Question

When sulphur in the form of $S_2\left(g\right)$ is heated at $900K$, the initial partial pressure of $S_8\left(g\right)$ which was $1atm$ falls by $29\%$ at equilibrium. This is because of conversion of some $S_8\left(g\right)$ to $S_2\left(g\right)$. Find the $K_p$ for reaction, $S_8\left(g\right)\rightleftharpoons 4S_2\left(g\right)$.

Answer 1

The given reaction is :-

$S_8\left(g\right)\rightleftharpoons 4S_2\left(g\right)$

Initial P : $1$ $0$

At eqm : $1-\alpha$ $4\alpha$ ($\alpha$ is degree of dissociation)

$=1-0.29$ $=4\times 0.29$

$=0.71$ $=1.16$

Now, $K_P=\frac{{\left(p_{S_2}\right)}^4}{p_{S_8}}=\frac{{\left(1.16\right)}^4}{0.71}=2.55$