Chemistry 30 concepts: chemical equilibrium

What equilibrium actually is

The idea: both reactions are still running. They are just running at the same rate, so the concentrations stop changing.

Dynamic, not stopped. Molecules keep converting in both directions; the traffic is equal in each, so nothing appears to happen.
Constant, not equal. The amounts hold steady wherever the reaction settles, which is almost never a 50/50 split.
Closed system only. If a product escapes, the reverse reaction can never keep up and the system never settles.
Reached from either direction. Start with pure reactants or pure products and, at the same temperature, you land at the same equilibrium position.

How the rates get equal

At the start there is plenty of reactant and no product, so the forward reaction is fast and the reverse is nonexistent. As product builds, the reverse speeds up and the forward slows, until the two meet. Everything after that is a stalemate that never stops moving.

The equilibrium constant

The idea: K is the ratio of products to reactants at equilibrium, with each concentration raised to its coefficient. Its size tells you which side the reaction favours.

For aA + bB ⇌ cC + dD:

K = ([C]ᶜ[D]ᵈ) ÷ ([A]ᵃ[B]ᵇ)

If K is	At equilibrium the mixture is
Much greater than 1	Mostly products — the reaction goes nearly to completion
Near 1	A genuine mixture of both
Much less than 1	Mostly reactants — barely any reaction

Coefficients become exponents, never multipliers.
Pure solids and pure liquids are left out, because their concentration does not change as the reaction proceeds.
K depends only on temperature. Adding reactant, changing volume or adding a catalyst shifts concentrations around but leaves K alone.
The reverse reaction has K' = 1/K, and doubling an equation squares its K.

ICE tables

The idea: Initial, Change, Equilibrium. The change row is where the coefficients do their work, and everything else is arithmetic.

Worked: hydrogen and iodine

1.00 mol of H₂ and 1.00 mol of I₂ are sealed in a 1.00 L flask. At equilibrium, [HI] = 1.56 mol/L. Find K for H₂ + I₂ ⇌ 2HI.

	H₂	I₂	HI
Initial	1.00	1.00	0
Change	−x	−x	+2x
Equilibrium	1.00 − x	1.00 − x	2x = 1.56

2x = 1.56, so x = 0.78 and each reactant is 1.00 − 0.78 = 0.22 mol/L.

K = (1.56)² ÷ [(0.22)(0.22)] = 2.43 ÷ 0.0484 ≈ 50

The change row follows the coefficients. HI has a coefficient of 2, so it changes by 2x while the reactants change by x.
Work in mol/L. If the question gives moles, divide by the volume first — with a 1.00 L container they happen to be the same number, which hides the step.
Check your answer for sense: a K of 50 says products are favoured, and indeed most of the hydrogen and iodine reacted.

Le Chatelier's principle

The idea: disturb a system at equilibrium and it shifts in the direction that partly undoes the disturbance.

Change	Shift	Does K change?
Add a reactant	Toward products	No
Remove a product	Toward products	No
Decrease the volume (raise pressure)	Toward the side with fewer moles of gas	No
Raise the temperature, exothermic forward	Toward reactants	Yes — K decreases
Raise the temperature, endothermic forward	Toward products	Yes — K increases
Add a catalyst	No shift; equilibrium is reached sooner	No
Add an inert gas at constant volume	No shift	No

Worked: the Haber process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ

High pressure favours ammonia: four moles of gas become two.
Low temperature favours ammonia, since the forward reaction is exothermic — but low temperature also makes the reaction slow, which is why industry compromises at around 450 °C and uses a catalyst.
Removing ammonia as it forms keeps pulling the system right.

That compromise is the standard exam question: the conditions that give the best yield are not the conditions that give it fastest.

Watch out: temperature is the only change that alters K. Everything else moves concentrations until the same ratio is restored.

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