3.2.3 Chemical equilibrium

At dynamic equilibrium, the forward and reverse reactions are both still happening. In a closed system their rates are equal, so you see no overall change in the amounts or concentrations of reactants and products.

The concentrations are constant, not necessarily equal. That distinction is worth guarding because “the forward and reverse reactions are the same” and “the concentrations are equal” are both tempting but wrong shortcuts.

Start by deciding what change has been imposed, then ask which direction of reaction partly opposes it. If you add a reactant, the equilibrium moves in the direction that uses some of it. Increasing pressure favours the side with fewer moles of gas; if the gas mole numbers are the same, pressure does not shift the equilibrium.

For temperature, treat heat as part of the reaction: raising the temperature favours the endothermic direction. Name the direction or the products/reactants explicitly; “shifts to the side with fewer moles” is not quite finished.

If you add a catalyst to an equilibrium mixture, it lowers the activation energy for both the forward and reverse reactions. Both rates increase, so the system reaches equilibrium faster.

You do not change the equilibrium position because the catalyst does not favour one direction over the other. At a fixed temperature, Kc is unchanged as well.

Choose an equilibrium where a change in composition gives a visible colour change, then alter one condition while keeping the others controlled. For concentration, add or remove one species and compare the new equilibrium colour. For temperature, warm or cool the same equilibrium mixture and let it re-establish equilibrium.

Your conclusion should connect the observed colour to the relative amounts of the coloured species. If you need to preserve a sample for analysis, think carefully about whether the equilibrium could keep shifting while you measure it.

Industrial conditions are usually a compromise. A temperature that gives the best equilibrium yield may give an unacceptably slow rate; a very high pressure may improve yield for some gas equilibria but cost more to build and run safely.

Separate the arguments. First say what the condition does to equilibrium position, then what it does to rate, and then bring in energy cost, equipment, safety or catalyst performance. That makes your final choice much easier to justify.

When you write Kc for a homogeneous equilibrium, put product concentrations over reactant concentrations and raise each concentration to the power of its stoichiometric coefficient. Use square brackets for equilibrium concentrations.

For aA + bB ⇌ cC + dD, the pattern is [C]^c[D]^d / ([A]^a[B]^b). I write the symbolic expression before putting any numbers into it; that catches inverted expressions and missing powers early.

Once the expression is correct, substitute equilibrium concentrations, not starting concentrations. If the question gives equilibrium moles and a volume, convert each required amount into mol dm⁻³ first.

Then calculate Kc and work out the units from the concentration powers that remain after cancellation. Keep the expression visible above the substitution so you can see exactly which value belongs to which species.

You can use the size of Kc as a quick guide to which side is favoured at equilibrium. A very large Kc means the equilibrium mixture is dominated by products; a very small Kc means reactants dominate.

If Kc is around 1, neither side overwhelmingly dominates. This is a qualitative judgement, so do not try to turn one Kc value into an exact percentage composition without more information.

• Kc ≫ 1: products favoured
• Kc ≪ 1: reactants favoured
• Kc around 1: appreciable amounts of both sides