Eduqas A-level Chemistry (A410QS) · C2 — Chemical change
Mini-Lesson
C2 · Chemical change
Eduqas C2 is where chemistry starts to change: periodicity and the chemistry of Groups 2 and 7, tests for ions, thermochemistry (q = mcΔT, Hess cycles and bond enthalpies), rates (collision theory and the Maxwell-Boltzmann distribution) and simple equilibria and acid-base reactions.
Work through each screen, answer every question as you go — the multiple-choice checks, the calculations and the sorting games — and collect ⭐ stars. Press Start when you are ready.
C2 · periodicity
Periodicity across Period 3
Reading left to right across a period, the nuclear charge rises while the electrons enter the same shell, so shielding barely changes.
Atomic radius decreases — the greater nuclear charge pulls the same shell in tighter.
First ionisation energy increases (with the Al and S dips).
Melting point rises Na → Mg → Al (metallic bonding strengthens: more delocalised electrons, higher charge), peaks at Si (giant covalent — strong covalent bonds must be broken), then collapses at P₄, S₈, Cl₂ (simple molecular, only van der Waals forces) and is lowest for Ar.
Explain, do not describe. "Silicon has the highest melting point" earns nothing; "silicon is giant covalent, so melting breaks many strong covalent bonds" earns the mark.
Quick check
Why does the radius shrink?
?Atomic radius decreases from sodium to chlorine. What is the best explanation?
C2 · Group 2
Group 2 — the alkaline earth metals
Going down Group 2 the atoms get bigger and more shielded, so the outer electrons are lost more easily — reactivity increases and first ionisation energy falls.
Hydroxide solubility increases down the group: Mg(OH)₂ is only sparingly soluble (milk of magnesia); Ba(OH)₂ is quite soluble and strongly alkaline.
Sulfate solubility decreases down the group: MgSO₄ is soluble, but BaSO₄ is insoluble — which is exactly why acidified BaCl₂ is the test for sulfate ions, and why a "barium meal" is safe to swallow.
Thermal stability of carbonates and nitrates increases down the group — the larger cation has a lower charge density and polarises the anion less.
Quick check
Two opposite trends
?Which statement correctly describes solubility in Group 2?
C2 · Group 7
Group 7 — the halogens
Down Group 7 the atoms are larger and more shielded, so an incoming electron is attracted less strongly: oxidising power decreases (Cl₂ > Br₂ > I₂) while the halide ions become better reducing agents (I⁻ > Br⁻ > Cl⁻). Boiling points rise down the group because the molecules have more electrons and stronger van der Waals forces.
Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq)a more powerful oxidising agent displaces a less powerful one
Halides with concentrated sulfuric acid show the reducing trend beautifully:
Cl⁻ — no redox; just misty HCl fumes (H₂SO₄ acting as an acid only).
Br⁻ — reduces S from +6 to +4: brown Br₂ and choking SO₂.
I⁻ — the strongest reducer: S goes all the way to −2, giving H₂S (rotten eggs), plus solid sulfur and purple I₂.
Disproportionation — one element both oxidised and reduced:
Cl₂ + 2NaOH → NaCl + NaClO + H₂Ochlorine goes 0 → −1 and 0 → +1 in the same reaction (cold, dilute NaOH → bleach)
Sort it
Oxidised, reduced or disproportionated?
Tap the named species, then the box that describes what happens to it in that reaction.
Carbonate: add dilute acid → effervescence; the gas turns limewater milky (CO₂).
Sulfate: acidify with dilute HCl, then add BaCl₂(aq) → white precipitate of BaSO₄. The acid first removes carbonate, which would also give a white precipitate.
Halides: acidify with dilute HNO₃, then add AgNO₃(aq). Cl⁻ → white AgCl, dissolves in dilute ammonia. Br⁻ → cream AgBr, dissolves only in concentrated ammonia. I⁻ → yellow AgI, insoluble even in concentrated ammonia.
Ammonium: warm with NaOH(aq) → NH₃ gas turns damp red litmus blue.
Order matters: always test for carbonate → sulfate → halide. Testing for halide first would fail, because Ag₂CO₃ and Ag₂SO₄ are also precipitates.
Quick check
Identify the halide
?A solution is acidified with dilute nitric acid and silver nitrate is added. A cream precipitate forms, which dissolves in concentrated ammonia but not in dilute ammonia. Which ion is present?
C2 · thermochemistry
Enthalpy changes and q = mcΔT
ΔH is the heat change at constant pressure. Exothermic reactions release heat (ΔH negative, the surroundings warm up); endothermic reactions absorb it (ΔH positive).
q = mcΔTq in J · m = mass of SOLUTION in g · c = 4.18 J g⁻¹ K⁻¹ · ΔT in K (or °C — the size of a degree is the same)
ΔH = −q ÷ ndivide by the moles of the LIMITING reagent · the minus sign converts a temperature RISE into a NEGATIVE ΔH
Two traps: (1) m is the mass of the solution absorbing the heat, not the mass of the solid you added. (2) Convert q from J to kJ before quoting ΔH in kJ mol⁻¹.
Calculate
Your turn — heat released
150.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises by 6.50 K. Calculate the heat released, in kJ. (Take the density of the mixture as 1.00 g cm⁻³ and c = 4.18 J g⁻¹ K⁻¹.)
kJ
Hint: total mass = 100 g. q = 100 × 4.18 × 6.50 = 2717 J. Now convert to kJ.
Calculate
Your turn — enthalpy of neutralisation
2In the same experiment, 0.0500 mol of water is formed. Using q = 2.72 kJ, calculate the enthalpy of neutralisation, in kJ mol⁻¹. Include the sign (type it like -50.0).
kJ mol⁻¹
Hint: ΔH = −q ÷ n = −2.717 ÷ 0.0500. The temperature ROSE, so ΔH must be negative.
C2 · Hess
Hess's law
Hess's law: the total enthalpy change is independent of the route taken. That lets us calculate ΔH values we cannot measure directly.
ΔHr = Σ ΔHf(products) − Σ ΔHf(reactants)formation data: arrows point UP from the elements, so products minus reactants
ΔHr = Σ ΔHc(reactants) − Σ ΔHc(products)combustion data: arrows point DOWN to the oxides, so reactants minus products
Remember: ΔHf of any element in its standard state is zero, and don't forget to multiply by the coefficients in the equation.
Calculate
Your turn — Hess cycle
3Calculate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), using ΔHf values in kJ mol⁻¹: CH₄ = −74.8, CO₂ = −393.5, H₂O(l) = −285.8, O₂ = 0. Include the sign.
Bond breaking is endothermic; bond making is exothermic.
ΔH = Σ(bonds broken) − Σ(bonds made)always use the value for the bonds in the actual molecules
Why answers from bond enthalpies are only approximate: a mean bond enthalpy is averaged over many different compounds, and the values apply to gaseous species — so any liquid or solid in the equation makes the estimate worse.
Calculate
Your turn — bond enthalpies
4Calculate ΔH for H₂(g) + Cl₂(g) → 2HCl(g). Mean bond enthalpies in kJ mol⁻¹: H–H = 436, Cl–Cl = 242, H–Cl = 431. Include the sign.
Collision theory and the Maxwell-Boltzmann distribution
A reaction happens only when particles collide with enough energy (at least the activation energy, Ea) and the correct orientation.
The Maxwell-Boltzmann distribution shows how molecular energies are spread. It starts at the origin (no molecule has zero energy), rises to a peak (the most probable energy) and has a long tail to the right. The area under the curve is the total number of molecules — so it cannot change.
Higher temperature: the peak moves right and gets lower; the area is unchanged, but a much larger fraction of molecules now exceeds Ea.
Catalyst: the curve does not move at all. The catalyst provides an alternative route of lower Ea, so the Ea line shifts left and more molecules lie beyond it.
Higher concentration or pressure: more particles per unit volume → more frequent collisions (the distribution of energies is unchanged).
Quick check
What a catalyst really does
?Which statement about adding a catalyst is correct?
C2 · equilibria
Dynamic equilibrium and Le Chatelier
A reversible reaction reaches dynamic equilibrium in a closed system when the forward and reverse rates are equal and the concentrations are constant (not equal).
Le Chatelier's principle: if a change is imposed, the position of equilibrium shifts to oppose it.
Kc = [products]coefficients ÷ [reactants]coefficientsequilibrium concentrations only · solids and pure liquids are omitted
The key A-level point: pressure, concentration and catalysts can move the position of equilibrium, but only temperature changes the value of Kc. A catalyst simply gets you there faster.
Calculate
Your turn — calculate Kc
5For H₂(g) + I₂(g) ⇌ 2HI(g) the equilibrium concentrations are [H₂] = 0.20, [I₂] = 0.20 and [HI] = 1.60 mol dm⁻³. Calculate Kc.
Tap a change on the left, then its effect on N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹.
Change imposed
Effect on the equilibrium
C2 · acids and bases
Acids, bases and the pH scale
A Brønsted-Lowry acid is a proton donor; a base is a proton acceptor. A strong acid is fully dissociated; a weak acid only partially.
pH = −log₁₀[H⁺]and [H⁺] = 10−pH
Strong is not the same as concentrated. A 0.001 mol dm⁻³ solution of HCl is a strong acid at a dilute concentration; concentrated ethanoic acid is a weak acid at a high concentration.
Calculate
Your turn — pH of a strong acid
6Calculate the pH of 0.0500 mol dm⁻³ hydrochloric acid, to 2 decimal places. (HCl is fully dissociated, so [H⁺] = 0.0500 mol dm⁻³.)
pH
Hint: pH = −log₁₀(0.0500).
Recap
The big ideas to know
Periodicity: radius falls and IE rises across a period; m.p. peaks at Si (giant covalent)
Group 2: reactivity increases down; hydroxides MORE soluble, sulfates LESS soluble
Group 7: oxidising power falls down; halide reducing power rises (I⁻ gives H₂S with conc. H₂SO₄)
Ion tests: carbonate → sulfate → halide, in that order