PCl3 (g) + Cl2 (g) ⇋ PCl5 (g) Kc = 91.0 at 400 K. What is the [Cl2] at equilibrium if the initial concentrations were 0.24 M for PCl3 and 1.50 M for Cl2 and 0.12 M PCl5.

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃ (g) + Cl₂ (g) ⇋ PCl₅ (g)

Kc = 91

So let's analyse, all the process:

PCl₃ (g) + Cl₂ (g) ⇋ PCl₅ (g)

Initially 0.24 M 1.50M 0.12 M

React x x x

Some amount of compound has reacted during the process.

In equilibrium we have

0.24 - x 1.50 - x 0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂]. [PCl₃]

91 = (0.12 + x) / (0.24 - x) (1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 + x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b = - 159.34; c = 32.64

(-b + - √ (b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M