Equilibrium constants and calculations: WACE Year 12 Chemistry

Syllabus dot point

“Write equilibrium constant expressions, interpret their magnitude, and calculate equilibrium concentrations using an ICE table”

9 min readUpdated 2 Jun 2026

Quick answer

The equilibrium constant $K_c$ is the ratio of product concentrations to reactant concentrations, each raised to its coefficient, at equilibrium. A large $K_c$ means products are favoured; comparing the reaction quotient $Q$ to $K_c$ tells you which way a reaction will shift.

What this dot point is asking

For a general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ , the equilibrium constant in terms of concentration is

K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}

Products go on top, reactants on the bottom, and each concentration is raised to the power of its stoichiometric coefficient. Concentrations are in $\text{mol L}^{-1}$ and are the equilibrium values, not the starting values.

Two rules about what to include. Pure solids and pure liquids are left out of the expression because their concentration is effectively constant. For example, for $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ the expression is just $K_c = [\text{CO}_2]$ . Water as a solvent in dilute aqueous reactions is also omitted.

Equilibrium constant expression

For $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ :

K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}

Omit pure solids and pure liquids.

Interpreting the magnitude of K

The size of $K_c$ tells you where the position of equilibrium lies.

$K_c \gg 1$ (large): the equilibrium mixture is mostly products; the forward reaction is favoured.
$K_c \ll 1$ (small): the equilibrium mixture is mostly reactants; the reverse reaction is favoured.
$K_c \approx 1$ : significant amounts of both reactants and products are present.

$K_c$ depends only on temperature. It does not change when you alter concentrations, pressure, or add a catalyst, but it does change when the temperature changes. This is why a question that changes the volume or adds more reactant asks about the new position of equilibrium (which shifts), not about $K_c$ (which is fixed unless the temperature is altered).

The reaction quotient Q

The reaction quotient $Q$ has exactly the same form as $K_c$ but uses the concentrations at any moment, not necessarily at equilibrium. Comparing $Q$ to $K_c$ predicts the direction of net reaction.

$Q < K_c$ : too few products, so the reaction proceeds forwards (to the right).
$Q > K_c$ : too many products, so the reaction proceeds backwards (to the left).
$Q = K_c$ : the system is already at equilibrium; no net change.

Finding Kc with an ICE table

At a certain temperature, $1.00\ \text{mol}$ of $\text{H}_2$ and $1.00\ \text{mol}$ of $\text{I}_2$ are placed in a $2.00\ \text{L}$ flask. The reaction is $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ . At equilibrium $1.56\ \text{mol}$ of $\text{HI}$ has formed. Find $K_c$ .

Step 1: Convert to concentrations

All entries in $K_c$ must be in $\text{mol L}^{-1}$ , so divide moles by the $2.00\ \text{L}$ volume. Initial: $[\text{H}_2] = [\text{I}_2] = 1.00 / 2.00 = 0.500\ \text{mol L}^{-1}$ and $[\text{HI}] = 0$ .

Step 2: Fill the change row from stoichiometry

The HI formed is $1.56 / 2.00 = 0.780\ \text{mol L}^{-1}$ , so the change in HI is $+0.780$ . From the 1:1:2 stoichiometry, $\text{H}_2$ and $\text{I}_2$ each change by $-0.390$ (half the HI change).

Step 3: Read off the equilibrium row

$[\text{H}_2] = 0.500 - 0.390 = 0.110\ \text{mol L}^{-1}$
$[\text{I}_2] = 0.500 - 0.390 = 0.110\ \text{mol L}^{-1}$
$[\text{HI}] = 0.780\ \text{mol L}^{-1}$

Step 4: Substitute into the expression

K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.780)^2}{(0.110)(0.110)} = \frac{0.6084}{0.0121} = 50.3.

Because the moles of gas are equal on both sides the volume cancels, so you could have used moles directly, but converting to concentration as a habit avoids slips on reactions where the volumes do not cancel. $K_c$ here is dimensionless; always quote the temperature, because $K_c$ is temperature-dependent.

Final answer: $K_c = 50.3$

Using initial instead of equilibrium values

The single most common error is substituting starting concentrations into the $K_c$ expression. $K_c$ is defined only at equilibrium, so you must use the equilibrium-row concentrations from the ICE table. Other frequent slips: forgetting to raise concentrations to their coefficients (the squared term for HI above), forgetting to divide moles by the volume to get concentration, and including pure solids or liquids in the expression. Also remember to apply the stoichiometric ratio when filling the change row, not just copy the same number for every species.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks

448\ ^\circ\text{C}

0.500\ \text{mol}

\text{PCl}_5

is placed in a

1.00\ \text{L}

sealed flask and allowed to reach equilibrium:

\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)

. At equilibrium

[\text{Cl}_2] = 0.150\ \text{mol L}^{-1}

. (a) Use an ICE table to find the equilibrium concentrations of all three gases. (b) Calculate

K_c

. (c) State, with a reason, the effect on

K_c

of halving the volume.

Show worked answer →

A 7 mark question rewards the completed ICE table, the $K_c$ value, and the conceptual answer.

(a) Initial: $[\text{PCl}_5] = 0.500$ , $[\text{PCl}_3] = [\text{Cl}_2] = 0$ . The $\text{Cl}_2$ formed is $0.150$ , and by the 1:1:1 stoichiometry $\text{PCl}_3$ also forms $0.150$ and $\text{PCl}_5$ falls by $0.150$ .

$[\text{PCl}_5] = 0.500 - 0.150 = 0.350\ \text{mol L}^{-1}$
$[\text{PCl}_3] = 0.150\ \text{mol L}^{-1}$
$[\text{Cl}_2] = 0.150\ \text{mol L}^{-1}$

(b)

K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{(0.150)(0.150)}{0.350} = 0.0643\ \text{mol L}^{-1}.

(c) Halving the volume increases the pressure but does not change $K_c$ ; only a temperature change alters $K_c$ . The position would shift left (towards fewer moles of gas), but $K_c$ stays at $0.0643$ .

Markers reward the three equilibrium concentrations, $K_c = 0.0643\ \text{mol L}^{-1}$ , and the statement that $K_c$ is unchanged by volume.

WACE 2020-style5 marks

For the reaction

\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)

K_c = 4.0

at a particular temperature. A mixture contains

[\text{N}_2\text{O}_4] = 0.50\ \text{mol L}^{-1}

and

[\text{NO}_2] = 2.0\ \text{mol L}^{-1}

. (a) Calculate the reaction quotient

Q

. (b) Determine whether the mixture is at equilibrium, and if not, predict the direction of net reaction with justification.

Show worked answer →

A 5 mark question rewards the correct $Q$ and the $Q$ versus $K_c$ comparison.

(a) $Q$ has the same form as $K_c$ but uses current concentrations:

Q = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(2.0)^2}{0.50} = \frac{4.0}{0.50} = 8.0.

(b) Compare $Q = 8.0$ with $K_c = 4.0$ . Since $Q > K_c$ there are too many products relative to equilibrium, so the net reaction proceeds in the reverse direction (to the left), converting $\text{NO}_2$ back to $\text{N}_2\text{O}_4$ until $Q$ falls to $4.0$ .

Markers reward $Q = 8.0$ with the squared term, the recognition $Q > K_c$ , and the leftward (reverse) prediction.

Practise this

Sources & how we know this

Chemistry ATAR course: syllabus and support materials — SCSA
Chemistry: past ATAR course examinations — SCSA