Presence of which reagent will affect the reversibility of the following reaction to make it irreversible reaction:

Step 1: Understand the equilibrium

The reaction shown appears to be: $\ce{CH3-I + H2O <=> CH3-OH + HI}$

(The LaTeX in question seems incomplete, but this is likely the intended reaction)

This is reversible because HI can react back with $\ce{CH3OH}$ to reform $\ce{CH3I}$.

Step 2: Making it irreversible

To make irreversible, we need to remove one of the products (Le Chatelier's principle).

Options:

• Remove HI
• Remove $\ce{CH3OH}$

Step 3: Analyze each option

(a) HOCl:

• Oxidizing agent
• Could oxidize HI to $\ce{I2}$
• Might work

(b) Liquid $\ce{NH3}$:

• Base, neutralizes HI
• $\ce{HI + NH3 -> NH4I}$
• Removes HI from equilibrium
• Could work

(c) Dilute $\ce{HNO3}$:

• Weak oxidizer (dilute)
• May not effectively remove HI

(d) Concentrated $\ce{HIO3}$:

• Strong oxidizing agent
• Oxidizes HI to $\ce{I2}$: $\ce{HIO3 + 5HI -> 3I2 + 3H2O}$
• Completely removes HI
• Makes reaction irreversible

Answer: (d) Concentrated $\ce{HIO3}$

Key Principle:

Le Chatelier's Principle:

• Removing a product shifts equilibrium forward
• Reaction becomes irreversible if product is continuously removed

Common methods:

• Precipitation (forms insoluble salt)
• Gas evolution (escapes from solution)
• Oxidation/reduction (converts product to different species)
• Neutralization (acid-base reaction)

In this case:

• $\ce{HIO3}$ oxidizes HI to $\ce{I2}$
• HI concentration drops to zero
• Reverse reaction cannot occur
• Forward reaction goes to completion