Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction:
$CH_4 + I_2 \rightleftharpoons CH_3-I + HI$
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According to Le Chatelier's principle, removing a product from a reversible reaction will shift the equilibrium to favor the formation of more products. In the iodination of alkanes, an oxidizing agent like $HIO_3$ or $HNO_3$ is used to remove the HI formed.
The iodination of methane is a reversible reaction.
$CH_4 + I_2 \overset{h\nu}{\rightleftharpoons} CH_3I + HI$
The hydrogen iodide (HI) formed as a product is a strong reducing agent and can reduce methyl iodide ($CH_3I$) back to methane ($CH_4$), shifting the equilibrium to the left.
To make the reaction proceed in the forward direction (irreversible), the HI product must be removed from the reaction mixture as it is formed. This can be achieved by adding a strong oxidizing agent that reacts with HI.
Let's examine the options:
(A) Concentrated HIO$_3$ (Iodic acid) is a strong oxidizing agent. It readily oxidizes HI to I$_2$.
$5HI + HIO_3 \rightarrow 3I_2 + 3H_2O$
By consuming HI, this reagent shifts the equilibrium of the main reaction to the right, making it effectively irreversible. This is the standard reagent used for this purpose.
The other options are either not strong enough or not typically used for this specific purpose.
