The units of K_c depend on the equation, because the powers may not cancel. Work them out from the expression: each concentration is in mol L^-1, so substitute the units and simplify. Many syllabus problems quote K_c without units, but you should be ready to derive them.

TASChemistrySyllabus dot point

How do we quantify the position of an equilibrium?

Write equilibrium expressions and calculate Kc, interpreting its size and units.

Writing the equilibrium constant expression, calculating Kc from equilibrium concentrations, interpreting its magnitude, and the effect of temperature on Kc.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-30

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

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What this dot point is asking

You must write the expression for $K_c$ , calculate it from equilibrium concentrations, and interpret what its size tells you.

Writing the expression

For a general reaction $aA + bB \rightleftharpoons cC + dD$ , the equilibrium constant in terms of concentrations is:

Pure solids and pure liquids are left out of the expression because their concentrations are effectively constant. Only aqueous and gaseous species appear.

Interpreting the magnitude

The size of $K_c$ tells you how far the reaction proceeds.

A large $K_c$ (much greater than 1) means the numerator dominates, so at equilibrium there are mostly products. The reaction goes nearly to completion.
A small $K_c$ (much less than 1) means the denominator dominates, so there are mostly reactants. Little product forms.
A $K_c$ near 1 means appreciable amounts of both are present.

Units

The units of $K_c$ depend on the equation, because the powers may not cancel. Work them out from the expression: each concentration is in $\text{mol L}^{-1}$ , so substitute the units and simplify. Many syllabus problems quote $K_c$ without units, but you should be ready to derive them.

example

Calculating Kc

For $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ , an equilibrium mixture in a $1.0\ \text{L}$ vessel contains $0.20\ \text{mol H}_2$ , $0.20\ \text{mol I}_2$ and $1.60\ \text{mol HI}$ . Find $K_c$ .

Step 1: Convert to concentrations (volume is $1.0\ \text{L}$ ): $[\text{H}_2] = 0.20$ , $[\text{I}_2] = 0.20$ , $[\text{HI}] = 1.60\ \text{mol L}^{-1}$ .

Step 2: Write the expression: $K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$ .

Step 3: Substitute: $K_c = \dfrac{(1.60)^2}{(0.20)(0.20)} = \dfrac{2.56}{0.040} = 64$ .

The units cancel here, and $K_c = 64$ (greater than 1) shows the equilibrium lies toward the product HI.

In the exam, always write the balanced equation, build the expression with the correct powers, omit pure solids and liquids, and comment on whether the value of $K_c$ favours reactants or products.

Exam-style practice questions

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

2022 TASC4 marksCompound A dissociates: 3A(g) reversible 2B(g) + C(g). At a particular temperature, a closed 2.0 L container initially contains 1.2 moles of A. At equilibrium 0.75 moles of A remain. Determine the value of the equilibrium constant at this temperature.

Show worked answer →

Set up an ICE table in moles. A reacted = 1.2 - 0.75 = 0.45 mol. From the 3:2:1 ratio, B formed = (2/3)(0.45) = 0.30 mol and C formed = (1/3)(0.45) = 0.15 mol.

Convert to concentrations (divide by 2.0 L): [A] = 0.75/2.0 = 0.375; [B] = 0.30/2.0 = 0.150; [C] = 0.15/2.0 = 0.075 mol L-1.

Kc = [B]^2 [C] / [A]^3 = (0.150)^2 (0.075) / (0.375)^3 = (0.0225 x 0.075) / 0.05273 = 0.001688 / 0.05273 = 0.032. (4 marks for ICE, concentrations, correct expression and value.)

2021 TASC2 marksFor the equilibrium E(g) + 2F(g) reversible 2G(g), an equilibrium mixture in a 1.50 L sealed container has 2.50 mol of gas E, 1.20 mol of gas F and 0.85 mol of gas G. Calculate the value of the equilibrium constant Kc at this temperature.

Show worked answer →

Convert each amount to a concentration by dividing by 1.50 L: [E] = 2.50/1.50 = 1.667; [F] = 1.20/1.50 = 0.800; [G] = 0.85/1.50 = 0.567 mol L-1.

Write the expression: Kc = [G]^2 / ([E][F]^2).

Kc = (0.567)^2 / (1.667 x (0.800)^2) = 0.321 / (1.667 x 0.640) = 0.321 / 1.067 = 0.301. So Kc is about 0.30 (mol L-1)^-1. (2 marks: correct concentrations and value.)

2023 TASC5 marksTo determine Kc for N2(g) + 3H2(g) reversible 2NH3(g), 63.0 g of nitrogen and 15.0 g of hydrogen were placed in a 100 L vessel at constant temperature. At equilibrium, 58.0 g of ammonia was present. Determine the equilibrium constant for this reaction. (M(H2) = 2.016, M(N2) = 28.02, M(NH3) = 17.034 g mol-1.)

Show worked answer →

Initial moles: N2 = 63.0/28.02 = 2.249 mol; H2 = 15.0/2.016 = 7.440 mol. Equilibrium NH3 = 58.0/17.034 = 3.405 mol.

From the stoichiometry, forming 3.405 mol NH3 uses N2 = 3.405/2 = 1.702 mol and H2 = (3/2)(3.405) = 5.107 mol. Equilibrium amounts: N2 = 2.249 - 1.702 = 0.547 mol; H2 = 7.440 - 5.107 = 2.333 mol; NH3 = 3.405 mol.

Concentrations (divide by 100 L): [N2] = 0.00547; [H2] = 0.02333; [NH3] = 0.03405 mol L-1.

Kc = [NH3]^2 / ([N2][H2]^3) = (0.03405)^2 / (0.00547 x (0.02333)^3) = 0.001159 / (0.00547 x 1.270 x 10^-5) = 0.001159 / 6.95 x 10^-8 = 1.67 x 10^4. (5 marks: ICE in moles, concentrations, expression and value.)

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