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. $0.75\,atm^3$
• B. $2.55\,atm^3$
• C. $25.0\,atm^3$
• D. $1.33\,atm^3$

Answer 1

Answer: (D)

$1.33\,atm^3$

Answer 2

$\begin{matrix}&S_{a\left(g\right)}& {<=>}&4S_{2\left(g\right)}\\\ \text{Initial pressure}&1&&0\\\ \text{At equilibrium}&1-\frac{25}{100}&&4\times\frac{25}{100}\\\ &=0.75&&=1\end{matrix}$ $K_{P}=\frac{\left(P_{S_2}\right)^{4}}{P_{S_8}}$ $=\frac{\left(1\right)^{4}}{0.75}$ $=1.33\,atm^{3}$