Question

On heating a mixture of ${SO}_2{Cl}_2$ and $CO$, two equilibria are simultaneously established:
${SO}_2{Cl}_2\rightleftharpoons {SO}_2+{Cl}_2$
$CO\left(g\right)+{Cl}_2\left(g\right)\rightleftharpoons CO{Cl}_2\left(g\right)$
On adding more ${SO}_2$ at equilibrium what will happen?

• A. Amount of $CO$ will decrease
• B. Amount of ${SO}_2{Cl}_2$ and $CO{Cl}_2$ will increase
• C. Amount of $CO$ will remain unaffected
• D. Amount of ${SO}_2{Cl}_2$ and $CO$ will increase

Answer 1

Answer: (D)

Amount of ${SO}_2{Cl}_2$ and $CO$ will increase

The given equilibrium are :-

${SO}_2{Cl}_2\rightleftharpoons {SO}_2+{Cl}_2\to \left(I\right)$

$CO\left(g\right)+{Cl}_2\left(g\right)\rightleftharpoons CO{Cl}_2\left(g\right)$

If more ${SO}_2$ is added at equilibrium then by Le Chatelier's principle reaction $\left(I\right)$ will proceed backwards and ${SO}_2{Cl}_2$ formation will be favoured and ${Cl}_2$ formation is less. Now, Since ${Cl}_2$ formation is reduced so in equation $\left(II\right)$ by Le Chatelier's principle, the reaction proceeds backwards side and enhances the formation of $CO$.