AS 7 Equilibria | A Level Chemistry 9701

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🎯What you need to be able to do

Explain reversible reactions and dynamic equilibrium, and why a closed system is needed.
State Le Chatelier’s principle and use it to predict the effect of temperature, concentration, pressure and a catalyst on the position of equilibrium.
Write expressions for K_c and, using mole fractions and partial pressures, for K_p; calculate them, including their units, and calculate equilibrium amounts.
State which changes alter the value of an equilibrium constant.
Describe and explain the conditions of the Haber process and the Contact process.
Name and give the formulas of the common acids and alkalis, and describe the Brønsted–Lowry theory.
Distinguish strong and weak acids and bases by dissociation, and by their pH, conductivity and reactions.
Describe neutralisation and salt formation, sketch pH titration curves for all four strong/weak combinations, and choose a suitable indicator.

📚The chemistry

7.1 Chemical equilibria

A reversible reaction can go both ways: products can react to re-form the reactants. The double-harpoon ⇌ shows it. Start with only reactants, and the forward reaction is fast at first and slows as they are used up; the reverse reaction starts at zero and speeds up as products build up. Eventually the two rates are equal.

That state is dynamic equilibrium: the rates of the forward and reverse reactions are equal, so the concentrations of reactants and products stay constant. It is dynamic because both reactions are still happening — molecules keep converting both ways — but no overall change can be seen. It needs a closed system, one that no substance can enter or leave: if a gaseous product escapes, the reverse reaction cannot keep pace and the reaction goes to completion instead.

Constant concentrations do not mean equal concentrations. The equilibrium mixture may be almost all products or almost all reactants; that balance is called the position of equilibrium.

Rate against time for a reversible reaction started with reactants only. The forward rate starts high and falls; the reverse rate starts at zero and rises. The two curves meet and continue together as one flat line, labelled equilibrium, where the rates are equal. — The forward rate falls and the reverse rate rises until they are equal: dynamic equilibrium.

Le Chatelier’s principle

If a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise that change. Applied to each kind of change:

Concentration. Add a reactant and the position moves to the right, using some of it up. Remove a product and it also moves right, to replace it.
Pressure (reactions involving gases). Increase the pressure and the position moves to the side with fewer gas molecules, which reduces the pressure. If both sides have the same number of gas molecules, as in H₂ + I₂ ⇌ 2HI, pressure has no effect on the position.
Temperature. Increase the temperature and the position moves in the endothermic direction, which absorbs heat. Decrease it and the position moves in the exothermic direction. The sign of ΔH is quoted for the forward reaction.
Catalyst. A catalyst speeds up the forward and reverse reactions equally, so it has no effect on the position. Equilibrium is simply reached sooner.

K_c

For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentrations is

\[ K_\mathrm{c} = \frac{[\mathrm{C}]^{c}\,[\mathrm{D}]^{d}}{[\mathrm{A}]^{a}\,[\mathrm{B}]^{b}} \]

Square brackets mean equilibrium concentration in mol dm⁻³; products go on top; each is raised to the power of its coefficient in the equation. Units are worked out by cancelling: for N₂ + 3H₂ ⇌ 2NH₃, (mol dm⁻³)² / (mol dm⁻³)⁴ = mol⁻² dm⁶. When the powers cancel completely, as for H₂ + I₂ ⇌ 2HI, K_c has no units.

A large K means the equilibrium lies to the right (mostly products); a small K means it lies to the left.

Mole fractions, partial pressures and K_p

For gases it is often more natural to use pressures. In a mixture of gases, the mole fraction of a gas is its share of the total moles, and its partial pressure is the pressure it would exert on its own — its share of the total pressure:

\[ x_\mathrm{A} = \frac{n_\mathrm{A}}{n_\mathrm{total}} \qquad p_\mathrm{A} = x_\mathrm{A} \times p_\mathrm{total} \]

The mole fractions add up to 1, and the partial pressures add up to the total pressure. The equilibrium constant in terms of partial pressures has the same shape as K_c:

\[ K_\mathrm{p} = \frac{p_\mathrm{C}^{\,c}\;p_\mathrm{D}^{\,d}}{p_\mathrm{A}^{\,a}\;p_\mathrm{B}^{\,b}} \]

Only gases appear in K_p. Its units come from the pressure unit used (Pa, kPa or atm), cancelled in the same way. You do not need the relationship between K_p and K_c, and calculations will never need a quadratic equation.

Calculating equilibrium amounts

Most calculations follow one layout. Write three rows under the equation: initial moles, change (in the ratio of the equation), and equilibrium moles. Then convert equilibrium moles into what the expression needs — concentrations (divide by volume) for K_c, partial pressures (mole fraction × total pressure) for K_p — and substitute.

What changes the value of K

This is examined as a statement, and it is simple: only temperature changes the value of an equilibrium constant. Changing a concentration or pressure moves the position, but the system re-settles with the same K. A catalyst changes neither the position nor K. For an exothermic forward reaction, raising the temperature decreases K; for an endothermic one it increases K.

Equilibria in industry

The Haber process makes ammonia:

\[ \mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)} \qquad \Delta H = -92\ \mathrm{kJ\ mol^{-1}} \]

Pressure about 20 000 kPa (200 atm). Four gas molecules become two, so high pressure moves the position right and also increases the rate. Higher still would give more ammonia, but the cost of compressors and thick-walled plant, and the safety risk, outweigh the gain.
Temperature about 450 °C. The forward reaction is exothermic, so a low temperature would give a higher yield — but the rate would be far too slow. The temperature is a compromise between yield and rate.
Iron catalyst to increase the rate. It does not change the yield.
Unreacted nitrogen and hydrogen are recycled, so although only about 15% is converted on each pass, almost all of it becomes ammonia in the end.

The Contact process makes sulfuric acid; its key step is

\[ \mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \qquad \Delta H = -197\ \mathrm{kJ\ mol^{-1}} \]

Temperature about 450 °C — the same yield-against-rate compromise for an exothermic reaction.
Pressure only slightly above atmospheric (about 100–200 kPa). Three gas molecules become two, so high pressure would help, but the conversion is already very high (over 99% after several passes over the catalyst), so the expense is not worth it.
Vanadium(V) oxide, V₂O₅, catalyst.
Excess air (oxygen) moves the position right and makes the most of the more valuable SO₂.

7.2 Acids and bases

The acids and alkalis you must know: hydrochloric acid, HCl; sulfuric acid, H₂SO₄; nitric acid, HNO₃; ethanoic acid, CH₃COOH; sodium hydroxide, NaOH; potassium hydroxide, KOH; ammonia, NH₃.

Brønsted–Lowry theory

A Brønsted–Lowry acid is a proton (H⁺) donor; a base is a proton acceptor. An acid can only behave as an acid if something accepts its proton, so every acid–base reaction is a proton transfer:

\[ \mathrm{HCl + H_2O \rightarrow H_3O^{+} + Cl^{-}} \] \[ \mathrm{NH_3 + H_2O \rightleftharpoons NH_4^{+} + OH^{-}} \]

In the first, water is a base; in the second, it is an acid. Water can do both.

Strong and weak

A strong acid is fully dissociated into ions in aqueous solution (HCl, HNO₃, H₂SO₄). A weak acid is only partially dissociated — an equilibrium lies well to the left:

\[ \mathrm{CH_3COOH(aq) \rightleftharpoons CH_3COO^{-}(aq) + H^{+}(aq)} \]

Strong bases (NaOH, KOH) are fully dissociated; ammonia is a weak base. “Strong” describes the extent of dissociation, not the concentration — a dilute solution of a strong acid is still a strong acid.

Water has a pH of 7; acidic solutions are below 7 and alkaline solutions above. Compare solutions of the same concentration, say 0.10 mol dm⁻³, and a weak acid has a lower [H⁺] than a strong one, which shows up three ways:

Higher pH, measured with a pH meter or universal indicator: about 2.9 for ethanoic acid against 1.0 for hydrochloric acid.
Lower electrical conductivity, because there are fewer ions.
Slower reaction with a reactive metal such as magnesium, or with a carbonate: fewer bubbles per second. The total volume of hydrogen is the same, because as H⁺ is used up the weak acid keeps dissociating until all of it has reacted.

Neutralisation, titration curves and indicators

Neutralisation is H⁺(aq) + OH⁻(aq) → H₂O(l), and it forms a salt from the remaining ions: NaOH + HCl gives NaCl, 2KOH + H₂SO₄ gives K₂SO₄, NH₃ + HNO₃ gives NH₄NO₃.

A pH titration curve plots pH against the volume of alkali added to an acid (or the reverse). Every curve has a steep, near-vertical section at the equivalence point, where a tiny volume changes the pH sharply. What differs is where that section starts and stops:

Strong acid, strong alkali — starts near pH 1, vertical section from about pH 3 to 11, equivalence at pH 7.
Weak acid, strong alkali — starts higher (about pH 3), rises gently, then a shorter vertical section from about pH 7 to 11; equivalence above 7 (about 9).
Strong acid, weak alkali — vertical section from about pH 3 to 7; equivalence below 7 (about 5); finishes lower (about pH 11).
Weak acid, weak alkali — no vertical section, just a gradual change in slope around pH 7.

Four pH curves for adding 0.100 mol per cubic decimetre alkali to 25.0 cubic centimetres of 0.100 mol per cubic decimetre acid, calculated from the equilibria. Each has a steep section at 25 cubic centimetres. Strong acid with strong alkali rises from about pH 1 to 12 with a long vertical section; weak acid with strong alkali starts near pH 3 and its vertical section sits higher; strong acid with weak alkali has its vertical section lower; weak acid with weak alkali has no vertical section. Bands mark methyl orange, 3.1 to 4.4, and phenolphthalein, 8.3 to 10.0. — Calculated pH curves for all four combinations, with the two indicator ranges shaded.

An indicator is suitable if its colour-change range lies entirely within the vertical section, because then a single drop takes it right through its change. The question will give you the ranges; for example methyl orange changes over about 3.1–4.4 and phenolphthalein over about 8.3–10.0. So:

strong acid + strong alkali: either indicator;
weak acid + strong alkali: phenolphthalein;
strong acid + weak alkali: methyl orange;
weak acid + weak alkali: no indicator gives a sharp end-point — use a pH meter.

✏️Worked example

2.00 mol of sulfur dioxide and 1.00 mol of oxygen are allowed to reach equilibrium at a fixed temperature. At equilibrium 1.80 mol of sulfur trioxide is present and the total pressure is 200 kPa. \[ \mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \qquad \Delta H = -197\ \mathrm{kJ\ mol^{-1}} \] (a) Calculate the partial pressure of each gas at equilibrium. (b) Write the expression for K_p and calculate its value, with units. (c) State and explain the effect on the value of K_p of (i) increasing the total pressure and (ii) increasing the temperature.

(a) Build the table. 1.80 mol of SO₃ formed, so by the 2 : 1 : 2 ratio 1.80 mol of SO₂ and 0.90 mol of O₂ were used.

SO₂: 2.00 − 1.80 = 0.20 mol

O₂: 1.00 − 0.90 = 0.10 mol

SO₃: 0 + 1.80 = 1.80 mol

total: 2.10 mol

Partial pressure = mole fraction × 200 kPa:

p(SO₂) = (0.20/2.10) × 200 = 19.0 kPa

p(O₂) = (0.10/2.10) × 200 = 9.52 kPa

p(SO₃) = (1.80/2.10) × 200 = 171 kPa

They add up to 200 kPa, as they must.

(b)

\[ K_\mathrm{p} = \frac{p_{\mathrm{SO_3}}^{\,2}}{p_{\mathrm{SO_2}}^{\,2}\;p_{\mathrm{O_2}}} = \frac{(171.4)^2}{(19.05)^2 \times 9.524} = 8.50\ \mathrm{kPa^{-1}} \]

Units: kPa² / (kPa² × kPa) = kPa⁻¹. (Full calculator values were carried through; rounding the partial pressures first would shift the third figure.)

(c)(i) No change. Increasing the pressure moves the position of equilibrium to the right, towards fewer gas molecules, but the mixture re-settles with the same value of K_p; only temperature changes an equilibrium constant. (ii) The forward reaction is exothermic, so raising the temperature moves the position to the left (the endothermic direction). There is less SO₃ and more SO₂ and O₂ at equilibrium, so K_p decreases.

Check it. Almost all the SO₂ was converted (90%), so the equilibrium lies well to the right and K should be comfortably greater than 1 in these units — 8.50 fits. The partial pressures summing to the total is a free check on the mole-fraction arithmetic; use it every time.

Putting initial amounts, or moles, into the expression. K_p needs equilibrium partial pressures. Using 2.00 and 1.00 (the starting amounts) or the moles themselves gives a meaningless number. The other classic loss is (c)(i): “higher pressure gives more SO₃, so K_p increases” confuses the position with the constant.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain what is meant by dynamic equilibrium, and why the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g) only reaches equilibrium in a closed container.

Dynamic equilibrium: the forward and reverse reactions continue at equal rates, so the concentrations of reactants and products remain constant. In an open container the CO₂ escapes as it is formed. The reverse reaction needs CO₂, so it cannot occur at the rate of the forward reaction, and the calcium carbonate decomposes completely. In a closed system the CO₂ stays, its pressure builds up until the reverse rate equals the forward rate, and equilibrium is established.

2. For 2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ mol⁻¹, brown NO₂ is in equilibrium with colourless N₂O₄. Predict and explain what you would see if the mixture is (a) cooled; (b) compressed at constant temperature.

(a) The forward reaction is exothermic. Cooling lowers the temperature, so by Le Chatelier’s principle the position moves in the exothermic direction, to the right, releasing heat. More colourless N₂O₄ forms: the mixture becomes paler. (b) There are 2 gas molecules on the left and 1 on the right. Increasing the pressure moves the position to the side with fewer gas molecules, the right, so the proportion of NO₂ falls. (The colour first darkens because compressing it raises the concentration of everything, then fades somewhat as the position shifts.)

3. 1.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed and left to reach equilibrium: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. At equilibrium, 0.667 mol of ester is present. Calculate K_c.

Equilibrium moles: ester 0.667, water 0.667, acid 1.00 − 0.667 = 0.333, ethanol 0.333. K_c = [ester][water] / ([acid][ethanol]). All four are divided by the same volume V, and there are two concentration terms top and bottom, so V cancels and moles can be used directly: K_c = (0.667 × 0.667)/(0.333 × 0.333) = 4.0, with no units. The volume only cancels because the numbers of terms match; for N₂ + 3H₂ it would not.

4. 1.00 mol of H₂ and 1.00 mol of I₂ are sealed in a 1.00 dm³ flask. At the temperature used, K_c = 49 for H₂ + I₂ ⇌ 2HI. Calculate the equilibrium amount of HI.

Let x mol of H₂ react. Equilibrium: H₂ (1 − x), I₂ (1 − x), HI 2x. K_c = (2x)² / (1 − x)² = 49. The expression is a perfect square, so take the square root of both sides (no quadratic needed): 2x/(1 − x) = 7, so 2x = 7 − 7x, x = 7/9 = 0.778. HI = 2x = 1.56 mol (and 0.222 mol each of H₂ and I₂ remain). Check: 1.556²/0.222² = 49.

5. Solutions of hydrochloric acid and ethanoic acid are both 0.10 mol dm⁻³. Describe three observations that would distinguish them, and explain the difference.

HCl is a strong acid, fully dissociated; CH₃COOH is weak, only partially dissociated (CH₃COOH ⇌ CH₃COO⁻ + H⁺, well to the left), so at the same concentration ethanoic acid has a much lower [H⁺]. Therefore: (1) with a pH meter or universal indicator, ethanoic acid has the higher pH (about 3, against 1); (2) it has a lower conductivity, having fewer ions; (3) with magnesium ribbon it fizzes more slowly. The total volume of hydrogen eventually produced from equal volumes is the same.

6. Sketch the titration curve for adding 0.10 mol dm⁻³ NaOH to 25.0 cm³ of 0.10 mol dm⁻³ CH₃COOH, and choose between methyl orange (3.1–4.4) and phenolphthalein (8.3–10.0).

The curve starts at about pH 3 (weak acid), rises gently as alkali is added, then has a steep section at 25.0 cm³ (equivalence, since the concentrations and the 1 : 1 ratio are equal) running from about pH 7 to 11, and levels off near pH 13 in excess NaOH. The equivalence point is above 7 (about 9). Phenolphthalein changes colour within the steep section, so it is suitable. Methyl orange would change during the gentle rise, long before the equivalence point, giving a wrong, gradual end-point.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

PhET — Reversible Reactions and Acid-Base Solutions, for dynamic equilibrium and strong versus weak acids at the particle level
Chemguide (Jim Clark) — the equilibria and acid–base sections, including a page of titration curves drawn out for every combination
Royal Society of Chemistry — resources on the Haber and Contact processes in industry