Question

consider the following system at equilibrium where $delta h^{circ}=-87.9 \text{kj}$ and $k_{c}=83.3$, at $500 \text{k}$.
$\text{pcl}_{3}(g)+\text{cl}_{2}(g)
ightleftharpoons\text{pcl}_{5}(g)$
if the volume of the equilibrium system is suddenly increased at constant temperature:
the value of $k_{c}$
$\bigcirc$ increases
$\bigcirc$ decreases
$\bigcirc$ remains the same
the value of $q_{c}$
$\bigcirc$ is greater than $k_{c}$
$\bigcirc$ is equal to $k_{c}$
$\bigcirc$ is less than $k_{c}$
the reaction must
$\bigcirc$ run in the forward direction to reestablish equilibrium.
$\bigcirc$ run in the reverse direction to reestablish equilibrium.
$\bigcirc$ remain the same. it is already at equilibrium.
the number of moles of $\text{cl}_{2}$ will
$\bigcirc$ increase
$\bigcirc$ decrease
$\bigcirc$ remain the same

Explanation:

• For \(K_c\): The equilibrium constant \(K_c\) depends only on temperature. Since the temperature is constant, \(K_c\) remains the same.
• For \(Q_c\): The reaction quotient \(Q_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}\). When the volume is increased, the concentrations of all species \([PCl_5]\), \([PCl_3]\), and \([Cl_2]\) decrease. The denominator \([PCl_3][Cl_2]\) (sum of moles of reactants: \(n_{reactants}=n_{PCl_3}+n_{Cl_2}\)) has a greater number of moles compared to the numerator (\(n_{PCl_5}\)). Using the formula \(c = \frac{n}{V}\), when \(V\) increases, \(Q_c=\frac{\frac{n_{PCl_5}}{V}}{\frac{n_{PCl_3}}{V}\times\frac{n_{Cl_2}}{V}}=\frac{n_{PCl_5}V}{n_{PCl_3}n_{Cl_2}}\). Since \(n_{PCl_3}+n_{Cl_2}>n_{PCl_5}\), \(Q_c < K_c\).
• For the reaction direction: When \(Q_c
• For moles of \(Cl_2\): As the reaction shifts forward (\(PCl_3(g)+Cl_2(g)

ightleftharpoons PCl_5(g)\)), \(Cl_2\) is consumed, so the number of moles of \(Cl_2\) decreases.

Answer:

• The value of \(K_c\): remains the same
• The value of \(Q_c\): is less than \(K_c\)
• The reaction must: run in the forward direction to reestablish equilibrium
• The number of moles of \(Cl_2\) will: decrease