Question

On heating a mixture of $S{O}_{2}C{l}_2$, and CO two equilibria are simultaneously established :-
$S{O}_{2}C{l}_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right)+C{l}_2\left(g\right)$
$CO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons COC{l}_2\left(g\right)$
on adding more $S{O}_2$ at equilibrium what will happen?

• A. Amount of $CO$ will decrease
• B. Amount of $S{O}_{2}C{l}_2$ and $COC{l}_2$ will increase
• C. Amount of CO will remain unaffected
• D. Amount of $S{O}_{2}C{l}_2$ and CO will increase

Answer 1

The correct option is A Amount of $CO$ will decrease

The given two equilibrium are :-

${SO}_2{Cl}_2\left(g\right)\rightleftharpoons {SO}_2\left(g\right)+{Cl}_2\left(g\right)\to \left(1\right)$

$CO\left(g\right)+{Cl}_2\left(g\right)\rightleftharpoons Co{Cl}_2\left(g\right)\to \left(2\right)$

Now, if more ${SO}_2$ is added at equilibrium, then by applying Le Chaterlier's principle to equilibrium $\left(1\right)$, we can see that the reaction will proceed towards ${SO}_2{Cl}_2\left(g\right)$ formation. So, in a way we can say that formation of ${Cl}_2\left(g\right)$ will get decreased.

Now, if we apply same principle to equilibrium $\left(2\right)$ we can see that, if ${Cl}_2\left(g\right)$ is decreased then the reaction will proceed in the direction of $Ca{Cl}_2\left(g\right)$. So, the formation of $Co\left(g\right)$ is increased.