Question

Two systems, $P{Cl}_5\left(g\right)\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$ and $CO{Cl}_2\left(g\right)\rightleftharpoons CO\left(g\right)+{Cl}_2\left(g\right)$ are simultaneously in equilibrium in a vessel at constant volume. If some $CO\left(g\right)$ is introduced in the vessel at constant volume, then at new equilibrium, the concentration of:

• A. $P{Cl}_5$ is greater
• B. $P{Cl}_3$ remains unchanged
• C. $P{Cl}_5$ is less
• D. ${Cl}_2$ greater

Answer 1

The correct option is C $P{Cl}_5$ is less

Explanation :

Given equation:

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

$COC{l}_2\left(g\right)\rightleftharpoons CO\left(g\right)+C{l}_2\left(g\right)$

It is give that both the above reactions run simultaneously in a vessel at constant volume.

If we add some amount of $CO\left(g\right)$ in the vessel at constant volume,

then at new equilibrium the concentration is $=?$

When $CO$ is added to the above system the second reaction will go backward by Le Chatelier's principle and Hence, the concentration of the common molecule $C{l}_2$​ will also reduce.

So, to counteract the decreased concentration of chlorine, first equilibrium will go forward and hence, the concentration of $PC{l}_5$​ will decrease.

$\therefore$ when $CO\left(g\right)$ is introduced in the vessel at constant volume, then at a new equilibrium, the concentration of $PC{l}_5$ get lesser.

Hence the correct answer is option $C$.