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IGCSE · Cambridge (CIE) · 0971

Dynamic equilibrium and changing conditions

Keep equal reaction rates separate from equal concentrations, then predict shifts from the supplied equation.

Chemical reactions · 0971 (9–1) · 6.3
Exam scope and assessment notes

Extended only (Supplement) · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.

Before you begin

  • Read a reversible equation and balanced gas coefficients.
  • Distinguish an exothermic forward reaction from its endothermic reverse.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

For an exothermic forward reaction, which direction is endothermic?

What you will learn

  • Describe equilibrium in a closed system using equal forward and reverse rates and constant concentrations.
  • Predict and explain temperature, pressure, concentration and catalyst effects from supplied reaction information.
Forward rateReverse rateEqual, non-zeroBoth continueReaction rateTime →
A schematic closed-system example initially containing reactants and no products. The forward rate starts higher and the reverse rate rises as products form. Both approach the same non-zero rate and remain equal at equilibrium. The lines overlap after equilibration: both reactions continue. The vertical axis is reaction rate, not concentration; equal concentrations are not shown or implied. No numerical scale is supplied.

Dynamic means both reactions continue

In a closed system at equilibrium, forward and reverse reaction rates are equal and reactant and product concentrations no longer change. Closed here means substances do not enter or leave; energy can still be transferred when temperature is changed. The reacting particles continue to interconvert, so constant composition does not mean the reaction has stopped.

The concentrations need not be equal to each other. Equal rates can maintain, for example, a mixture with much more reactant than product or the opposite. When a condition changes, the system can have a temporary imbalance of rates and then reach a new equilibrium. Distinguish the new position, which describes the mixture, from how quickly the system gets there.

Explain each condition through the response

Increasing temperature favours the endothermic direction; decreasing it favours the exothermic direction. Increasing pressure by reducing the volume of a reacting gas mixture favours the side with fewer gas particles in the balanced equation. If both sides have the same total gaseous coefficients, this pressure change does not favour either side in the syllabus model. Count gaseous species rather than every solid or liquid in an equation.

Adding a reactant or removing a product tends to favour the direction that uses the added reactant or replaces the removed product. Use the actual species and equation to state the direction. A suitable catalyst speeds both directions and helps equilibrium be reached sooner, but it does not change the equilibrium composition at unchanged temperature. It must not be described as increasing the equilibrium yield just because it speeds a reaction.

Equilibrium: equal forward/reverse rates; constant concentrations, not necessarily equal concentrations.

Pause and explain

What does a catalyst change for a closed reversible system at fixed temperature?

Put the idea to work

Worked example

For 2A(g)+B(g)⇌2C(g), the forward reaction is exothermic. Predict the equilibrium direction after separately raising temperature, increasing pressure by compression, removing C, and adding a suitable catalyst at unchanged temperature.

Show the worked solution
  1. Higher temperature favours the endothermic reverse direction, shifting towards A and B.
  2. Higher pressure favours the two gas particles on the C side rather than three on the reactant side; removing C also favours forming more C.
  3. A catalyst changes the speed of reaching equilibrium, but not its final position at the unchanged temperature.

Answer Temperature increase: left. Compression: right. Removing C: right. Catalyst: no position change.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

For N₂(g)+3H₂(g)⇌2NH₃(g), predict the pressure-compression effect on position and explain why a catalyst cannot provide that same position shift.

Give me a hint

Count gas coefficients, then distinguish position from rate.

Compare my reasoning
  1. Reactants have four gas-particle ratio parts and products two.
  2. Compression favours the fewer-particle product side, increasing the equilibrium proportion of ammonia under the stated model.
  3. A catalyst speeds approach in both directions but leaves equilibrium position unchanged at fixed temperature.

Compression favours ammonia; a catalyst speeds equilibration without the same position shift.

Look for these in your work

  • Gas counts justify the pressure prediction.
  • Catalyst and pressure effects remain distinct.
2 · Independent

For H₂(g)+I₂(g)⇌2HI(g), predict the pressure-compression effect on equilibrium position. Separately predict the response when HI is removed.

Give me a hint

Both gas sides total two coefficient parts, but product removal is a different change.

Compare my reasoning
  1. Both sides contain equal total gas coefficients, so compression does not favour either side in this model.
  2. Removing HI makes the mixture respond by forming more HI.
  3. The removal therefore favours the forward direction, even though the pressure comparison did not shift position.

Compression: no favoured side. Removing HI: forward direction favoured.

Look for these in your work

  • Equal gaseous coefficients are checked.
  • Concentration and pressure changes are not treated as equivalent.
3 · Transfer

A forward reaction is endothermic, and products have fewer gas coefficient parts than reactants. Predict the separate effects of cooling and compression, then explain why constant concentrations after equilibration do not mean particles stopped reacting.

Give me a hint

Temperature follows heat direction; pressure follows gas counts.

Compare my reasoning
  1. Cooling favours the exothermic reverse direction.
  2. Compression favours the fewer-gas-particle product side, so the two changes favour opposite directions.
  3. At each new equilibrium, equal forward and reverse rates keep concentrations constant while both reactions continue.

Cooling favours reverse; compression favours forward; equilibrium remains dynamic.

Look for these in your work

  • Each prediction uses the correct separate evidence.
  • Constant composition is explained with continuing equal rates.

Common mistakes

  • Describing equilibrium as no reactions occurring.
  • Requiring all concentrations to be equal.
  • Saying higher temperature always gives more products without using the reaction's heat direction.
  • Saying a catalyst changes equilibrium yield.
  • Counting solid coefficients when deciding a gas-pressure shift.
Recall without your notes

What can you explain now?

A closed equilibrium has constant reactant and product concentrations. Must those concentrations be equal to each other?

Compare with the explanation

No

The forward and reverse rates are equal; the different substances' concentrations can remain constant at unequal values.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Use a new supplied reversible equation to predict all four condition changes. Explain why equal rates, constant concentrations and equilibrium position are related but different statements.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Chemistry (9-1)

Course scope and assessment

143 sequenced lessons cover all twelve syllabus areas with 429 guided, independent and application problems. 49 of the 49 syllabus sections have linked teaching and practice. This measures availability of mapped content, not full programme completion or guaranteed exam readiness. Three earlier starter lessons remain available. Extend your formula notebook as new reagents and ions appear; the starter notebook is not a universal formula reference. Paper-based method exercises support laboratory learning but do not demonstrate practical competence or complete either practical-assessment route. Written practice is self-checked. This is the separate 9–1 qualification 0971; Cambridge states that its 2026–2028 subject content is otherwise the same as 0620. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. The 2029 syllabus is separately published at https://www.cambridgeinternational.org/Images/764365-2029-syllabus.pdf and is not yet mapped here. This course supports exams in 2026, 2027 and 2028 only.

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