Question

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

Answer 1

Initially, the partial pressure of $S_2=1.42$ atm and the partial pressure of $C{S}_2=0$ atm.
x atm of $S_2$ will react to reach equilibrium. $1.42-x$ atm of $S_2$ will remain at equilibrium. x atm of $C{S}_2$ will be formed.
$K_p=\frac{P_{C{S}_2}}{P_{S_2}}$
$9.40=\frac{x}{1.42-x}$
$1.42-x=\frac{x}{9.40}$
$1.42-x=0.106x$
$1.42=1.106x$
$P_{C{S}_2}=x=1.28atm$
$P_{S_2}=1.42-x=1.42-1.28=0.14atm$