Dynamic equilibrium and changing conditions
Keep equal reaction rates separate from equal concentrations, then predict shifts from the supplied equation.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
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.
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.
Sources for this lesson
Pause and explain
What does a catalyst change for a closed reversible system at fixed temperature?
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
- Higher temperature favours the endothermic reverse direction, shifting towards A and B.
- 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.
- 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.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →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.
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
- Reactants have four gas-particle ratio parts and products two.
- Compression favours the fewer-particle product side, increasing the equilibrium proportion of ammonia under the stated model.
- 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.
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
- Both sides contain equal total gas coefficients, so compression does not favour either side in this model.
- Removing HI makes the mixture respond by forming more HI.
- 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.
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
- Cooling favours the exothermic reverse direction.
- Compression favours the fewer-gas-particle product side, so the two changes favour opposite directions.
- 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.
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.
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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.