E. Chemical Systems and Equilibrium · Chemical Equilibrium

The Equilibrium Law and Writing K Expressions

E3.2 E3.4 E2.1 E2.4 A1.12About 70 minutes

Learning goals

I can state the equilibrium law and write the equilibrium constant expression K for a balanced equation.
I can write K for heterogeneous equilibria, leaving out pure solids and liquids.
I can calculate K from equilibrium concentrations.
I can use the size of K to say whether reactants or products are favoured at equilibrium.

Words to know

equilibrium law (law of mass action)
For aA + bB ⇌ cC + dD at equilibrium, the ratio [C]ᶜ[D]ᵈ ÷ ([A]ᵃ[B]ᵇ) has a constant value at a given temperature.

equilibrium constant (K or Kc)
The constant value of the equilibrium law ratio, calculated from equilibrium concentrations in mol/L. It depends only on the reaction and the temperature.

equilibrium constant expression
The ratio of product concentrations to reactant concentrations, each raised to the power of its coefficient in the balanced equation.

homogeneous equilibrium
An equilibrium in which all reactants and products are in the same phase.

heterogeneous equilibrium
An equilibrium involving more than one phase; pure solids and pure liquids are left out of K.

One number to describe a mixture

Flasks of H₂, I₂ and HI can end up with very different equilibrium mixtures depending on what you put in. That sounds messy. In 1864 two Norwegian chemists, Cato Guldberg and Peter Waage, found the order hiding inside the mess: for any reaction at a fixed temperature, one particular ratio of the equilibrium concentrations always comes out the same. That ratio is the equilibrium constant, K, and it is the most useful number in this unit.

Discovering the pattern

Here are three experiments with H₂(g) + I₂(g) ⇌ 2 HI(g), all at the same temperature (about 450 °C). All concentrations are equilibrium values in mol/L.

Trial	[H₂]	[I₂]	[HI]	[HI] ÷ ([H₂][I₂])	[HI]² ÷ ([H₂][I₂])
1	0.220	0.220	1.56	32.2	50.3
2	0.533	0.0328	0.934	53.4	49.9
3	0.110	0.110	0.780	64.5	50.3

Only the ratio with [HI] squared (its coefficient is 2) is constant.

Three different equilibrium mixtures of H₂, I₂ and HI at the same temperature. The concentrations differ, but [HI]² ÷ ([H₂][I₂]) is about 50 in every flask.

The equilibrium law

For the general balanced equation

aA + bB ⇌ cC + dD

the equilibrium constant expression is

K = [C]ᶜ[D]ᵈ ÷ ([A]ᵃ[B]ᵇ)

Products go on top, reactants on the bottom.
Each concentration is raised to the power of its coefficient in the balanced equation.
The concentrations are equilibrium values in mol/L.
K has one value for a reaction at a given temperature. Changing the starting amounts, the volume or the pressure does not change K; only a change in temperature does.

In this course, K is written without units. (In university chemistry you will see that K is properly defined from unitless ratios called activities, which is why this works.) Some books write Kc to show that it uses concentrations.

Worked example

Example 1: Writing K expressions

Write the equilibrium constant expression for each reaction.

(a) N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) (b) 2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g) (c) 4 NH₃(g) + 5 O₂(g) ⇌ 4 NO(g) + 6 H₂O(g)

(a) Product NH₃ with power 2; reactants N₂ (power 1) and H₂ (power 3): K = [NH₃]² ÷ ([N₂][H₂]³).
(b) K = [SO₂]²[O₂] ÷ [SO₃]².
(c) Water here is a gas, so it is included: K = [NO]⁴[H₂O]⁶ ÷ ([NH₃]⁴[O₂]⁵).

(a) K = [NH₃]²/([N₂][H₂]³); (b) K = [SO₂]²[O₂]/[SO₃]²; (c) K = [NO]⁴[H₂O]⁶/([NH₃]⁴[O₂]⁵).

Heterogeneous equilibria: leave out solids and liquids

The "concentration" of a pure solid or pure liquid is fixed by its density: a lump of CaCO₃ has the same number of moles per litre whether the lump is big or small. Because it cannot change, it is built into the value of K and left out of the expression. Water is also left out when it is the solvent in a dilute aqueous solution, because its concentration hardly changes.

CaCO₃(s) ⇌ CaO(s) + CO₂(g): K = [CO₂]
H₂O(l) ⇌ H₂O(g): K = [H₂O(g)]
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq): K = [Ag⁺][Cl⁻] (this special K is the solubility product, Ksp, studied in Acid-Base and Solubility Equilibria)

Include (g) and (aq) species. Leave out (s), (l), and water acting as the solvent.

Worked example

Example 2: Calculating K from equilibrium concentrations

At 25 °C, an equilibrium mixture of N₂O₄(g) ⇌ 2 NO₂(g) contains [N₂O₄] = 0.0898 mol/L and [NO₂] = 0.0203 mol/L. Calculate K.

Write the expression: K = [NO₂]² ÷ [N₂O₄].
Substitute equilibrium values: K = (0.0203)² ÷ 0.0898.
K = 4.121 × 10⁻⁴ ÷ 0.0898 = 4.59 × 10⁻³.

K = 4.59 × 10⁻³ (3 significant figures). Because K is much less than 1, the reactant N₂O₄ is favoured at 25 °C.

Worked example

Example 3: K with a heterogeneous system

For C(s) + CO₂(g) ⇌ 2 CO(g) at a high temperature, an equilibrium mixture contains [CO₂] = 0.20 mol/L and [CO] = 0.60 mol/L, with some solid carbon left. Find K.

Carbon is a solid, so it is left out: K = [CO]² ÷ [CO₂].
K = (0.60)² ÷ 0.20 = 0.36 ÷ 0.20 = 1.8.

K = 1.8. The amount of solid carbon does not matter, as long as some is present.

What the size of K tells you

Because products are on top, a big K means a lot of product at equilibrium.

K > 1: products are favoured. If K is enormous (say 10¹⁰ or more), the reaction goes essentially to completion, and we usually just write →.
K ≈ 1: reactants and products are present in comparable amounts.
K < 1: reactants are favoured. If K is tiny, hardly any product forms.

K tells you how far a reaction goes, not how fast. A reaction with a huge K can still be very slow (rates are studied in Energy Changes and Rates of Reaction).

The size of K on a scale where each tick is 100 times bigger than the last. N₂O₄ ⇌ 2 NO₂ at 25 °C lies on the reactant side; H₂ + I₂ ⇌ 2 HI at about 450 °C lies on the product side.

Just for fun

Guldberg and Waage were brothers-in-law as well as research partners. Their law of mass action proved that some relationships really do stay constant at a fixed temperature, even around a family dinner table.

Hands-on activity

Testing the pattern on a spreadsheet

You need: a computer or phone with a free spreadsheet app (or paper and a calculator).

Enter the three trials from the table at the top of this lesson in columns: [H₂], [I₂], [HI].
In new columns, calculate four different ratios for each trial: [HI]/([H₂][I₂]), [HI]²/([H₂][I₂]), [HI]/([H₂] + [I₂]) and ([H₂][I₂])/[HI]².
Which ratio is the same (within rounding) for all three trials? Which other ratio is also constant, and how is it related?

What to look for: [HI]²/([H₂][I₂]) ≈ 50 every time, and ([H₂][I₂])/[HI]² ≈ 0.020 every time. The second is just the first turned upside down (1 ÷ 50 = 0.020): it is K for the reverse reaction. Sums and ratios without the exponent do not stay constant. This is exactly the kind of data analysis Guldberg and Waage did by hand.

Your turn

Write K for 2 NO(g) + O₂(g) ⇌ 2 NO₂(g).

Write K for H₂(g) + Cl₂(g) ⇌ 2 HCl(g).

Write K for NH₄Cl(s) ⇌ NH₃(g) + HCl(g).

Write K for Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq).

For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), an equilibrium mixture has [PCl₅] = 1.60 mol/L, [PCl₃] = 0.40 mol/L and [Cl₂] = 0.40 mol/L. Find K.

For which reaction are products favoured: one with K = 3.2 × 10⁻⁵ or one with K = 2.1 × 10⁴?

Why is the amount of solid CaCO₃ left out of K for CaCO₃(s) ⇌ CaO(s) + CO₂(g)?

More practice: warm-up

In a K expression, do products go on the top or the bottom?

What do the exponents in a K expression come from?

What is the only thing that changes the value of K for a given reaction?

Write K for CO(g) + Cl₂(g) ⇌ COCl₂(g).

A reaction has K = 1.0 × 10⁻¹². Are products or reactants favoured?

More practice: on your own

Write K for 2 H₂S(g) + 3 O₂(g) ⇌ 2 H₂O(g) + 2 SO₂(g).

Write K for Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s).

For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), an equilibrium mixture contains [N₂] = 0.400 mol/L, [H₂] = 1.20 mol/L and [NH₃] = 0.200 mol/L. Find K.

For 2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g), [SO₃] = 1.40 mol/L, [SO₂] = 0.60 mol/L and [O₂] = 0.30 mol/L at equilibrium. Find K.

For CaCO₃(s) ⇌ CaO(s) + CO₂(g), K = 0.010 at a certain temperature. A sealed container holds some CaCO₃ and CaO. What is [CO₂] at equilibrium? Does it depend on how much CaCO₃ is present?

Challenge

For 2 HI(g) ⇌ H₂(g) + I₂(g) at about 450 °C, use the trial 1 data in this lesson to calculate K. How is it related to K for H₂ + I₂ ⇌ 2 HI?

For H₂(g) + I₂(g) ⇌ 2 HI(g), K = 50 at about 450 °C. An equilibrium mixture has [H₂] = 0.100 mol/L and [I₂] = 0.200 mol/L. What is [HI]?

A student writes K = [CO₂][H₂O] ÷ [NaHCO₃]² for 2 NaHCO₃(s) ⇌ Na₂CO₃(s) + CO₂(g) + H₂O(g). Find and fix the error.

Check your understanding

Quiz

What is K for 2 A(g) + B(g) ⇌ 3 C(g)?

For Fe₃O₄(s) + 4 H₂(g) ⇌ 3 Fe(s) + 4 H₂O(g), the K expression is:

Which change alters the value of K for a reaction?

A reaction has K = 6.0 × 10⁵ at 25 °C. At equilibrium:

For PCl₅ ⇌ PCl₃ + Cl₂ (all gases), [PCl₅] = 0.50, [PCl₃] = 0.20 and [Cl₂] = 0.10 mol/L at equilibrium. K is:

Why is liquid water left out of K for a reaction in dilute aqueous solution?

For parents and teachers

Big idea (E3.2, E3.4, E2.4): at equilibrium, K = [products]^coefficients ÷ [reactants]^coefficients is a constant for a given reaction and temperature. Pure solids and liquids are left out. K > 1 favours products; K < 1 favours reactants.

How to help: give your teen a balanced equation and ask them to say the K expression out loud ("products over reactants, coefficients become powers, skip solids and liquids"). Ask them to explain why the size of K says nothing about speed.

Common difficulties:

Multiplying by the coefficient instead of using it as an exponent.
Including solids or liquid water in K, or leaving out water vapour.
Using starting concentrations instead of equilibrium concentrations (the next two lessons fix this with ICE tables).
Thinking K changes when more reactant is added. Only temperature changes K.