Equilibria — A-Level Chemistry Study Guide

For: A-Level Chemistry candidates sitting A-Level Chemistry.

Covers: Reversible reactions and dynamic equilibrium, Le Chatelier's principle, equilibrium constants Kc and Kp, acid-base equilibria including pH, Kw and Ka, and buffer solutions with the Henderson-Hasselbalch equation.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Chemistry style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Equilibria?

Equilibria describes the state of a closed chemical system where the rate of the forward reaction equals the rate of the reverse reaction, so macroscopic properties (concentration, colour, pressure, pH) stay constant over time with no net change in reactant or product quantities. Reversible equilibrium reactions are written with the double harpoon symbol ⇌ to distinguish them from irreversible one-way reactions. This is a core 8-12 mark topic across AS and A Level A-Level Chemistry papers, appearing in multiple choice, structured calculation, and practical analysis questions.

2. Reversible reactions and dynamic equilibrium

A reversible reaction is a reaction that can proceed in both forward (reactants → products) and reverse (products → reactants) directions under the same reaction conditions. Dynamic equilibrium is the specific state of a closed reversible system where both forward and reverse reactions continue to occur at equal rates, so no net change in observable properties is seen. Three non-negotiable features of dynamic equilibrium:

1. The system is closed: no matter is added or removed, and no gas escapes
2. Rate of forward reaction = rate of reverse reaction
3. All macroscopic properties (concentration, colour, pressure) remain constant

Worked example

For the gas-phase decomposition of dinitrogen tetroxide: $$N_2{O}_4(g)\rightleftharpoons 2N{O}_2(g)$$ N₂O₄ is colourless, while NO₂ is a dark brown gas. If you seal pure N₂O₄ in a closed gas syringe at 298K, the initial pale colour will darken as NO₂ forms, then stop changing once equilibrium is reached. If you start with pure NO₂ instead, you will reach the same equilibrium ratio of N₂O₄:NO₂ at the same temperature, confirming equilibrium can be approached from either direction.

3. Le Chatelier's principle

Le Chatelier's principle predicts how an equilibrium system responds to external changes in conditions: if a change in concentration, temperature, or pressure is applied to a system at equilibrium, the system shifts in the direction that opposes the change to establish a new equilibrium state. Breakdown of common condition changes:

• Concentration: Increasing reactant concentration shifts equilibrium right to consume extra reactant; increasing product concentration shifts equilibrium left to consume extra product.
• Pressure (gas-phase reactions only): Increasing pressure shifts equilibrium to the side with fewer moles of gas to reduce total pressure; decreasing pressure shifts to the side with more moles of gas.
• Temperature: If the forward reaction is exothermic (ΔH = negative), increasing temperature shifts equilibrium left to absorb extra heat; if the forward reaction is endothermic (ΔH = positive), increasing temperature shifts equilibrium right.
• Catalysts: Do not shift equilibrium position or change yield, only reduce the time taken to reach equilibrium by lowering activation energy for both forward and reverse reactions equally.

Worked example

For the industrial Haber process for ammonia production: $$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)\Delta H=-92{\text{ kJ mol}}^{-1}$$ Increasing pressure shifts equilibrium right (4 moles of gas on the left, 2 on the right) to raise ammonia yield, while increasing temperature shifts equilibrium left (forward reaction is exothermic) to reduce yield. Examiners regularly ask for this industrial application, so you should also be able to explain why a compromise temperature of 450°C is used to balance yield and reaction rate.

4. $K_c$ and $K_p$ — definitions and units

Equilibrium constants quantify the ratio of product to reactant concentrations (Kc) or partial pressures (Kp) at equilibrium, at a fixed temperature. Only temperature changes the value of Kc or Kp; changes in concentration, pressure, or catalysts do not alter K values, only shift equilibrium position.

$K_c$ (concentration equilibrium constant)

For a general homogeneous reaction (all species same state) $aA+bB\rightleftharpoons cC+dD$: $$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$$ Where $[X]$ = equilibrium concentration of X in mol dm⁻³. Solids and pure liquids are omitted from Kc expressions, as their concentration is constant and assigned a value of 1. Units of Kc are calculated by cancelling units of concentration: for the Haber process, $K_c$ units are $(mold{m}^{-3}{)}^2/[(mold{m}^{-3})(mold{m}^{-3}{)}^3]=mo{l}^{-2}d{m}^6$.

$K_p$ (gas-phase equilibrium constant)

For gas-phase reactions, Kp uses partial pressures instead of concentrations. The partial pressure of a gas $p(X)=\text{mole fraction of X}\times \text{total pressure }{P}_{total}$, where mole fraction $X(X)=\text{moles of X}/\text{total moles of gas in the system}$. Kp is written as: $$K_p=\frac{p(C{)}^{c}p(D{)}^d}{p(A{)}^{a}p(B{)}^b}$$ Units of Kp match the units of pressure used (kPa, Pa, atm) raised to the difference between total moles of product and reactant gas.

Worked example

For the reaction $2S{O}_2(g)+O_2(g)\rightleftharpoons 2S{O}_3(g)$ at 1000K, equilibrium concentrations are $[S{O}_2]=0.20mold{m}^{-3}$, $[O_2]=0.10mold{m}^{-3}$, $[S{O}_3]=0.60mold{m}^{-3}$. Calculate Kc and its units: $$K_c=\frac{(0.60{)}^2}{(0.20{)}^2\times 0.10}=\frac{0.36}{0.004}=90mo{l}^{-1}d{m}^3$$

5. pH, $K_w$, $K_a$ — strong vs weak acids

Acid-base equilibria describe the dissociation of acids and bases in aqueous solution, quantified using pH, Kw, and Ka values.

• pH: Defined as $pH=-{\log }_{10}[H^{+}(aq)]$, where $[H^{+}]$ is hydrogen ion concentration in mol dm⁻³.
• Strong acids: Fully dissociate in aqueous solution, so $[H^{+}]=\text{initial acid concentration}$ for monoprotic strong acids (e.g. HCl, HNO₃).
• Weak acids: Only partially dissociate, so $[H^{+}]<<\text{initial acid concentration}$.
• $K_w$ (ionic product of water): Water dissociates very slightly: $H_{2}O(l)\rightleftharpoons H^{+}(aq)+O{H}^{-}(aq)$. $K_w=[H^{+}][O{H}^{-}]=1.0\times 10^{-14}mo{l}^{2}d{m}^{-6}$ at 298K, so neutral solution has $[H^{+}]=[O{H}^{-}]=1.0\times 10^{-7}mold{m}^{-3}$ and pH =7 at 298K.
• $K_a$ (acid dissociation constant): For a weak acid HA: $HA(aq)\rightleftharpoons H^{+}(aq)+A^{-}(aq)$, $K_a=\frac{[H^{+}][A^{-}]}{[HA]}$. $p{K}_a=-{\log }_{10}{K}_a$, so smaller pKa = stronger weak acid.

Worked example

Calculate the pH of 0.01 mol dm⁻³ ethanoic acid at 298K, given $K_a=1.7\times 10^{-5}mold{m}^{-3}$. Use two standard approximations: 1) all H⁺ comes from acid dissociation, so $[H^{+}]=[A^{-}]$, 2) dissociation is negligible so equilibrium $[HA]=\text{initial }[HA]=0.01mold{m}^{-3}$. $$K_a=\frac{[H^{+}{]}^2}{[HA]}⟹[H^{+}]=\sqrt{K_a\times [HA]}=\sqrt{1.7\times 10^{-5}\times 0.01}=4.12\times 10^{-4}mold{m}^{-3}$$ $$pH=-{\log }_{10}(4.12\times 10^{-4})=3.38$$ Examiners regularly ask you to state these two approximations for weak acid pH calculations, so memorize them explicitly.

6. Buffer solutions — Henderson-Hasselbalch

A buffer solution resists changes in pH when small amounts of acid, base, or dilution are added. There are two types:

1. Acidic buffers: Made of a weak acid and its conjugate base (e.g. ethanoic acid + sodium ethanoate), pH <7
2. Basic buffers: Made of a weak base and its conjugate acid (e.g. ammonia + ammonium chloride), pH >7 Buffers work by having a large reservoir of both the weak acid (HA) and conjugate base (A⁻):

• Added H⁺ reacts with A⁻: $A^{-}+H^{+}\to HA$, removing excess H⁺
• Added OH⁻ reacts with HA: $HA+O{H}^{-}\to A^{-}+H_{2}O$, removing excess OH⁻

The Henderson-Hasselbalch equation, derived directly from the Ka expression, lets you calculate buffer pH quickly: $$pH=p{K}_a+{\log }_{10}\left(\frac{[A^{-}]}{[HA]}\right)$$ Note that you can use the mole ratio of A⁻:HA instead of concentration ratio, since total volume cancels out for both species.

Worked example

Calculate the pH of a buffer made by mixing 50 cm³ of 0.1 mol dm⁻³ ethanoic acid (pKa = 4.76) and 25 cm³ of 0.1 mol dm⁻³ sodium ethanoate. $$\text{Moles of ethanoic acid (HA)}=0.05\times 0.1=0.005mol$$ $${\text{Moles of ethanoate (A}}^{-})=0.025\times 0.1=0.0025mol$$ $$pH=4.76+{\log }_{10}\left(\frac{0.0025}{0.005}\right)=4.76-0.30=4.46$$

7. Common Pitfalls (and how to avoid them)

• Wrong move: Including solids or pure liquids in Kc/Kp expressions. Why: Students assume all reactants and products are included. Correct move: Only include gaseous and aqueous species; solids and pure liquids have constant activity so they are omitted, assigned a value of 1.
• Wrong move: Using initial concentrations instead of equilibrium concentrations to calculate Kc. Why: Students forget Kc only describes the equilibrium state. Correct move: Use ICE (Initial, Change, Equilibrium) tables to calculate equilibrium concentrations if they are not given directly.
• Wrong move: Stating that pressure or concentration changes alter Kc/Kp values. Why: Students confuse shifts in equilibrium position with changes to the equilibrium constant. Correct move: Only temperature changes Kc/Kp; all other factors only shift equilibrium position without changing K.
• Wrong move: Assuming neutral pH is always 7, regardless of temperature. Why: Students memorize pH 7 = neutral without context. Correct move: Neutral is defined as $[H^{+}]=[O{H}^{-}]$, so at temperatures above 298K where Kw increases, neutral pH is less than 7.
• Wrong move: Using concentration instead of mole ratio for Henderson-Hasselbalch calculations when volumes are mixed. Why: Students forget total volume is the same for both conjugate acid and base. Correct move: Use mole ratio to save time, as volume cancels out for both species.

8. Practice Questions (A-Level Chemistry Style)

Question 1

The reaction between carbon monoxide and hydrogen to make methanol is: $$CO(g)+2H_2(g)\rightleftharpoons C{H}_{3}OH(g)\Delta H=-91{\text{ kJ mol}}^{-1}$$ (a) State and explain the effect of increasing total pressure on the equilibrium yield of methanol. (2 marks) (b) State and explain the effect of increasing temperature on the value of Kp for this reaction. (2 marks)

Solution

(a) Yield of methanol increases. There are 3 moles of gas on the left side of the equation and 1 mole on the right, so increasing pressure shifts equilibrium to the right to reduce total gas moles and oppose the pressure increase. (b) Kp decreases. The forward reaction is exothermic, so increasing temperature shifts equilibrium to the reverse endothermic direction to absorb extra heat, reducing the ratio of products to reactants at equilibrium.

Question 2

Calculate the pH of 0.02 mol dm⁻³ hydrochloric acid (strong monoprotic acid) and 0.02 mol dm⁻³ nitrous acid (weak acid, $K_a=4.0\times 10^{-4}mold{m}^{-3}$) at 298K. (4 marks)

Solution

Hydrochloric acid is fully dissociated: $[H^{+}]=0.02mold{m}^{-3}$ $$pH=-{\log }_{10}(0.02)=1.70$$ For nitrous acid, use weak acid approximations: $$[H^{+}]=\sqrt{K_a\times [HA]}=\sqrt{4.0\times 10^{-4}\times 0.02}=2.83\times 10^{-3}mold{m}^{-3}$$ $$pH=-{\log }_{10}(2.83\times 10^{-3})=2.55$$

Question 3

A buffer is made by mixing 0.3 mol dm⁻³ propanoic acid (pKa = 4.87) and 0.2 mol dm⁻³ sodium propanoate. Calculate the change in pH when 0.02 mol of NaOH is added to 1 dm³ of this buffer, assuming no volume change. (3 marks)

Solution

Initial pH: $$pH=4.87+{\log }_{10}\left(\frac{0.2}{0.3}\right)=4.87-0.176=4.69$$ After adding NaOH, OH⁻ reacts with propanoic acid to form propanoate: New $[HA]=0.3-0.02=0.28mold{m}^{-3}$ New $[A^{-}]=0.2+0.02=0.22mold{m}^{-3}$ $$\text{New pH}=4.87+{\log }_{10}\left(\frac{0.22}{0.28}\right)=4.87-0.105=4.76$$ $$\text{pH change}=4.76-4.69=+0.07$$

9. Quick Reference Cheatsheet

10. What's Next

Equilibria is a foundational topic that connects to multiple later areas of the A-Level Chemistry syllabus. You will use Kc and Kp calculations for reaction kinetics, organic reaction mechanism analysis, and solubility product (Ksp) calculations for sparingly soluble ionic compounds. Acid-base equilibria is required for titration curve interpretation, transition metal complex stability constant calculations, and biological chemistry questions covering blood pH regulation. Mastery of this topic is non-negotiable to score well on Paper 4 (A Level Structured Questions) and Paper 5 (Planning, Analysis and Evaluation) questions that require equilibrium-based practical design. If you struggle with any of the concepts, calculations, or exam-style questions covered in this guide, you can ask Ollie your personal AI tutor for personalised explanations, extra practice problems, or step-by-step walkthroughs of tricky calculations at any time on the homepage. You can also access our full library of A-Level Chemistry topic guides, past paper walkthroughs, and revision quizzes to reinforce your learning ahead of your exams.

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