The equilibrium constant $K_{p}$ of the reaction: $2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$ is 900 atm at 800 K. A mixture containing $S O_{3}$ and $O_{2}$ having initial partial pressure of 1 and 2 atm. respectively is heated at constant volume to equilibriate. Calculate the partial pressure of $O_{2}$ gas at 800 K.

2.47

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4.01

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0.97

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1.1

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Solution

The correct option is A 2.47
Consider the reaction,
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g) initial pressure 102 At equilibrium (1 - x) x (2 + \frac{x}{2})$ ,

$K_{p} = 900 a t m$
$∴ K_{p} = \frac{(P_{S O_{2}}^{'})^{2} (P_{O_{2}}^{'})}{(P_{S O_{3}}^{'})^{2}}$
$∴ 900 = \frac{x^{2} (2 + \frac{x}{2})}{(1 - x)^{2}}$
$∴$ Since $K_{p}$ of this reaction is small and thus, $x < 1$
$∴ 900 = \frac{x^{2} (2)}{(1 - x)^{2}} (∵ 2 + \frac{x}{2} \approx 2)$
$= \frac{2 x^{2}}{(1 - x)^{2}}$
$∴ 30 = \frac{\sqrt{2} x}{(1 - x)}$
$∴ x = \frac{30}{\sqrt{2} + 30}$
$∴ x = 0.95$
$∴$ At equilibrium,
$P_{S O_{3}}^{'} = 1 - x = 1 - 0.95 = 0.05$ atm
$P_{S O_{2}}^{'} = x = 0.95$ atm
$P_{O_{2}}^{'} = 2 + \frac{x}{2} = 2.475$ atm

Suggest Corrections

Similar questions

Q. The equilibrium constant

K_{p}

for the reaction,

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

, is

900 {a t m}^{- 1}

800 K

. A mixture containing

{S O}_{3}

and

O_{2}

having initial partial pressure of

1

and

2 a t m

respectively is heated at constant volume to equilibriate. Calculate the partial pressure of each gas at

800 K

at equilibrium.

The equilibrium constant Kp of the reaction, 2 $S O_{2}$ (g) + $O_{2}$ (g) $\leftrightharpoons$ 2 $S O_{3}$ (g) is 900 atm. at 800 K. A mixture containing $S O_{3}$ and $O_{2}$ having initial pressure of 1 and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of $S O_{2}$ gas at 800 K.