For the reaction $I_{2} (g) ⇌ 2 I (g), K_{c} = 0.01 {molL}^{- 1}$ The volume of the vessel if $1.0 mol$ of $I_{2}$ and 0.5 mol of I are present at equilibrium is

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15 L

25 L

12.5 L

40 L

answer is C.

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Detailed Solution

$\begin{array}{l} I_{2} (g) ⇌ 2 I (g) \\ 1.0 mol 0.5 mol \end{array}$ $K_{c} = \frac{[I]^{2}}{[I_{2}]} = \frac{(0.5 mol / V)^{2}}{(1.0 mol / V)} = 0.01 {molL}^{- 1}$

This give $V = 25 L$

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