Calculate K from Initial + One Final Concentration

Question

A student ran the following reaction in the laboratory at 447 K:

PCl₅(g) PCl₃(g) + Cl₂(g)

When she introduced 1.38 moles of PCl₅(g) into a 1.00 liter container, she found the equilibrium concentration of Cl₂(g) to be 3.85×10^-2 M.

Calculate the equilibrium constant, K_c, she obtained for this reaction.

K_c = 

Answer 1

PCl5(g) <==> PCl3(g) + Cl2(g)

1.38.................0................0.........Initial

-x....................+x..............+x........Change

1.38-x..............x.................x........Equilibrium

At equilibrium, x = 3.85x10^-2 M, therefore the following concentrations are present @equilibrium:

[PCl5] = 1.38 - 0.0385 = 1.34 M

[PCl3] = 0.0385 M

[Cl2] = 0.0385 M

Kc = [PCl3][Cl2] / [PCl5] = (0.0385)(0.0385) / (1.34)

Kc = 1.1x10^-3