The equilibrium constant $K_{p}$ for the reaction, $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ , is $900 {a t m}^{- 1}$ at $800 K$ . 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.

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Solution

The system in the initial stage does not contain ${S O}_{2}, {S O}_{3}$ will thus, decompose to form ${S O}_{2}$ and $O_{2}$ until equilibrium is reached. The partial pressure of ${S O}_{3}$ will decrease.

Let the decrease in partial pressure be $2 x$ .

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

At equi. $(2 x)$ $(2 + x)$ $(1 - 2 x)$

Applying law of mass action,

$K_{p} = \frac{{(1 - 2 x)}^{2}}{{(2 x)}^{2} (2 + x)}$

$900 = \frac{{(1 - 2 x)}^{2}}{{8 x}^{2}}$

or $x = 0.0115 a t m$

Thus, the partial pressures at equilibrium are:

$p_{S O_{2}} = 2 \times 0.0115 = 0.023 a t m$

$p_{S O_{2}} = 2 + 0.0115 = 2.0115 a t m$

$p_{S O_{3}} = 1 - 2 \times 0.0115 = 0.977 a t m$

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Similar questions

Q. 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.

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.