Dynamic equilibrium and Kc (SACE Stage 2 Chemistry)

Syllabus dot point

“Describe dynamic equilibrium in closed systems and use the equilibrium constant expression Kc to relate equilibrium concentrations and reaction extent.”

9 min readUpdated 2 Jun 2026

Quick answer

Dynamic equilibrium is reached in a closed system when the forward and reverse reactions occur at equal rates, so concentrations stay constant even though both reactions continue. The equilibrium constant $K_c$ is the ratio of product concentrations to reactant concentrations, each raised to its coefficient. A large $K_c$ favours products; a small $K_c$ favours reactants. Calculate it from equilibrium concentrations (often via an ICE table). Only temperature changes the value of $K_c$ .

Jump to a section

What this dot point is asking
Lead worked calculation
Dynamic equilibrium
The equilibrium constant expression
The ICE-table method
Interpreting the magnitude
Units
Why it matters for managing processes
Exam-style questions

What this dot point is asking

SACE expects you to describe dynamic equilibrium, write the $K_c$ expression, calculate $K_c$ or an unknown concentration, find units, compare $Q$ to $K_c$ , and explain why only temperature alters $K_c$ .

Lead worked calculation

Calculating Kc from an equilibrium mixture

For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ , a $1.0\ \text{L}$ flask at equilibrium contains $0.040\ \text{mol}$ of $\text{N}_2\text{O}_4$ and $0.080\ \text{mol}$ of $\text{NO}_2$ . Find $K_c$ and its units.

Step 1: Convert amounts in moles to equilibrium concentrations

The $K_c$ expression requires concentrations in $\text{mol L}^{-1}$ , not raw mole amounts. Concentration is defined as moles divided by the volume of the container, so you divide every equilibrium amount by the flask volume. This step is essential: substituting mole amounts directly into the $K_c$ expression would give the wrong numerical value whenever the volume is not exactly $1\ \text{L}$ .

[\text{N}_2\text{O}_4] = \frac{0.040\ \text{mol}}{1.0\ \text{L}} = 0.040\ \text{mol L}^{-1}

[\text{NO}_2] = \frac{0.080\ \text{mol}}{1.0\ \text{L}} = 0.080\ \text{mol L}^{-1}

Step 2: Write the Kc expression from the balanced equation

The $K_c$ expression is built directly from the balanced equation: products go in the numerator, reactants in the denominator, and each concentration is raised to the power of its stoichiometric coefficient. For this equation the coefficient of $\text{NO}_2$ is $2$ , so its concentration is squared. The coefficient of $\text{N}_2\text{O}_4$ is $1$ , so it appears to the first power.

K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(0.080)^2}{0.040} = \frac{6.4 \times 10^{-3}}{0.040} = 0.16

Step 3: Find the units by tracking concentration units through the expression

The units of $K_c$ depend on how the powers in the expression work out, and they must be determined by substituting units algebraically. Here, the numerator has units of $(\text{mol L}^{-1})^2$ because of the squared coefficient, and the denominator has units of $(\text{mol L}^{-1})^1$ . Cancelling one power from the numerator against the denominator leaves one power of $\text{mol L}^{-1}$ .

\text{units} = \frac{(\text{mol L}^{-1})^2}{\text{mol L}^{-1}} = \text{mol L}^{-1}

Final answer: $K_c = 0.16\ \text{mol L}^{-1}$ . Because $K_c < 1$ , the equilibrium mixture contains more $\text{N}_2\text{O}_4$ than $\text{NO}_2$ , meaning the equilibrium position lies toward the reactant side.

Dynamic equilibrium

In a closed system, dynamic equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction. Macroscopic properties (concentrations, colour, pressure) stay constant, but the reactions have not stopped: reactants are still forming products and products still forming reactants, at equal and opposite rates. Equilibrium can be approached from either side and is only possible in a closed system where nothing escapes.

The equilibrium constant expression

Kc 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}

Products go on top, reactants on the bottom, each raised to the power of its coefficient. Pure solids and pure liquids are omitted because their concentration is effectively constant; only aqueous and gaseous species appear.

The ICE-table method

Most $K_c$ calculations follow one routine: build an Initial, Change, Equilibrium table in moles, use the stoichiometry to complete the change row, convert to concentrations by dividing by the volume, then substitute. Check the volume each time, because the powers in the expression may not cancel, so the volume does not always disappear.

Interpreting the magnitude

$K_c \gg 1$ : the mixture is mostly products; the reaction goes nearly to completion.
$K_c \ll 1$ : the mixture is mostly reactants; little product forms.
$K_c \approx 1$ : appreciable amounts of both are present.

The reaction quotient $Q$ has the same form as $K_c$ but uses current (not necessarily equilibrium) concentrations. Comparing them predicts the direction of change: if $Q < K_c$ the reaction goes forward; if $Q > K_c$ it goes backward; if $Q = K_c$ it is at equilibrium.

Only temperature changes Kc

Changing concentration, pressure or volume, or adding a catalyst, shifts the equilibrium position but leaves $K_c$ unchanged. Only a change in temperature alters the value of $K_c$ , because it changes the forward and reverse rate constants by different amounts. For an exothermic forward reaction, raising the temperature lowers $K_c$ .

Units

The units of $K_c$ depend on the equation, because the concentration powers may not cancel. Substitute $\text{mol L}^{-1}$ for each concentration and simplify. For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ the units are $\dfrac{(\text{mol L}^{-1})^2}{(\text{mol L}^{-1})(\text{mol L}^{-1})^3} = (\text{mol L}^{-1})^{-2}$ .

Using initial concentrations or inverting the expression

The two classic errors are substituting initial instead of equilibrium concentrations, and writing the expression upside down (reactants on top), which gives the reciprocal. Always complete the ICE table to equilibrium values first, put products in the numerator, and raise each concentration to its coefficient (not multiply by it). Also remember to divide moles by the volume to get concentration before substituting.

Why it matters for managing processes

$K_c$ quantifies how far a reaction proceeds and, with $Q$ , predicts which way a mixture will shift. Engineers use it alongside Le Chatelier reasoning to choose conditions that push industrial equilibria, such as ammonia synthesis, toward a high yield of product.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2022-style5 marks

For

\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)

, a

2.0\ \text{L}

flask initially holds

1.0\ \text{mol}

\text{H}_2

and

1.0\ \text{mol}

\text{I}_2

. At equilibrium

0.20\ \text{mol}

\text{H}_2

remains. Determine

K_c

for the reaction at this temperature.

Show worked answer →

Step 1: $\text{H}_2$ reacted $= 1.0 - 0.20 = 0.80\ \text{mol}$ . From the $1:1:2$ ratio, $\text{I}_2$ reacted $= 0.80\ \text{mol}$ (so $0.20\ \text{mol}$ remains) and $\text{HI}$ formed $= 1.60\ \text{mol}$ . (2 marks)

Step 2: concentrations (divide by $2.0\ \text{L}$ ): $[\text{H}_2] = 0.10$ ; $[\text{I}_2] = 0.10$ ; $[\text{HI}] = 0.80\ \text{mol L}^{-1}$ . (1 mark)

Step 3: $K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(0.80)^2}{(0.10)(0.10)} = \dfrac{0.64}{0.010} = 64$ . (2 marks)

SACE 2020-style4 marks

At a certain temperature,

K_c = 4.0

for

\text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) + \text{D}(g)

. A mixture in a

1.0\ \text{L}

flask has

[\text{A}] = [\text{B}] = 0.50

and

[\text{C}] = [\text{D}] = 0.20\ \text{mol L}^{-1}

. By calculating the reaction quotient

Q

, determine whether the system is at equilibrium and, if not, which direction it will shift.

Show worked answer →

Step 1: $Q = \dfrac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} = \dfrac{(0.20)(0.20)}{(0.50)(0.50)} = \dfrac{0.040}{0.25} = 0.16$ . (2 marks)

Step 2: $Q = 0.16$ is less than $K_c = 4.0$ , so the system is not at equilibrium. (1 mark)

Step 3: because $Q < K_c$ , there are too few products relative to equilibrium, so the reaction shifts to the right (forward), making more $\text{C}$ and $\text{D}$ until $Q = K_c$ . (1 mark)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board