Lesson 1 of 3
Dynamic equilibrium
Understand what stays constant and how a disturbance is opposed.
A-Level 8873, revised syllabus (2026-2027)
I can explain reversible reactions and dynamic equilibrium.Syllabus
Syllabus 8873, 8(a). Explain reversible reactions and dynamic equilibrium.
I can apply Le Chatelier's Principle qualitatively.Syllabus
Syllabus 8873, 8(b). Apply Le Chatelier's Principle qualitatively.
I can identify changes that affect the equilibrium constant.Syllabus
Syllabus 8873, 8(c). Identify changes that affect the equilibrium constant.
Equilibrium is equal rates, not equal amounts
A closed system can look unchanged while molecules continue reacting.
A reversible reaction can proceed in both directions under suitable conditions. For H2(g) + I2(g) ⇌ 2HI(g), hydrogen and iodine form hydrogen iodide, while hydrogen iodide can decompose. Initially, a mixture containing only H2 and I2 has a forward reaction but no reverse reaction: no HI is present yet.
As products accumulate, the reverse reaction becomes possible and its rate increases. Meanwhile, reactants are used up and the forward rate decreases. At dynamic equilibrium, the forward and reverse rates are equal. In a closed system at constant temperature, each concentration then stays constant because each species is made and removed at the same rate. Both reactions continue.
Two different graph signatures of equilibrium
The forward and reverse rates converge to the same nonzero value. Below, reactant and product concentrations become constant at different values. These are qualitative sketches, not the same vertical quantity.
Check your understanding
A sealed mixture keeps constant concentrations of 0.80 mol dm-3 reactant and 0.15 mol dm-3 product. Do the unequal values rule out equilibrium?
Show the answer
No. Equilibrium means equal forward and reverse rates, not equal concentrations.
If both reactions continue at equal rates, the concentrations stay constant at whatever values Kc allows. These unequal values can be equilibrium values.
Predict the response to a disturbance
Track the imposed change separately from the subsequent reaction.
Make a guess. You are not marked.
For 2SO2(g) + O2(g) <=> 2SO3(g), the forward reaction is exothermic. The mixture is heated at constant pressure. What happens to the amount of SO3 at the new equilibrium?
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It decreases, as heating favours the endothermic reverse reaction
Heating favours the endothermic direction, which absorbs the added energy. Here that is the reverse reaction, so some SO3 breaks down.
Le Chatelier's Principle: when a system at equilibrium is disturbed, its equilibrium position changes in a direction that tends to oppose the disturbance. The response partly offsets the imposed change; it does not promise to restore every original concentration.
| Change | Immediate disturbance | Subsequent net reaction |
|---|---|---|
| Add H2 at fixed volume and temperature | [H2] rises immediately. | Forward reaction consumes some added H2; more NH3 forms. |
| Remove NH3 | [NH3] falls immediately. | Forward reaction replaces some removed product. |
| Compress the gas mixture at constant temperature | All gas concentrations rise; total pressure rises. | Forward reaction reduces gas amount: 4 mol gas on the left become 2 mol on the right. |
| Increase temperature | Heat is supplied. | The endothermic reverse direction is favoured; the equilibrium NH3 yield decreases. |
| Add a catalyst | Both reaction directions gain a faster pathway. | No change in equilibrium composition at the same temperature; equilibrium is reached faster. |
For a pressure question, state how pressure changes. Compression changes the reacting gas concentrations. For H2(g) + I2(g) ⇌ 2HI(g), equal gas amounts occur on each side, so compression produces no equilibrium shift in this ideal-gas treatment. Adding an inert gas at fixed volume does not change the concentrations of the reacting gases, even though total pressure rises. Do not use total pressure alone to infer a shift.
Treat heat as a bookkeeping aid for the temperature response: an exothermic forward reaction releases heat, so heating favours the reverse direction. Heat is not a species to insert into a Kc expression. A temperature increase can make the forward reaction faster while making the equilibrium proportion of product smaller; rate and final composition are different questions.
Worked exampleFollow a concentration jump
For A(aq) ⇌ B(aq), a little concentrated A solution is added to an equilibrium mixture, with negligible volume change. Describe [A] and [B] immediately and as equilibrium is re-established.
Try it first. Then show the worked stepsHide the worked steps
- [A] jumps upwards because A was added. [B] is unchanged at the instant of addition.
- The forward rate increases, so net conversion of A to B occurs. [A] falls from its immediate peak; [B] rises.
- At the new equilibrium the rates are equal again. Both concentrations can differ from their original values, while their equilibrium ratio is fixed by the same temperature.
Distinguish the instantaneous physical addition from the later chemical readjustment. Do not sketch both concentrations jumping when only A was added.
Check your understanding
For 2SO2(g) + O2(g) <=> 2SO3(g), the forward reaction is exothermic. Which two changes increase the equilibrium proportion of SO3?
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Lowering the temperature and raising the pressure
Cooling favours the exothermic forward direction. Higher pressure favours the side with fewer gas moles, 2 rather than 3. Cooling also slows the reaction, so it is not automatically the best industrial choice.