, $HI$ is removed from the reaction mixture. As a result, the backward reaction is prevented and the iodination proceeds effectively in the forward direction.

4. Conclusion

The reagent that makes the reaction irreversible is:

$\boxed{\text{D: Concentrated } HIO_3}$

5. Comparison with stored answer

Stored correct answer: D

My derived answer: D

So, they agree.