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The reaction $I_{2} (g) ⇌ 2 I (g)$ is studied at 1200 Kat constant volume with initial pressure of $I_{2}$ equal to 0.07 atm. If the pressure at equilibrium is 0.12 atm, the equilibrium constant of the reaction is

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0.01 atm

1.0 atm

0.5 atm

0.25 atm

answer is C.

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Detailed Solution

$\underset{2}{I_{2} (g)} \underset{p_{0} - p}{2 I (g)}$ Total pressure $= p_{0} + p$

Hence $p_{0} + p = 0.07 atm + p = 0.12 atm \Rightarrow p = 0.05 atm$

$K_{p} = \frac{{(p_{I})}^{2}}{(p_{l})} = \frac{(2 p)^{2}}{p_{0} - p} = \frac{(2 \times 0.05 atm)^{2}}{(0.07 - 0.05) atm} = 0.5 atm$

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