Reversible Reactions and Equilibria

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.

1. Haber process: N₂ + 3H₂ ⇌ 2NH₃. Industrial ammonia production.
2. Contact process (key step): 2SO₂ + O₂ ⇌ 2SO₃. Industrial sulfuric acid production.
3. Hydration of copper(II) sulfate: CuSO₄·5H₂O ⇌ CuSO₄ + 5H₂O. Blue hydrated ⇌ white anhydrous. Easy visible demo.
4. 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.

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.

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.

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.

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.

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).

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.

1. Burn sulfur: $\text{S} + \text{O}_2 \rightarrow \text{SO}_2$. (Sulfur from elemental sulfur deposits or as a byproduct of oil/gas desulfurisation.)

2. 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.

3. 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').

4. 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).

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:

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

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