Coaching Centre
The equilibrium constant
Writing equilibrium expressions for homogeneous and heterogeneous systems, what the size of K tells you, the reaction quotient Q and predicting the direction of change, manipulating K when equations are reversed or multiplied, and how temperature changes K.
Mix hydrogen and iodine in a sealed flask at 450 °C, or start with pure hydrogen iodide, or use any random mixture of all three. Wait, measure the equilibrium concentrations, and plug them into one particular fraction. Every time, the answer is about 50. That number, the equilibrium constant, is the single most useful fact about an equilibrium.
The equilibrium expression
For a general reaction at equilibrium
$$a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$$
the equilibrium constant is
$$K_{eq} = \frac{[\text{C}]^c\,[\text{D}]^d}{[\text{A}]^a\,[\text{B}]^b}$$
Square brackets mean concentration in mol L⁻¹ at equilibrium. Products go on top, reactants underneath, and each concentration is raised to the power of its coefficient in the balanced equation. This relationship is called the equilibrium law. It was discovered by experiment, by measuring many equilibrium mixtures, long before anyone could explain it.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g): $\;K_{eq} = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq): $\;K_{eq} = \dfrac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]}$
2SO₂(g) + O₂(g) ⇌ 2SO₃(g): $\;K_{eq} = \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$
Homogeneous and heterogeneous equilibria
In a homogeneous equilibrium all species are in the same phase, such as all gases or all aqueous. In a heterogeneous equilibrium there is more than one phase, such as a solid in contact with a gas or a solution.
Pure solids and pure liquids are left out of the expression. Their "concentration" is just their density divided by molar mass, which is fixed no matter how much of the substance is present. Adding more marble chips to a sealed container doesn't change how many moles of CaCO₃ are packed into each litre of marble. Those constant values are absorbed into K.
| Equilibrium | Expression |
|---|---|
| CaCO₃(s) ⇌ CaO(s) + CO₂(g) | $K_{eq} = [\text{CO}_2]$ |
| H₂O(l) ⇌ H₂O(g) | $K_{eq} = [\text{H}_2\text{O(g)}]$ |
| CO₂(g) ⇌ CO₂(aq) | $K_{eq} = \dfrac{[\text{CO}_2\text{(aq)}]}{[\text{CO}_2\text{(g)}]}$ |
| AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) | $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$ |
| CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) | $K_a = \dfrac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]}$ |
The last two rows show that many named constants are just K for a particular kind of reaction. Ksp, the solubility product, is K for a solid dissolving into its ions (lesson 5I). Ka and Kb are K for an acid or base reacting with water, and Kw = [H₃O⁺][OH⁻] is K for water ionising itself. Water is left out of Ka because it is the solvent, present in huge, effectively constant amount. You'll use these in Module 6.
What the size of K tells you
| Value of K | Position of equilibrium | Example |
|---|---|---|
| K > 10⁴ (roughly) | Lies far to the right; mostly products; reaction is essentially complete | 2H₂ + O₂ ⇌ 2H₂O, K ≈ 10⁸⁰ at 25 °C |
| 10⁻⁴ < K < 10⁴ | Significant amounts of both reactants and products | H₂ + I₂ ⇌ 2HI, K = 50 at 450 °C |
| K < 10⁻⁴ (roughly) | Lies far to the left; mostly reactants; very little reaction | N₂ + O₂ ⇌ 2NO, K ≈ 10⁻³⁰ at 25 °C |
The cut-offs are rough guides, not sharp rules. K says nothing about rate: hydrogen and oxygen can sit together in a balloon for years despite an enormous K, because the activation energy is high.
A note on units
Because the powers top and bottom are often different, K can have units, such as mol⁻² L² for the Haber process. Strictly, K is defined with each concentration divided by a standard 1 mol L⁻¹, which makes it a pure number, and NESA data and marking guidelines usually give K without units. Check what your exam question does, and don't let units distract you.
The reaction quotient, Q
The reaction quotient Q has exactly the same form as K, but uses the concentrations at any moment, not only at equilibrium. Comparing Q with K tells you which way a mixture will react:
- Q < K: there are too few products compared with equilibrium. Net forward reaction until Q = K.
- Q = K: the system is at equilibrium. No net change.
- Q > K: there are too many products. Net reverse reaction until Q = K.
Q is what Le Chatelier's principle is really measuring. Adding a reactant makes the denominator of Q bigger, so Q drops below K and the reaction moves forward. Halving the volume of N₂O₄ ⇌ 2NO₂ doubles both concentrations, so $Q = \frac{(2[\text{NO}_2])^2}{2[\text{N}_2\text{O}_4]} = 2K$, and the reaction moves backwards.
At 25 °C, K = 4.6 × 10⁻³ for N₂O₄(g) ⇌ 2NO₂(g). A flask contains [N₂O₄] = 0.040 mol L⁻¹ and [NO₂] = 0.020 mol L⁻¹. Is it at equilibrium? If not, which way will it react?
$Q = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \dfrac{(0.020)^2}{0.040} = 1.0\times10^{-2}$
Q > K, so the mixture is not at equilibrium. There is a net reverse reaction: NO₂ combines to form N₂O₄ and the brown colour fades until Q falls to 4.6 × 10⁻³.
- For H₂ + I₂ ⇌ 2HI with 1.0 mol L⁻¹ of each reactant, press Let it reach equilibrium. What is [HI] at equilibrium? Check that the final Q equals 50.
- Reset, set [H₂] and [I₂] to 0 and [HI] to 1.0. Where is the Q marker now, and which way does the reaction go?
- Find starting concentrations for which nothing happens when you press the button.
- Choose N₂O₄ ⇌ 2NO₂ starting with 0.10 mol L⁻¹ N₂O₄. What fraction of the N₂O₄ dissociates? Then try 1.0 mol L⁻¹. Does a larger fraction or a smaller fraction dissociate?
Changing the equation changes K
K belongs to a particular equation, written a particular way. For 2SO₂ + O₂ ⇌ 2SO₃, K = 280 at 1000 K.
| Change to the equation | Effect on K | Example |
|---|---|---|
| Reverse it | $K' = \dfrac{1}{K}$ | 2SO₃ ⇌ 2SO₂ + O₂: K′ = 1/280 = 3.6 × 10⁻³ |
| Multiply all coefficients by n | $K' = K^n$ | SO₂ + ½O₂ ⇌ SO₃ (n = ½): K′ = √280 = 16.7 |
| Add two equations | $K' = K_1 \times K_2$ | — |
So whenever you quote a K value, the equation it refers to must be stated too.
Temperature is the only thing that changes K
Changing a concentration or the pressure moves the system away from equilibrium by changing Q; the system then reacts until Q equals the same K. A catalyst changes neither. Only temperature changes the value of K itself, because it changes the forward and reverse rates by different amounts (lesson 5D).
| Forward reaction | Temperature increased | Temperature decreased |
|---|---|---|
| Exothermic (ΔH < 0) | K decreases; equilibrium shifts left | K increases; equilibrium shifts right |
| Endothermic (ΔH > 0) | K increases; equilibrium shifts right | K decreases; equilibrium shifts left |
For example, K for the Haber process is about 6 × 10⁵ at 25 °C but only about 2 × 10⁻⁵ at 500 °C, so the yield of ammonia falls sharply when the reactor is heated. The plant runs hot anyway for the sake of rate.
Test yourself
What is the equilibrium expression for 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)?
Water is a gas here, so it is included.
What is the expression for C(s) + H₂O(g) ⇌ CO(g) + H₂(g)?
Solid carbon is omitted.
For a reaction, K = 2.5 × 10⁻³. A mixture has Q = 0.15. What happens?
Q > K, so there are too many products.
For H₂(g) + I₂(g) ⇌ 2HI(g), K = 50 at 450 °C. What is K for HI(g) ⇌ ½H₂(g) + ½I₂(g) at the same temperature?
Reverse (1/50 = 0.020) and halve (√0.020 = 0.14).
For an exothermic reaction, which change will increase the value of K?
Only temperature changes K. Cooling favours the exothermic forward reaction.
Sort each change by its effect on K.
The equilibrium constant is calculated from concentrations. Pure and liquids are left out of the expression. When concentrations at any moment are used instead, the result is called the reaction . If this is less than K, the reaction proceeds in the direction. The only factor that changes K is .
For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), K = 0.042 at 250 °C. A 2.0 L vessel contains 0.20 mol PCl₅, 0.10 mol PCl₃ and 0.10 mol Cl₂ at 250 °C. Determine whether the system is at equilibrium and predict any change in the colour of the mixture (Cl₂ is pale green; the other gases are colourless).
Concentrations: [PCl₅] = 0.20/2.0 = 0.10 mol L⁻¹; [PCl₃] = [Cl₂] = 0.10/2.0 = 0.050 mol L⁻¹.
$Q = \dfrac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \dfrac{0.050\times0.050}{0.10} = 0.025$
Q (0.025) is less than K (0.042), so the system is not at equilibrium. There will be a net forward reaction, converting PCl₅ into PCl₃ and Cl₂ until Q rises to 0.042. Because [Cl₂] increases, the pale green colour of the mixture will become more intense.