PLAY · A-Level · AQA 7405

Titrate it.

Drag a burette and watch a real titration curve redraw, shift the Haber equilibrium and reshape a Maxwell-Boltzmann distribution — every number computed properly, not sketched.

The burette, on a slider

FIG. 01 · ACIDS & BASES · 3.1.12.5
pH titration curve: pH on the vertical axis against volume of base added on the horizontal axis pH V(base)/cm³ equivalence ½ eq · pH = pKa
Acid in the flask · 25.0 cm³
Base in the burette
0.100 mol dm⁻³
0.100 mol dm⁻³
4.76

Ethanoic acid is 4.76. Only bites when the flask holds a weak acid.

0.0 cm³
Indicator
pH now
1.00
Equivalence at
25.0
Pick an indicator to test it.

Why it is marked: the pH is solved from the full charge-balance equation, so every part of the curve is real — the buffer plateau, the half-equivalence point where pH = pKa, and the near-vertical jump. An indicator only works if its whole colour-change range sits inside that vertical section, which is exactly why phenolphthalein passes a weak-acid / strong-base titration and methyl orange does not.

Push the equilibrium

FIG. 02 · EQUILIBRIA · 3.1.6.1
Equilibrium yield of ammonia against temperature, with the chosen pressure NH₃/% T/°C

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   ΔH = −92 kJ mol⁻¹  ·  1:3 feed, ideal-gas model

450 °C

Forward reaction is exothermic, so heating shifts it back — and cuts the yield.

200 atm

4 mol of gas become 2, so squeezing pushes the equilibrium to the right.

Catalyst · Fe

Watch the yield when you add it. Then read the note.

Equilibrium mixture · mole %
NH₃N₂H₂

Live mole fractions of the mixture at equilibrium.

NH₃ yield
23%
Kp / atm⁻²
3.8e−5

Why it is marked: Kp changes with temperature only — raise the pressure and the number in the readout does not move, but the position of equilibrium does, because the mole fractions rearrange to keep Kp constant. The catalyst changes neither. That is the whole three-mark answer, and it is why industry settles on a compromise near 450 °C and 200 atm: a colder reactor would yield more but take far too long.

The tail that does the work

FIG. 03 · KINETICS · 3.1.5.2
Maxwell-Boltzmann distribution of molecular energies with the activation energy marked n(E) E/kJ mol⁻¹ Ea Ea(cat)
500 K

The dashed curve stays at 500 K so you can see the shift. Same area under both — the molecules do not go anywhere, they just redistribute.

50 kJ mol⁻¹
0 kJ mol⁻¹

A catalyst does not move the curve. It moves the line — an alternative route with a lower activation energy.

Molecules with E ≥ Ea
Rate vs 500 K, no cat.
×1.0

Why it is marked: a 10 °C rise barely moves the peak, yet it can roughly double the rate — because the rate depends on the fraction of molecules in the tail beyond Ea, and that fraction is exponential in −Ea/RT. Examiners want the shaded-area language: more molecules with energy greater than or equal to the activation energy, so more collisions are successful per second.

Six real questions

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