When sulphur is heated at $900 \mathrm{~K}, \mathrm{~S}_8$ is converted to $S_2$. What will be the e

Question

When sulphur is heated at $900 \mathrm{~K}, \mathrm{~S}_8$ 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?

Professor PC Thomas Classes · Answer

(d) $K_p=\frac{\left(P_{\mathrm{S}_2}\right)^4}{P_{\mathrm{S}_8}}=\frac{(1)^4}{0.75}=1.33 \mathrm{~atm}^3$