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Dalton's Law of Partial Pressures

In a 13L ve...

Question

In a $13 L$ vessel initially following reaction occur $C (s) + S_{2} (g) ⇌ {C S}_{2} (g)$ by $12 g$ $C$ , $64 g$ $S_{2}$ , $76 g$ ${C S}_{2}$ at $1027^{o} C$ . Then total pressure is:

200 R

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158 R

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100 R

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79 R

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Solution

The correct option is A $200 R$
$C (s) + S_{2} (g) ⇌ C S_{2} (g)$
no.of moles of $S_{2} (g) = \frac{64}{64} = 1$ mole
no.of moles of $C S_{2} (g) = \frac{76}{76} = 1$ mole
$P_{t o t a l} = P (C S_{2}) (g) + P S_{2} (g)$
$(∵ C (s)$ is not a gas)
$(∵ P V = n R T)$
$P = \frac{2 \times R \times T}{V}$
$= \frac{2 \times R \times (1027 + 273)}{13}$
$= \frac{2 R \times 1300}{13} = 200 R$

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