Loav · VCE Chemistry · Unit 3 / Area of Study 2
U3A2-03 · 2026 curriculum · Expanded original teaching guide · Study guide
1. Turn an equilibrium into a mathematical relationship
For the homogeneous equilibrium aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ). Square brackets denote concentration in mol L⁻¹; coefficients become powers, not multipliers placed before concentrations. All concentrations in Kc must be equilibrium values from the same mixture at one temperature. The expression applies to the particular equation as written. Labels A–D are placeholders; use the real balanced equation, including states, in a chemical answer.
Kc describes an equilibrium concentration ratio, not a reaction rate. A large value tends to indicate a product-favoured equilibrium for that equation, while a small value tends to indicate a reactant-favoured one. It is not a percentage yield or a guarantee that every reactant molecule is consumed. The initial mixture and stoichiometry still determine the actual amounts. Comparing the numerical Kc values of unrelated equations can be misleading, particularly where powers and units differ.
This course asks for concentration-based equilibrium expressions and their units. Work them out algebraically. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = [NH₃]² / ([N₂][H₂]³). The unit is (mol L⁻¹)²/(mol L⁻¹)⁴ = L² mol⁻². For H₂(g) + I₂(g) ⇌ 2HI(g), powers cancel and Kc has no units. More advanced thermodynamic equilibrium constants use dimensionless activities; do not let that different convention replace the concentration-unit treatment requested here.
2. Q uses the same expression at any instant
The reaction quotient Q has exactly the same algebraic form as Kc but uses current concentrations, whether or not equilibrium has been reached. Compare Q with Kc for the same equation at the same temperature. If Q < Kc, the product/reactant ratio is too small relative to equilibrium and net forward reaction occurs. If Q > Kc, net reverse reaction occurs. If Q = Kc, the composition satisfies the equilibrium condition.
This is more reliable than counting which side “has more particles” in a mixed-composition disturbance. Adding both a reactant and a product can have competing effects: calculate Q rather than guessing. If no product is initially present while denominator concentrations are positive, Q can be zero. If a denominator species is zero, do not blindly divide by zero; reason from the available species and the limiting behaviour.
3. Build an initial–change–equilibrium calculation
Begin by converting amounts to concentrations using c = n/V, with volume in litres. List initial concentrations, stoichiometric changes and equilibrium concentrations. Choose one change variable and multiply it by the equation’s coefficients. A change of x for N₂ corresponds to 3x for H₂ and 2x for NH₃. Negative concentrations are physically impossible and can expose an incorrect root or a mistaken assumed direction.
Worked example 1: find Kc from an observed equilibrium
Original synthetic data: a 1.00 L vessel at a fixed temperature initially contains 0.500 mol H₂(g), 0.500 mol I₂(g), and no HI(g). At equilibrium, [HI] = 0.600 mol L⁻¹. For H₂(g) + I₂(g) ⇌ 2HI(g), product increase is 2x = 0.600, hence x = 0.300 mol L⁻¹. Each reactant concentration is 0.500 − 0.300 = 0.200 mol L⁻¹. Kc = 0.600²/(0.200 × 0.200) = 9.00, with no units. The equilibrium amount of HI is cV = 0.600 × 1.00 = 0.600 mol. Distinguish this amount from its concentration even though the numerical values happen to match.
Worked example 2: predict a new mixture’s direction
At the same temperature, another mixture has [H₂] = 0.400, [I₂] = 0.200 and [HI] = 0.600 mol L⁻¹. Q = 0.600²/(0.400 × 0.200) = 4.50. Since 4.50 < 9.00, net forward reaction makes HI while consuming H₂ and I₂. The fact that HI already has the highest individual concentration does not imply reverse reaction. Direction depends on the complete ratio with the correct powers.
4. Kc belongs to an equation and temperature
Reversing an equation inverts Kc. Multiplying every coefficient by a factor raises Kc to that power. For the HI example, 2HI(g) ⇌ H₂(g) + I₂(g) has Kc = 1/9.00 = 0.111. For ½H₂(g) + ½I₂(g) ⇌ HI(g), Kc = √9.00 = 3.00. These are different mathematical descriptions of the same equilibrium chemistry, not three contradictory experimental results. Units must be transformed with the expression as well.
Changing temperature can change Kc for a fixed equation. Heating favours the endothermic direction: Kc increases when the forward reaction is endothermic and decreases when it is exothermic. Changes of starting concentrations or vessel volume change Q immediately, but Kc stays fixed if temperature is unchanged. The subsequent reaction restores Q = Kc. A catalyst changes the time required, not the equilibrium ratio.
As a useful check, uniform doubling of gas concentrations multiplies Q by 2 raised to the difference between total product and reactant coefficients. For the HI equation that difference is zero, so Q is unchanged. For ammonia formation it is −2, so Q becomes one quarter of its previous value; starting at equilibrium, this favours net forward reaction. This mathematical explanation and the gas-mole Le Chatelier prediction agree.
5. Work backwards from incomplete equilibrium information
Many questions give an equilibrium constant and one measured quantity, then ask for an initial amount. Do not put an unknown initial concentration directly into Kc. First use the balanced equation to express how much of that species changed. The concentration appearing in Kc is its initial concentration plus or minus that change. After finding the equilibrium concentration, reverse the stoichiometric change to recover the initial concentration, then multiply by volume if the target is an amount.
An initial–change–equilibrium record can consistently use moles throughout the stoichiometric stage, then convert every equilibrium amount to concentration. Alternatively, use concentrations throughout when volume is fixed. Both methods work. Mixing an initial amount in mol with a change in mol L⁻¹ does not. If volume changes during a disturbance, finish the instantaneous amount-to-concentration conversion at the new volume before setting up the chemical adjustment.
Worked example 3: recover an unknown initial amount
Challenge, original synthetic data. At a chosen temperature, CO(g) + 2H₂(g) ⇌ CH₃OH(g) has Kc = 2.00 L² mol⁻². A 2.00 L vessel initially contains 1.20 mol H₂, no methanol, and an unknown amount of CO. At equilibrium there is 0.200 mol CH₃OH. Find the initial CO amount.
Initial [H₂] = 1.20/2.00 = 0.600 mol L⁻¹. Methanol increases from zero to 0.200/2.00 = 0.100 mol L⁻¹, so the concentration changes are −0.100 for CO, −0.200 for H₂ and +0.100 for CH₃OH. Thus equilibrium [H₂] = 0.400 mol L⁻¹. Let equilibrium [CO] be y, rather than confusing it with its initial value.
Kc = [CH₃OH]/([CO][H₂]²), hence 2.00 = 0.100/(y × 0.400²). Rearranging yields y = 0.100/(2.00 × 0.160) = 0.3125 mol L⁻¹. Initial [CO] = 0.3125 + 0.100 = 0.4125 mol L⁻¹. Therefore initial n(CO) = 0.4125 × 2.00 = 0.825 mol. Check: consuming 0.200 mol CO and 0.400 mol H₂ makes the observed 0.200 mol methanol; the remaining concentrations reproduce Kc = 2.00.
6. Test changes quantitatively before choosing a direction
When multiple species are changed, use the ratio Qnew/Qold to simplify the reasoning. At fixed volume for the methanol equation, doubling CO multiplies the denominator by 2; doubling H₂ multiplies it by 2² = 4; doubling CH₃OH multiplies the numerator by 2. If all three happen together from an equilibrium mixture, Qnew = (2/8)Kc = Kc/4, so net forward reaction follows. “Both sides were added, so nothing happens” ignores exponents.
Q and Kc must correspond to the same temperature and equation convention. If a problem supplies Kc for the reverse equation, either invert it or write Q for that reverse equation. Do not compare one forward ratio to a reverse constant. Raising the whole equation to twice its coefficients squares the entire concentration ratio and its units. It does not double Kc. Enthalpy values, in contrast, scale linearly with equation coefficients; keep these transformation rules separate.
Some homogeneous aqueous examples contain charged species. Charge does not become the exponent: the stoichiometric coefficient does. For 2Fe³⁺(aq) + Sn²⁺(aq) ⇌ 2Fe²⁺(aq) + Sn⁴⁺(aq), Kc = [Fe²⁺]²[Sn⁴⁺]/([Fe³⁺]²[Sn²⁺]). The units cancel because the total concentration power is three on each side. The equation is charge-balanced: both sides total +8. Do not count “3+” as a coefficient of three.
A calculated equilibrium composition must meet more than the Kc expression. It must conserve atoms, use the stated volume and remain nonnegative. If one measured concentration is inaccurate, a calculation can produce an apparently neat number that violates these physical checks. Units and conservation are independent checks, not decoration at the end. When a conclusion is based on a graph, record the precision available from that graph and avoid overinterpreting the last calculator digit.
Practice
- For 2NO₂(g) ⇌ N₂O₄(g), equilibrium concentrations are 0.200 and 0.300 mol L⁻¹ respectively. Calculate Kc and units.
- For that same temperature, calculate Q and direction when [NO₂] = 0.400 and [N₂O₄] = 0.200 mol L⁻¹.
- Find Kc and units for N₂O₄(g) ⇌ 2NO₂(g) using question 1.
- A 2.00 L vessel has equilibrium [HI] = 0.600 mol L⁻¹. Find n(HI). Explain why substituting 1.20 into a concentration expression would be wrong.
- For CO(g) + 2H₂(g) ⇌ CH₃OH(g), an equilibrium mixture has [CO] = 0.250, [H₂] = 0.400 and [CH₃OH] = 0.0800 mol L⁻¹. Calculate Kc, then the Kc for 2CH₃OH(g) ⇌ 2CO(g) + 4H₂(g).
- Starting at equilibrium for the methanol equation, [CO] is doubled and [CH₃OH] tripled while [H₂], volume and temperature remain unchanged. Predict direction without knowing the original concentrations.
- For 2Fe³⁺(aq) + Sn²⁺(aq) ⇌ 2Fe²⁺(aq) + Sn⁴⁺(aq), Kc = 4.00 and an equilibrium mixture has [Fe³⁺] = 0.100, [Sn²⁺] = 0.200, [Sn⁴⁺] = 0.0500 mol L⁻¹. Find [Fe²⁺].
Worked answers
- Kc = 0.300/0.200² = 7.50 L mol⁻¹. Product concentration is first power; reactant concentration is squared.
- Q = 0.200/0.400² = 1.25 L mol⁻¹. Q < 7.50, so net forward reaction produces N₂O₄. Use concentrations from the same instant.
- Kc(reverse) = 1/7.50 = 0.133 mol L⁻¹. Both the numerical value and units invert.
- n = cV = 0.600 × 2.00 = 1.20 mol. This is an amount; the equilibrium expression requires 0.600 mol L⁻¹. Amounts cannot replace concentrations unless all volume factors are explicitly included.
- Forward Kc = 0.0800/(0.250 × 0.400²) = 2.00 L² mol⁻². The new equation is the reverse doubled, so its constant is (1/2.00)² = 0.250 mol⁴ L⁻⁴. Invert first, then square both value and units.
- Qnew/Qold = 3/2 = 1.50. Therefore Qnew = 1.50Kc and net reverse reaction is required. Both a reactant and a product were added; only the full ratio resolves their competing effects.
- 4.00 = [Fe²⁺]² × 0.0500/(0.100² × 0.200). Thus [Fe²⁺]² = 0.160 (mol L⁻¹)² and [Fe²⁺] = 0.400 mol L⁻¹. Select the positive square root because concentration cannot be negative. The exponent two comes from the coefficient, not the ion charge.
