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Law of Mass Action

At 800 K a re...

Question

At 800 K a reaction mixture contained 0.5 mole of $S O_{2}$ . 0.12 mole of $O_{2}$ and 5 mole of $S O_{3}$ at equilibrium. $K_{c}$ for the equilibrium $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ is 833 lit/ mole. If the voiume of the container is 1 litre, calculate how much $O_{2}$ is to be added at this equilibrium in order to get 5.2 moles of $S O_{3}$ at the same temperature.

A

2.34 mole

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B

0.34 mole

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C

1.43 mole

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D

3.23 mole

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Solution

The correct option is B 0.34 mole
Here $K_{c} = \frac{[S O_{3}]^{2}}{[O_{2}] [S O_{2}]^{2}} = 833$
$∵$ volume=1litre
$∵$ $\frac{(5.2)^{2}}{[O_{2}] [S O_{2}]^{2}} - = 833$
$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$
$n e q (1)$ $0.5$ $0.12$ $5$
added $x$
$x i$ $0.5$ $0.12 + x$ $5$
neq $0.5 - 0.2$ $0.12 + x - 0.1$ $5.2
$∴$ from(1)
$= \frac{(5.2)^{2}}{(0.02 + x) (0.3)^{2}} = 833$
$= \frac{27.04}{0.09 \times 933} = 0.02 + x$
$x = \frac{27.04}{74.97} - 0.02 = 0.34 m o l e$

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