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Question

The equilibrium constant for the reaction is $$9.40$$ at $${900}^{o}C$$.  $${S}_{2}(g)+C(s)\rightleftharpoons {CS}_{2}(g)$$. Calculate the pressure of two gases at equilibrium when $$1.42$$ atm of $${S}_{2}$$ and excess of $$C(s)$$ come to equilibrium.


Solution

Given equation
$$S_2(g)+C(s)\rightarrow CS_2(g)$$
Initial pressure of $$S_2$$ is $$1.42$$atm
Equilibrium pressure of $$S_2$$ & $$CS_2$$ is $$1.42-x$$ & $$x$$ respectively
Equilibrium constant$$=\dfrac{P_{CS_2}}{P_{S_2}}$$
$$9.40=\dfrac{x}{1.42-x}$$
Solving for $$x$$
$$13.35-9.40x=x$$
$$x=1.284$$ atm
so $$P_{CS_2}=1.284 $$atm
$$P_{S_2}=1.42-1.284=0.136$$atm

Chemistry

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