Question

use the references to access important values if needed for this question.
consider the following system at equilibrium where \\( \delta h^{\circ}=87.9 \mathrm{~kj} \\), and \\( k_{c}=0.0120 \\), at 500 k:
\\( \mathrm{pcl}_{5}(g) \
ightleftharpoons \mathrm{pcl}_{3}(g)+\mathrm{cl}_{2}(g) \\)
when 0.34 moles of \\( \mathrm{pcl}_{3}(g) \\) are added to the equilibrium system at constant temperature:
the value of \\( k_{c} \\)
o increases
o decreases
o remains the same
the value of \\( q_{c} \\)
o is greater than \\( k_{c} \\)
o is equal to \\( k_{c} \\)
o is less than \\( k_{c} \\)
the reaction must
o run in the forward direction to reestablish equilibrium
o run in the reverse direction to reestablish equilibrium
o remain in the current position, since it is already at equilibrium
the concentration of \\( \mathrm{pcl}_{3} \\) will
o increase
o decrease
o remain the same

Explanation:

Step1: Effect on \(K_{c}\)

The equilibrium constant \(K_{c}\) depends only on temperature. Since the temperature is constant, adding \(PCl_{3}(g)\) does not change \(K_{c}\).

Step2: Effect on \(Q_{c}\)

The reaction quotient \(Q_{c}=\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}\). When \(PCl_{3}(g)\) is added, the numerator of \(Q_{c}\) increases. So \(Q_{c}>K_{c}\).

Step3: Direction of reaction

According to Le - Chatelier's principle, when \(Q_{c}>K_{c}\), the reaction will shift in the reverse direction to re - establish equilibrium.

Step4: Effect on \([PCl_{3}]\)

As the reaction shifts in the reverse direction (\(PCl_{3}(g)+Cl_{2}(g)
ightarrow PCl_{5}(g)\)), the concentration of \(PCl_{3}\) will decrease.

Answer:

• The value of \(K_{c}\): remains the same
• The value of \(Q_{c}\): is greater than \(K_{c}\)
• The reaction must: run in the reverse direction to reestablish equilibrium
• The concentration of \(PCl_{3}\) will: decrease