Topic 13 · page 1 of 12

Writing Kc

At a fixed temperature the equilibrium concentrations of a reaction always fit one expression, and the value of that expression is the equilibrium constant Kc.

12 min read13.1

A number for every equilibrium

In the Unit 2 equilibria topic you met dynamic equilibrium: the forward and backward rates are equal, so the concentrations stay constant. You could say which way a change pushes the mixture. You could not say how far the reaction goes, or what stays the same while it moves.

The equilibrium constant answers both. It is one number that describes a reaction at one temperature. This page writes the expression for it using concentrations, which gives $K_\text{c}$ . The next pages add pressures, units, calculations, and then what changes the number.

The equilibrium constant 13.1

Here are three equilibrium mixtures of hydrogen, iodine and hydrogen iodide, all at the same temperature. The data are illustrative. Each started from different amounts, so each settled at different concentrations.

\ce{H2(g) + I2(g) <=> 2HI(g)}

Mixture	[H₂] / mol dm⁻³	[I₂] / mol dm⁻³	[HI] / mol dm⁻³	[HI]² ÷ ([H₂][I₂])
1	0.0020	0.0050	0.0224	50
2	0.0040	0.0040	0.0283	50
3	0.0100	0.0010	0.0224	50

The concentrations differ, but the combination in the last column is the same every time. That combination, with each concentration raised to the power of its coefficient in the equation, is the equilibrium constant.

Square brackets mean "concentration of", in mol dm⁻³. Concentrations during the reaction, before equilibrium is reached, give a different value. The expression equals $K_\text{c}$ only once the rates are equal.

Writing the expression 13.1

MethodWriting a Kc expression

Start from a balanced equation. The coefficients become the powers.
Products go on top, reactants on the bottom.
Write each substance as [X], raised to the power of its coefficient.
Multiply the terms on the top, and multiply the terms on the bottom.
Drop the power when the coefficient is 1.

Worked exampleThree expressions

\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad K_\text{c} = \frac{[\ce{NH3}]^2}{[\ce{N2}]\,[\ce{H2}]^3}

\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \qquad K_\text{c} = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^2\,[\ce{O2}]}

\ce{2HI(g) <=> H2(g) + I2(g)} \qquad K_\text{c} = \frac{[\ce{H2}]\,[\ce{I2}]}{[\ce{HI}]^2}

Homogeneous systems 13.1

In a homogeneous system every species appears in the expression. The three examples above are all gases. Two more:

\ce{Fe^3+(aq) + SCN-(aq) <=> [Fe(SCN)]^2+(aq)} \qquad K_\text{c} = \frac{[\ce{[Fe(SCN)]^2+}]}{[\ce{Fe^3+}]\,[\ce{SCN-}]}

\ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)} \qquad K_\text{c} = \frac{[\ce{CH3COOC2H5}]\,[\ce{H2O}]}{[\ce{CH3COOH}]\,[\ce{C2H5OH}]}

In the second equation all four substances are liquids mixed together, and water is a product like the others, so it stays in. Page 3 deals with solids and with water as a solvent.

The same reaction, written differently 13.1

The value of $K_\text{c}$ belongs to the equation as you write it. Change the equation and you change $K_\text{c}$ in a predictable way.

Worked exampleSame reaction, three equations

Suppose $K_\text{c}$ = 0.040 (illustrative) for $\ce{A <=> B}$ .

Reverse: $\ce{B <=> A}$ has $K_\text{c}$ = 1 ÷ 0.040 = 25.
Double: $\ce{2A <=> 2B}$ has $K_\text{c}$ = 0.040² = 1.6 × 10⁻³.

This is why an exam question always gives you the equation, and why your expression must match it.

Check yourself

Write the expression for $K_\text{c}$ for each reaction. (a) $\ce{2NO(g) + O2(g) <=> 2NO2(g)}$ (b) $\ce{N2O4(g) <=> 2NO2(g)}$ (c) $\ce{CH3COOH(l) + C2H5OH(l) <=> CH3COOC2H5(l) + H2O(l)}$
A student writes $K_\text{c} = \dfrac{[2\ce{NH3}]}{[\ce{N2}] + 3[\ce{H2}]}$ for the Haber process. Give the correct expression and name the two mistakes.
At one temperature an equilibrium mixture contains 0.0030 mol dm⁻³ H₂, 0.0060 mol dm⁻³ I₂ and 0.0300 mol dm⁻³ HI. Calculate $K_\text{c}$ for $\ce{H2(g) + I2(g) <=> 2HI(g)}$ (value only; units come on page 4).
Use your answer to question 3 to find $K_\text{c}$ for $\ce{2HI(g) <=> H2(g) + I2(g)}$ at the same temperature.
At the same temperature a second equilibrium mixture has [H₂] = 0.0100 mol dm⁻³ and [I₂] = 0.0020 mol dm⁻³. Calculate [HI].
Why must the concentrations in $K_\text{c}$ be equilibrium concentrations?

Answers cover these until you have tried the questions

(a) $K_\text{c} = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2\,[\ce{O2}]}$ . (b) $K_\text{c} = \dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$ . (c) $K_\text{c} = \dfrac{[\ce{CH3COOC2H5}]\,[\ce{H2O}]}{[\ce{CH3COOH}]\,[\ce{C2H5OH}]}$ . All four species are liquids in one mixture, so water stays in.
Correct: $K_\text{c} = \dfrac{[\ce{NH3}]^2}{[\ce{N2}]\,[\ce{H2}]^3}$ . The first mistake is adding the reactant terms instead of multiplying them. The second is putting the coefficient in front of the bracket instead of as a power (2 in [2NH₃], and 3 in 3[H₂], should be the powers [NH₃]² and [H₂]³).
$K_\text{c} = \dfrac{0.0300^2}{0.0030 \times 0.0060} = \dfrac{9.00 \times 10^{-4}}{1.8 \times 10^{-5}} =$ 50
Reversing the equation gives the reciprocal: 1 ÷ 50 = 0.020.
$K_\text{c}$ = 50 at this temperature, so $[\ce{HI}]^2$ = 50 × 0.0100 × 0.0020 = 1.00 × 10⁻³, and [HI] = √(1.00 × 10⁻³) = 0.032 mol dm⁻³ (0.0316).
Only at equilibrium are the forward and backward rates equal and the concentrations constant. Before that the concentrations are still changing, and the expression gives a different value that is not $K_\text{c}$ .