The equilibrium constant Kc: TCE Chemistry (Tasmania)

Syllabus dot point

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

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; a small $K_c$ means reactants are favoured. Calculate it by finding equilibrium concentrations (often via an ICE table) and substituting into the expression. Only a change in temperature changes the value of $K_c$ .

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

Equilibrium constant expression

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

Products 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 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.

Only temperature changes Kc

Adding more reactant, changing pressure or volume, or adding a catalyst shifts the position of equilibrium but does not change the value of $K_c$ . Only a change in temperature alters $K_c$ , 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 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.

Putting reactants on top

The expression always has products in the numerator and reactants in the denominator; writing it upside down gives the reciprocal and a wildly wrong answer. Other slips: forgetting to raise each concentration to its coefficient (not multiply by it), forgetting to divide moles by the volume to get concentration, and using initial instead of equilibrium concentrations.

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 questions

Questions in the style of TASC 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.

Practice4 marks

Compound

\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.)

Practice5 marks

To 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)

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC