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CIE 9701 Chemistry · AS · Topic 7

Equilibria

Clear, syllabus-mapped CIE 9701 Chemistry revision notes on equilibria: explanations, worked examples and exam technique, then a free targeted practice drill.

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Contents: 7 sections

Syllabus points

Dynamic equilibrium

A reversible reaction can proceed in both directions. In a closed system it reaches dynamic equilibrium, and both words carry meaning:

Saying the reaction has stopped is wrong, and saying the concentrations are equal is wrong. Nothing appears to change because the two rates match.

Le Chatelier's principle

If a change is made to a system at equilibrium, the position of equilibrium shifts so as to oppose that change.
ChangeShift
Increase concentration of a reactantTowards the products
Increase total pressureTowards the side with fewer gas molecules
Increase temperatureIn the endothermic direction
Add a catalystNo shift at all

Two of those need care.

Pressure only matters if the number of gas molecules differs between the two sides. In the Haber process, N₂ + 3H₂ ⇌ 2NH₃, there are four gas molecules on the left and two on the right, so higher pressure favours ammonia. In H₂ + I₂ ⇌ 2HI there are two on each side, so pressure has no effect on the position at all.

Temperature is the only change that alters the value of the equilibrium constant. Everything else shifts the position while leaving K unchanged.

A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but at the same position. This is worth stating precisely, because "a catalyst increases the yield" is a common and costly error.

The industrial compromise

The Haber process is exothermic, so a low temperature would give a higher yield. In practice about 450 °C is used, because at genuinely low temperatures the rate is uneconomically slow. The chosen conditions are a compromise between yield and rate, and that phrase is what the mark scheme wants.

Equilibrium constants

For the general reaction aA + bB ⇌ cC + dD,

Kc = ([C]^c[D]^d) ÷ ([A]^a[B]^b)

Products on top, each concentration raised to its balancing number. Kp is the same expression using partial pressures.

The units are worked out from the expression, and they differ from reaction to reaction. If the powers cancel, Kc has no units.

A large K means the position lies well to the right; a small K means it lies to the left. K changes only with temperature.

Worked example

For H₂ + I₂ ⇌ 2HI at equilibrium, a 1.00 dm³ vessel contains 0.200 mol H₂, 0.200 mol I₂ and 1.60 mol HI.

Kc = (1.60^2) ÷ (0.200 × 0.200) = (2.56) ÷ (0.0400) = 64.0

The powers cancel here, one on top against two below of the same kind, so Kc has no units.

Acids and bases

A Brønsted-Lowry acid is a proton donor; a base is a proton acceptor.

When an acid donates a proton it becomes its conjugate base. The pair differ by exactly one H⁺.

For HCl + H₂O → Cl⁻ + H₃O⁺, the conjugate pairs are HCl and Cl⁻, and H₂O and H₃O⁺. Water accepts a proton here, so it is acting as a base; with ammonia it donates one and acts as an acid.

Strong and concentrated are different words. A strong acid is fully dissociated in water; a weak acid only partially dissociates. Concentration is about how much acid is dissolved. A concentrated weak acid and a dilute strong acid are entirely possible.

The pH scale

pH = -\log[H^+]

The scale is logarithmic, so each whole unit is a factor of ten in hydrogen ion concentration. A solution of pH 2 has one hundred times the hydrogen ion concentration of one at pH 4.

For a strong monobasic acid, [H⁺] equals the concentration of the acid, because dissociation is complete. 0.0100 mol dm⁻³ HCl has [H⁺] = 0.0100, and

pH = -log(0.0100) = 2.00

A weak acid of the same concentration has a higher pH, because only a fraction of its molecules have dissociated.

Reversing the calculation, [H⁺] = 10⁻ᵖᴴ.

Common mistakes

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