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 $$1.42$$atmEquilibrium pressure of $$S_2$$ & $$CS_2$$ is $$1.42-x$$ & $$x$$ respectivelyEquilibrium constant$$=\dfrac{P_{CS_2}}{P_{S_2}}$$$$9.40=\dfrac{x}{1.42-x}$$Solving for $$x$$$$13.35-9.40x=x$$$$x=1.284$$ atmso $$P_{CS_2}=1.284$$atm$$P_{S_2}=1.42-1.284=0.136$$atmChemistry

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