Question

At equilibrium, the reaction below has an equilibrium constant (Kc) of 4.2 x 10-2 and concentrations of products at equilibrium are: [PCl3] = 0.10 M and [Cl2] = 0.10 M. What is equilibrium concentration of PCl5?
PCl5(g)%25%255C--------- 5CPCl3(g) +Cl2(g)
a) 0.00042 M
b) 0.20 M
c) 0.10 M
d) 2.4 M
e) 0.24 M

Answer 1

Answer:

The equilibrium concentration of [PCl₅] = 0.24M

Explanation:

This is the reaction.

PCl₅(g) → PCl₃(g) +Cl₂(g)

Now let's make, the expression for Kc

Remember that concentrations must be in M

Kc = ( [PCl₃] . [Cl₂] ) / [PCl₅]

4.2x10⁻² = [0.1] . [0.1] / [PCl₅]

[PCl₅] =  [0.1] . [0.1] / 4.2x10⁻²

[PCl₅] = 0.24M