Many reactions proceed until one reactant is exhausted — these go to completion and are called irreversible. For example, magnesium reacting with hydrochloric acid goes to completion.
Other reactions are reversible — the products can re-react to reform the original reactants. The reaction can proceed in both the forward and reverse directions simultaneously.
$$\ce{CuSO4 . 5H2O(s) <=> CuSO4(s) + 5H2O(g)}$$
Heating blue hydrated copper(II) sulphate gives white anhydrous copper(II) sulphate. Adding water reverses this. We use the equilibrium sign ⇌ to denote reversible reactions.
Consider: $$\ce{H2(g) + I2(g) <=> 2HI(g)}$$
Starting with only H₂ and I₂, initially only the forward reaction occurs. As HI is produced, the reverse reaction begins. Gradually, the forward rate falls (reactants are consumed) and the reverse rate rises (products increase) until the two rates become equal. At this point, dynamic equilibrium is established.
Importantly, the same equilibrium can also be reached starting from pure HI — the system always converges to the same equilibrium concentrations at the same temperature.
Left: starting from H₂ + I₂. Right: starting from HI. Both reach the same equilibrium concentrations.
At dynamic equilibrium, both forward and reverse reactions continue at equal rates. The system appears static (no net change in composition) but molecules are constantly reacting in both directions.
A book on a desk is in static equilibrium — nothing is moving. Chemical equilibrium is fundamentally different because reactions are still occurring at the molecular level.
For the general reversible reaction: $$\ce{A + B <=> C + D}$$
Forward rate: \(r_f = k_1[\text{A}][\text{B}]\) Reverse rate: \(r_r = k_2[\text{C}][\text{D}]\)
At equilibrium, \(r_f = r_r\), so: \(k_1[\text{A}][\text{B}] = k_2[\text{C}][\text{D}]\)
$$\frac{k_1}{k_2} = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} = K_c$$Rate vs time: forward rate decreases, reverse rate increases, until both are equal at equilibrium.
Constant concentrations do NOT mean equal concentrations. At equilibrium [H₂] may be 0.060 mol dm⁻³ while [HI] is 0.157 mol dm⁻³ — they are constant, not equal.
A simple example: water in a sealed container: $$\ce{H2O(l) <=> H2O(g)}$$
Molecules with high kinetic energy escape from the liquid surface (evaporation). As vapour builds up, condensation begins. Eventually the rate of evaporation = rate of condensation — liquid–vapour equilibrium is established. The vapour pressure at this point is the vapour pressure of water at that temperature.
When a system reaches equilibrium, there is always a fixed mathematical relationship between the concentrations of reactants and products (at constant temperature). For the general reaction:
$$m\text{A} + n\text{B} \rightleftharpoons p\text{C} + q\text{D}$$[X] = concentration of X in mol dm⁻³. Exponents = stoichiometric coefficients. Products go in the numerator; reactants in the denominator.
Pure solids and pure liquids have constant concentrations and are excluded from equilibrium expressions. $$\ce{Fe2O3(s) + 3CO(g) <=> 2Fe(s) + 3CO2(g)} \quad K_c = \frac{[\ce{CO2}]^3}{[\ce{CO}]^3}$$
1. \(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\)
$$K_c = \frac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}$$2. \(\ce{2SO2(g) + O2(g) <=> 2SO3(g)}\)
$$K_c = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^2[\ce{O2}]}$$3. \(\ce{H2(g) + I2(g) <=> 2HI(g)}\)
$$K_c = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}$$Substitute mol dm⁻³ for each concentration term and cancel. For example:
If the number of moles of gas is equal on both sides (e.g. \(\ce{H2 + I2 <=> 2HI}\)), the units cancel → Kc is dimensionless.
The value of Kc depends on how the equation is written. If the equation is reversed, the new Kc = 1/(original Kc). If the equation is multiplied by factor n, the new Kc = (original Kc)n.
Mole fraction of gas A = moles of A ÷ total moles of all gases. Mole fractions are dimensionless and always sum to 1.
The partial pressure of gas A = its mole fraction × total pressure. The sum of all partial pressures equals the total pressure.
Example: 1.2 mol N₂, 0.5 mol H₂, 0.3 mol O₂, total pressure \(8.0\times10^5\) Pa. Partial pressure of H₂:
$$p_{\ce{H2}} = \frac{0.5}{2.0} \times 8.0\times10^5 = 2.0\times10^5\text{ Pa}$$For gas-phase reactions, Kp is defined using partial pressures instead of concentrations. No square brackets are used.
Units: \(\dfrac{\text{Pa}^2}{\text{Pa}\times\text{Pa}^3} = \text{Pa}^{-2}\)
Changes in concentration or pressure shift the position of equilibrium but do not change the value of Kc or Kp. Only a change in temperature changes K.
Equilibrium calculations always follow a systematic three-step approach using the ICE table method. See the Worked Examples and ICE Calculator tabs for full interactive solutions.
Q has the same form as Kc but uses current (non-equilibrium) concentrations. Comparing Q with Kc tells us the direction of reaction:
"If one or more factors that affect a system in equilibrium are changed, the position of equilibrium shifts in the direction which opposes the change."
Le Chatelier's principle is a qualitative tool. It lets us predict how an equilibrium will respond to disturbances without doing any calculation.
The position of equilibrium describes the relative amounts of reactants and products in the equilibrium mixture.
A catalyst speeds up the rate at which equilibrium is reached (increases both forward and reverse rates equally) but has no effect on the position of equilibrium or the value of Kc.
Changing concentration shifts the position but does NOT change the value of Kc or Kp.
Pressure changes only affect reactions where there is a difference in the total number of gas moles on each side.
Changing pressure shifts the position but does NOT change the value of Kc or Kp.
Temperature is the only factor that changes the value of Kc.
↑ T → equilibrium shifts left → less SO₃ → Kc decreases.
↑ T → equilibrium shifts right → more H₂ and I₂ → Kc increases.
| Change Applied | Effect on Position | Effect on Kc |
|---|---|---|
| ↑ [Reactant] | Shifts right (more products) | No change |
| ↑ [Product] | Shifts left (more reactants) | No change |
| ↑ Pressure (unequal gas moles) | Shifts toward fewer gas moles | No change |
| ↑ Temperature (exothermic rxn) | Shifts left (fewer products) | Decreases |
| ↑ Temperature (endothermic rxn) | Shifts right (more products) | Increases |
| Add catalyst | No change (reaches eq. faster) | No change |
Temperature Only Changes K
Problem: \(\ce{CH3COCH3 + HCN <=> CH3C(OH)(CN)CH3}\). At equilibrium in 500 cm³: 0.013 mol propanone, 0.013 mol HCN, 0.011 mol product. Calculate Kc.
Problem: 0.29 mol CH₃COOH + 0.25 mol C₂H₅OH mixed. At equilibrium 0.18 mol ester present. Calculate Kc.
$$\ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O}$$| Species | CH₃COOH | C₂H₅OH | CH₃COOC₂H₅ | H₂O |
|---|---|---|---|---|
| I | 0.29 | 0.25 | 0 | 0 |
| C | −0.18 | −0.18 | +0.18 | +0.18 |
| E | 0.11 | 0.07 | 0.18 | 0.18 |
Problem: 2.40×10⁻² mol H₂ and 1.38×10⁻² mol I₂ placed in a sealed tube. At equilibrium, 1.20×10⁻³ mol I₂ remains. Calculate Kc.
$$\ce{H2(g) + I2(g) <=> 2HI(g)}$$I₂ consumed = \(1.38\times10^{-2} - 1.20\times10^{-3} = 1.26\times10^{-2}\) mol
H₂ : I₂ = 1:1, so H₂ consumed = \(1.26\times10^{-2}\) mol
HI : I₂ = 2:1, so HI formed = \(2.52\times10^{-2}\) mol
Problem: At 430°C, Kc = 54.3 for \(\ce{H2(g) + I2(g) <=> 2HI(g)}\). A mixture has [H₂] = 0.050, [I₂] = 0.050, [HI] = 0.40 mol dm⁻³. In which direction will the reaction proceed?
Q = 64 > Kc = 54.3
∴ The reaction proceeds in the reverse direction (left) to decrease [HI] and increase [H₂] and [I₂] until Q = Kc.
Problem: At equilibrium: 12.5 mol SO₂, 87.5 mol O₂, 100 mol SO₃. Total pressure = 1.6×10⁷ Pa. Calculate Kp.
$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$Problem: 2 mol N₂ + 6 mol H₂ at 680°C, \(P = 2\times10^7\) Pa. At equilibrium, 3 mol NH₃ present. Calculate Kp.
$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$\(n_{\ce{N2}}=0.5,\; n_{\ce{H2}}=1.5,\; n_{\ce{NH3}}=3,\; n_{total}=5\)
Select a reaction above to see the ICE table.
| Equilibrium Reaction | Temperature | Kc | Units | Notes |
|---|
| Equilibrium Reaction | Temperature | Kp | Units | Notes |
|---|