Question

When sulphur is heated at $900$ K, $S_8$ is converted to $S_2$. What will be the equilibrium constant for the reaction if initial pressure of $1$ atm falls by $25\%$ at equilibrium?

• A. $0.75at{m}^3$
• B. $2.55at{m}^3$
• C. $25.0at{m}^3$
• D. $1.33at{m}^3$

Answer 1

The correct option is D $1.33at{m}^3$

The given reaction is :-

$S_8\rightleftharpoons 4S_2$

Initial pressure : $1atm$ $0$ (given)

At equilibrium : $1-0.25$ $4\times 0.25=1atm$ (At eqm $P$ of $S_8$ falls by $25$%)

$=0.75atm$

So, $K_P=\frac{{\left(P_{S_2}\right)}^4}{P_{S_8}}$

$=\frac{{\left(1\right)}^4}{0.75}=\frac{4}{3}=1.33at{m}^3$