6. Chemical reactions

Classify a change by comparing the substances before and after it. If their chemical identities are unchanged, the change is physical. If one or more new substances with different compositions and properties form, the change is chemical.

When the amount of limiting reactant is unchanged, a faster gas-forming reaction gives a steeper curve and reaches the same final gas volume sooner.

A catalyst participates in steps of the reaction but is regenerated, so it is not used up overall. It can therefore be recovered with the same chemical identity after the reaction.

Measure the reactants, assemble the apparatus, add the final reactant and start the timer together, then record gas volume or mass at fixed time intervals until the reading becomes constant. Calculate a rate from change in volume or mass divided by time.

Average rate over an interval = change in measured quantity ÷ time interval. Use the graph scale and include units, such as cm³/s for gas volume or g/s for mass change.

Use this reasoning chain: condition changes particle behaviour → successful collisions per unit time change → reaction rate changes.

For temperature answers, include both effects: particles collide more often and more collisions have sufficient energy. Temperature does not lower the activation energy.

At the same temperature, particle kinetic energies are unchanged, but the lower Ea means a larger fraction of collisions have sufficient energy to react. There are therefore more successful collisions per unit time.

Choose a method by linking it to the reaction and data needed. Consider whether gas forms, whether it is safe to release, the expected volume or mass change, reading frequency, measurement resolution, and the main systematic losses.

For A + B ⇌ C + D, the forward reaction forms C and D from A and B, while the reverse reaction forms A and B from C and D.

Use the condition to choose direction: heat drives dehydration; water drives hydration. The observed colour identifies which form is present.

A closed system prevents substances entering or leaving, allowing the opposing reactions to establish and maintain equilibrium.

For pressure, count gaseous coefficients only. For temperature, label the forward direction exothermic or endothermic. For concentration, identify which side consumes the changed species. Then state the shift and the resulting yield change.

One mole of nitrogen reacts reversibly with three moles of hydrogen to form two moles of ammonia. All three substances are gases under the reaction conditions.

Nitrogen is separated from air. Hydrogen is produced industrially from methane, commonly by reaction with steam before purification.

Keep the values attached to the correct process: Haber uses a very high pressure of 200 atm and an iron catalyst.

Two moles of sulfur dioxide react reversibly with one mole of oxygen to form two moles of sulfur trioxide. Each substance is gaseous under the reaction conditions.

Burning sulfur gives S + O₂ → SO₂. Roasting a metal sulfide ore in air also produces sulfur dioxide as the sulfur is oxidised.

Contact and Haber both use 450 °C, but Contact uses only 2 atm and the catalyst V₂O₅.

Industrial conditions maximise neither rate nor single-pass equilibrium yield alone. They balance production speed, equilibrium composition, plant cost, energy use, equipment strength, and safety; unreacted gases can be recycled.

Determine the oxidation number from the formula and ion charges, then write it in Roman numerals immediately after the element name when required.

One species is oxidised while another is reduced. The two processes are linked because oxygen or electrons transferred from one species are gained by another.

In CuO + H₂ → Cu + H₂O, hydrogen gains oxygen and is oxidised; copper(II) oxide loses oxygen and is reduced.

1. Compare oxygen in each reactant and product. 2. Identify a species that gains oxygen. 3. Identify another species that loses oxygen. 4. If both occur in the same reaction, classify it as redox.

In Zn + CuO → ZnO + Cu, zinc gains oxygen and is oxidised; copper(II) oxide loses oxygen and is reduced.

Zn → Zn²⁺ + 2e⁻ shows oxidation: zinc loses two electrons and its oxidation number increases from 0 to +2.

Cu²⁺ + 2e⁻ → Cu shows reduction: copper ions gain two electrons and the oxidation number decreases from +2 to 0.

For Zn + Cu²⁺ → Zn²⁺ + Cu: Zn → Zn²⁺ + 2e⁻ is oxidation, while Cu²⁺ + 2e⁻ → Cu is reduction. The electrons cancel when the half-equations are combined.

In SO₄²⁻, oxygen is −2: S + 4(−2) = −2, so sulfur is +6. Show the total equation before solving the unknown oxidation number.

Use the reagent's own change to identify the unknown: manganate(VII) is reduced, so the tested substance is reducing; iodide is oxidised to iodine, so the tested substance is oxidising.

It accepts electrons from the other species, so its own oxidation number decreases.

It donates electrons to the other species, so its own oxidation number increases.

1. Assign oxidation numbers or write half-equations. 2. Find the species reduced: it is the oxidising agent. 3. Find the species oxidised: it is the reducing agent. 4. Check that electrons lost equal electrons gained.