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Characteristics of Equilibrium Constant

If Kp for t...

Question

If $K_{p}$ for the reaction $N_{2} O_{4} ⇌ 2 {N O}_{2}$ is $0.66$ , then what is the equilibrium pressure of $N_{2} O_{4}$ ?
(Total pressure at equilibrium is $0.5 a t m$ )

0.168

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0.322

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Solution

The correct option is A $0.168$
$N_{2} O_{4} ⇌ {2 N O}_{2} K_{P} = 0.66$
$t = 0$ $P_{0}$ $0$ Given total pressure $0.5$ atm
$t = t_{e q}$ $P_{0} (1 - α)$ $2 P_{0} α$ $P_{0} (1 + α) = 0.5$
$K_{P} = \frac{{4 P_{0}}^{2} α^{2}}{P_{0} (1 - α)} = 0.66$
$\frac{4 P_{0} α^{2}}{1 - α} = 0.66 \frac{2 α^{2}}{1 - α^{2}} = 0.66 P_{0} (3 / 2) = 0.5$
${3 α}^{2} = 1 - α^{2}$ $P_{0} = 1 / 3$
${4 α}^{2} = 1$
$\Rightarrow α = 1 / 2$
So, $N_{2} O_{4}$ Eq'b pressure
$\Rightarrow \frac{1}{3} \times \frac{1}{2} \Rightarrow\approx 0.168$

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