Question

The reactions $PC{l}_5\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ and $COC{l}_2\left(g\right)\rightleftharpoons CO\left(g\right)+C{l}_2\left(g\right)$ are simultaneously in equilibrium at constat volume. A few moles of CO(g) are introduced into the vessel. After some time, the new equlibrium concentration of:

• A. $PC{l}_5$ will remain unchanged
• B. $C{l}_2$ will be greater
• C. $PC{l}_5$ will become less
• D. $PC{l}_5$ will become greater

Answer 1

The correct option is C $PC{l}_5$ will become less

${PCl}_5\left(g\right)\rightleftharpoons {PCl}_3\left(g\right)+{Cl}_2\left(g\right)\to \left(1\right)$

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

The above reactions are performed simultaneously in equilibrium at constant volume.

If few moles $CO$ are added, the equilibrium of reaction (2) is shifted backward and ${Cl}_2$ quantity will reduce.

Reaction in ${Cl}_2$ quantity effects the equilibrium of reaction (1) in forward direction.

Hence, quantity of ${PCl}_5\left(g\right)$ decreases.

Option C is correct.