# Exercise 9 | Chemical Equilibrium

Here is a reaction: A (g) + B (g) $⇄$ C (g)

At equilibrium, a 1.5 L container was found to contain 0.10 moles of A, 0.10 moles of B and 0.20 moles of C.

If 0.05 moles of A and 0.05 moles of B are added to this system, what is the new equilibrium concentration of C?

Let’s calculate KC of this system:

KC

⇒ KC = 30 M-1

[A]eq = [B]eq = $\frac{0.10}{1.5}$ = 6.7 x 10-2 mol.L-1

[C]eq = $\frac{0.20}{1.5}$ = 1.3 x 10-1 mol.L-1

Let’s calculate the new equilibrium concentration [C]eq2:

[A] and [B] increase by adding 0.05 moles of A and B.

According to Le Chatelier’s principle, the equilibrium moves to the right:

[C]eq2 = [C]eq + x = 1.3 x 10-1 + x   with x = change in the number of moles of C per liter

[A]eq2 = [A]eq + $\frac{0.05}{1.5}$ – x = 1.0 x 10-1 – x

[B]eq2 = [A]eq2 = 1.0 x 10-1 – x

KC = 30

= 30

⇒ 30 x2 – 6.0 x + 0.30 = 1.3 x 10-1 + x

⇒ 30 x2 – 5.0 x + 0.17 = 0

⇒ x = 0.12 or x = 4.8 x 10-2

If x = 0.12, [A]eq2 = 0.10 - 0.12  < 0 ⇒ this is impossible

So x = 4.8 x 10-2

[C]eq2 = 1.3 x 10-1 + x = 1.3 x 10-1 + 4.8 x 10-2

[C]eq2 = 0.18 mol.L-1