Question

The equilibrium constant for the reaction is $9.40$ at ${900}^{o}C$. $S_2\left(g\right)+C\left(s\right)\rightleftharpoons {CS}_2\left(g\right)$. Calculate the pressure of two gases at equilibrium when $1.42$ atm of $S_2$ and excess of $C\left(s\right)$ come to equilibrium.

Answer 1

Given equation

$S_2\left(g\right)+C\left(s\right)\to C{S}_2\left(g\right)$

Initial pressure of $S_2$ is $1.42$atm

Equilibrium pressure of $S_2$ & $C{S}_2$ is $1.42-x$ & $x$ respectively

Equilibrium constant$=\frac{P_{C{S}_2}}{P_{S_2}}$

$9.40=\frac{x}{1.42-x}$

Solving for $x$

$13.35-9.40x=x$

$x=1.284$ atm

so $P_{C{S}_2}=1.284$atm

$P_{S_2}=1.42-1.284=0.136$atm