The units of K_c depend on the equation, because the powers may not cancel. Substitute mol L^-1 for each concentration and simplify. For N_2 + 3H_2 2NH_3 the units are (mol L^-1)^2(mol L^-1)(mol L^-1)^3 = (mol L^-1)^-2.

TASC TCE 2022-style practice: Compound A dissociates: 3A(g) 2B(g) + C(g). At a particular temperature a closed 2.0\ L container initially holds 1.2\ mol of A. At equilibrium 0.75\ mol of A remains. Determine the value of the equilibrium constant at this temperature.

Set up an ICE table in moles. A reacted = 1.2 - 0.75 = 0.45\ mol. From the 3:2:1 ratio, B formed = 23(0.45) = 0.30\ mol and C formed = 13(0.45) = 0.15\ mol. (1 mark) Convert to concentrations (divide by 2.0\ L): [A] = 0.375; [B] = 0.150; [C] = 0.075\ mol L^-1. (1 mark) $K_c = [B]^2[C][A]^3 = ((0.150)^2(0.075))/((0.375)^3) = 1.69 × 10^-35.27 × 10^-2 = 0.032.$ (2 marks for the correct expression and value.)

TASC TCE 2023-style practice: To determine K_c for N_2(g) + 3H_2(g) 2NH_3(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. (M(H_2) = 2.016, M(N_2) = 28.02, M(NH_3) = 17.03\ g mol^-1.) Determine K_c.

Initial moles: n(N_2) = 63.0/28.02 = 2.249; n(H_2) = 15.0/2.016 = 7.440. Equilibrium n(NH_3) = 58.0/17.03 = 3.405\ mol. (1 mark) Forming 3.405\ mol of NH_3 uses n(N_2) = 3.405/2 = 1.702 and n(H_2) = 32(3.405) = 5.107\ mol. Equilibrium: n(N_2) = 0.547; n(H_2) = 2.333; n(NH_3) = 3.405\ mol. (2 marks) Concentrations (divide by 100\ L): [N_2] = 5.47 × 10^-3; [H_2] = 2.333 × 10^-2; [NH_3] = 3.405 × 10^-2\ mol L^-1. (1 mark) $K_c = [NH_3]^2[N_2][H_2]^3 = (3.405 × 10^-2)^2(5.47 × 10^-3)(2.333 × 10^-2)^3 = 1.67 × 10^4.$ (1 mark)

§-Syllabus dot point

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, using an ICE table to calculate Kc from equilibrium concentrations, interpreting its magnitude and units, and the effect of temperature, with fully worked TASC-style examples.

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

TASC expects you to write the $K_c$ expression, calculate $K_c$ from equilibrium data using an ICE table, work out its units, and interpret what its size tells you.

Writing the expression

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

The ICE table method

Most $K_c$ calculations follow the same routine: build an Initial, Change, Equilibrium table in moles, use the stoichiometry to fill the change row, convert to concentrations by dividing by the volume, then substitute. Always check the volume because the powers in the expression may not cancel.

Calculating Kc from an equilibrium mixture

Calculate the equilibrium constant $K_c$ for the reaction $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ given that at equilibrium a $1.0\ \text{L}$ vessel contains $0.20\ \text{mol}$ of $\text{H}_2$ , $0.20\ \text{mol}$ of $\text{I}_2$ and $1.60\ \text{mol}$ of $\text{HI}$ .

Step 1: Convert moles to molar concentrations

The equilibrium constant expression requires concentrations in $\text{mol L}^{-1}$ , not amounts in moles. Divide each equilibrium amount by the volume of the vessel. Because the volume is $1.0\ \text{L}$ , the numerical values are unchanged here, but the unit matters:

[\text{H}_2] = \frac{0.20\ \text{mol}}{1.0\ \text{L}} = 0.20\ \text{mol L}^{-1}, \quad [\text{I}_2] = 0.20\ \text{mol L}^{-1}, \quad [\text{HI}] = 1.60\ \text{mol L}^{-1}.

Step 2: Write the equilibrium constant expression

Products appear in the numerator and reactants in the denominator, each raised to the power of its stoichiometric coefficient. The coefficient of HI is $2$ , so $[\text{HI}]$ is squared:

K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}.

Step 3: Substitute the equilibrium concentrations and evaluate

Replace each symbol with its numerical value and simplify. The units cancel because there are two concentration terms in the numerator and two in the denominator:

K_c = \frac{(1.60)^2}{(0.20)(0.20)} = \frac{2.56}{0.040} = 64.

Since $K_c = 64 \gg 1$ , the equilibrium position lies strongly towards the product HI; the forward reaction is favoured.

Final answer: $K_c = 64$ (dimensionless for this reaction)

Interpreting the magnitude

The size of $K_c$ shows how far the reaction proceeds:

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

Units

The units of $K_c$ depend on the equation, because the 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 $\dfrac{(\text{mol L}^{-1})^2}{(\text{mol L}^{-1})(\text{mol L}^{-1})^3} = (\text{mol L}^{-1})^{-2}$ .

Manipulating Kc when the equation changes

The form of $K_c$ depends on how the equation is written, so altering the equation alters the constant in predictable ways. If you reverse an equation, the new constant is the reciprocal, $K' = 1/K_c$ . If you multiply every coefficient by a factor $n$ , the new constant is $K_c$ raised to the power $n$ , $K' = (K_c)^n$ . If you add two equilibria, the overall constant is the product of the individual constants. These rules let you find a constant for a target reaction from constants for related reactions, and they parallel the way Hess's law combines enthalpies.

Kc for solutions and gases

For reactions in solution, the concentrations in $K_c$ are in $\text{mol L}^{-1}$ as usual. For gaseous equilibria, an equivalent constant $K_p$ can be written in terms of partial pressures, but $K_c$ remains valid provided concentrations are used. In every case, pure solids and pure liquids are omitted because their effective concentration does not change, and water as a solvent in dilute aqueous reactions is also left out. Always state the temperature with any $K_c$ value, because the constant is temperature dependent.

In the exam, write the balanced equation, build an ICE table where amounts change, convert to concentrations, substitute into the correctly powered expression, derive units if asked, 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.

TCE 20224 marksCompound

\text{A}

dissociates:

3\text{A}(g) \rightleftharpoons 2\text{B}(g) + \text{C}(g)

. At a particular temperature a closed

2.0\ \text{L}

container initially holds

1.2\ \text{mol}

\text{A}

. At equilibrium

0.75\ \text{mol}

\text{A}

remains. Determine the value of the equilibrium constant at this temperature.

Show worked answer →

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

Convert to concentrations (divide by $2.0\ \text{L}$ ): $[\text{A}] = 0.375$ ; $[\text{B}] = 0.150$ ; $[\text{C}] = 0.075\ \text{mol L}^{-1}$ . (1 mark)

K_c = \frac{[\text{B}]^2[\text{C}]}{[\text{A}]^3} = \frac{(0.150)^2(0.075)}{(0.375)^3} = \frac{1.69 \times 10^{-3}}{5.27 \times 10^{-2}} = 0.032.

(2 marks for the correct expression and value.)

TCE 20235 marksTo determine

K_c

for

\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

63.0\ \text{g}

of nitrogen and

15.0\ \text{g}

of hydrogen were placed in a

100\ \text{L}

vessel at constant temperature. At equilibrium,

58.0\ \text{g}

of ammonia was present. (

M(\text{H}_2) = 2.016

M(\text{N}_2) = 28.02

M(\text{NH}_3) = 17.03\ \text{g mol}^{-1}

.) Determine

K_c

Show worked answer →

Initial moles: $n(\text{N}_2) = 63.0/28.02 = 2.249$ ; $n(\text{H}_2) = 15.0/2.016 = 7.440$ . Equilibrium $n(\text{NH}_3) = 58.0/17.03 = 3.405\ \text{mol}$ . (1 mark)

Forming $3.405\ \text{mol}$ of $\text{NH}_3$ uses $n(\text{N}_2) = 3.405/2 = 1.702$ and $n(\text{H}_2) = \tfrac{3}{2}(3.405) = 5.107\ \text{mol}$ . Equilibrium: $n(\text{N}_2) = 0.547$ ; $n(\text{H}_2) = 2.333$ ; $n(\text{NH}_3) = 3.405\ \text{mol}$ . (2 marks)

Concentrations (divide by $100\ \text{L}$ ): $[\text{N}_2] = 5.47 \times 10^{-3}$ ; $[\text{H}_2] = 2.333 \times 10^{-2}$ ; $[\text{NH}_3] = 3.405 \times 10^{-2}\ \text{mol L}^{-1}$ . (1 mark)

K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(3.405 \times 10^{-2})^2}{(5.47 \times 10^{-3})(2.333 \times 10^{-2})^3} = 1.67 \times 10^{4}.

(1 mark)