Question

When sulphur in the form of $S_8$ is heated at 900K, the initial pressure of one atm of $S_{(}8)$ falls by 29% at equilibrium. This is because of conversion of some $S_8$ to $S_2$. Find the value of equilibrium constant for this reaction.

• A. $1.16at{m}^3$
• B. $0.71at{m}^3$
• C. $2.55at{m}^3$
• D. $5.1at{m}^3$

Answer 1

The correct option is C $2.55at{m}^3$

The equilibrium reaction is $S_8\left(g\right)\iff 4S_2\left(g\right).$

	$S_8$	$S_2$
Initial	1	0
Equilibrium	$1-0.29=0.71$	$4\times 0.29$

The expression for the equilibrium constant is $K_p=\frac{(P_{S_2}{)}^4}{P_{S_8}}=\frac{(4\times 2.9{)}^4}{0.71}=2.55at{m}^3.$