In a sealed flask a reversible reaction settles to a point where nothing seems to change, yet molecules are still reacting in both directions. Disturb it and the system pushes back, which is how industry decides the conditions for making ammonia and sulfuric acid.
In a closed system a reversible reaction reaches dynamic equilibrium: the forward and reverse reactions continue at equal rates, so the concentrations stay constant. Le Chatelier's principle: if a condition of a system at equilibrium is changed, the position of equilibrium shifts to minimise the change. Adding a reactant shifts it right; raising the pressure shifts it towards fewer gas molecules; raising the temperature shifts it in the endothermic direction. A catalyst speeds up both reactions equally, so it does not move the position; it only gets there faster.
Start the simulator with only reactant molecules and watch the forward and reverse rates. Note the moment the two rate bars become equal: that is equilibrium, even though particles are still reacting.
Change one condition at a time in the Le Chatelier simulator, using the Haber process N2 + 3H2 ⇌ 2NH3, ΔH = −92 kJ mol−1.
| Change | Shift | Reason |
|---|---|---|
| add N2 or H2 | right, more NH3 | uses up some of the added reactant |
| remove NH3 | right | replaces the removed product |
| increase pressure | right | 4 mol of gas on the left, 2 on the right: fewer molecules lower the pressure |
| increase temperature | left, less NH3 | the reverse reaction is endothermic and absorbs the heat |
| add a catalyst | none | both rates rise equally; equilibrium reached sooner |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
At dynamic equilibrium:
Dynamic equilibrium can only be established in:
What is the effect of a catalyst on a system at equilibrium?
For N2 + 3H2 ⇌ 2NH3, increasing the pressure shifts the position of equilibrium:
Theory says low temperature and high pressure give the best yield of ammonia. Industry has to balance yield against rate and cost.
| Process | Equation | Conditions |
|---|---|---|
| Haber | N2 + 3H2 ⇌ 2NH3, ΔH = −92 | about 450 °C, 200 atm (20 MPa), iron catalyst |
| Contact | 2SO2 + O2 ⇌ 2SO3, ΔH = −197 | about 450 °C, 1 to 2 atm, V2O5 catalyst |
A catalyst does not change the position of equilibrium or the yield; it increases the rate of the forward and reverse reactions equally. And when explaining a pressure change, always count the moles of gas on each side of the equation; if they are equal, pressure has no effect on the position.
Unlocks once the four checks above are done. Worth more XP, written to AS Paper 1 and Paper 2 standard.
For 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −197 kJ mol−1. Which change increases the equilibrium yield of SO3?
For H2(g) + I2(g) ⇌ 2HI(g), what is the effect of increasing the pressure on the position of equilibrium?
Why is the Haber process run at about 450 °C rather than a lower temperature?
Cr2O72− (orange) + H2O ⇌ 2CrO42− (yellow) + 2H+. What colour change is seen when acid is added?
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