What is a reversible reaction in the context of Edexcel IGCSE Chemistry?

In Edexcel IGCSE Chemistry, a reversible reaction is a chemical reaction where the reactants form products, which can then react together to give the reactants back. This means the reaction can proceed in both forward and backward directions. An example is the reaction between nitrogen and hydrogen to form ammonia, which can decompose back into nitrogen and hydrogen.

How is dynamic equilibrium defined in reversible reactions?

Dynamic equilibrium in reversible reactions occurs when the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in the concentrations of reactants and products. At this point, the system is stable, and the concentrations remain constant, although both reactions continue to occur.

What factors can affect the position of equilibrium in a chemical reaction?

The position of equilibrium in a chemical reaction can be affected by changes in concentration, temperature, and pressure. According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in these conditions, the system will adjust to counteract the change and restore a new equilibrium state.

Why is Le Chatelier's Principle important in understanding equilibria?

Le Chatelier's Principle is crucial for understanding how equilibria respond to external changes. It predicts how the position of equilibrium will shift when there is a change in concentration, temperature, or pressure. This principle helps in predicting the outcome of reactions under different conditions, which is essential for industrial applications and laboratory experiments.

Can you explain the concept of equilibrium constant (Kc) in reversible reactions?

The equilibrium constant (Kc) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients. It provides insight into the position of equilibrium; a large Kc indicates a greater concentration of products, while a small Kc suggests a greater concentration of reactants. Kc is specific to a particular reaction at a given temperature.

Why is "Saying 'the reaction has stopped' at equilibrium" a common mistake in Reversible Reactions and Equilibria?

Why it happens: The concentrations are constant, so it LOOKS stopped. How to avoid it: At equilibrium, the reaction is STILL OCCURRING in both directions — just at EQUAL rates. The concentrations don't change because forward + backward perfectly cancel. Always state: 'rate of forward = rate of backward' (or 'continues in both directions at equal rates'). Source: 4CH1 Examiner Reports 2022-2024

Why is "Predicting the wrong direction for temperature change" a common mistake in Reversible Reactions and Equilibria?

Why it happens: Confusion about exo/endo direction. How to avoid it: T INCREASE → equilibrium shifts in the ENDOTHERMIC direction (absorbs heat → opposes increase). T DECREASE → shifts in EXOTHERMIC direction (releases heat → opposes decrease). Identify whether the FORWARD reaction is exo or endo from the given ΔH; the backward is the opposite. Source: 4CH1 Examiner Reports 2023-2024

Why is "Including solid or liquid species when counting moles for pressure analysis" a common mistake in Reversible Reactions and Equilibria?

Why it happens: Counting everything in the equation. How to avoid it: Pressure changes only affect GAS species (state symbol (g)). Solids and liquids have negligible volume and are unaffected by pressure. Count ONLY moles of GAS on each side.

Why is "Saying a catalyst increases the yield of products" a common mistake in Reversible Reactions and Equilibria?

Why it happens: Equating 'faster' with 'more'. How to avoid it: A catalyst does NOT change the equilibrium position → does NOT change the yield. It only changes the SPEED of approach to equilibrium. So yield is unchanged; same NH₃ at equilibrium, just reached faster. Source: 4CH1 Examiner Reports 2022-2024

Why is "Explaining only ONE side of the compromise (e.g. 'high T gives high rate' without mentioning yield)" a common mistake in Reversible Reactions and Equilibria?

Why it happens: Forgetting it's a TRADE-OFF question. How to avoid it: A 'compromise' question expects BOTH sides. For T: low T good for yield BUT slow rate → 450 °C is the middle ground. For P: high P good for yield BUT expensive → 200 atm is the middle ground. State 'compromise' explicitly.

Why is "Saying 'doubling concentration doubles the yield' or quantitative-sounding claims" a common mistake in Reversible Reactions and Equilibria?

Why it happens: Over-extending qualitative Le Chatelier predictions. How to avoid it: Le Chatelier is QUALITATIVE only at IGCSE — predict the DIRECTION of shift (forwards / backwards / no change), not the SIZE. 'Doubling [N₂] shifts equilibrium forwards, INCREASING yield of NH₃' — but you cannot say BY HOW MUCH.

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Reversible Reactions and Equilibria

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Reversible Reactions and Equilibria — Pearson Edexcel International GCSE Chemistry 4CH1 Study Notes (2026 onwards, Spec Issue 4)

Reactions that proceed in both directions (⇌). Dynamic equilibrium: forward rate = backward rate, concentrations constant, reaction still occurring on molecular scale, closed system required. Le Chatelier: system disturbed shifts to oppose change. T ↑ → endo direction. P ↑ → fewer moles gas. [Reactant] ↑ → forward shift. Catalyst doesn't shift equilibrium, only speeds approach. Industrial: Haber (450 °C / 200 atm / Fe) and Contact (450 °C / 1-2 atm / V₂O₅) — compromise conditions between yield, rate and cost.

At a glance

Reversible reaction: products can re-form reactants. Written with ⇌ (e.g. N₂ + 3H₂ ⇌ 2NH₃).
Dynamic equilibrium: forward rate = backward rate → concentrations CONSTANT → reaction still occurring. Needs CLOSED system.
Le Chatelier: system disturbed → equilibrium SHIFTS to OPPOSE the change.
T ↑ → shifts in ENDOTHERMIC direction. T ↓ → shifts in EXOTHERMIC direction.
P ↑ (gas) → shifts to side with FEWER moles of gas. P ↓ → more moles.
[Reactant] ↑ → shifts FORWARDS (consumes added reactant). [Product] ↑ → shifts BACKWARDS.
Catalyst: does NOT shift equilibrium — only reaches it FASTER. Both forward + backward rates ↑ equally.
Haber: N₂ + 3H₂ ⇌ 2NH₃ (exo, 4→2 mol gas). 450 °C, 200 atm, Fe — compromise.
Contact: 2SO₂ + O₂ ⇌ 2SO₃ (exo, 3→2 mol gas). 450 °C, 1-2 atm, V₂O₅ — high yield without high P.

What you’ll learn

Mapped to the Cambridge IGCSE 4CH1 syllabus (2026 onwards).

3.16 — Know that some reactions can be reversed and can reach a dynamic equilibrium in a closed system.
3.17 — Understand the concept of dynamic equilibrium: forward and backward reactions occurring at equal rates; concentrations of reactants and products remain constant.
3.18 — (Paper 2) Apply the principle of Le Chatelier: when a system at equilibrium is disturbed, the position of equilibrium shifts to oppose the change.
3.19 — (Paper 2) Predict the effect of changes in temperature, pressure and concentration on the position of equilibrium for given reversible reactions.
3.20 — (Paper 2) Explain the conditions used in the Haber process for ammonia production (450 °C, 200 atm, iron catalyst) as a compromise between yield and rate.
3.21 — (Paper 2) Explain the conditions used in the Contact process for sulfuric acid production (450 °C, 1-2 atm, V₂O₅ catalyst).

Reversible reactions and dynamic equilibrium (spec 3.16, 3.17)

⇌ means both directions. Equilibrium: equal rates, constant concentrations, closed system.

Reversible reactions. Some chemical reactions can proceed in BOTH directions — the products can react together to re-form the reactants. We use the symbol ⇌ instead of → to indicate this:

$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$

The left-to-right ('forward') reaction is N₂ + H₂ combining to make NH₃. The right-to-left ('backward') reaction is NH₃ decomposing back to N₂ + H₂. Both happen simultaneously in the reaction vessel.

Classic examples of reversible reactions.

Haber process: N₂ + 3H₂ ⇌ 2NH₃. Industrial ammonia production.
Contact process (key step): 2SO₂ + O₂ ⇌ 2SO₃. Industrial sulfuric acid production.
Hydration of copper(II) sulfate: CuSO₄·5H₂O ⇌ CuSO₄ + 5H₂O. Blue hydrated ⇌ white anhydrous. Easy visible demo.
Hydrogen + iodine: H₂(g) + I₂(g) ⇌ 2HI(g). Studied as a 'model' equilibrium in chemistry research.

Dynamic equilibrium. When a reversible reaction is allowed to proceed in a CLOSED system (no atoms can escape), it eventually reaches a state called dynamic equilibrium:

The rate of the FORWARD reaction = the rate of the BACKWARD reaction.
The concentrations of all species REMAIN CONSTANT over time (no NET change).
The reaction has NOT STOPPED — both directions are still occurring at the molecular level.

The system is 'dynamic' (constantly changing on the molecular scale) but 'static' (no net change on the macroscopic scale).

Why this happens.

Initially, when we mix N₂ and H₂, only the forward reaction can proceed (no NH₃ yet to decompose). As NH₃ accumulates, the BACKWARD reaction starts to occur. Over time:

The forward rate DECREASES (reactants being used up).
The backward rate INCREASES (more product to react).
Eventually they MEET AT THE SAME VALUE → equilibrium.

At equilibrium, NH₃ is still forming AND breaking down — just at the same speed. So the AMOUNT of each species stays the same.

Dynamic equilibrium is reached when the forward and backward rates become equal — concentrations then stay constant.

Why a closed system is essential.

If the system were OPEN (e.g. NH₃ escaping into the atmosphere), the products could never accumulate enough to drive the backward reaction. Equilibrium would NEVER establish — the forward reaction would just keep going (slowly) until reactants ran out. In an open container, you'd never see dynamic equilibrium.

For the Haber process, this is why the reactor is SEALED at high pressure: it keeps everything inside so equilibrium can be reached and maintained.

Molecular-level picture.

Imagine a sealed flask of H₂ + I₂ + HI at equilibrium. At any instant:

About 50% of the molecules are HI; 25% are H₂; 25% are I₂ (depending on T).
1 million HI molecules per second decompose to H₂ + I₂.
1 million H₂ + I₂ pairs per second collide and form HI.
Net change: zero. Macroscopically constant.

Each INDIVIDUAL molecule is constantly being converted back and forth, but the POPULATIONS are stable. That's dynamic equilibrium.

Analogy — running on a treadmill. You're running fast (legs moving), but you stay in the same place (no net change in position). At dynamic equilibrium, the molecules are 'running' fast (forward and backward reactions occurring rapidly), but the composition stays constant.

Reversible reaction: products can re-form reactants. Symbol ⇌.
Dynamic equilibrium: forward rate = backward rate.
Concentrations CONSTANT; reaction has NOT stopped (still occurring on molecular level).
Requires CLOSED system (no products escape).
Examples: Haber (NH₃), Contact (SO₃), hydrated CuSO₄.

See the full worked example for reversible reactions and equilibria →

Le Chatelier's principle (Paper 2, spec 3.18, 3.19)

System disturbed → shifts to OPPOSE the change.

Le Chatelier's principle is one of the most powerful predictive tools in chemistry. It states:

If a system at equilibrium is subjected to a change in TEMPERATURE, PRESSURE, or CONCENTRATION, the equilibrium will SHIFT in the direction that OPPOSES (counteracts) the change.

The system 'pushes back' against the disturbance. The new equilibrium that establishes has different concentrations, but it's still characterised by 'forward rate = backward rate' — just at different absolute rates.

This is a Paper 2 topic (BOLD P content) — be ready to use it qualitatively to predict the direction of shift.

Three types of disturbance you must be able to handle:

(1) Temperature change.

T INCREASE → equilibrium shifts in the ENDOTHERMIC direction.
T DECREASE → equilibrium shifts in the EXOTHERMIC direction.

Why? Raising temperature adds heat to the system; Le Chatelier predicts the system will move to ABSORB the added heat. The endothermic direction absorbs heat — so that's the direction of shift. (And vice versa for cooling.)

Application — Haber. N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ/mol (forward exothermic). If T is increased, the backward direction is endothermic → equilibrium shifts BACKWARDS (towards reactants) → YIELD OF NH₃ DECREASES.

(2) Pressure change (for gas reactions only).

P INCREASE → equilibrium shifts to the side with FEWER moles of gas.
P DECREASE → equilibrium shifts to the side with MORE moles of gas.

Why? Raising pressure compresses gases. The system shifts to REDUCE the pressure. Pressure is reduced by having FEWER GAS MOLES. So shift towards whichever side has fewer moles of gas.

Count moles of GAS ONLY. Solids and liquids don't change much with pressure — ignore them.

Whatever the disturbance, the equilibrium position shifts in the direction that counteracts it.

Application — Haber. N₂ + 3H₂ ⇌ 2NH₃. Reactants: 1 + 3 = 4 mol gas. Products: 2 mol gas. Forward direction reduces moles of gas (4 → 2). So increasing P shifts FORWARDS → MORE NH₃ → YIELD INCREASES.

Application — Contact. 2SO₂ + O₂ ⇌ 2SO₃. Reactants: 2 + 1 = 3 mol gas. Products: 2 mol gas. Forward reduces moles of gas (3 → 2). Increasing P → more SO₃.

When pressure has NO effect. If both sides have equal moles of gas. e.g. H₂(g) + I₂(g) ⇌ 2HI(g): 2 mol gas → 2 mol gas. Increasing P → NO shift; equilibrium position unchanged.

(3) Concentration change.

[Reactant] INCREASE → equilibrium shifts FORWARDS (towards products — to consume the added reactant).
[Product] INCREASE → equilibrium shifts BACKWARDS (towards reactants — to consume the added product).
[Reactant] DECREASE → equilibrium shifts BACKWARDS (to replenish the reactant).
[Product] DECREASE → equilibrium shifts FORWARDS (to make more of the product).

Application — removing a product. If we continuously REMOVE NH₃ from the Haber reactor as it forms (by cooling and condensing it), the [NH₃] decreases → equilibrium shifts FORWARDS to replace it → MORE NH₃ formed over time. This is a key trick in industrial chemistry to maximise overall yield.

Mnemonic for Le Chatelier.

Whatever change you make, the system DOES THE OPPOSITE. Add heat → it absorbs heat (endo direction). Squash it (raise P) → it shrinks (fewer moles of gas). Add reactant → it consumes it (forward shift). The system always 'fights back'.

Why Le Chatelier works.

A more rigorous explanation (not required at IGCSE, but useful): the rates of the forward and backward reactions depend on the concentrations of reactants and products. When you disturb the system:

Adding more N₂ → forward rate increases (more reactant collisions). Backward rate unchanged at first.
Net effect: forward > backward → equilibrium shifts forwards.
As NH₃ builds up, backward rate increases too.
Eventually they meet again (at a new equilibrium with higher [NH₃] and lower [N₂] than before, but different from the original).

Effect of catalyst on equilibrium — NONE.

A catalyst speeds up BOTH the forward AND the backward reaction by the SAME factor (because both go through the same lowered Ea pathway). So:

Forward rate × f, backward rate × f → equilibrium STILL forward = backward.
Concentrations at equilibrium UNCHANGED → POSITION of equilibrium UNCHANGED.
BUT the system reaches equilibrium MUCH FASTER.

Industrially this is invaluable: same NH₃ yield per pass, but in minutes rather than hours → many more passes per day → much more total NH₃ produced.

Le Chatelier: system disturbed → shifts to OPPOSE the change.
T ↑ → endo direction. T ↓ → exo direction.
P ↑ (gas) → side with fewer moles of gas. P ↓ → more moles.
[Reactant] ↑ → forward shift. [Product] removed → forward shift.
Catalyst: NO shift in equilibrium position; just reaches it faster.
Always identify exo/endo direction and count gas moles before predicting.

See the full worked example for reversible reactions and equilibria →

The Haber process — ammonia production (Paper 2, spec 3.20)

N₂ + 3H₂ ⇌ 2NH₃. 450 °C / 200 atm / Fe. Compromise conditions.

The Haber process is the industrial method for manufacturing AMMONIA from atmospheric nitrogen and hydrogen:

$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92\ \text{kJ/mol}$

Developed by Fritz Haber in 1909 and scaled up by Carl Bosch. One of the most important industrial chemical processes — produces ~ 200 million tonnes of NH₃ globally per year. Ammonia is the basis of artificial fertilisers (NH₄NO₃, (NH₄)₂SO₄, urea) that feed about HALF the world's population. Also used to make nitric acid (HNO₃), explosives, cleaning products.

Industrial conditions.

Variable	Value	Why
Temperature	~ 450 °C	Compromise — rate vs yield
Pressure	~ 200 atm	Compromise — yield vs cost / safety
Catalyst	Iron (Fe) with K₂O / Al₂O₃ promoters	Lowers Ea → equilibrium reached faster

Source of reactants.

Nitrogen (N₂): from the atmosphere (air is ~ 78% N₂). Air is liquefied and separated to give pure N₂.
Hydrogen (H₂): usually from CH₄ + H₂O → CO + 3H₂ (steam reforming of methane in natural gas). Sometimes from electrolysis of water (zero-carbon route).

Temperature — 450 °C compromise.

Forward reaction is EXOTHERMIC (ΔH = −92 kJ/mol). By Le Chatelier:

LOW T → shifts forward → HIGH yield. ✓ But: rate is very slow → reaction takes days → uneconomic. ✗
HIGH T → shifts backward → LOW yield. ✗ But: rate is fast. ✓

Compromise: 450 °C. Yield is only ~ 15% per pass at 450 °C, but the rate is high enough that this happens in minutes. Overall: high throughput.

450 °C balances a workable rate against an acceptable ammonia yield — the classic compromise condition.

Why not 300 °C? Yield would be ~ 50% but rate would be too slow for industrial throughput. Why not 600 °C? Rate higher but yield only ~ 5% — too little NH₃ per pass.

Pressure — 200 atm compromise.

Forward: 4 moles of gas → 2 moles of gas. By Le Chatelier:

HIGH P → shifts forward → HIGH yield. ✓ But: huge equipment cost (thick pipes, large reactors). Safety risk. Compressing gas is energy-intensive. ✗
LOW P → shifts backward → LOW yield. ✗ But: cheap equipment. ✓

Compromise: 200 atm. High enough to get decent yield (~ 15%), low enough that equipment + energy costs are manageable. Higher pressure (e.g. 1000 atm) would improve yield, but the extra capital + energy cost outweighs the benefit.

Iron catalyst.

The IRON catalyst is essential. Without it, the reaction at 450 °C would take hours or days. With iron, equilibrium is reached in minutes.

The catalyst is usually 'promoted' (mixed with small amounts of K₂O and Al₂O₃) to improve its activity. It's a SOLID — placed as small pellets in a 'fixed-bed' reactor. The gases flow over the catalyst surface.

Critically: the catalyst does NOT change the YIELD (position of equilibrium). It only changes the SPEED. Without the catalyst, you'd still get ~ 15% yield per pass — just much more slowly.

Recycling unreacted gases.

After one pass through the reactor, only ~ 15% of N₂ + H₂ has been converted. The hot mixture is COOLED in heat exchangers:

NH₃ liquefies (bp −33 °C) → drained off as liquid ammonia.
Unreacted N₂ + H₂ stay as gas → pumped back to the reactor for another pass.

Over many passes, OVERALL CONVERSION reaches ~ 97% — most of the N₂ + H₂ ends up as NH₃. Continuous removal of NH₃ also helps shift equilibrium forwards (Le Chatelier — removing product makes the system 'try to replace' it).

Summary of compromises.

Condition	Best for yield	Best for rate / cost	Compromise
Temperature	LOW (exo forward)	HIGH (rate)	450 °C
Pressure	HIGH (4 → 2 mol gas)	LOW (cost / safety)	200 atm
Catalyst	irrelevant	needed	Iron (Fe)

The Haber process is THE classic 4CH1 'compromise conditions' question — be ready to explain WHY each value is chosen, not just to state the value.

Uses of ammonia.

~ 80% of NH₃ produced is converted to fertilisers:

Ammonium nitrate (NH₄NO₃): NH₃ + HNO₃ → NH₄NO₃. Major fertiliser.
Ammonium sulfate ((NH₄)₂SO₄): 2NH₃ + H₂SO₄ → (NH₄)₂SO₄. Cheap fertiliser.
Urea (CO(NH₂)₂): NH₃ + CO₂ → urea. Most-used solid fertiliser globally.

Other uses: nitric acid (oxidation of NH₃), explosives (RDX, TNT), cleaning products, refrigeration (anhydrous NH₃ as a refrigerant).

N₂ + 3H₂ ⇌ 2NH₃; ΔH = −92 kJ/mol; 4 mol gas → 2 mol gas.
Conditions: 450 °C, 200 atm, iron catalyst.
T compromise: low T good for yield (exo forward) but slow rate.
P compromise: high P good for yield (fewer gas moles) but very expensive.
Iron catalyst: speeds rate; does NOT change equilibrium position.
Recycle unreacted N₂ + H₂: overall conversion ~ 97%; NH₃ condenses out at low T.
Use: fertilisers (NH₄NO₃, urea), nitric acid, explosives.

See the full worked example for reversible reactions and equilibria →

The Contact process — sulfuric acid (Paper 2, spec 3.21)

2SO₂ + O₂ ⇌ 2SO₃. 450 °C / 1-2 atm / V₂O₅. High yield without high P.

The Contact process is the industrial method for manufacturing SULFURIC ACID (H₂SO₄). The critical equilibrium step is:

$2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H = -197\ \text{kJ/mol}$

H₂SO₄ is the world's most-produced industrial chemical by mass (~ 230 million tonnes/year). Used for fertilisers ((NH₄)₂SO₄, ammonium phosphates), detergents, paint pigments, lead-acid car batteries, oil refining, dyes, plastics, and general acid duty.

Full process — four stages.

Burn sulfur: $\text{S} + \text{O}_2 \rightarrow \text{SO}_2$ . (Sulfur from elemental sulfur deposits or as a byproduct of oil/gas desulfurisation.)
CATALYTIC OXIDATION (the equilibrium step): $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ . This is the key step we focus on for 4CH1.
Absorption in concentrated H₂SO₄ (NOT in water — too violent): $\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7$ (oleum or 'fuming sulfuric acid').
Dilution: oleum + water → 2 H₂SO₄. Now we have ordinary concentrated sulfuric acid (98%).

Industrial conditions for the equilibrium step.

Variable	Value	Why
Temperature	~ 450 °C	Compromise — rate vs yield
Pressure	~ 1-2 atm (essentially atmospheric)	Low P sufficient
Catalyst	Vanadium(V) oxide (V₂O₅)	Lowers Ea

Temperature — 450 °C compromise.

Forward reaction is HIGHLY EXOTHERMIC (ΔH = −197 kJ/mol — even more than the Haber process). By Le Chatelier:

LOW T → shifts forward → HIGH yield (favours SO₃).
HIGH T → shifts backward → low yield + faster rate.

Compromise at 450 °C: yield is ~ 95-99% (much higher than the Haber process at the same T because the equilibrium constant is much larger). Rate is also acceptable with the V₂O₅ catalyst.

Why not lower T (e.g. 300 °C)? Yield would be even higher (closer to 100%), but the rate would be slower and the V₂O₅ catalyst is less active.

Pressure — only 1-2 atm needed.

Forward: 2 SO₂ + 1 O₂ = 3 moles of gas → 2 SO₃ = 2 moles of gas. By Le Chatelier:

HIGH P → shifts forward → MORE SO₃.

BUT: the equilibrium ALREADY lies far to the right at 1-2 atm (yield > 95%). Going to higher P would only give a marginal improvement, at much higher capital + energy cost. Not economic.

So the Contact process uses only ATMOSPHERIC PRESSURE (or slightly above to push gases through the converter).

Contrast with Haber. The Haber process needs 200 atm because its equilibrium position is unfavourable at 1 atm (only ~ 0.1% NH₃). The Contact process doesn't need high P because its equilibrium ALREADY favours products even at 1 atm. Different reactions have different equilibrium constants → different P requirements.

Vanadium(V) oxide catalyst (V₂O₅).

V₂O₅ is the catalyst of choice for the Contact process. Why?

Effective at 450 °C (works at the operating temperature).
Cheaper than platinum (the older alternative).
Resistant to 'poisoning' by trace impurities.

Often promoted with K₂SO₄ to increase activity.

Why high yield, low P, moderate T?

The equilibrium constant K is LARGE for this reaction at 450 °C (about 4 × 10⁴ atm⁻¹). This means equilibrium strongly favours products even without high pressure. Compare with the Haber process where K is much smaller (~ 10⁻² atm⁻² at 450 °C), so we need high P to push it forward.

(Don't worry about the math — at IGCSE the qualitative idea is sufficient: 'the equilibrium already lies far right, so high P is unnecessary'.)

Final step — SO₃ to H₂SO₄ — DO NOT use water directly.

You might expect: SO₃ + H₂O → H₂SO₄. Why not just do this?

Because this reaction is HIGHLY EXOTHERMIC and produces a CHOKING MIST of sulfuric acid that's hard to control. Instead:

Dissolve SO₃ in concentrated H₂SO₄ → oleum (H₂S₂O₇), a manageable liquid.
Dilute oleum carefully with water → 2 H₂SO₄.

Net: SO₃ + H₂O → H₂SO₄, but done in two stages for control.

Side-by-side: Haber vs Contact.

Feature	Haber	Contact
Reaction	N₂ + 3H₂ ⇌ 2NH₃	2SO₂ + O₂ ⇌ 2SO₃
ΔH	−92 kJ/mol	−197 kJ/mol
Mol gas (R → P)	4 → 2	3 → 2
Temperature	450 °C	450 °C
Pressure	200 atm	1-2 atm
Catalyst	Iron (Fe)	Vanadium(V) oxide
Yield per pass	~ 15%	~ 95-99%
Product use	Fertilisers	Sulfuric acid

Both use 450 °C compromise (same logic — exo forward + need acceptable rate). But Haber needs HIGH P (less favourable equilibrium at 1 atm) whereas Contact only needs ~ 1 atm (equilibrium already favourable).

2SO₂ + O₂ ⇌ 2SO₃; ΔH = −197 kJ/mol (very exothermic).
Conditions: 450 °C, 1-2 atm, V₂O₅ catalyst.
Yield per pass ~ 95-99% (much higher than Haber).
Low P sufficient because equilibrium already favours products at 1 atm.
SO₃ + H₂SO₄ → H₂S₂O₇ (oleum); then dilute with water → H₂SO₄ (not direct because too violent).
Uses of H₂SO₄: fertilisers, detergents, batteries, dyes — world's most-produced industrial chemical.

See the full worked example for reversible reactions and equilibria →

Catalysts and equilibrium — speeding without shifting (spec 3.18, 3.20)

Catalyst lowers Ea for BOTH directions equally → no shift, only faster.

A common Paper 2 question: 'Does a catalyst shift the equilibrium position?'

The answer is NO — a catalyst does NOT shift the equilibrium position. It only speeds up the rate at which equilibrium is reached.

Why NOT.

A catalyst provides an alternative pathway with a LOWER activation energy. This applies to BOTH the forward AND the backward reactions — they both go through the same lowered Ea (the catalyst doesn't 'prefer' one direction over the other).

Forward rate × f (where f > 1, the speedup factor).
Backward rate × f (same factor).

Therefore:

If they were equal before (at equilibrium), they're still equal after (just both faster).
The RATIO forward:backward = 1:1 throughout.
Concentrations at equilibrium UNCHANGED.

So the catalyst SPEEDS UP THE APPROACH TO EQUILIBRIUM, but doesn't change WHERE the equilibrium sits.

Why this is industrially valuable.

In the Haber process, equilibrium at 450 °C gives ~ 15% NH₃. With the iron catalyst, you reach this 15% in MINUTES. Without the catalyst, you'd reach the same 15% only after HOURS or DAYS.

Industrially: same yield per pass + much faster pass → many more passes per day → MUCH MORE NH₃ produced per year. Without the catalyst, the Haber process would be uneconomic.

Mark-scheme phrasing to memorise.

For full marks on a 'does a catalyst affect equilibrium' question, use these phrases:

'A catalyst lowers the activation energy of BOTH the forward and the backward reactions.'
'Both rates are increased by the same factor.'
'The position of equilibrium is UNCHANGED.'
'Equilibrium is reached FASTER.'

Visualisation on a reaction profile.

For a reversible reaction:

Uncatalysed: tall hump (Ea forward) over the reactants line, with a slightly different drop to products.
Catalysed: shorter hump for both directions.
Same reactants and products levels (ΔH unchanged) → same equilibrium position.

Other things that do NOT shift equilibrium.

Adding more INERT gas (a gas not involved in the reaction, e.g. argon to the Haber reactor): increases TOTAL pressure but doesn't change the partial pressures or moles of reactant/product gases → no shift.
Changing the size of the container at constant T and constant total moles of gas: redistributes by Le Chatelier (volume change ≡ pressure change in opposite direction).

Things that DO shift equilibrium (recap).

Temperature.
Total pressure (gas reactions, unequal mol gas).
Concentration of a reactant or product.

Putting it all together — Haber process control.

To maximise NH₃ production, industrial chemists:

Use the COMPROMISE temperature (450 °C) — accept lower yield to gain rate.
Use a COMPROMISE pressure (200 atm) — accept extra cost for higher yield.
Use the iron CATALYST — reach equilibrium faster (more passes per day).
REMOVE NH₃ continuously (condense out at low T) → Le Chatelier shifts forward → more NH₃ formed.
RECYCLE unreacted gases → overall conversion ~ 97%.

Each of these steps uses the principles we've covered: Le Chatelier, catalysis, and continuous removal of product.

Catalyst lowers Ea for BOTH forward and backward reactions equally.
Both rates × same factor → position of equilibrium UNCHANGED.
Equilibrium is reached FASTER (the practical benefit).
Does NOT change ΔH or yield, only speed.
Mark-scheme phrase: 'alternative pathway with lower Ea; both rates equally affected'.
Other things that DON'T shift: adding inert gas; the catalyst itself.

See the full worked example for reversible reactions and equilibria →

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Reversible-reaction symbol: ⇌ (NOT → or ⇆ in mark scheme — the double-arrow ⇌ is standard).
Dynamic equilibrium: 'rate of FORWARD reaction = rate of BACKWARD reaction' (the mark-scheme phrase).
Le Chatelier statement: 'if a system at equilibrium is disturbed, the position of equilibrium shifts to OPPOSE the change'.
Temperature: T ↑ → endo direction; T ↓ → exo direction. Always identify ΔH direction first.
Pressure: P ↑ → side with fewer moles of GAS. Only counts gas species; ignore solids and liquids.
Catalyst: 'lowers activation energy of BOTH forward and backward reactions EQUALLY' → no shift, only faster.
Haber conditions: 450 °C, 200 atm, iron catalyst. Both T and P are COMPROMISES (use the word compromise).
Contact conditions: 450 °C, 1-2 atm, V₂O₅. Low P sufficient because equilibrium already favours products.
Industrial trick: continuously REMOVE product (NH₃ in Haber) → equilibrium shifts forward → higher overall conversion.

How it’s examined

Reversible reactions appears on most 4CH1 papers, typically worth 5-8 marks per session. Paper 2 (4CH1/2C) is the main location since Le Chatelier (3.18-3.21) is bold 'P' content. Common question types: (a) define dynamic equilibrium (4 marks); (b) apply Le Chatelier to predict effect of T / P / concentration on a given equilibrium (4-5 marks each); (c) explain the conditions used in the Haber process (often a full 6-8 mark essay — temperature compromise + pressure compromise + catalyst); (d) explain why a catalyst doesn't shift equilibrium (4 marks). The Haber-process essay is the SINGLE most common Paper 2 question on this topic — be fluent with the compromises. Paper 1 may include simpler reversibility questions (hydrated/anhydrous CuSO₄ demo) without Le Chatelier depth.

Sources: Pearson Edexcel International GCSE Chemistry 4CH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4CH1 Examiner Reports 2022-2024; 4CH1/2C May/Jun 2024 question paper and mark scheme; 4CH1/2C January 2024 question paper and mark scheme; 4CH1/1C May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Reversible Reactions and Equilibria

Step-by-step solutions to past-paper-style questions on reversible reactions and equilibria, written exactly the way a tutor would explain them at the board.

1Explain what is meant by a dynamic equilibrium (5 marks, Paper 2)
Extended• Adapted from 4CH1/2C May/Jun 2024 Q4• dynamic equilibrium, reversible reactions, spec-3.17, spec-3.18
Question
The reaction $\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})$ can reach a dynamic equilibrium in a sealed flask. (a) State what is meant by a 'reversible reaction'. (b) State what is meant by a 'dynamic equilibrium'. (c) Explain why a dynamic equilibrium can only be reached in a CLOSED system. (5 marks)
Step-by-step solution
1. Step 1
  (a) Reversible reaction (1 M). A reaction in which the products can react together to re-form the reactants — the reaction can proceed in BOTH directions. Written with the equilibrium arrow ⇌ instead of →.
  $\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI(g)}$
2. Step 2
  (b) Dynamic equilibrium — definition (1 A). A state reached in a CLOSED system where the FORWARD and BACKWARD reactions are still occurring (the reaction has NOT stopped), but at EQUAL RATES, so that there is NO NET CHANGE in the concentrations of reactants and products.
3. Step 3
  (b) 'Dynamic' part explained (1 A). Although the macroscopic properties (concentrations, colour, pressure) appear constant, on the molecular scale particles are constantly reacting in both directions: H₂ and I₂ react to form HI just as fast as HI breaks down into H₂ and I₂. The system is 'dynamic' (moving) at the molecular level but 'static' at the macroscopic level.
4. Step 4
  (c) Why a closed system is needed (1 B). A CLOSED system is one where NO reactant or product can ESCAPE (and no new substance can enter). If the system were OPEN: any gaseous product could escape, lowering its concentration, so the BACKWARD reaction could never occur at the same rate → equilibrium never establishes.
5. Step 5
  (c) Specific example (1 B). For H₂ + I₂ ⇌ 2HI, if the flask were open, HI gas would escape into the atmosphere — the system could not 'feed back' to make H₂ and I₂ again. So you'd just see one-way conversion of H₂ + I₂ → HI until reactants ran out. In a sealed flask, all species remain in contact → forward and backward rates can become equal.
Answer
(a) A reversible reaction can proceed in both directions: products can react to re-form reactants. Written with ⇌. (b) Dynamic equilibrium = forward and backward rates EQUAL → concentrations CONSTANT → reaction has NOT stopped (still occurring on molecular scale). (c) A closed system stops products escaping → both reactants and products remain in contact → the backward reaction can occur. In an open system, products escape and equilibrium cannot establish.
Examiner tip
Edexcel 5-mark mark scheme: 1 for reversible reaction definition; 1 for 'forward and backward rates equal'; 1 for 'concentrations remain constant' / 'reaction has not stopped'; 1 for closed system; 1 for explaining why an open system would not work. Common error: saying 'the reaction stops' — wrong, both directions still occur, just at equal rates. The phrase 'rate of forward reaction = rate of backward reaction' is mark-scheme gold.
2Le Chatelier — predict the effect of temperature (5 marks, Paper 2)
Extended• Adapted from 4CH1/2C Jan 2024 Q9• Le Chatelier, temperature, Haber, spec-3.20
Question
The Haber process is a reversible reaction used industrially to produce ammonia: $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92\ \text{kJ/mol}$ . (a) State Le Chatelier's principle. (b) Predict and EXPLAIN what happens to the position of equilibrium if the temperature is INCREASED. (c) Predict and EXPLAIN what happens to the YIELD of NH₃ if the temperature is INCREASED. (5 marks)
Step-by-step solution
1. Step 1
  (a) Le Chatelier's principle (1 M). If a system at equilibrium is disturbed by changing the temperature, pressure, or concentration, the equilibrium will SHIFT to OPPOSE (counteract) the change. In effect, the system 'pushes back' against the disturbance.
2. Step 2
  (b) Identify endothermic / exothermic directions (1 A). Given: forward reaction (N₂ + 3H₂ → 2NH₃) is EXOTHERMIC (ΔH = −92 kJ/mol, negative). Therefore the BACKWARD reaction (2NH₃ → N₂ + 3H₂) is ENDOTHERMIC (+92 kJ/mol, positive).
3. Step 3
  (b) Apply Le Chatelier (1 A). Increasing temperature ADDS heat to the system. The system shifts to OPPOSE this — to ABSORB the added heat. The endothermic direction is the BACKWARD direction. So the equilibrium shifts BACKWARD (to the LEFT — towards reactants).
4. Step 4
  (c) Effect on yield of NH₃ (1 B). Equilibrium shifts BACKWARDS → MORE NH₃ decomposes → MORE N₂ and H₂ form → LESS NH₃ at equilibrium. So increasing temperature DECREASES the yield of NH₃.
5. Step 5
  (c) Industrial compromise (1 B). Logically, the Haber process should use LOW temperature for maximum yield. But at low T, the reaction is too SLOW (low rate of reaction). The industrial compromise: 450 °C — high enough to give a useful rate, low enough to keep yield acceptable (~15% per pass; unreacted gases are recycled). This is a classic 'rate vs yield' compromise — see Topic 3.2 on rates.
Answer
(a) Le Chatelier: if a system at equilibrium is disturbed, the equilibrium shifts to OPPOSE the change. (b) Forward reaction is exothermic, so backward is endothermic. Increasing T → system shifts to absorb heat → shifts BACKWARDS (towards reactants, N₂ + H₂). (c) Yield of NH₃ DECREASES. Industrially, T = 450 °C is a compromise: lower T would give higher yield but lower rate; higher T would give faster rate but lower yield.
Examiner tip
Edexcel 5-mark mark scheme: 1 for Le Chatelier statement (opposing/counteracting); 1 for identifying exo/endo directions; 1 for applying it (shifts backwards / endothermic direction); 1 for effect on yield (decreases); 1 for compromise explanation. Mark scheme accepts 'equilibrium moves to the LEFT' OR 'shifts backwards / towards reactants'. Common error: confusing the direction of shift when ΔH is given for the FORWARD reaction.
3Le Chatelier — predict the effect of pressure (5 marks, Paper 2)
Extended• Adapted from 4CH1/2C May/Jun 2024 Q5• Le Chatelier, pressure, Haber, spec-3.20
Question
For the Haber process $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$ , predict and explain the effect of INCREASING the pressure on (a) the position of equilibrium and (b) the yield of NH₃. (c) State why the Haber process is operated at 200 atm and not at 1000 atm. (5 marks)
Step-by-step solution
1. Step 1
  (a) Count moles of gas on each side (1 M). Left side (reactants): 1 mol N₂ + 3 mol H₂ = 4 moles of gas. Right side (products): 2 mol NH₃ = 2 moles of gas. The forward reaction REDUCES the number of moles of gas (4 → 2).
2. Step 2
  (a) Apply Le Chatelier (1 A). Increasing pressure DISTURBS the equilibrium by raising the pressure. The system shifts to OPPOSE this — to REDUCE the pressure. Pressure is reduced by having FEWER GAS MOLES → equilibrium shifts to the side with FEWER moles of gas. Here: shifts to the RIGHT (products, 2 mol gas).
3. Step 3
  (b) Effect on yield (1 A). Equilibrium shifts to the RIGHT → MORE NH₃ formed → YIELD of NH₃ INCREASES. (Higher pressure favours NH₃ production.)
4. Step 4
  (c) Why 200 atm not 1000 atm (1 B). Higher pressure would give EVEN HIGHER yield. But there are practical drawbacks: • Equipment cost: pipes, reactors, compressors must be much thicker and stronger at 1000 atm → very expensive plant. • Safety: extremely high-pressure gases are dangerous — risk of explosion / leaks. • Energy cost: compressing gas to 1000 atm requires a huge amount of electrical energy → high running costs.
5. Step 5
  (c) Compromise (1 B). 200 atm is a COMPROMISE between yield and cost/safety: high enough to give a useful yield, low enough that equipment and energy costs are manageable. Modern Haber plants often run at 150-250 atm depending on optimisation.
Answer
(a) Reactant side has 4 mol gas; product side has 2 mol gas. Increasing P → equilibrium shifts to the side with FEWER moles of gas → shifts to the RIGHT (products). (b) Yield of NH₃ INCREASES. (c) Higher P (e.g. 1000 atm) would give even higher yield but the cost of pressure-resistant equipment, the energy to compress, and the safety risks make it uneconomic. 200 atm is the compromise.
Examiner tip
Edexcel 5-mark mark scheme: 1 for counting moles of gas on each side (4 vs 2); 1 for 'shifts to fewer moles of gas' / 'shifts to the right'; 1 for yield increases; 1 for cost/equipment/safety reason; 1 for stating it's a compromise. Common error: forgetting to count moles of gas (always count gas moles, not solid/aqueous). Reactions with equal moles of gas on both sides are NOT affected by pressure changes.
4Why a catalyst does NOT shift equilibrium (4 marks, Paper 2)
Extended• Adapted from 4CH1/2C Jan 2024 Q8• Le Chatelier, catalyst, Haber, spec-3.20, spec-3.21
Question
Iron is used as a catalyst in the Haber process. Explain why adding the iron catalyst does NOT shift the position of equilibrium but DOES affect how quickly equilibrium is reached. (4 marks)
Step-by-step solution
1. Step 1
  Catalyst's effect on rate (1 M). A catalyst provides an ALTERNATIVE PATHWAY with a LOWER activation energy → more particles have E ≥ Ea → more successful collisions per second → faster rate. For a reversible reaction, the catalyst lowers Ea for BOTH the forward AND the backward reaction (same activation barrier, same pathway).
2. Step 2
  Equal speeding-up of both directions (1 A). The forward rate INCREASES by some factor f. The backward rate also increases by the SAME factor f. Therefore the RATIO of forward rate to backward rate is unchanged.
3. Step 3
  Position of equilibrium unchanged (1 A). At equilibrium, forward rate = backward rate. If the catalyst multiplies BOTH by the same factor f, they remain equal — so the system is still at equilibrium. The CONCENTRATIONS of reactants and products at equilibrium are UNCHANGED → position of equilibrium is UNCHANGED.
4. Step 4
  But equilibrium reached FASTER (1 B). Without a catalyst, the Haber reaction at 450 °C would take hours or days to reach equilibrium. With iron, both the forward and backward reactions are sped up → the system reaches the equilibrium composition in minutes. This makes the process commercially viable. Industrially, this is critical: the same NH₃ yield, but in much less time → more NH₃ per day → more profit.
Answer
A catalyst lowers the activation energy for BOTH the forward and backward reactions by the SAME factor. Forward rate × f; backward rate × f. The ratio is unchanged → equilibrium concentrations unchanged → position of equilibrium unchanged. But both rates are increased → equilibrium is REACHED FASTER. So a catalyst is desirable industrially because it gives the same yield in less time → higher throughput.
Examiner tip
Edexcel 4-mark mark scheme: 1 for catalyst lowers Ea / provides alternative pathway; 1 for affects BOTH forward AND backward rates equally; 1 for position of equilibrium UNCHANGED; 1 for equilibrium reached FASTER. Common error: saying the catalyst 'pushes the reaction forward' — it doesn't, only speeds both directions equally.
5Conditions for the Haber process — full compromise explanation (8 marks, Paper 2)
Extended• Adapted from 4CH1/2C May/Jun 2024 Q9• Haber, compromise, industrial, spec-3.20, spec-3.21
Question
The Haber process is operated at approximately 450 °C, 200 atm, with an iron catalyst. The reaction is $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92\ \text{kJ/mol}$ . Explain WHY each of these three conditions is chosen — including any compromises made between yield, rate and cost. (8 marks)
Step-by-step solution
1. Step 1
  Temperature analysis — equilibrium (1 M). The forward reaction is EXOTHERMIC (ΔH = −92 kJ/mol). By Le Chatelier, INCREASING T shifts equilibrium BACKWARDS (endothermic direction) → DECREASES yield. So for max yield, T should be LOW.
2. Step 2
  Temperature analysis — rate (1 A). But at LOW T, the reaction RATE is very slow (few particles have energy ≥ Ea). So low T gives high yield but takes forever — not commercially useful.
3. Step 3
  Temperature compromise — 450 °C (1 A). 450 °C is a COMPROMISE: low enough to give an acceptable yield (~15% per pass) but high enough for the reaction to proceed at a USEFUL RATE. Lower T would maximise yield but take too long; higher T would be fast but give very low yield (and waste energy).
4. Step 4
  Pressure analysis — equilibrium (1 B). Reactant side: 4 moles of gas (1 N₂ + 3 H₂). Product side: 2 moles of gas (2 NH₃). Increasing P shifts equilibrium to the side with FEWER moles of gas → to the RIGHT → INCREASES yield of NH₃. So for max yield, P should be HIGH.
5. Step 5
  Pressure analysis — cost / safety (1 B). Very high P requires extremely thick, expensive pipes and reactors → high capital cost; large amounts of electrical energy to compress the gas → high running cost; safety risks (leaks, explosion).
6. Step 6
  Pressure compromise — 200 atm (1 B). 200 atm gives a reasonable yield (~15-20% per pass) while keeping equipment and energy costs at a manageable level. Higher P (e.g. 1000 atm) would give better yield but the extra equipment cost is not justified by the gain.
7. Step 7
  Iron catalyst — what it does (1 B). Iron provides an alternative pathway with LOWER activation energy → BOTH forward and backward rates increase EQUALLY → equilibrium is reached MUCH FASTER. It does NOT change the position of equilibrium (yield is unchanged).
8. Step 8
  Catalyst — why iron (1 B). Iron is cheap, abundant, and effective at the operating temperature. Without the catalyst, the Haber process would be impossibly slow at 450 °C (Ea too high). The iron speeds up the approach to equilibrium → same NH₃ yield per pass, but in minutes rather than hours → more production per day → economic.
Answer
Temperature (450 °C): compromise. Forward exo → low T favours yield BUT slows rate. 450 °C balances acceptable yield (~15%) with useful rate. Pressure (200 atm): compromise. 4 mol gas → 2 mol gas → high P favours yield BUT high P is expensive and dangerous. 200 atm balances good yield with practical equipment/cost. Iron catalyst: lowers Ea → speeds BOTH directions equally → equilibrium reached faster (NOT shifted). Iron chosen for low cost + effectiveness. Without catalyst the process would be too slow.
Examiner tip
Edexcel 8-mark mark scheme: 2 marks each for temperature compromise (yield + rate sides); 2 marks for pressure compromise (yield + cost sides); 2 marks for the catalyst (rate effect + no effect on yield); 2 marks for explicitly noting it's all a 'compromise'. This is a classic Paper 2 essay-style question — write in clear paragraphs. Common error: not explaining the OPPOSING factor that creates the compromise (just saying 'high T favours rate' misses that low T favours yield).

Model Answers — Reversible Reactions and Equilibria

High-scoring sample answers for reversible reactions and equilibria on the Pearson Edexcel IGCSE 4CH1 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4CH1/2C-style — dynamic equilibrium4 marks
Define a dynamic equilibrium. State the conditions necessary for a dynamic equilibrium to be established. Explain what is happening at the molecular level even though the bulk concentrations appear constant. (4 marks)
Model answer
Definition of a dynamic equilibrium.

A dynamic equilibrium is a state reached in a reversible reaction in a CLOSED system where:

The rate of the FORWARD reaction = the rate of the BACKWARD reaction.

As a result, the concentrations (and other macroscopic properties like colour, pressure, total mass) of all reactants and products REMAIN CONSTANT over time.

The reaction has NOT stopped — it is still occurring in both directions.

Conditions necessary for a dynamic equilibrium.

The reaction must be REVERSIBLE (can proceed in both directions). Written with ⇌.

The system must be CLOSED — neither reactants nor products can escape (no matter is added or removed from outside). For gas-phase reactions, this means a sealed container; for solution reactions, it usually means a flask with no evaporation.

Constant temperature must be maintained (changing T disturbs the equilibrium — by Le Chatelier).

Sufficient TIME must have passed for the system to settle. Initially, if you start with only reactants, the forward rate is fast and backward rate zero. As products form, backward rate increases; as reactants are consumed, forward rate decreases. Eventually the two rates EQUAL each other → equilibrium.

At the molecular level — what's happening.

Even though the concentrations APPEAR constant, on a molecular scale particles are constantly reacting:

For each forward reaction that occurs (reactants → products), an equivalent backward reaction also occurs (products → reactants).

Individual molecules are constantly being converted, but the NET amount of each species stays the same because the two opposing processes balance.

Example: in H₂ + I₂ ⇌ 2HI, at any instant, many H₂ + I₂ pairs are colliding and forming HI, AND many HI pairs are colliding and breaking back into H₂ and I₂. These two rates are equal at equilibrium.

Analogy — escalators. Imagine two escalators side by side in a department store: one going up, one going down. If 100 people per minute go UP and 100 per minute go DOWN, the number of people on each floor stays the same (concentrations constant) — but the system is DYNAMIC: people are constantly moving between floors. If you photographed it, no one floor is empty; everyone is constantly in flux. That's dynamic equilibrium.

Why a closed system is essential. In an OPEN system, products can escape. If HI escaped from the flask, the backward reaction (HI → H₂ + I₂) couldn't proceed, and the forward reaction would just keep going until reactants ran out — no equilibrium would be established. Closing the system traps both reactants and products together so both directions can occur.
Why this scores
Edexcel 4-mark mark scheme: 1 for 'forward and backward rates equal'; 1 for 'concentrations constant' / 'reaction has not stopped'; 1 for stating closed system requirement; 1 for explaining molecular-level dynamic behaviour (still occurring in both directions). Mark scheme keyword phrase: 'rate of forward reaction = rate of backward reaction'. Use exactly.

Question 2

4CH1/2C-style — Le Chatelier overview6 marks

State Le Chatelier's principle. Use it to explain the effect of (a) increasing temperature, (b) increasing pressure, and (c) increasing the concentration of a reactant on the position of equilibrium of a generic reversible reaction. (6 marks)

Model answer

Le Chatelier's principle.

If a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the position of equilibrium will shift in the direction that OPPOSES (counteracts) the change.

The system 'pushes back' against the disturbance to minimise its effect. The new equilibrium that establishes has different concentrations, but the relationship 'forward rate = backward rate' is restored.

(a) Effect of increasing temperature.

INCREASING T adds heat to the system.
The equilibrium shifts to ABSORB heat → in the endothermic direction.
If the forward reaction is EXOTHERMIC (ΔH negative) → backward is endothermic → equilibrium shifts BACKWARDS (towards reactants) → yield of products DECREASES.
If the forward reaction is ENDOTHERMIC (ΔH positive) → forward is endothermic → equilibrium shifts FORWARDS (towards products) → yield INCREASES.

Example: N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ/mol, exothermic). Increasing T → shifts BACKWARDS → less NH₃ at equilibrium.

Decreasing temperature has the OPPOSITE effect — shifts in the exothermic direction (releases heat to oppose the cooling).

(b) Effect of increasing pressure (gas reactions only).

INCREASING P forces the gas into a smaller volume → raises particle density.
Equilibrium shifts to REDUCE pressure → to the side with FEWER moles of gas.
If there are equal moles of gas on both sides → NO effect (e.g. H₂ + I₂ ⇌ 2HI: 2 mol on each side).

Examples:

N₂ + 3H₂ ⇌ 2NH₃ — 4 mol gas → 2 mol gas. Increasing P shifts FORWARDS → MORE NH₃.
2SO₂ + O₂ ⇌ 2SO₃ — 3 mol gas → 2 mol gas. Increasing P shifts FORWARDS → MORE SO₃.
N₂O₄ ⇌ 2NO₂ — 1 mol gas → 2 mol gas. Increasing P shifts BACKWARDS → MORE N₂O₄.

Decreasing pressure has the OPPOSITE effect — shifts to the side with more moles of gas.

(c) Effect of increasing the concentration of a reactant.

INCREASING [reactant] disturbs the system by having too much reactant.
Equilibrium shifts to USE UP the added reactant → shifts in the FORWARD direction (towards products).
Yield of products INCREASES.

Example: in N₂ + 3H₂ ⇌ 2NH₃, adding more N₂ shifts equilibrium FORWARDS, producing more NH₃ (and consuming some of the added N₂).

Increasing the concentration of a PRODUCT has the opposite effect — equilibrium shifts BACKWARDS to use up the added product. So removing a product (e.g. by condensing NH₃ and tapping it off in the Haber process) shifts equilibrium FORWARDS → more product forms → higher overall yield.

(d) Bonus: effect of a catalyst.

A catalyst does NOT shift the position of equilibrium — it speeds up BOTH the forward and backward reactions equally → equilibrium is reached FASTER but at the SAME composition.

Summary table.

Change	Effect on equilibrium
T ↑	Shifts in endothermic direction
T ↓	Shifts in exothermic direction
P ↑ (gas)	Shifts to fewer moles of gas
P ↓ (gas)	Shifts to more moles of gas
[Reactant] ↑	Shifts forwards (away from added reactant)
[Product] ↑	Shifts backwards (away from added product)
Catalyst	NO shift; reaches equilibrium faster

Why this scores

Edexcel 6-mark mark scheme: 1 for Le Chatelier statement (opposes change); 1 each for the three predictions (temperature, pressure, concentration); 1 for the explanatory phrase for each (endothermic / fewer moles / consume added). Mark scheme accepts 'moves' or 'shifts' interchangeably. Catalyst non-effect bonus point if covered.

Question 3
4CH1/2C-style — Haber compromise6 marks
Describe the conditions used in the Haber process for the manufacture of ammonia and explain why each is a COMPROMISE. The reaction is $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92\ \text{kJ/mol}$ . (6 marks)
Model answer
Industrial conditions for the Haber process:

Variable Value Notes
Temperature ~ 450 °C Compromise (see below)
Pressure ~ 200 atm Compromise (see below)
Catalyst Iron (often promoted with K₂O / Al₂O₃) Speeds up approach to equilibrium

(1) Temperature: 450 °C — a compromise.

The forward reaction is EXOTHERMIC (ΔH = −92 kJ/mol).

By Le Chatelier, INCREASING T shifts equilibrium BACKWARDS → DECREASES yield of NH₃.

So for maximum YIELD, T should be LOW. (At room temperature, the equilibrium would lie almost entirely on the products side.)

BUT at LOW T, the RATE of reaction is very slow — most particles don't have energy ≥ Ea.

For commercial production, you need a useful RATE (kg of NH₃ per hour) AND a useful YIELD.

450 °C is a COMPROMISE: high enough that the rate is acceptable, low enough that the yield (~15% per pass) is still useful. Higher T gives faster reaction but much less NH₃; lower T gives more NH₃ but takes too long.

(2) Pressure: 200 atm — a compromise.

Reactant side: 4 moles of gas (1 N₂ + 3 H₂). Product side: 2 moles of gas (2 NH₃).

By Le Chatelier, INCREASING P shifts equilibrium FORWARDS (towards fewer moles of gas) → INCREASES yield of NH₃.

So for maximum YIELD, P should be HIGH.

BUT very high P (e.g. 1000 atm) requires:

extremely thick, expensive pipes and reactors (capital cost ↑↑)

large amounts of electrical energy to compress gases (running cost ↑↑)

much greater safety risks (high pressure can cause explosions)

200 atm is a COMPROMISE: high enough to give a useful yield, low enough that equipment and energy costs are manageable. Higher P would give more NH₃ per pass but at much higher capital + energy cost.

(3) Iron catalyst — speeds up but doesn't shift.

Iron is used as a catalyst (often 'promoted' with K₂O and Al₂O₃ to increase activity).

It provides an ALTERNATIVE PATHWAY with LOWER activation energy → BOTH forward and backward rates increase equally → equilibrium is reached FASTER but at the SAME composition.

Without the catalyst, the Haber process at 450 °C would take many hours; with iron, equilibrium is reached in minutes → much greater throughput.

Iron is CHEAP and EFFECTIVE at these conditions, hence chosen.

(4) Recycling unreacted gases.

Because each pass converts only 15% of N₂ + H₂ to NH₃, the unreacted gases are SEPARATED from the NH₃ (by cooling the mixture — NH₃ liquefies at higher T than N₂ or H₂) and RECYCLED back into the reactor. Over many passes, the OVERALL conversion is much higher ( 97-99%). The constant removal of NH₃ also helps shift the equilibrium forwards (by Le Chatelier — removing product makes the system 'try to replace' it).

Use of ammonia. ~ 80% of NH₃ produced is used to make fertilisers: ammonium nitrate (NH₄NO₃), ammonium sulfate ((NH₄)₂SO₄), and urea (CO(NH₂)₂). Other uses: nitric acid manufacture, explosives, cleaning products.

Summary of compromises.

Condition Best for yield Best for rate / cost Compromise
Temperature LOW (exo forward) HIGH (rate) 450 °C
Pressure HIGH (4 → 2 mol) LOW (cost / safety) 200 atm
Catalyst irrelevant needed iron
Why this scores
Edexcel 6-mark mark scheme: 2 each for explaining T (exo + rate), P (moles of gas + cost), and catalyst (Ea + same equilibrium). Use the word 'compromise' explicitly — it's a mark-scheme keyword. Bonus credit for mentioning recycling unreacted gases (shifts equilibrium further forward by Le Chatelier as NH₃ is removed). 4CH1/2C asks this question almost every year.
Question 4
4CH1/2C-style — Contact process5 marks
The Contact process is used to make sulfuric acid. The key equilibrium step is $2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H = -197\ \text{kJ/mol}$ . State the conditions used and explain (using Le Chatelier) why each is chosen. (5 marks)
Model answer
Conditions used in the Contact process (for the key equilibrium step SO₂ → SO₃):

Variable Value
Temperature ~ 450 °C
Pressure ~ 1-2 atm (essentially atmospheric)
Catalyst Vanadium(V) oxide (V₂O₅)

(1) Temperature: 450 °C — a compromise.

The forward reaction is EXOTHERMIC (ΔH = −197 kJ/mol, very exothermic — even more than the Haber process!).

By Le Chatelier, LOW T would maximise yield (shifts forwards in the exothermic direction).

BUT at low T, the rate is too slow — even with a catalyst.

450 °C is a COMPROMISE: high enough to give a useful rate, low enough to keep yield acceptable (~95-99% per pass — Contact has a much higher yield than Haber because the equilibrium lies far to the right at moderate T).

(2) Pressure: 1-2 atm — only slightly above atmospheric.

Reactant side: 2 mol SO₂ + 1 mol O₂ = 3 moles of gas. Product side: 2 mol SO₃ = 2 moles of gas.

By Le Chatelier, increasing P shifts forwards (3 → 2 mol gas) → favours SO₃ formation.

BUT the equilibrium ALREADY lies far to the right at 1-2 atm (yield > 95%), so there's little to gain from going higher.

Higher P would not significantly improve yield but would cost more in equipment + energy → NOT economic.

1-2 atm is essentially just enough to push gas through the converter. The Contact process does NOT need high pressure (unlike the Haber process which DOES — different equilibrium position).

(3) Vanadium(V) oxide catalyst — V₂O₅.

Provides an alternative pathway with lower Ea → faster forward AND backward rates → equilibrium reached FASTER.

Does NOT shift the position of equilibrium (yield unchanged by catalyst).

V₂O₅ chosen because it's effective at 450 °C, doesn't poison easily, and is cheaper than Pt (an alternative).

Promoter: K₂SO₄ improves the activity of V₂O₅.

(4) Use of the products.

SO₃ is dissolved in concentrated H₂SO₄ to give oleum (H₂S₂O₇), then diluted with water to give H₂SO₄. (Direct dissolving in water is too violent.) H₂SO₄ is used in fertilisers (NH₄)₂SO₄, detergents, lead-acid car batteries, and as a general industrial acid.

Comparison with the Haber process.

Haber: 4 → 2 mol gas → high pressure needed. Contact: 3 → 2 mol gas → small change → only 1-2 atm.

Haber: yield only ~15% per pass → needs recycling. Contact: yield > 95% per pass → high single-pass conversion.

Both: exothermic forward, so use moderate T (~ 450 °C) as a compromise.

Both: use a catalyst (Fe vs V₂O₅) to speed approach to equilibrium.
Why this scores
Edexcel 5-mark mark scheme: 1 each for naming the three conditions (T, P, catalyst); 1 for explaining T compromise (exo + rate); 1 for explaining low P sufficient (equilibrium already favourable); 1 for catalyst lowers Ea, doesn't shift equilibrium. Bonus credit for noting the comparison with Haber. The point that LOW pressure is used in Contact despite Le Chatelier favouring high P is a common discriminator question.

Variable	Value	Notes
Temperature	~ 450 °C	Compromise (see below)
Pressure	~ 200 atm	Compromise (see below)
Catalyst	Iron (often promoted with K₂O / Al₂O₃)	Speeds up approach to equilibrium

Variable	Value
Temperature	~ 450 °C
Pressure	~ 1-2 atm (essentially atmospheric)
Catalyst	Vanadium(V) oxide (V₂O₅)

Key Definitions and Keywords — Reversible Reactions and Equilibria

Definitions to memorise and the exact keywords mark schemes credit for reversible reactions and equilibria answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4CH1 sitting.

Reversible reaction
Examiner keyword
A reaction in which the products can react together to re-form the reactants — the reaction can proceed in both directions. Written with the symbol ⇌ instead of →.
Dynamic equilibrium
Examiner keyword
A state reached in a closed system where the rate of the FORWARD reaction = the rate of the BACKWARD reaction. The concentrations of all species REMAIN CONSTANT. The reaction has NOT stopped — it is still occurring in both directions at the molecular level.
Closed system
Examiner keyword
A system in which no reactants or products can enter or leave. Required for a dynamic equilibrium — otherwise products would escape and the backward reaction couldn't occur. For gases, a sealed container.
Le Chatelier's principle
Examiner keyword
If a system at equilibrium is disturbed by a change in temperature, pressure, or concentration, the position of equilibrium shifts in the direction that OPPOSES (counteracts) the change.
Position of equilibrium
Examiner keyword
The relative amounts of reactants and products present at equilibrium. 'Lying to the right' = mostly products; 'lying to the left' = mostly reactants. A change in T, P, or concentration can SHIFT the position of equilibrium (Le Chatelier).
Haber process
Examiner keyword
The industrial manufacture of ammonia: N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ/mol). Conditions: 450 °C, 200 atm, iron catalyst. Both T and P are compromises between yield and rate/cost. Ammonia is used to make fertilisers, nitric acid, and explosives.
Contact process
Examiner keyword
The industrial manufacture of sulfuric acid. The key equilibrium: 2SO₂ + O₂ ⇌ 2SO₃ (ΔH = −197 kJ/mol). Conditions: 450 °C, 1-2 atm, V₂O₅ catalyst. Equilibrium lies far to the right → > 95% yield per pass.
Catalyst (effect on equilibrium)
Examiner keyword
A catalyst lowers the activation energy of BOTH the forward and backward reactions by the SAME amount → both rates increase equally → equilibrium is reached FASTER. The position of equilibrium is UNCHANGED.
Compromise conditions (industrial)
Examiner keyword
Operating conditions (T, P) chosen to balance MAXIMUM YIELD (favoured by Le Chatelier) against MAXIMUM RATE and PRACTICAL COST/SAFETY. Neither extreme is chosen; a middle ground gives acceptable values of both.
Anhydrous ⇌ hydrated copper(II) sulfate
A visible demonstration of reversibility: $\text{CuSO}_4 \cdot 5\text{H}_2\text{O (blue)} \rightleftharpoons \text{CuSO}_4 \text{(white)} + 5\text{H}_2\text{O}$ . Heating the blue hydrated form drives off water giving the white anhydrous form (forward, endothermic). Adding water reverses the reaction (backward, exothermic). Also a test for water.

Common Mistakes and Misconceptions — Reversible Reactions and Equilibria

The traps other students keep falling into on reversible reactions and equilibria questions — taken from recent Pearson Edexcel IGCSE 4CH1 examiner reports and mark schemes — and how to avoid them.

✕Saying 'the reaction has stopped' at equilibrium
4CH1 Examiner Reports 2022-2024
Why it happens
The concentrations are constant, so it LOOKS stopped.
How to avoid it
At equilibrium, the reaction is STILL OCCURRING in both directions — just at EQUAL rates. The concentrations don't change because forward + backward perfectly cancel. Always state: 'rate of forward = rate of backward' (or 'continues in both directions at equal rates').
✕Predicting the wrong direction for temperature change
4CH1 Examiner Reports 2023-2024
Why it happens
Confusion about exo/endo direction.
How to avoid it
T INCREASE → equilibrium shifts in the ENDOTHERMIC direction (absorbs heat → opposes increase). T DECREASE → shifts in EXOTHERMIC direction (releases heat → opposes decrease). Identify whether the FORWARD reaction is exo or endo from the given ΔH; the backward is the opposite.
✕Including solid or liquid species when counting moles for pressure analysis
Why it happens
Counting everything in the equation.
How to avoid it
Pressure changes only affect GAS species (state symbol (g)). Solids and liquids have negligible volume and are unaffected by pressure. Count ONLY moles of GAS on each side.
✕Saying a catalyst increases the yield of products
4CH1 Examiner Reports 2022-2024
Why it happens
Equating 'faster' with 'more'.
How to avoid it
A catalyst does NOT change the equilibrium position → does NOT change the yield. It only changes the SPEED of approach to equilibrium. So yield is unchanged; same NH₃ at equilibrium, just reached faster.
✕Explaining only ONE side of the compromise (e.g. 'high T gives high rate' without mentioning yield)
Why it happens
Forgetting it's a TRADE-OFF question.
How to avoid it
A 'compromise' question expects BOTH sides. For T: low T good for yield BUT slow rate → 450 °C is the middle ground. For P: high P good for yield BUT expensive → 200 atm is the middle ground. State 'compromise' explicitly.
✕Saying 'doubling concentration doubles the yield' or quantitative-sounding claims
Why it happens
Over-extending qualitative Le Chatelier predictions.
How to avoid it
Le Chatelier is QUALITATIVE only at IGCSE — predict the DIRECTION of shift (forwards / backwards / no change), not the SIZE. 'Doubling [N₂] shifts equilibrium forwards, INCREASING yield of NH₃' — but you cannot say BY HOW MUCH.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Reversible Reactions and Equilibria — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Reversible Reactions and Equilibria

Short Study Notes in the form of Slides

Short Study Notes — Reversible Reactions and Equilibria

Detailed Notes

What you’ll learn

How it’s examined

1Explain what is meant by a dynamic equilibrium (5 marks, Paper 2)

2Le Chatelier — predict the effect of temperature (5 marks, Paper 2)

3Le Chatelier — predict the effect of pressure (5 marks, Paper 2)

4Why a catalyst does NOT shift equilibrium (4 marks, Paper 2)

5Conditions for the Haber process — full compromise explanation (8 marks, Paper 2)

Reversible reaction

Dynamic equilibrium

Closed system

Le Chatelier's principle

Position of equilibrium

Haber process

Contact process

Catalyst (effect on equilibrium)

Compromise conditions (industrial)

Anhydrous ⇌ hydrated copper(II) sulfate

✕Saying 'the reaction has stopped' at equilibrium

✕Predicting the wrong direction for temperature change

✕Including solid or liquid species when counting moles for pressure analysis

✕Saying a catalyst increases the yield of products

✕Explaining only ONE side of the compromise (e.g. 'high T gives high rate' without mentioning yield)

✕Saying 'doubling concentration doubles the yield' or quantitative-sounding claims

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Reversible Reactions and Equilibria — frequently asked questions