Calculating the Equilibrium Constant: Study Guide (Chemistry 7.4)

From measurements to a number

In Topic 7.3 you learned to write the expression for K. In this topic you put real measurements into it. The idea is simple:

Measure the concentrations (or partial pressures) of the species at equilibrium, substitute them into the expression from the balanced equation, and calculate.

Chemists measure equilibrium amounts in several ways. A coloured species can be measured with a spectrophotometer, because absorbance is proportional to concentration (Topic 3.13). The pressure of a gas mixture can be read from a gauge. A sample can be removed and titrated. Whatever the method, the numbers are only useful if they are taken at equilibrium.

How do you know it is at equilibrium?

A system is at equilibrium when its concentrations or partial pressures stop changing. In an experiment, you take readings over time and wait until they are constant. Figure 1 shows a reaction approaching equilibrium. The readings to use for K are the ones on the flat part of the curves.

Notice two things in the figure. First, the reactants do not run out: both are still present at equilibrium. Second, the changes are linked by the equation. For every 1 mol L⁻¹ of CH₄ used up, 2 mol L⁻¹ of H₂S are used up and 4 mol L⁻¹ of H₂ are made. You will use that link in Worked example 2.

The method in four steps

1. Write the balanced equation with state symbols.
2. Write the expression for Kc or Kp, leaving out pure solids and liquids.
3. Get every equilibrium value in the right form: mol L⁻¹ for Kc, atm for Kp. Convert moles to concentrations (divide by volume), or a total pressure to partial pressures, if you need to.
4. Substitute and calculate. Give K to the number of significant figures in the data, with no unit.

Worked example 1: substituting measured values

Question. (a) For 2NOCl(g) ⇌ 2NO(g) + Cl₂(g), an equilibrium mixture contains [NOCl] = 0.0500 mol L⁻¹, [NO] = 0.0100 mol L⁻¹ and [Cl₂] = 0.00500 mol L⁻¹. Calculate Kc. (b) For 2NO₂(g) ⇌ 2NO(g) + O₂(g), the equilibrium partial pressures in a vessel are P_NO₂ = 0.400 atm, P_NO = 0.100 atm and P_O₂ = 0.0500 atm. Calculate Kp.

(a)

1. Kc = [NO]²[Cl₂] / [NOCl]².
2. Substitute: Kc = (0.0100)² × 0.00500 / (0.0500)².
3. Numerator: 1.00 × 10⁻⁴ × 0.00500 = 5.00 × 10⁻⁷. Denominator: 2.50 × 10⁻³.
4. Kc = 5.00 × 10⁻⁷ ÷ 2.50 × 10⁻³ = 2.00 × 10⁻⁴.

(b)

1. Kp = (P_NO)² × P_O₂ / (P_NO₂)².
2. Substitute: Kp = (0.100)² × 0.0500 / (0.400)² = 5.00 × 10⁻⁴ / 0.160.
3. Kp = 3.13 × 10⁻³.

Check. Both answers are three significant figures, like the data. Neither has a unit. Part (b) used only pressures, because it asks for Kp.

When only one species is measured: the ICE table

Often you know what you put in at the start, and you measure just one species at equilibrium. The balanced equation then tells you how much of everything else changed. An ICE table (Initial, Change, Equilibrium) keeps this tidy:

• I: starting concentrations (or pressures).
• C: the change for each species, in the ratio of the coefficients. Reactants go down (−), products go up (+).
• E: I + C for each species.

Work in concentrations from the start. If you are given moles, divide by the volume first.

Worked example 2: starting amounts plus one measurement

Question. 0.400 mol of CH₄ and 0.800 mol of H₂S are sealed in a 2.00 L vessel and heated. The reaction is CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g). At equilibrium the vessel contains 0.0400 mol of CS₂. Calculate Kc at this temperature.

1. Concentrations. [CH₄]₀ = 0.400 ÷ 2.00 = 0.200 mol L⁻¹; [H₂S]₀ = 0.800 ÷ 2.00 = 0.400 mol L⁻¹; [CS₂]ₑ = 0.0400 ÷ 2.00 = 0.0200 mol L⁻¹.
2. ICE table (mol L⁻¹). Let the change in CS₂ be +0.0200; the other changes follow the 1 : 2 : 1 : 4 ratio.

	CH₄	H₂S	CS₂	H₂
Initial	0.200	0.400	0	0
Change	−0.0200	−0.0400	+0.0200	+0.0800
Equilibrium	0.180	0.360	0.0200	0.0800

3. Expression. Kc = [CS₂][H₂]⁴ / ([CH₄][H₂S]²).
4. Substitute. Numerator: 0.0200 × (0.0800)⁴ = 0.0200 × 4.096 × 10⁻⁵ = 8.192 × 10⁻⁷. Denominator: 0.180 × (0.360)² = 0.180 × 0.1296 = 0.023328.
5. Kc = 8.192 × 10⁻⁷ ÷ 0.023328 = 3.51 × 10⁻⁵.

Why each step matters. Using moles instead of concentrations gives 1.40 × 10⁻⁴, four times too big, because the powers on top (1 + 4 = 5) and bottom (1 + 2 = 3) do not cancel the volume. Using the starting values for CH₄ and H₂S gives 2.56 × 10⁻⁵, which mixes a starting mixture with an equilibrium one and is not K at all.

Worked example 3: a total pressure with a solid

Question. Solid ammonium carbamate is placed in an evacuated flask and heated. It decomposes: NH₂COONH₄(s) ⇌ 2NH₃(g) + CO₂(g). When equilibrium is reached, the total pressure is 0.300 atm and some solid remains. Calculate Kp.

1. Expression. The solid is left out: Kp = (P_NH₃)² × P_CO₂.
2. Partial pressures. All the gas comes from the solid, in the ratio 2 NH₃ : 1 CO₂. So NH₃ is 2/3 of the total pressure and CO₂ is 1/3: P_NH₃ = 0.200 atm, P_CO₂ = 0.100 atm. (Check: 0.200 + 0.100 = 0.300 atm.)
3. Substitute. Kp = (0.200)² × 0.100 = 0.0400 × 0.100 = 4.00 × 10⁻³.

Common slip. Splitting the total pressure in half for each gas (0.150 atm each) ignores the 2 : 1 ratio and gives 3.38 × 10⁻³.

Reading K as a relationship between variables

K is fixed at a given temperature. So if one equilibrium concentration is different in another experiment, the others must be different in a way that keeps K the same. The powers tell you by how much.

For 2A(g) ⇌ B(g), Kc = [B] / [A]². If a second equilibrium mixture at the same temperature has [A] twice as large, then [A]² is four times as large, so [B] must also be four times as large to keep the ratio equal to Kc.

This is also a good test of your own data. If two experiments at the same temperature give values of K that differ by much more than the measurement uncertainty, one of the calculations (or measurements) is probably wrong.

Common misconceptions

• Using starting amounts in K. Only equilibrium values go into K. Starting values give Q, not K.
• Using moles instead of concentrations. Divide by the volume first. It only “works” in special cases.
• Ignoring the mole ratio in the change row. If CS₂ goes up by 0.0200, H₂ goes up by 4 × 0.0200, not 0.0200.
• Getting the signs wrong. Reactants decrease and products increase when the reaction moves forward from a start with no products.
• Splitting a total pressure equally. Partial pressures follow the mole ratio of the gases.
• Including the solid. A solid must be present, but it is not in the expression.
• “More reactant gives a bigger K.” Different starting amounts give different equilibrium concentrations but the same K at the same temperature.
• Giving K a unit. In this course K is written as a plain number.

Where this leads

Next, in Topic 7.5, you will use the size of K to judge whether a reaction goes almost to completion or barely happens. Later, Topic 7.7 runs this topic’s method backwards: you will be given K and starting amounts and asked to find the equilibrium concentrations. If writing the expression still feels shaky, review Topic 7.3.

Try the practice questions now, then use the revision notes and the checklist to consolidate.