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Equilibrium and acid reactions review
A summary of the whole Equilibrium and Acid Reactions module, key words and formulas, and a mixed HSC-style test covering dynamic equilibrium, ΔG, Le Chatelier's principle, collision theory, K and Q, ICE calculations, dissolving, cycad processing and Ksp.
Use this page to check the whole module. Read the summary, make sure you can use every formula and explain every key word, then do the test without looking back.
Module summary
| Topic | What you need to know |
|---|---|
| Dynamic equilibrium (5A) | In a closed system a reversible reaction reaches a state where forward and reverse rates are equal and concentrations are constant (not equal). Open systems and static equilibria are different. Combustion and photosynthesis are not equilibrium systems. |
| ΔG, ΔH and ΔS (5B) | ΔG = ΔH − TΔS (ΔS in kJ K⁻¹ mol⁻¹). ΔG < 0 means spontaneous. Four sign combinations; crossover temperature T = ΔH/ΔS. Equilibrium is reached because enthalpy and entropy pull in opposite directions. |
| Le Chatelier's principle (5C) | A system at equilibrium responds to a disturbance by partially counteracting it. Concentration, pressure/volume (gases only) and temperature shift the position; a catalyst and an inert gas at constant volume don't. Cobalt chloride and Fe³⁺/SCN⁻ practicals; reading concentration–time graphs. |
| Collision theory (5D) | Disturbances make the forward and reverse rates temporarily unequal. Eₐ(fwd) − Eₐ(rev) = ΔH; the endothermic direction has the higher Eₐ and speeds up more on heating (Maxwell–Boltzmann). A catalyst lowers both Eₐ equally. |
| K and Q (5E) | K = [products]^coefficients ÷ [reactants]^coefficients, omitting solids and pure liquids. Large K → products favoured. Q < K → forward; Q > K → reverse. Reverse the equation → 1/K; multiply by n → Kⁿ. Only temperature changes K. |
| Calculations (5F) | ICE tables with the mole ratio in the Change row; concentrations not moles; perfect squares, quadratics and the small-x approximation. Colorimetry with a calibration curve to measure K for FeSCN²⁺. |
| Dissolving (5G) | Ion–dipole attraction and hydration; dissociation equations; saturated solutions are dynamic equilibria (tracer evidence). ΔH(soln) = lattice + hydration; endothermic dissolving is driven by entropy. Solubility curves and solubility rules. |
| Removing toxins (5H) | Aboriginal peoples removed water-soluble toxins (cycasin, macrozamin, BMAA) from cycad seeds by crushing and leaching in running water, which keeps Q < K, plus ageing and roasting. |
| Ksp (5I) | Ksp expressions; s from Ksp (s², 4s³); compare Ksp only for the same formula type; Q > Ksp → precipitate (dilute first); net ionic equations; common ion effect. |
Key formulas
| Formula | Use |
|---|---|
| $\Delta G = \Delta H - T\Delta S$ | Predicting spontaneity; convert ΔS to kJ |
| $T = \dfrac{\Delta H}{\Delta S}$ | Temperature at which spontaneity changes |
| $K_{eq} = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$ | Equilibrium constant (Q has the same form) |
| $E_a(\text{fwd}) - E_a(\text{rev}) = \Delta H$ | Linking activation energies |
| $c = \dfrac{n}{V}$, $\;c_1V_1 = c_2V_2$ | Concentrations and dilution on mixing |
| $K_{sp} = s^2$ (AB), $4s^3$ (AB₂) | Solubility from Ksp |
| $A = \varepsilon l c$ | Colorimetry (absorbance proportional to concentration) |
Putting it together
The module is built on one idea: a reversible reaction in a closed system settles to a dynamic equilibrium, where forward and reverse reactions continue at equal rates. Thermodynamics explains why it stops where it does: the balance of enthalpy and entropy in ΔG. K puts a number on where it stops, and Q tells you which way a mixture will move to get there. Le Chatelier's principle is the qualitative version of comparing Q with K, and collision theory explains it at the level of particles. Only temperature changes K itself, because the endothermic direction, with the larger activation energy, responds more strongly to heating. The same ideas apply to a solid dissolving: a saturated solution is an equilibrium with constant Ksp, Q tells you whether a precipitate will form, and Aboriginal peoples' leaching of cycad seeds in running water is a long-tested application of keeping Q below K. In Module 6 you'll apply all of this to acids and bases, where Ka, Kb and Kw are just K under new names.
Key words
absorbance · activation energy · ageing · BMAA · calibration curve · catalyst · closed system · colorimeter · common ion effect · cycasin · dissociation · dynamic equilibrium · endothermic · enthalpy · entropy · equilibrium constant · equilibrium law · equilibrium position · exothermic · Gibbs free energy · heterogeneous equilibrium · homogeneous equilibrium · hydration · ICE table · ion–dipole attraction · isolated system · Keq · Ksp · lattice energy · Le Chatelier's principle · leaching · macrozamin · Maxwell–Boltzmann distribution · molar solubility · net ionic equation · open system · precipitate · reaction quotient · reversible reaction · saturated · small-x approximation · solubility · solubility product · solubility rules · spectator ion · spontaneous · static equilibrium · supersaturated · unsaturated
Test yourself
For a reaction, ΔH = +40 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Under which conditions is it spontaneous?
T = 40 ÷ 0.100 = 400 K; above this, TΔS > ΔH.
Which statement about a system at dynamic equilibrium is correct?
Rates are equal, so concentrations, and properties that depend on them, are constant.
[Co(H₂O)₆]²⁺(aq, pink) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq, blue) + 6H₂O(l), ΔH > 0. A purple equilibrium mixture is placed in an ice bath. What is observed?
Cooling favours the exothermic reverse reaction, which forms the pink complex.
At 25 °C, K = 4.6 × 10⁻³ for N₂O₄(g) ⇌ 2NO₂(g). What is K for 2NO₂(g) ⇌ N₂O₄(g)?
1 ÷ 4.6 × 10⁻³ = 217.
Which change increases the yield of products for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹, and also increases K?
Pressure and extra O₂ raise the yield but not K. Only temperature changes K.
What is the molar solubility of Ag₂CO₃? (Ksp = 8.46 × 10⁻¹²)
Ksp = (2s)²s = 4s³, so s = ∛(8.46 × 10⁻¹² ÷ 4).
Equal volumes of 2.0 × 10⁻⁴ mol L⁻¹ BaCl₂ and 2.0 × 10⁻⁴ mol L⁻¹ Na₂SO₄ are mixed. Ksp(BaSO₄) = 1.08 × 10⁻¹⁰. Which is correct?
After mixing each is 1.0 × 10⁻⁴; Q = 1.0 × 10⁻⁸ > Ksp.
Why did Aboriginal peoples leach crushed cycad kernels in running water rather than a still pool?
[toxin(aq)] stays low, so leaching continues.
In a Maxwell–Boltzmann distribution, raising the temperature
The curve flattens and shifts right; the area stays the same.
Label the Q versus K diagram.
For Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq), ΔH < 0, how does the equilibrium respond?
Put the steps for predicting a precipitate in order.
At equilibrium in a system, the forward and reverse rates are . The only factor that changes the value of K is . If Q is greater than K, a net reaction occurs. Adding a common ion the solubility of a salt.
1.00 mol of PCl₅ is placed in a 2.0 L flask at 250 °C, where K = 0.042 for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).
(a) Calculate the equilibrium concentrations of all three gases.
(b) The volume of the flask is then suddenly halved at constant temperature. Use Q to predict the direction of the shift, and explain your answer using Le Chatelier's principle.
(a) [PCl₅]₀ = 1.00 ÷ 2.0 = 0.50 mol L⁻¹. E row: [PCl₅] = 0.50 − x, [PCl₃] = [Cl₂] = x.
$0.042 = \dfrac{x^2}{0.50-x} \Rightarrow x^2 + 0.042x - 0.021 = 0 \Rightarrow x = \dfrac{-0.042 + \sqrt{0.042^2 + 4(0.021)}}{2} = 0.125$
[PCl₃] = [Cl₂] = 0.125 mol L⁻¹; [PCl₅] = 0.375 mol L⁻¹. (Check: 0.125² ÷ 0.375 = 0.042 ✓)
(b) Halving the volume doubles every concentration: [PCl₃] = [Cl₂] = 0.25, [PCl₅] = 0.75 mol L⁻¹. Q = 0.25² ÷ 0.75 = 0.083, which is greater than K (0.042), so there is a net reverse reaction until Q falls back to 0.042. This agrees with Le Chatelier's principle: halving the volume increases the pressure, and the system partially counteracts this by shifting towards the side with fewer gas particles (1 mol on the left compared with 2 mol on the right).