A vessel contains $N_{2} O_{4} & {NO}_{2}$ in 2:3 molar ratio at 10 atm under equilibrium. Now, $K_{P}$ for $N_{2} O_{4} ⇌ 2 {NO}_{2} is$

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$4.5 {atm}^{2}$

10 atm

$9 {atm}^{- 1}$

9 atm

answer is A.

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Detailed Solution

$N_{2} O_{4} ⇌ 2 {NO}_{2}$

$P_{N_{2} O_{4}} = 10 \times \frac{2}{5} = 4 (P_{N_{2} O_{4}} = P_{T} . X_{N_{2} O_{4}})$ $P_{{NO}_{2}} = P_{T} . X_{{NO}_{2}} = 10 \times \frac{3}{5} = 6$

$so K_{p} = \frac{P_{{NO}_{2}^{2}}}{P_{N_{2} O_{4}}} = \frac{6 \times 6}{4} = 9 atm$

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