Chemical
Equilibria

H2 Chemistry (9476), JC1

By the end of this topic, you can answer

1

In a sealed bottle of fizzy drink, carbon dioxide keeps leaving and re-entering the drink. So why does nothing seem to change?

2

How can one number, K, tell you how far a reaction goes before it stops?

3

Mix 1 mol of ethanoic acid with 1 mol of ethanol. How much ester is there once equilibrium is reached?

4

Why does a fizzy drink bubble when you open it, and why does a cold drink stay fizzy for longer?

5

The Haber process runs at about 450 °C, though a lower temperature gives more ammonia. Why?

Recall: what you know already

1Heating blue hydrated CuSO₄ makes it white. Adding water turns it blue again. What kind of reaction is this?

A reversible reaction.[1]

2From Reaction Kinetics: does a catalyst change the amount of product made?

No. It only speeds up the reaction.[1]

3From Chemical Energetics: what does a negative ΔH tell you about the forward reaction?

It is exothermic: heat is released to the surroundings.[1]
1

Dynamic Equilibrium

  • DefinitionReversible reaction: a reaction that takes place in both the forward and backward directions (⇌)
DefinitionDynamic equilibrium: the state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction, such that there is no net change in the concentrations of reactants and products, in a closed system.

AnalogyTwo crowded shops joined by a door: people walk both ways all the time, but once as many go each way per minute, the number in each shop stops changing.

1

Dynamic Equilibrium

  • At equilibrium, [reactants] and [products] are constant, but usually not equal, and both reactions are still occurring
  • The same equilibrium is reached whether we start from the reactants or the products
  • DefinitionClosed system: no matter is lost to or gained from the surroundings. In an open system, products escape, so the reaction goes to completion.
1

Dynamic Equilibrium

Quick checkWhich is true at dynamic equilibrium?

AThe forward and backward reactions have stopped.
B[reactants] = [products]
CThe rates of the forward and backward reactions are equal.
DIt can only be reached from the reactants.
NoteC: both reactions continue at equal rates, so concentrations stay constant (but usually not equal).
2

Equilibrium Constants

For aA + bB ⇌ cC + dD:

ExpressionUnitsUse for
Kc[C]c[D]d[A]a[B]b(mol dm⁻³)(c + d) − (a + b)species in solution or gas, as concentrations
KppCc pDdpAa pBb(atm or Pa)(c + d) − (a + b)gases only, as partial pressures

Partial pressure: pA = xA × ptotal, where the mole fraction xA = nA ÷ ntotal

2

Equilibrium Constants

  • Heterogeneous equilibria: leave out pure solids, pure liquids and the solvent, as their concentrations are constant
  • CaCO₃(s) ⇌ CaO(s) + CO₂(g): Kc = [CO₂], Kp = pCO₂
NoteK is changed only by temperature. Concentration, pressure and catalysts do not change K. K says nothing about how fast equilibrium is reached.
Size of KPosition of equilibriumMixture
K ≫ 1lies to the rightmostly products
K ≪ 1lies to the leftmostly reactants
2

Equilibrium Constants

Your turnWrite Kc and Kp, with units, for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

Kc = [NH₃]²[N₂][H₂]³  mol⁻² dm⁶[1]
Kp = pNH₃²pN₂ pH₂³  atm⁻²[1]

ChallengeWrite Kc for Fe³⁺(aq) + 3H₂O(l) ⇌ Fe(OH)₃(s) + 3H⁺(aq).

Kc = [H⁺]³[Fe³⁺] mol² dm⁻⁶: leave out the solid and the solvent, water.[1]
3

Calculations with K

3.1Find Kc from equilibrium amounts

Worked exampleCH₃CO₂H(l) + C₂H₅OH(l) ⇌ CH₃CO₂C₂H₅(l) + H₂O(l). To make the ester ethyl ethanoate, 2.0 mol of ethanoic acid and 2.0 mol of ethanol are mixed. At equilibrium, 1.33 mol of water is present and the total volume is 0.500 dm³. Find Kc.

ICE table: Initial, Change, Equilibrium. Convert equilibrium moles to concentration (÷ V) before substituting into K.

CH₃CO₂HC₂H₅OHesterH₂O
I / mol2.02.000
C / mol−1.33−1.33+1.33+1.33
E / mol0.670.671.331.33

Kc = (1.33/0.500)(1.33/0.500)(0.67/0.500)(0.67/0.500) = 3.94

No units. Water is a product here, not the solvent, so it is included.

3

Calculations with K

3.2Find Kp from initial partial pressures

Worked example2SO₂(g) + O₂(g) ⇌ 2SO₃(g). In the Contact process, SO₂ (1.00 atm) and O₂ (0.50 atm) reach equilibrium in a vessel of fixed volume at 675 K. The total pressure at equilibrium is 1.01 atm. Find Kp.

At fixed V and T, partial pressure ∝ moles: let x atm of O₂ react, so SO₂ falls by 2x and SO₃ rises by 2x.

SO₂O₂SO₃
I / atm1.000.500
C / atm−2x−x+2x
E / atm0.020.010.98

(1.00 − 2x) + (0.50 − x) + 2x = 1.01 → x = 0.49

Kp = pSO₃²pSO₂² pO₂ = (0.98)²(0.02)²(0.01)

Kp = 2.40 × 10⁵ atm⁻¹

3

Calculations with K

Your turn1.00 mol of H₂ and 1.00 mol of I₂ are heated in a 1.00 dm³ flask: H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium there is 1.56 mol of HI. Find Kc.

Since 1.56 mol of HI is formed, the amounts of H₂ and I₂ each fall by 0.78 mol, leaving 0.22 mol of each.[1]
Kc = (1.56)²(0.22)(0.22) = 50.3 (no units)[1]
4

Le Chatelier's Principle

DefinitionLe Chatelier's Principle: if a change is made to a system in dynamic equilibrium, the position of equilibrium shifts so as to counteract (minimise) that change and re-establish equilibrium.
Qc < Kc: too few products → forward reaction favoured, POE shifts right
Qc = Kc: system is at equilibrium, no net change
Qc > Kc: too many products → backward reaction favoured, POE shifts left

Q has the same expression as K but uses concentrations at any moment.

4

Le Chatelier's Principle

4.1Concentration: A + B ⇌ C + D

Adding A

  1. POE shifts right to consume some of the added A
  2. Forward reaction is favoured
  3. [A] ↓ (but stays higher than before), [B] ↓, [C] and [D] ↑
  4. Kc unchanged

Removing a product, e.g. D: POE shifts right to produce more D.

4

Le Chatelier's Principle

4.2Pressure: 2NO₂(g) ⇌ N₂O₄(g)

PredictNO₂ is brown, N₂O₄ is colourless. The syringe is pushed in. What happens to the colour?

  1. Pressure ↑ → POE shifts to the side with fewer gas particles to reduce the pressure
  2. 2 mol gas → 1 mol gas: POE shifts right
  3. Brown colour becomes paler (after first darkening as all gases are compressed)
4

Le Chatelier's Principle

NoteSpecial cases
  • Equal moles of gas on both sides (2HI ⇌ H₂ + I₂): no shift
  • Inert gas at constant volume: partial pressures unchanged → no shift
  • Inert gas at constant pressure: volume ↑, partial pressures ↓ → POE shifts to more gas particles

Your turnPredict the effect of increasing pressure on each POE: (a) N₂ + 3H₂ ⇌ 2NH₃, (b) H₂ + I₂ ⇌ 2HI, (c) PCl₅ ⇌ PCl₃ + Cl₂ (all gases).

(a)
The position of equilibrium shifts to the right, as there are fewer moles of gas on the right (2 mol, against 4 mol on the left).[1]
(b)
The position of equilibrium does not shift, as there are equal numbers of moles of gas on both sides (2 mol each).[1]
(c)
The position of equilibrium shifts to the left, as there are fewer moles of gas on the left (1 mol, against 2 mol on the right).[1]
4

Le Chatelier's Principle

4.3Temperature: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ mol⁻¹

Increasing temperature

  1. POE shifts left
  2. The endothermic backward reaction is favoured to absorb the excess heat
  3. [NH₃] ↓ → Kc decreases

T ↑ favours the endothermic direction, T ↓ favours the exothermic direction. Only temperature changes K.

4.4Catalyst

  1. Rates of forward and backward reactions increase equally
  2. Ea of both directions lowered by the same amount, so kf and kb increase by the same factor
  3. Kc = kfkb → no effect on POE or K
  4. Equilibrium is only reached faster
4

Le Chatelier's Principle

Quick checkFor an endothermic forward reaction, raising the temperature

Ashifts the POE left and decreases K.
Bshifts the POE right and increases K.
Cshifts the POE right but leaves K unchanged.
Dhas no effect on the POE.
NoteB: T ↑ favours the endothermic (forward) direction to absorb the heat, so more products form and K increases.
5

The Haber Process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)   ΔH = −92 kJ mol⁻¹

5

The Haber Process

Temperature: 450 °C

  • Forward reaction exothermic: low T shifts POE right → higher yield
  • But at low T, rate is too slow
  • Moderate T: reasonable yield in a reasonable time

Pressure: 250 atm

  • 4 mol gas → 2 mol gas: high p shifts POE right → higher yield, also faster rate
  • But very high p needs costly, thick-walled plant

Catalyst: finely divided Fe

  • Shortens the time to reach equilibrium, yield unchanged
  • Cheaper than other catalysts (e.g. Ru)
  • NH₃ is cooled and condensed and removed: POE shifts right, unreacted N₂ and H₂ recycled
6

K and Gibbs Free Energy

Supplementary knowledgeΔG⦵ = −RT ln K is not required by the 9476 syllabus. It links Chemical Energetics and Chemical Equilibria.

ΔG⦵ = −RT ln K

G of the mixture falls to a minimum at equilibrium, where ΔG = 0.

ΔG⦵KPosition of equilibrium
negative> 1lies right, mostly products
zero= 1
positive< 1lies left, mostly reactants

Exit ticket

1Define dynamic equilibrium.

Rate of forward reaction equals rate of backward reaction, so there is no net change in concentrations, in a closed system.[1]

2Write Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), with units.

Kc = [SO₃]²[SO₂]²[O₂],  mol⁻¹ dm³[1]

3Why does a catalyst not change the yield?

It increases the rates of the forward and backward reactions equally, so K and the POE are unchanged.[1]

4Why is 450 °C used in the Haber process, not a lower temperature?

A lower T gives a higher yield, but the rate is too slow[1]
450 °C gives a reasonable yield in a reasonable time.[1]