Calculating K and equilibrium concentrations

Using ICE tables to calculate K from experimental data and to find equilibrium concentrations from K, including perfect-square and quadratic problems and the small-x approximation, plus the colorimetry investigation that measures K for the iron(III) thiocyanate equilibrium.

You usually can't measure every concentration in an equilibrium mixture. Often you know what you put in and can measure just one thing at the end, perhaps the depth of a colour. The stoichiometry of the balanced equation lets you work out everything else. A tidy bookkeeping grid called an ICE table does it for you.

ICE tables

An ICE table has three rows for each species:

Initial concentration (or moles),
Change as the system reacts to equilibrium, which must follow the mole ratio in the equation (reactants lose, products gain),
Equilibrium concentration = Initial + Change.

Only the E row goes into the equilibrium expression. Work in mol L⁻¹, or work in moles and divide by the volume before substituting into K.

Type 1: finding K from data

K from initial amounts and one equilibrium measurement

0.400 mol of SO₂ and 0.200 mol of O₂ are placed in a 1.00 L vessel at 1000 K. At equilibrium, analysis shows 0.296 mol of SO₃ is present. Calculate K for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

SO₃ gained 0.296 mol L⁻¹. By the 2 : 1 : 2 ratio, SO₂ lost 0.296 and O₂ lost 0.148.

	2SO₂	O₂	2SO₃
I	0.400	0.200	0
C	−0.296	−0.148	+0.296
E	0.104	0.052	0.296

$K = \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \dfrac{(0.296)^2}{(0.104)^2(0.052)} = 156$

Type 2: finding concentrations from K

When K and the starting amounts are known, let the unknown change be $x$, write the E row in terms of $x$, substitute into K and solve.

A perfect square

0.50 mol of H₂ and 0.50 mol of I₂ are sealed in a 2.0 L flask at 450 °C, where K = 50 for H₂(g) + I₂(g) ⇌ 2HI(g). Find the equilibrium concentrations.

Initial concentrations are 0.50 ÷ 2.0 = 0.25 mol L⁻¹.

	H₂	I₂	2HI
I	0.25	0.25	0
C	−x	−x	+2x
E	0.25 − x	0.25 − x	2x

$50 = \dfrac{(2x)^2}{(0.25-x)^2}$. Both sides are perfect squares, so take the square root: $\dfrac{2x}{0.25-x} = \sqrt{50} = 7.07$

$2x = 1.768 - 7.07x \Rightarrow 9.07x = 1.768 \Rightarrow x = 0.195$

[HI] = 2x = 0.39 mol L⁻¹; [H₂] = [I₂] = 0.25 − 0.195 = 0.055 mol L⁻¹. Check: 0.39² ÷ 0.055² = 50 ✓

Using the quadratic formula

2.00 mol L⁻¹ CO and 1.00 mol L⁻¹ H₂O are heated to 1000 K, where K = 1.4 for CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). Find the equilibrium concentrations.

E row: [CO] = 2.00 − x, [H₂O] = 1.00 − x, [CO₂] = [H₂] = x

$1.4 = \dfrac{x^2}{(2.00-x)(1.00-x)}$, so $x^2 = 1.4(2.00 - 3.00x + x^2)$, which rearranges to $0.4x^2 - 4.2x + 2.8 = 0$

$x = \dfrac{4.2 \pm \sqrt{4.2^2 - 4(0.4)(2.8)}}{2(0.4)} = \dfrac{4.2 \pm 3.63}{0.8}$, giving $x = 0.715$ or $x = 9.79$

The second root is impossible: you can't use up 9.79 mol L⁻¹ of a reactant when only 1.00 is present. So [CO₂] = [H₂] = 0.715, [CO] = 1.285 and [H₂O] = 0.285 mol L⁻¹.

Always reject a root that is negative or that would make any equilibrium concentration negative.

The small-x approximation

When K is very small, very little reacts, so $x$ is tiny compared with the initial concentration and $(c - x) \approx c$. That turns a quadratic into a one-step calculation.

When K is small

0.100 mol L⁻¹ N₂O₄ is placed in a flask at 25 °C, where K = 4.6 × 10⁻³ for N₂O₄ ⇌ 2NO₂. Estimate [NO₂] at equilibrium.

E row: [N₂O₄] = 0.100 − x, [NO₂] = 2x. Assume 0.100 − x ≈ 0.100.

$4.6\times10^{-3} = \dfrac{(2x)^2}{0.100} \Rightarrow 4x^2 = 4.6\times10^{-4} \Rightarrow x = 0.0107$, so [NO₂] ≈ 0.021 mol L⁻¹

Check the assumption: x is 10.7% of 0.100, which is more than the usual 5% limit. Solving the full quadratic gives x = 0.0102 and [NO₂] = 0.020 mol L⁻¹, so the approximation is about 5% too high here. It works well when K is less than about 10⁻⁴ times the starting concentration.

You'll use this approximation constantly for weak acids in Module 6, where K values are often 10⁻⁵ or smaller.

Measuring K by colorimetry

Iron(III) ions and thiocyanate ions form a blood-red complex:

$$\text{Fe}^{3+}\text{(aq)} + \text{SCN}^-\text{(aq)} \rightleftharpoons \text{FeSCN}^{2+}\text{(aq)}$$

Fe³⁺ is very pale yellow and SCN⁻ is colourless at these concentrations, so the intensity of the red colour depends only on [FeSCN²⁺]. A colorimeter (or UV–visible spectrophotometer) measures how much light of a chosen colour a solution absorbs.

A colorimeter passes light of one colour through the sample and measures how much gets through.

The absorbance A is directly proportional to concentration (the Beer–Lambert law, $A = \varepsilon l c$, where ε is a constant for the substance and wavelength and l is the path length). Choose the colour that the sample absorbs most strongly, the complementary colour: a red solution absorbs blue-green light strongly.

Calibration. Make standards with a large excess of Fe³⁺ and small known amounts of SCN⁻. The excess pushes the equilibrium so far right that practically all the SCN⁻ becomes FeSCN²⁺, so [FeSCN²⁺] equals the SCN⁻ added. Measure their absorbances and plot a calibration graph.
Equilibrium mixtures. Mix comparable amounts of Fe³⁺ and SCN⁻, so that a real equilibrium mixture forms. Measure each absorbance and read [FeSCN²⁺] from the calibration line.
ICE. Initial concentrations come from the dilution ($c_1V_1 = c_2V_2$). Subtract [FeSCN²⁺] from each to find [Fe³⁺] and [SCN⁻] at equilibrium, then calculate K for each mixture and average.

Reading an unknown concentration from a calibration line.

InteractiveMeasuring K for FeSCN²⁺

Try this

With the Blue filter, look at the calibration graph. Now try Red. Why is red light useless for this solution?
Switch to 2. Equilibrium mixtures. Which mixture has the highest absorbance? Why?
Press Show the K calculation for each mixture. How close are the five values to each other?
Using mixture 1, check the sim's working by hand: the initial concentrations are 1.00 × 10⁻³ mol L⁻¹ Fe³⁺ and 2.0 × 10⁻⁴ mol L⁻¹ SCN⁻.

The absorbances are generated from a model with K = 138 plus small random errors. Published values for this equilibrium vary with ionic strength and temperature.

Investigation: determining K for iron(III) thiocyanate

Aim: to determine K for Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) by colorimetry.

Materials: 0.200 mol L⁻¹ Fe(NO₃)₃ in 0.5 mol L⁻¹ HNO₃ (for the standards), 2.00 × 10⁻³ mol L⁻¹ Fe(NO₃)₃ in 0.5 mol L⁻¹ HNO₃, 2.00 × 10⁻³ mol L⁻¹ KSCN, 0.5 mol L⁻¹ HNO₃, graduated pipettes or burettes, colorimeter with blue (about 450 nm) filter, cuvettes.

Method:

Prepare five standards, each containing 5.00 mL of 0.200 mol L⁻¹ Fe(NO₃)₃, 0 to 1.00 mL of 2.00 × 10⁻³ mol L⁻¹ KSCN, made up to 10.00 mL with HNO₃. Measure the absorbance of each and draw the calibration line.
Prepare five equilibrium mixtures, each with 5.00 mL of 2.00 × 10⁻³ mol L⁻¹ Fe(NO₃)₃ and 1.00, 2.00, 3.00, 4.00 or 5.00 mL of KSCN, made up to 10.00 mL with HNO₃.
Measure each absorbance, read [FeSCN²⁺] from the calibration line, and calculate K for each mixture.

Notes: the nitric acid stops Fe³⁺ from forming brown hydroxo complexes, which would also absorb. Measure everything at the same temperature, because K depends on temperature. Rinse the cuvette with the next solution and wipe it so no fingerprints are in the light path.

Safety: wear safety glasses; nitric acid is corrosive. Thiocyanate solutions must not be mixed with concentrated acid. Dispose of solutions as directed by your teacher.

Test yourself

1.0 mol of N₂ and 3.0 mol of H₂ are placed in a 2.0 L vessel. At equilibrium 0.40 mol of NH₃ is present. How many moles of H₂ remain?
For A(g) ⇌ 2B(g), the initial [A] = 0.50 mol L⁻¹ and no B is present. At equilibrium [B] = 0.20 mol L⁻¹. What is K?
Why are the calibration standards for the Fe³⁺/SCN⁻ experiment made with a large excess of Fe³⁺?
Which filter colour should be chosen to measure the absorbance of a red solution?

Put the steps for finding equilibrium concentrations from K in order.

Label the parts of a colorimeter.

4 marks

At a certain temperature, K = 64 for H₂(g) + I₂(g) ⇌ 2HI(g). 0.30 mol of H₂ and 0.30 mol of I₂ are sealed in a 1.5 L flask at this temperature. Calculate the equilibrium concentration of each species.

Your answer