SACE Board SACE 2022-style practice: For H_2(g) + I_2(g) 2HI(g), a 2.0\ L flask initially holds 1.0\ mol of H_2 and 1.0\ mol of I_2. At equilibrium 0.20\ mol of H_2 remains. Determine K_c for the reaction at this temperature.

Step 1: H_2 reacted = 1.0 - 0.20 = 0.80\ mol. From the 1:1:2 ratio, I_2 reacted = 0.80\ mol (so 0.20\ mol remains) and HI formed = 1.60\ mol. (2 marks) Step 2: concentrations (divide by 2.0\ L): [H_2] = 0.10; [I_2] = 0.10; [HI] = 0.80\ mol L^-1. (1 mark) Step 3: K_c = [HI]^2[H_2][I_2] = (0.80)^2(0.10)(0.10) = 0.640.010 = 64. (2 marks)

§-Syllabus dot point

SAChemistrySyllabus dot point

What does it mean for a reaction to reach dynamic equilibrium, and how does Kc describe it?

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

Dynamic equilibrium in closed systems, writing and interpreting the Kc expression, the ICE-table method, units and the temperature dependence of Kc, with fully worked SACE-style equilibrium-constant calculations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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

The equilibrium constant expression

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.

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 $\dfrac{(\text{mol L}^{-1})^2}{(\text{mol L}^{-1})(\text{mol L}^{-1})^3} = (\text{mol L}^{-1})^{-2}$ .

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 practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20225 marksFor

\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 20204 marksAt 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)