The big idea: At equilibrium the concentrations of reactants and products stop changing. The equilibrium constant Kc captures where that balance sits as a single number.
Write it as products over reactants, each concentration raised to the power of its stoichiometric coefficient from the balanced equation. For the general reaction
aA + bB ⇌ cC + dD
the expression is the products' terms divided by the reactants' terms.
- equilibrium constant (in terms of concentrations)
- equilibrium concentrations of the products (mol dm⁻³)
- equilibrium concentrations of the reactants (mol dm⁻³)
- the stoichiometric coefficients, used as powers
What the magnitude of Kc tells you: Kc only depends on temperature — not on starting amounts, pressure or a catalyst.
- Large K_{c} (≫ 1): the top of the fraction dominates, so at equilibrium there are mostly products — the reaction lies far to the right. - Small K_{c} (≪ 1): the bottom dominates, so mostly reactants remain — the reaction lies far to the left. - K_{c} ≈ 1: comparable amounts of reactants and products.
Leave out pure solids and liquids: Only species whose concentration can actually change appear in Kc: gases (g) and aqueous species (aq).
Pure solids (s) and pure liquids (l) — including the solvent water in dilute solution — have an effectively constant concentration, so they are omitted from the expression.
For example, for CaCO3(s) ⇌ CaO(s) + CO2(g) the expression is just Kc = [CO2] — the two solids do not appear.
Exam questions usually give you the initial amounts and one measured equilibrium value, then ask for Kc. An ICE table (Initial / Change / Equilibrium) keeps the bookkeeping straight.
How to use an ICE table: (1) Write Initial amounts (in mol, or concentrations) for every species.
(2) Work out the Change from the one value you are told — changes are in the ratio of the coefficients, with reactants decreasing (−) and products increasing (+).
(3) Add down each column to get the Equilibrium amounts.
(4) Convert each to a concentration (÷ volume) before putting it into Kc.
| H2(g) + I2(g) ⇌ 2HI(g) | H2 | I2 | HI |
|---|---|---|---|
| Initial (mol) | 1.00 | 1.00 | 0 |
| Change (mol) | −0.80 | −0.80 | +1.60 |
| Equilibrium (mol) | 0.20 | 0.20 | 1.60 |
Worked example — Kc from an ICE table
1.00 mol of H2 and 1.00 mol of I2 are placed in a sealed 2.00 dm³ flask and reach equilibrium:
H2(g) + I2(g) ⇌ 2HI(g).
At equilibrium 1.60 mol of HI is present. Calculate Kc.
Solution
- ICE table — HI rose from 0 to 1.60 mol, so the change in HI is +1.60 mol. The coefficients are 1 : 1 : 2, so H2 and I2 each fall by half of 1.60 = 0.80 mol:
- Equilibrium amounts (add down each column): H2 = 1.00 − 0.80 = 0.20 mol; I2 = 0.20 mol; HI = 1.60 mol.
- Convert to concentrations by dividing each by the volume, 2.00 dm³:
- Formula first — write the Kc expression (HI has coefficient 2, so it is squared):
- Substitute the equilibrium concentrations:
- Work it out — Kc is dimensionless here (equal moles of gas top and bottom):
Final answer
Kc = 64. Because the volume cancels here (2 mol of gas → 2 mol of gas), you would get the same answer using amounts in mol directly.
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The reaction quotient Q has the same form as Kc, but you put in the concentrations at any moment — not just at equilibrium. Comparing Q with Kc tells you which way the reaction must go to reach equilibrium.
Q vs Kc — the direction rule: A system always moves so that Q heads towards Kc.
- Q < K_{c}: not enough product yet → the reaction shifts forward (makes more product). - Q = K_{c}: already at equilibrium → no net change. - Q > K_{c}: too much product → the reaction shifts backward (makes more reactant).
| Compare Q with Kc | What it means | Which way the reaction shifts |
|---|---|---|
| Q < Kc | too few products (numerator too small) | shifts forward (→ makes more products) |
| Q = Kc | ratio already matches Kc | at equilibrium — no net change |
| Q > Kc | too many products (numerator too big) | shifts backward (← makes more reactants) |
Worked example — predicting the direction
For N2(g) + 3H2(g) ⇌ 2NH3(g), Kc = 0.50 at a certain temperature. A mixture has [N2] = 1.0, [H2] = 1.0 and [NH3] = 2.0 mol dm⁻³. In which direction will the reaction proceed?
Solution
- Formula first — write Q with the same expression as Kc (H2 is cubed, NH3 squared):
- Substitute the current concentrations:
- Evaluate Q and compare with Kc = 0.50:
- Apply the rule — Q > Kc, so there is too much product:
Final answer
Q = 4.0 > Kc = 0.50, so the reaction shifts to the left (backward), consuming NH3 until Q falls to 0.50.
Thermodynamics meets equilibrium: How far a reaction goes (its K) is decided by its standard Gibbs energy change, ΔG°. The two are linked by the data-booklet equation
ΔG° = −RT ln K.
A spontaneous reaction (ΔG° < 0) has K > 1, so it favours products; a non-spontaneous one (ΔG° > 0) has K < 1, so it favours reactants.
- standard Gibbs energy change (J mol⁻¹)
- gas constant = 8.31 J K⁻¹ mol⁻¹
- absolute temperature (K)
- equilibrium constant at that temperature
Reading the relationship qualitatively: Because the natural log of a number bigger than 1 is positive and of a number smaller than 1 is negative:
- K > 1 → ln K > 0 → ΔG° = −RT ln K is negative → reaction is spontaneous, products favoured. - K = 1 → ln K = 0 → ΔG° = 0 → neither side favoured. - K < 1 → ln K < 0 → ΔG° is positive → reaction is non-spontaneous, reactants favoured.
So the sign of ΔG° and the size of K always tell the same story.
Watch the units: With R = 8.31 J K⁻¹ mol⁻¹, this equation gives ΔG° in J mol⁻¹ — divide by 1000 to quote it in kJ mol⁻¹.
Always use T in kelvin, and take ln (natural log), not log₁₀.
Worked example — ΔG° from K
A gas-phase reaction has K = 150 at 298 K. Calculate its standard Gibbs energy change, ΔG°. (R = 8.31 J K⁻¹ mol⁻¹.)
Solution
- Formula first — the given relationship:
- Substitute R, T and K (in kelvin, with ln of K):
- Evaluate ln 150 = 5.01, then multiply:
- Work it out in J mol⁻¹, then convert to kJ mol⁻¹:
Final answer
ΔG° = −12.4 kJ mol⁻¹. It is negative because K > 1, so the reaction is spontaneous and favours products.