Question

When sulphur in the form of $S_8$ is heated at 900$\text{K}$, the initial pressure of 1$\text{atm}$ falls by $20\%$ at equilibrium. This is because of conversion of some $S_8$ to $S_2$. Find the value of equilibrium constant for this reaction.

Answer 1

The equilibrium reaction is
$\begin{array}{cccc}{S}_8\left(g\right)& \rightleftharpoons & 4S_2\left(g\right)& \\ 1& & 0& \text{(Initially)}\\ 1-x& & 4x& \text{(Equilibrium)}\end{array}$
But at equilibrium, $x=20\%=0.20$
Hence, $p_{S_8}=1-x=1-0.20=0.80$
$p_{S_2}=4x=4\times 0.20=0.80$

$K_p=\frac{(p_{S_2}{)}^4}{p_{S_8}}=\frac{(0.80{)}^4}{0.8}=0.512$