3.1.6 Equilibria (Kc)

Picture a reversible reaction in a closed system. Both directions are available: the forward reaction makes products and the reverse reaction turns those products back into reactants. That is what allows a dynamic equilibrium to develop.

Dynamic equilibrium is reached when the forward and reverse rates become equal. Both reactions are still occurring; the macroscopic concentrations then remain constant.

For dynamic equilibrium, I’d learn the two essential ideas together: the forward and reverse reactions continue at equal rates in a closed system, and the macroscopic concentrations stay constant. They do not have to be equal concentrations.

If a concentration–time graph has levelled off, do not read the flat line as “reaction stopped”. At equilibrium particles are still reacting in both directions; equal rates make the macroscopic concentrations steady.

Le Chatelier’s principle predicts how an equilibrium responds to a change in concentration, pressure or temperature: the position shifts in the direction that tends to oppose the imposed change.

Use it as a prediction rule, then name the chemical consequence. If pressure rises, for example, favour the side with fewer moles of gas when the two sides differ in gaseous mole count.

For concentration, the system shifts to use up some of what you add or replace some of what you remove. For pressure, compare gaseous mole numbers. For temperature, treat heat as a reactant in the endothermic direction.

A catalyst is the exception students often try to force into Le Chatelier: it speeds both directions and reaches the same equilibrium composition sooner.

A catalyst lowers activation energy for both forward and reverse reactions. Both rates increase, so equilibrium is reached faster, but the equilibrium position and equilibrium constant at that temperature are unchanged.

If a question separates rate and yield, keep them separate in your answer: a catalyst can improve production rate without increasing the equilibrium yield.

Predict the direction that opposes the imposed change, then identify which side is favoured. A catalyst reaches the same equilibrium faster; it does not move the equilibrium position.

Always state the direction and the consequence for the named species. “Equilibrium shifts right” is incomplete if the question asks about yield, concentration or colour.

Industrial conditions are chosen using equilibrium yield, reaction rate, energy cost, equipment cost and safety together. The condition giving the highest theoretical yield may be too slow or too expensive to use.

This is why a “compromise” is a real chemical/economic choice, not a vague phrase. State what is gained and what is sacrificed when you justify the selected temperature or pressure.

Build Kc directly from the balanced equilibrium equation: equilibrium concentrations of products over reactants, each raised to its stoichiometric coefficient.

I would write the symbolic Kc expression before putting any numbers in. For aA + bB ⇌ cC + dD, the powers come straight from the balanced coefficients; once that structure is right, the substitution is much safer.

Square brackets in a Kc expression mean equilibrium concentration in mol dm⁻³. If you are given initial amounts, first use the reaction stoichiometry to find equilibrium amounts and then divide by the equilibrium volume.

Do not substitute starting concentrations into Kc unless the question explicitly tells you those are already equilibrium values.

If you change a concentration at fixed temperature, the equilibrium composition can shift, but Kc itself does not change. Adding a catalyst also leaves Kc unchanged because it changes the rates of both directions, not the thermodynamic equilibrium.

After a concentration disturbance the system settles to new equilibrium concentrations that satisfy the same Kc value at that temperature.

Use the balanced equation exactly as written when you construct Kc. Product concentrations go in the numerator, reactant concentrations in the denominator, and each concentration is raised to its equation coefficient.

If you double every coefficient in the equilibrium equation, you square the corresponding Kc expression. Do not casually rescale the equation after a numerical Kc has been defined for it.

Calculate Kc from equilibrium concentrations, not initial values. Substitute each concentration with the correct stoichiometric power and derive the units from the net concentration powers if they do not cancel.

Before pressing the calculator, write the complete expression with brackets and powers. It is much easier to see a missing square or inverted ratio there than in a line of numbers.

If one equilibrium concentration is unknown, express the concentration changes from the balanced-equation ratio first, then substitute the resulting equilibrium concentrations into Kc. That keeps the stoichiometry visible while you do the algebra.

If the algebra gives more than one mathematical root, test the answers chemically. A concentration cannot be negative or exceed what the starting amounts make possible.

When a question asks what changes the numerical value of Kc, the answer is temperature. Heating favours the endothermic direction, so Kc rises when the forward reaction is endothermic and falls when the forward reaction is exothermic.

Pressure and concentration can change equilibrium composition, and a catalyst changes how quickly equilibrium is reached, but none of those changes Kc at fixed temperature.