The Effect of Temperature and Pressure Changes on Equilibrium [HT only]

SPEC 4.6.2.6 – 4.6.2.7 + 4.10.4.1 • PAPER 2 • HIGHER TIER ONLY

Temperature, Pressure & The Haber Process

Two final factors that shift equilibrium — and the textbook industrial application that ties everything together. The 6-mark Haber question wants three strands: rate, yield, and economic compromise.

What You'll Learn

HIGHER TIER ONLY

All content in this lesson appears on the Higher Tier paper only (8462/2H). Spec sub-points 4.6.2.6 (temperature), 4.6.2.7 (pressure), and the HT-only Haber points in 4.10.4.1.

By the end of this lesson

Predict the effect of temperature changes using the endothermic / exothermic direction of the reaction (spec 4.6.2.6).

Predict the effect of pressure changes using the number of molecules of gas on each side of the equation (spec 4.6.2.7).

Recall the Haber Process conditions (~450°C, ~200 atm, iron catalyst) and the equation (spec 4.10.4.1).

Explain the compromise between rate, yield, and cost — the structure required for a Level 3 / 6-mark answer.

Builds on Lessons 6.05 and 6.06

This lesson assumes you can apply Le Chatelier's principle and define dynamic equilibrium. If "shifts to oppose the change" doesn't ring a bell, revisit Lesson 6.06 first.

Temperature Effect — The Principle

Every reversible reaction has one direction that is exothermic (releases energy) and the opposite direction that is endothermic (takes in energy). Temperature changes shift the equilibrium toward whichever direction opposes the change.

What the AQA spec says (4.6.2.6)

"If the temperature of a system at equilibrium is increased: the relative amount of products at equilibrium increases for an endothermic reaction [and] decreases for an exothermic reaction." The reverse holds for a temperature decrease.

Plain-English logic (Le Chatelier)

Increase the temperature → system absorbs the extra energy → shifts in the endothermic direction (which takes energy in). Decrease the temperature → system releases energy → shifts in the exothermic direction.

The two-step process for any temperature question

Step 1: identify which direction (forward or reverse) is endothermic. Step 2: increasing T favours the endothermic direction; decreasing T favours the exothermic one.

Temperature — The Four Cases

The spec lays out all four cases explicitly. Here they are mapped to shift direction and yield change:

Forward reaction is...Temperature changeEquilibrium shifts...Effect on yield of product

Exothermic (ΔH negative)Increase T← LEFT (endothermic = reverse)Yield decreases

Exothermic (ΔH negative)Decrease TRIGHT → (exothermic = forward)Yield increases

Endothermic (ΔH positive)Increase TRIGHT → (endothermic = forward)Yield increases

Endothermic (ΔH positive)Decrease T← LEFT (exothermic = reverse)Yield decreases

Quick rule

Increase T → endothermic direction. If the forward reaction is endothermic, that means the forward direction. If the forward reaction is exothermic, the reverse direction is endothermic — so equilibrium shifts back the other way.

Worked Example — Temperature on Exothermic Reaction

Temperature-effect question · Spec 4.6.2.6 2 MARKS

Consider a generic gas-phase equilibrium where the forward reaction is exothermic:

2A(g) + B(g) ⇌ C(g) (ΔH negative — exothermic)

Explain the effect of increasing the temperature on the yield of C.

M1

Identify which direction is endothermic. The forward reaction is exothermic (ΔH negative). So the reverse reaction is endothermic.

M2

Apply the rule. Increasing T shifts equilibrium in the endothermic direction → toward the left → less C is formed → yield of C decreases.

SPEC

Spec 4.6.2.6 verbatim: "the relative amount of products at equilibrium decreases for an exothermic reaction" when T is increased.

TIP

Always state both the direction and the yield change. "Shifts left" alone is not the full mark.

Pressure Effect — The Principle

Pressure changes only matter for gas-phase equilibria. The shift direction depends on counting the number of molecules of gas on each side of the symbol equation.

What the AQA spec says (4.6.2.7)

"For gaseous reactions at equilibrium: an increase in pressure causes the equilibrium position to shift towards the side with the smaller number of molecules as shown by the symbol equation. A decrease in pressure causes the equilibrium position to shift towards the side with the larger number of molecules."

Plain-English logic (Le Chatelier)

Higher pressure means molecules are squeezed together more. The system opposes this by shifting toward the side with fewer molecules — fewer molecules occupy less space, reducing the pressure. Lower pressure → system shifts toward more molecules to increase pressure.

STATE SYMBOLS MATTER

Pressure only affects equilibria where the number of gas molecules changes between sides. Solids, liquids, and dissolved species (aq) are not counted. Always check state symbols before counting.

Counting Molecules from the Equation

The mark scheme keyword is "number of molecules of gas" — taken directly from the symbol equation. Add up the coefficients on each side, but only for species with (g).

Worked example

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Left side: 1 + 3 = 4 molecules of gas

Right side:2 molecules of gas

Increasing pressure shifts equilibrium toward the side with fewer molecules → the right → more NH₃ is formed → yield increases.

Mark-scheme phrase

Use "fewer molecules of gas" or "smaller number of molecules" — both score. Avoid "shifts to lower pressure side" — that's the misconception examiners flag, even though it leads to the same answer.

Edge Case — Equal Molecules of Gas

If both sides of the equation have the same number of gas molecules, there is no "smaller" or "larger" side to shift toward. Pressure has no effect on the equilibrium position.

EXAMINER ALERT — RECURRING ERROR

Students see a pressure change in the question and assume there must be a shift. Always count molecules first. If they're equal, the answer is "no effect on equilibrium position" — even though pressure does still affect the rate.

Worked example — equal molecules

H₂(g) + I₂(g) ⇌ 2HI(g)

Left side: 1 + 1 = 2 molecules of gas

Right side:2 molecules of gas

Pressure has no effect on the position of equilibrium because the number of gas molecules is the same on both sides.

Mark-scheme answer

"No effect on equilibrium position because there are equal numbers of gas molecules on each side of the equation." Both halves score: the conclusion and the reason.

Practice — Pressure & Equal Molecules

Pressure-effect question · Spec 4.6.2.7 2 MARKS

The following gas-phase equilibrium is set up in a sealed container:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

State and explain the effect of increasing the pressure on the equilibrium position.

M1

Effect: No effect on the equilibrium position.

M2

Reason: There are equal numbers of gas molecules on each side of the equation (2 on the left, 2 on the right).

TRAP

Don't claim a shift. Don't discuss rate either — the question asks specifically about the equilibrium position. Stating "rate increases" earns no marks here, even though it's true.

EXTEND

Pressure still affects rate (more frequent collisions per unit volume) — but rate and position are independent. This is the same principle as the catalyst trap from Lesson 6.06.

The Haber Process — Spec Context

The most-examined industrial application of equilibrium. Used to manufacture ammonia (NH₃), which is then converted into nitrogen-based fertilisers.

Raw materials & their sources (spec 4.10.4.1 recallable)

Nitrogen — from the air (about 78% N₂)

Hydrogen — from natural gas (mostly methane, CH₄) by reaction with steam

What the AQA spec says (4.10.4.1)

"The purified gases are passed over a catalyst of iron at a high temperature (about 450°C) and a high pressure (about 200 atmospheres). Some of the hydrogen and nitrogen reacts to form ammonia. The reaction is reversible..."

"On cooling, the ammonia liquefies and is removed. The remaining hydrogen and nitrogen are recycled."

The reaction

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH negative (exothermic)

4 molecules of gas (left) → 2 molecules of gas (right)

Forward reaction is exothermic

Memorise these facts

Sources: N₂ from air; H₂ from natural gas. Conditions: ~450°C · ~200 atm · iron catalyst. Recovery: ammonia liquefies on cooling and is removed; unreacted N₂ and H₂ are recycled. Each fact can be a mark on a recall question.

Haber Temperature — Why ~450°C?

The forward reaction is exothermic. By Le Chatelier, lower temperature would give a higher yield of NH₃. So why doesn't industry just use a low temperature?

Low temperature problem

Low T → high yield but the rate is too slow. Industry can't wait days or weeks for each batch — equilibrium needs to be reached within useful timescales.

High temperature problem

High T → fast rate but equilibrium shifts toward reactants (endothermic direction) → low yield of NH₃. Wasted energy and wasted reactants.

The compromise: ~450°C

High enough for an acceptable rate; low enough that the yield is still commercially viable. The iron catalyst helps further by speeding up the rate without affecting the position of equilibrium.

Mark-scheme structure

Always present temperature as a compromise between rate and yield. Both halves of the trade-off must appear in your answer to score the second mark.

Reading Haber graphs (HT spec — interpret graphs of conditions vs rate/yield)

Past papers commonly show NH₃ yield vs temperature, with multiple curves at different pressures. The shape always follows the same logic:

Two patterns to read off: yield falls as T rises (forward reaction is exothermic — heat shifts equilibrium left); yield rises as P rises (4 → 2 molecules — high P shifts equilibrium right). The 450°C operating line is a compromise — accept lower yield to get acceptable rate.

Haber Pressure — Why ~200 atm?

The forward reaction goes from 4 molecules of gas to 2. By Le Chatelier, higher pressure shifts the equilibrium toward the right, increasing the yield. So why not 1000 atm?

Higher pressure → higher yield

Increasing pressure shifts the equilibrium toward the side with fewer molecules (the products). More NH₃ is formed.

Why not "as high as possible"?

Three practical limits that mark schemes credit:

Equipment cost: very high-pressure reactors require thick, reinforced vessels — expensive to build and maintain

Energy cost: compressing gas to extreme pressures requires large amounts of electricity

Safety: high-pressure equipment is hazardous and needs careful engineering to contain safely

The compromise: ~200 atm

High enough for a usefully shifted equilibrium and high yield; low enough that the equipment is economically and safely manageable.

Mark-scheme structure

Always present pressure as a compromise between yield and cost (equipment + energy). Pure "use highest possible pressure" reasoning is wrong — that's the recurring examiner-flagged error.

The Iron Catalyst — Rate Only

Industry uses a finely-divided iron catalyst in the Haber Process. From Lesson 6.06: a catalyst speeds up the rate of both forward and reverse reactions equally — it has no effect on the equilibrium position or yield.

Why use a catalyst at all?

Without one, the rate at 450°C would still be commercially too slow. The iron catalyst lowers the activation energy of both directions, so equilibrium is reached much faster — at the same equilibrium position.

Catalyst summary for the Haber Process

✓ Rate — speeds up both forward and reverse equally; equilibrium reached faster

✗ Yield — no effect on equilibrium position; yield unchanged

✓ Economics — allows lower operating temperature, saving energy costs

Don't conflate rate and yield in Haber answers

"The catalyst increases the yield of ammonia" → wrong, drops a mark. The catalyst increases the rate at which equilibrium is reached. Equilibrium position is unchanged.

Worked Example — 6-Mark Haber Conditions

Synoptic 6-mark question · Spec 4.6.2.6, 4.6.2.7, 4.10.4.1 6 MARKS · LEVEL OF RESPONSE

The Haber Process uses these conditions: ~450°C, ~200 atm, with an iron catalyst.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH negative

Evaluate the conditions used in the Haber Process. Explain why each condition is chosen.

M1

Raw materials: N₂ from air (abundant, free) and H₂ from natural gas. The availability and cost of these raw materials matter — but ammonia is still profitable to produce because nitrogen is essentially free from the atmosphere.

M2

Temperature (~450°C): A compromise. The forward reaction is exothermic, so a low T would shift equilibrium right and give higher yield — but rate would be too slow. High T would speed the rate but shift equilibrium left (endothermic direction) → low yield. 450°C balances rate and yield.

M3

Pressure (~200 atm): The forward reaction has 4 gas molecules → 2 gas molecules. Higher pressure shifts equilibrium toward the side with fewer molecules → more NH₃ formed → higher yield. But very high pressures need expensive reinforced equipment and large energy inputs to compress the gases. 200 atm is a compromise between yield and cost.

M4

Iron catalyst: Speeds up the forward and reverse reactions equally — equilibrium is reached faster but the position is unchanged. Lets the process operate at a lower temperature than would otherwise be needed (saving energy costs) without sacrificing rate.

M5

Recycling: Unreacted N₂ and H₂ are recycled back into the reactor; ammonia is removed by cooling and liquefying. This further improves the economics of the process — no reactant is wasted.

∴

Compromise conclusion: The conditions balance the spec's four factors: availability and cost of raw materials, cost of energy supplies, control of equilibrium position (yield), and rate of reaction. No single condition is optimal alone — every choice is a trade-off.

TIP

Level 3 (5–6 marks) requires: rate considerations + yield/equilibrium reasoning + economic factors (raw materials and energy) + an explicit "compromise" conclusion. Single-aspect answers (e.g. only discussing yield) cap at Level 2 (3–4 marks).

Common Errors — At-A-Glance

Run your answer past this table before moving on. Each row is a recurring examiner-flagged error.

What students write ✗What earns marks ✓Why it matters

"Shifts to the lower-pressure side""Shifts to the side with fewer molecules of gas "Spec wording uses molecules, not pressure

"Pressure shifts equilibrium" (when moles are equal)" No effect — equal molecules of gas on both sides"Always count molecules first

"Use the highest temperature possible"Low T favours yield; high T favours rate; 450°C is a compromiseCaps at Level 1 in 6-mark question

"Catalyst increases the yield""Catalyst increases rate ; yield is unchanged"Rate ≠ position; recurring trap

Discussing rate when asked about positionRead the question — rate and position are independentWrong focus = wasted marks

Haber answer with no economic reasoningAlways include cost / safety / energy in 6-mark answersLevel 3 requires economic trade-offs

Three Things to Remember

Three habits that protect Le Chatelier marks across every temperature, pressure, and Haber question.

1. Identify the endothermic direction first

Before answering any temperature question, work out which direction (forward or reverse) is endothermic. Then increasing T favours that direction; decreasing T favours the exothermic one. Two-step process — don't skip step one.

2. Count molecules of gas — including the equal-moles trap

For pressure questions, count molecules from the symbol equation. Use spec wording: "fewer molecules of gas" or "smaller number of molecules". If they're equal on both sides → no effect on position.

3. Haber Process: rate + yield + economic factors + compromise

Every 6-mark Haber question expects all four strands. Discuss how each condition affects rate AND yield (equilibrium position), the economic factors (raw materials, energy supplies, equipment cost), and conclude with the word "compromise". Single-aspect answers cap at Level 2.

Summary

Spec 4.6.2.6 — Temperature Effects

Increase T → equilibrium shifts in the endothermic direction → relative amount of products increases for endothermic, decreases for exothermic

Decrease T → equilibrium shifts in the exothermic direction → reverse holds

Two-step process: identify endothermic direction first, then apply the rule

Spec 4.6.2.7 — Pressure Effects (Gases Only)

Increase P → shifts to side with smaller number of molecules of gas

Decrease P → shifts to side with larger number of molecules of gas

Equal molecules → no effect on equilibrium position (although rate still affected)

The Haber Process — Spec 4.10.4.1

Reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) , ΔH negative (exothermic)

Conditions: ~450°C · ~200 atm · iron catalyst

Process: ammonia liquefies on cooling and is removed; unreacted N₂ and H₂ are recycled

Why Those Conditions? — Compromises

Temperature: 450°C balances rate (high T) vs yield (low T favours exothermic forward)

Pressure: 200 atm balances yield (4 → 2 molecules favours high P) vs equipment + energy cost

Catalyst: rate only, no yield change — allows lower-T operation

Raw materials: N₂ from air (free); H₂ from natural gas (cost varies)

Coming next — Module 7

Module 6 (Rate & Extent) is now complete. Next up: Module 7 — Organic Chemistry, starting with crude oil, hydrocarbons, and alkanes (Lesson 7.01).