Question

use the references to access important values if needed for this question.
consider the following reaction:
mathrm{pcl}_{5}(g)
ightleftharpoons mathrm{pcl}_{3}(g)+mathrm{cl}_{2}(g)
if 0.0260 moles of ( mathrm{pcl}_{5}(g), 0.603 ) moles of ( mathrm{pcl}_{3} ), and 0.249 moles of ( mathrm{cl}_{2} ) are at equilibrium in a 17.8 l container at 598 k, the value of the equilibrium constant, ( k_{c} ), is

Explanation:

Step1: Calculate the molar concentrations

The formula for molar concentration \(c=\frac{n}{V}\), where \(n\) is the number of moles and \(V\) is the volume of the container.
For \(PCl_{5}\): \(c_{PCl_{5}}=\frac{0.0260\space mol}{17.8\space L}\approx 0.00146\space M\)
For \(PCl_{3}\): \(c_{PCl_{3}}=\frac{0.603\space mol}{17.8\space L}\approx 0.034\space M\)
For \(Cl_{2}\): \(c_{Cl_{2}}=\frac{0.249\space mol}{17.8\space L}\approx 0.014\space M\)

Step2: Write the equilibrium constant expression

For the reaction \(PCl_{5}(g)
ightleftharpoons PCl_{3}(g)+Cl_{2}(g)\), the equilibrium constant \(K_{c}=\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}\)

Step3: Substitute the values into the equilibrium constant expression

\(K_{c}=\frac{0.034\times0.014}{0.00146}\)
\(K_{c}=\frac{0.000476}{0.00146}\approx 0.326\)

Answer:

\(0.326\)