# Exercise 9 | Chemical Equilibrium

## General Chemistry 3 - Exercise 9

Here is a reaction: A (g) + B (g) $\rightleftarrows $ 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 K_{C }of this system:

K_{C} = $\frac{{\left[\mathrm{C}\right]}_{\mathrm{eq}}}{{\left[\mathrm{A}\right]}_{\mathrm{eq}}{\left[\mathrm{B}\right]}_{\mathrm{eq}}}$

⇒ 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} = $\frac{{\left[\mathrm{C}\right]}_{\mathrm{eq}2}}{{\left[\mathrm{A}\right]}_{\mathrm{eq}2}{\left[\mathrm{B}\right]}_{\mathrm{eq}2}}$ = 30

$\frac{1.3\times {10}^{-1}+\mathrm{x}}{{\left(1.0\times {10}^{-1}-\mathrm{x}\right)}^{2}}$ = 30

⇒ 30 x^{2} – 6.0 x + 0.30 = 1.3 x 10^{-1} + x

⇒ 30 x^{2} – 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}