Question

Following two equilibrium is simultaneously established in a container.
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$CO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons COC{l}_2\left(g\right)$
If some $Ni\left(s\right)$ is introduced in the container forming $Ni\left(CO{)}_4\right(g)$ then at new equilibrium:

• A. $PC{l}_3$ concentration will increase
• B. $PC{l}_3$ concentration will decrease
• C. $C{l}_2$ concentration will remain same
• D. $CO$ concentration will remain same

Answer 1

The correct option is B $PC{l}_3$ concentration will decrease

According to the Le chatelier's principle when we add $Ni\left(s\right)$, it will react with $CO\left(g\right)$ to form $Ni(CO{)}_4$. This leads the second equilibrium to shift in backward direction as concentration of $CO$ is decreasing. Due to this, the concentration of $C{l}_2$ also increases. Hence, the first equilibrium also shifts backward so as to maintain the equilibrium or in other words to maintain the equilibrium, concentration of $PC{l}_3$ will decrease.
Option (a) is incorrect because first equilibrium shifts backward which leads to decrease in concentration of $PC{l}_3.$
Option (b) is correct because $PC{l}_3$ concentration will decrease.
Option (c) is incorrect because if we add or remove any thing in an equilibrium then according to Le chatelier's principle equilibrium will shift either towards forward direction or towards backward direction so as to minimize the effect of the changes.
Option (d) is incorrect because $CO$ is consumed in complex formation, means it will not have constant concentration.