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Chemical Equilibria

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Edexcel IAL Chemistry · Unit 4

Chemical Equilibria

Big idea: when a reversible reaction reaches equilibrium, there's a fixed "scorecard" ratio of products to reactants — called the equilibrium constant — and that number only changes if you change the temperature. Everything else (concentration, pressure, catalysts) can shift the equilibrium's *position*, but never the constant itself.

Summary — What This Chapter Covers

K_c — the equilibrium constant written in terms of concentrations (mol dm⁻³), used for reactions in solution.
K_p — the equilibrium constant written in terms of partial pressures (Pa or kPa), used for gaseous reactions.
Solids (and liquids, for K_p) are left out of equilibrium expressions entirely.
Calculating K_c or K_p often needs an ICE table (Initial, Change, Equilibrium) to find missing quantities.
K_p calculations sometimes require converting moles → mole fraction → partial pressure first.
Only a change in temperature changes the value of K_c or K_p. Concentration, pressure, and catalysts shift the position of equilibrium but leave K unchanged.
Entropy links to equilibrium through ΔS_total = R ln K — a big K means the reaction is strongly product-favoured entropically.

1. The Equilibrium Constant, K_c

1.1 Writing the K_c Expression

For any reversible reaction that has settled into equilibrium, we can write an expression that links the equilibrium constant to how much of each substance is hanging around. For a general reaction:

General Reaction

aA + bB ⇌ cC + dD

the equilibrium constant expression is:

K_c Expression

K_c = [C]^c[D]^d ÷ [A]^a[B]^b

Products (raised to their coefficients) on top, reactants on the bottom — always. The little letters a, b, c, d become powers.

Think of it like a recipe ratio at the finish line: once the reaction stops changing overall (even though forward and backward reactions are still happening — it's dynamic equilibrium), you measure how much of everything is present and plug it into this fraction.

Golden Rule

Solids are never included in the K_c expression. Only aqueous, liquid, and gaseous species that actually have a meaningful "concentration" that changes get a slot in the expression. A solid's concentration doesn't really change (its "concentration" is just a fixed density), so it's treated as a constant and folded into K_c itself.

Worked Example — Deducing K_c

Write the K_c expression for: Ag⁺(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq)

Answer:

K_c = [Fe³⁺(aq)] ÷ ([Fe²⁺(aq)][Ag⁺(aq)])

Notice Ag(s) is completely missing — it's a solid, so it never appears in the expression at all, not even as a "1".

Practice Question

Deduce the K_c expression for: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

1.2 Calculating K_c from Moles and Volume

Exam questions rarely hand you concentrations directly — instead they usually give you the number of moles of each substance at equilibrium plus the total volume of the mixture. You need to convert moles into concentration first:

Concentration from Moles

concentration (mol dm⁻³) = moles ÷ volume (dm³)

Worked Example — Calculating K_c

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

500 cm³ of the equilibrium mixture contains 0.235 mol ethanoic acid, 0.035 mol ethanol, 0.182 mol ethyl ethanoate, and 0.182 mol water. Find K_c.

Step 1 — concentrations (divide each by 0.500 dm³):

[CH₃COOH] = 0.235 / 0.500 = 0.470 mol dm⁻³
[C₂H₅OH] = 0.035 / 0.500 = 0.070 mol dm⁻³
[CH₃COOC₂H₅] = 0.182 / 0.500 = 0.364 mol dm⁻³
[H₂O] = 0.182 / 0.500 = 0.364 mol dm⁻³

Step 2 — expression: K_c = [H₂O][CH₃COOC₂H₅] ÷ ([C₂H₅OH][CH₃COOH])

Step 3 — substitute: K_c = (0.364 × 0.364) ÷ (0.070 × 0.470) = 4.03

Step 4 — units: every unit is mol dm⁻³, two on top and two on bottom, so they all cancel — K_c has no units here.

When You're Only Given Initial Amounts

If a question only gives you the starting amounts and the equilibrium amount of just one substance, build an ICE table (Initial, Change, Equilibrium). Use the reaction's mole ratio to work out how much every other substance must have changed by, since they all change in the same fixed proportions dictated by the balanced equation.

Worked Example — Using an ICE Table

CH₃COOC₂H₅(l) + H₂O(l) ⇌ CH₃COOH(l) + C₂H₅OH(l)

0.1000 mol ethyl ethanoate + 0.1000 mol water, made up to 1.00 dm³. At equilibrium, 0.0654 mol water remains.

	CH₃COOC₂H₅	H₂O	CH₃COOH	C₂H₅OH
Initial	0.1000	0.1000	0	0
Change	−0.0346	−0.0346	+0.0346	+0.0346
Equilibrium	0.0654	0.0654	0.0346	0.0346

Water dropped from 0.1000 to 0.0654, a change of −0.0346 mol. Because the ratio is 1:1:1:1, every other species changes by the same amount (0.0346), just in the direction the equation dictates.

K_c = (0.346 × 0.346) ÷ (0.654 × 0.654) = 0.28 (no units — they all cancel)

Practice Question

A 250 cm³ flask at equilibrium contains 0.40 mol of A, 0.20 mol of B, and 0.60 mol of C for the reaction A(aq) + B(aq) ⇌ C(aq). Calculate K_c including units.

2. The Equilibrium Constant, K_p

K_p is basically K_c's twin for gas-phase reactions — instead of concentrations, we use partial pressures (the pressure each individual gas would exert if it alone filled the container).

K_p Expression (for aA(g) + bB(g) ⇌ cC(g) + dD(g))

K_p = p_C^c · p_D^d ÷ (p_A^a · p_B^b)

Same shape as K_c, just swap concentrations for partial pressures (in kPa or Pa).

Heterogeneous Reactions

For reactions with a mix of states (e.g. a solid decomposing to give a gas), both solids and liquids are ignored in the K_p expression — only gases get a slot. E.g. for CaCO₃(s) ⇌ CaO(s) + CO₂(g), the whole expression collapses down to just K_p = p_CO₂.

Worked Example — Direct Partial Pressures

2SO₂(g) + O₂(g) ⇌ 2SO₃(g), with p(SO₂) = 1.0×10⁶ Pa, p(O₂) = 7.0×10⁶ Pa, p(SO₃) = 8.0×10⁶ Pa.

K_p = p²(SO₃) ÷ [p²(SO₂) × p(O₂)] = (8.0×10⁶)² ÷ [(1.0×10⁶)² × (7.0×10⁶)]

K_p = 9.1 × 10⁻⁶ Pa⁻¹

Units: Pa² ÷ (Pa² × Pa) = Pa⁻¹ — always work these out from the raw expression, don't guess.

2.1 When You're Only Given Moles and Total Pressure

Sometimes you don't get partial pressures directly — instead you're given the number of moles of each gas at equilibrium and the total pressure of the container. You need two extra steps first:

Mole fraction of each gas = (moles of that gas) ÷ (total moles of all gases)
Partial pressure of each gas = (mole fraction) × (total pressure)

Mole Fraction → Partial Pressure

p_gas = x_gas × P_total

Worked Example — Mole Fraction Route

H₂(g) + I₂(g) ⇌ 2HI(g) at 600 K. Moles at equilibrium: H₂ = 1.71×10⁻³, I₂ = 2.91×10⁻³, HI = 1.65×10⁻². Total pressure = 100 kPa.

Step 1 — total moles: 1.71×10⁻³ + 2.91×10⁻³ + 1.65×10⁻² = 2.112×10⁻²

Step 2 — mole fractions: H₂ = 0.0810, I₂ = 0.1378, HI = 0.7813

Step 3 — partial pressures (× 100 kPa): H₂ = 8.10 kPa, I₂ = 13.78 kPa, HI = 78.13 kPa

Step 4 — K_p expression: K_p = p²(HI) ÷ [p(H₂) × p(I₂)]

Step 5 — substitute: K_p = 78.13² ÷ (8.10 × 13.78) = 54.7

Step 6 — units: Pa² ÷ (Pa × Pa) — all units cancel, so K_p has no units here.

Practice Question

At equilibrium a container holds 0.60 mol N₂, 1.80 mol H₂, and 0.60 mol NH₃ at a total pressure of 200 kPa, for N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Calculate K_p.

3. Equilibrium Constant & Changing Conditions

This is the section students trip up on most, so let's be crystal clear about the difference between two things that sound similar but aren't:

Position of equilibrium — which side (reactants or products) currently has more "stuff" in it. This shifts around with concentration, pressure, and temperature changes (think Le Chatelier).
Equilibrium constant (K) — the actual numerical value of that products-over-reactants ratio. This is a fixed number for a given reaction at a given temperature, and only temperature can change it.

The One Rule to Remember

Concentration changes, pressure changes, and catalysts can all shift the position of equilibrium — but the system always settles back down to the same value of K. Only changing the temperature actually changes what K itself equals.

3.1 Effect of Temperature

Change	How the equilibrium shifts
Increase in temperature	Shifts in the endothermic direction, to absorb the extra heat and reverse the change
Decrease in temperature	Shifts in the exothermic direction, to release heat and reverse the change

For a reaction that's exothermic in the forward direction: raising the temperature pushes the equilibrium from right back to left (favouring reactants), so the ratio [products]/[reactants] gets smaller — meaning K decreases.

For an endothermic forward reaction: raising the temperature pushes equilibrium further right (favouring products), so [products]/[reactants] gets bigger — meaning K increases.

3.2 Effect of Pressure

Pressure changes only matter for reactions involving gases.

Change	How the equilibrium shifts
Increase in pressure	Shifts towards the side with the smaller number of gas moles (to reduce pressure again)
Decrease in pressure	Shifts towards the side with the larger number of gas moles (to increase pressure again)

Crucially: neither K_c nor K_p is affected by pressure changes. The system shifts to a brand new equilibrium position, but when you recalculate K from the new concentrations or partial pressures, you get the exact same number as before. It's the same idea as adding more of one reactant to a K_c equilibrium in solution — the position moves, but K stays put.

3.3 Effect of a Catalyst

A catalyst speeds up the forward and reverse reactions by exactly the same factor. That means the system reaches equilibrium faster, but it arrives at the same position and the same value of K it would have reached anyway, just quicker. Catalysts never appear in the equilibrium expression and never change K.

Worked Example

AB(aq) + CD(aq) ⇌ AC(aq) + BD(aq), ΔH = +180 kJ mol⁻¹. Which factors would change the value of K_c?

Answer: Only a change in temperature. Adding a catalyst speeds up reaching equilibrium but doesn't move its position or change K_c. Changing concentration would shift the position but the system would settle back to the same K_c.

Practice Question

For the exothermic reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), predict what happens to the value of K_c if (a) the pressure is increased, and (b) the temperature is increased.

4. Equilibrium Constant & Temperature — The Mechanics

It helps to actually watch the numbers move, not just memorise the rule. Take the endothermic decomposition of hydrogen iodide:

Endothermic Example

2HI(g) ⇌ H₂(g) + I₂(g)

with K_c = [H₂][I₂] ÷ [HI]². As temperature increases:

[H₂] and [I₂] increase (more HI breaks down)
[HI] decreases
Top of the fraction goes up, bottom goes down → K_c increases

Exothermic Example

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

with K_c = [SO₃]² ÷ ([SO₂]²[O₂]). As temperature increases:

[SO₃] decreases
[SO₂] and [O₂] increase
Top goes down, bottom goes up → K_c decreases

Quick Way to Remember It

Ask yourself: "Is the forward reaction endo or exo?" Then: heating always favours the endothermic direction. If forward is endothermic, heating pushes forward → more products → bigger K. If forward is exothermic, heating pushes backward → fewer products → smaller K.

5. Equilibrium Constant & Entropy

This links back to thermodynamics: every spontaneous change has a positive total entropy change (ΔS_total > 0). The total entropy change is made of two parts:

Total Entropy Change

ΔS_total = ΔS_sys + ΔS_surr

sys = the system (the reacting chemicals) · surr = the surroundings

ΔS_sys barely shifts with temperature (unless a state changes), but ΔS_surr shifts a lot, because it depends directly on temperature:

Entropy of Surroundings

ΔS_surr = −ΔH ÷ T

ΔH = enthalpy change (J mol⁻¹), T = absolute temperature (K)

Worked Example — Is Decomposition Spontaneous?

CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +177.9 kJ mol⁻¹, ΔS_sys = +160.4 J K⁻¹ mol⁻¹.

At 293 K (20°C):

ΔS_surr = −177,900 ÷ 293 = −607.2 J K⁻¹ mol⁻¹

ΔS_total = 160.4 + (−607.2) = −446.8 J K⁻¹ mol⁻¹ → negative → NOT spontaneous

At 1173 K (900°C):

ΔS_surr = −177,900 ÷ 1173 = −151.7 J K⁻¹ mol⁻¹

ΔS_total = 160.4 + (−151.7) = +8.7 J K⁻¹ mol⁻¹ → positive → spontaneous!

This is exactly why limestone only decomposes into quicklime when you heat it strongly in a kiln.

5.1 Why Equilibrium Reactions Are Special

In a normal one-way reaction, ΔS_total just needs to be positive to go forward. But an equilibrium reaction is reversible — it can be reached from either the pure reactant side or the pure product side. That means both the forward and backward journeys towards the equilibrium mixture must have positive ΔS_total (both are spontaneous moves towards the middle).

Picture entropy plotted against the % composition of a gas mixture (like N₂O₄ ⇌ 2NO₂). The entropy curve peaks somewhere in the middle — that peak is the equilibrium mixture. Moving from pure N₂O₄ towards that peak increases entropy (spontaneous). Moving from pure NO₂ towards that peak also increases entropy (also spontaneous). But once you're sitting at the peak, moving further in either direction would actually decrease total entropy — which is why the reaction doesn't run to completion either way. It just sits at the top of the hill.

Mental Picture

Think of entropy as a hill with the equilibrium mixture sitting right at the summit. Any ball placed anywhere on the slope (any starting mixture of reactants/products) will naturally roll up towards the peak — but once it's at the very top, it has nowhere spontaneous left to roll. That's equilibrium.

5.2 Connecting Entropy to K

At the exact equilibrium point, the forward and backward entropy changes cancel out to zero. This gives us a direct link between total entropy and the equilibrium constant:

Entropy–K Relationship

ΔS_total = R ln K

R = gas constant (8.31 J K⁻¹ mol⁻¹)

Rearranged to calculate K directly from a known entropy change:

Rearranged for K

ln K = ΔS_total ÷ R → K = e^{(ΔS_total ÷ R)}

As a rough guide (not a strict rule): a very large K suggests the equilibrium position sits heavily towards the products (right-hand side), while a very small K suggests it sits heavily towards the reactants (left-hand side).

Practice Question

A reaction at 500 K has ΔS_total = +42.5 J K⁻¹ mol⁻¹. Calculate K for this reaction (R = 8.31 J K⁻¹ mol⁻¹).

What to Memorise

K_c expression

K_c = [products]^powers ÷ [reactants]^powers, using equilibrium concentrations. Solids excluded.

K_p expression

Same shape as K_c but with partial pressures instead of concentrations. Solids and liquids excluded.

Concentration formula

concentration = moles ÷ volume (dm³)

Mole fraction

x_gas = moles of gas ÷ total moles of all gases present

Partial pressure

p_gas = mole fraction × total pressure

Only temperature changes K

Concentration, pressure, and catalysts shift the position of equilibrium, never the value of K itself.

Heating favours endothermic

Raising T always shifts equilibrium in the endothermic direction, regardless of which way that is.

Entropy of surroundings

ΔS_surr = −ΔH ÷ T. As T increases, this term shrinks (matters less).

Total entropy for spontaneity

ΔS_total = ΔS_sys + ΔS_surr must be positive for a change to happen spontaneously.

Entropy–K link

ΔS_total = R ln K, so K = e^(ΔS_total/R)

Concepts Checklist

I can write a correct K_c expression from a balanced equation, leaving out solids.
I can calculate concentrations from moles and volume, then substitute into K_c.
I can build an ICE table when only initial and one equilibrium amount are given.
I can write a correct K_p expression, leaving out solids and liquids.
I can convert moles → mole fraction → partial pressure before calculating K_p.
I can correctly deduce the units of K_c and K_p from the raw expression.
I understand that only temperature changes the value of K — not concentration, pressure, or catalysts.
I can predict whether K increases or decreases when temperature rises, for both endothermic and exothermic reactions.
I can explain why a catalyst doesn't change K, only the time taken to reach equilibrium.
I can calculate ΔS_surr = −ΔH/T and use it to judge spontaneity at different temperatures.
I understand why both forward and backward reactions must have positive ΔS_total at equilibrium.
I can use ΔS_total = R ln K to calculate K, and interpret whether K favours products or reactants.

Exam Tips & Common Mistakes

Forgetting to exclude solids/liquids Students often include CaCO₃(s) or H₂O(l) in expressions where they shouldn't be. Double-check every state symbol before writing K.

Mixing up "position of equilibrium" with "value of K" A very common mark-scheme trap: a question changes the pressure and asks "what happens to K_c?" — the answer is always "no change," even though the position clearly shifts. Read carefully whether it's asking about position or K.

Skipping the units step Units for K vary reaction to reaction — never assume "no units." Always substitute the actual units (mol dm⁻³ or Pa) into the expression and cancel them algebraically to find the real answer.

Forgetting powers on non-1 coefficients If a species has a coefficient of 2 or 3 in the balanced equation, that number becomes a power in the K expression — not a multiplier. [NH₃]² is very different from 2[NH₃].

Confusing endothermic/exothermic direction when temperature rises Always identify which direction (forward or backward) is endothermic first, then remember: heating always favours the endothermic direction, full stop — regardless of which way that pushes K.

Using ΔH in kJ instead of J in entropy calculations ΔS is measured in J K⁻¹ mol⁻¹, but ΔH is usually given in kJ mol⁻¹. Convert ΔH to J mol⁻¹ (× 1000) before plugging into ΔS_surr = −ΔH/T, or your answer will be out by a factor of 1000.

Forgetting significant figures Match your final answer's significant figures to the least precise data given in the question — examiners do check this.

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3. Equilibrium Constant & Changing Conditions
4. Equilibrium Constant & Temperature — The Mechanics
5. Equilibrium Constant & Entropy
Exam Tips & Common Mistakes

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Chemical Equilibria

Chemical Equilibria

Summary — What This Chapter Covers

1. The Equilibrium Constant, Kc

1.1 Writing the Kc Expression

1.2 Calculating Kc from Moles and Volume

2. The Equilibrium Constant, Kp

2.1 When You're Only Given Moles and Total Pressure

3. Equilibrium Constant & Changing Conditions

3.1 Effect of Temperature

3.2 Effect of Pressure

3.3 Effect of a Catalyst

4. Equilibrium Constant & Temperature — The Mechanics

5. Equilibrium Constant & Entropy

5.1 Why Equilibrium Reactions Are Special

5.2 Connecting Entropy to K

What to Memorise

Concepts Checklist

Exam Tips & Common Mistakes

1. The Equilibrium Constant, K_c

1.1 Writing the K_c Expression

1.2 Calculating K_c from Moles and Volume

2. The Equilibrium Constant, K_p