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Raoult's Law

The equilibri...

Question

The equilibrium constant for the reaction is 9.40 at $900^{\circ} C$ . $S_{2} (g) + C (s) ⇌ C S_{2} (g) .$
The pressure of two gases at equilibrium, when 1.42 atm of $S_{2}$ and excess of $C (s)$ come to equilibrium is :

P_{C S_{2}} = 1.284 a t m

P_{S_{2}} = 0.1365 a t m

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P_{C S_{2}} = 0.1365 a t m

P_{S_{2}} = 1.284 a t m

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P_{C S_{2}} = 1.345 a t m

P_{S_{2}} = 0.1456 a t m

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P_{C S_{2}} = 1.444 a t m

P_{S_{2}} = 0.1857 a t m

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Solution

The correct option is A $P_{C S_{2}} = 1.284 a t m$ , $P_{S_{2}} = 0.1365 a t m$
Initial pressure of $S_{2}$ is 1.42 atm. The equilibrium pressures of $S_{2}$ and $C S_{2}$ are 1.42-x and x respectively.
The expression for the equilibrium constant is $K = \frac{P_{C S_{2}}}{P_{S_{2}}} = \frac{x}{1.42 - x} = 9.4$
Thus $x = 1.284 a t m$ and $1.42 - x = 0.1365 a t m$ .
Thus $P_{C S_{2}} = 1.284 a t m$

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