Question

use the references to access important values if needed for this question. consider the following reaction where ( k_{c}=0.0120 ) at ( 500 mathrm{~k} ). ( mathrm{pcl}_{5}(g)
ightleftharpoons mathrm{pcl}_{3}(g)+mathrm{cl}_{2}(g) ) a reaction mixture was found to contain ( 0.119 ) moles of ( mathrm{pcl}_{5}(g), 0.0395 ) moles of ( mathrm{pcl}_{3}(g) ), and ( 0.0441 ) moles of ( mathrm{cl}_{2}(g) ), in a ( 1.00 ) liter container. calculate ( q_{c} ). ( q_{c}= ) is the reaction at equilibrium? if not, what direction must it run in order to reach equilibrium? the reaction must run in the forward direction to reach equilibrium. the reaction must run in the reverse direction to reach equilibrium. the reaction is at equilibrium.

Explanation:

Step1: Calculate the concentrations

Since concentration \(c=\frac{n}{V}\) and \(V = 1.00\space L\), the concentrations are:
\([\text{PCl}_5]=0.119\space M\), \([\text{PCl}_3]=0.0395\space M\), \([\text{Cl}_2]=0.0441\space M\)

Step2: Write the expression for \(Q_c\)

For the reaction \(\text{PCl}_5(g)
ightleftharpoons\text{PCl}_3(g)+\text{Cl}_2(g)\), the reaction - quotient \(Q_c\) is given by \(Q_c=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}\)

Step3: Substitute the values into the \(Q_c\) expression

\(Q_c=\frac{0.0395\times0.0441}{0.119}\)

$$
LATEXBLOCK0
$$

Step4: Compare \(Q_c\) and \(K_c\)

Given \(K_c = 0.0120\) at \(500\space K\). Since \(Q_c(0.0146)>K_c(0.0120)\)

Answer:

\(Q_c = 0.0146\). The reaction must run in the reverse direction to reach equilibrium.