# Question 3c6f1

May 28, 2017

The concentration of $C {l}_{2}$ would increase.

#### Explanation:

I published a note on this topic ~two weeks ago... Here is the link to my answer presentation. => https://socratic.org/questions/what-is-meant-by-a-stress-on-a-reaction-at-equilibrium?source=search

Quick Answer => Adding $P C {l}_{5}$ overloads the product side of the reaction much like extra kids getting onto a balance beam (seesaw). Visualize the reaction tilting toward the product side because of the extra $P C {l}_{5}$. To reestablish the equilibrium, the reaction needs to shift away from the applied stress. This means some of the $P C {l}_{5}$ must decompose into $P C {l}_{3} \mathmr{and} C {l}_{2}$ thereby increasing the concentration of $C {l}_{2}$ (and PCl_3)# causing the balance to shift to the reactant side to establish a new equilibrium.

Go to the link mentioned above for more support on this. It is really very simple concept. Good luck.

May 28, 2017

The equilibrium position shifts so as to nullify the change you have made. The change you made was to add $P C {l}_{5}$ so to nullify this the equilibrium shifts in the direction of reagents, thereby using up some of the $P C {l}_{5}$ by decomposing it into $P C {l}_{3}$ and $C {l}_{2}$, so $C {l}_{2}$ concentration increases.