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?

Question

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?

| General Chemistry 2

Accepted Answer

Let’s calculate K_Cof this system:

K_C =

\frac{{[C]}_{eq}}{{[A]}_{eq} {[B]}_{eq}}

⇒ K_C= 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

K_C =

\frac{{[C]}_{eq 2}}{{[A]}_{eq 2} {[B]}_{eq 2}}

= 30

\frac{1.3 \times 10^{- 1} + x}{{(1.0 \times 10^{- 1} - x)}^{2}}

= 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

Exercise 9 | Chemical Equilibrium