Question

consider the reaction below.
pcl₅(g)⇌pcl₃(g) + cl₂(g)
at 500 k, the reaction is at equilibrium with the following concentrations.
pcl₅= 0.0095 m
pcl₃ = 0.020 m
cl₂ = 0.020 m
what is the equilibrium constant for the given reaction?
○ 0.042
○ 0.42
○ 2.4
○ 24

Explanation:

Step1: Recall the equilibrium constant formula

For a reaction \( aA + bB
ightleftharpoons cC + dD \), the equilibrium constant \( K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b} \). For the reaction \( \text{PCl}_5(\text{g})
ightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \), the formula is \( K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \).

Step2: Substitute the given concentrations

We know \( [\text{PCl}_5] = 0.0095\ \text{M} \), \( [\text{PCl}_3] = 0.020\ \text{M} \), and \( [\text{Cl}_2] = 0.020\ \text{M} \). Substituting these values into the formula: \( K_c=\frac{(0.020)(0.020)}{0.0095} \).

Step3: Calculate the numerator and then the denominator

First, calculate the numerator: \( (0.020)(0.020)=0.0004 \). Then divide by the denominator: \( \frac{0.0004}{0.0095}\approx0.042 \).

Answer:

0.042