The reaction in which dissociation of PCl₅ (g) takes place B. The dissociation of PCl5 increases. Therefore, B. The dissociation of PCl5 increases is correct .
Le Chatelier's principle: When a stress is applied to a system in equilibrium, the system will shift its position to counteract the stress and re-establish equilibrium.
Adding an inert gas (helium) increases the total number of moles of gas in the system. This represents a stress on the equilibrium.
According to Le Chatelier's principle, the equilibrium will shift in the direction that consumes the added gas.
In this case, the dissociation of PCl5 consumes one mole of gas (PCl5) and produces two moles of gas (PCl3 and Cl2).
Therefore, the equilibrium will shift to the right, favoring the dissociation of PCl5.
Therefore, the dissociation of PCl5 increases when helium is added at constant pressure.
Here's why the other options are incorrect:
A. The dissociation of PCl5 decreases: This is the opposite of what happens according to Le Chatelier's principle.
C. Does not affect the degree of dissociation of PCl5: While inert gases do not directly participate in the reaction, they still affect the equilibrium through the total pressure.
This is why the dissociation of PCl5 increases in this case.
D. Decrease in the formation of PCl3: This is a consequence of the increase in the dissociation of PCl5, not the cause.
Question
Consider the following reaction in which dissociation of PCl₅ (g) takes place:
PCl₅ (g)⇌PCl₃ (g)+Cl₂ (g)
At a constant pressure, an inert gas Helium is added.
Which of the following statement is correct?
A. The dissociation of PCl5 decreases.
B. The dissociation of PCl5 increases.
C. Does not affect the degree of dissociation of PCl5
D. Decrease in the formation of PCl3