How far?

In a nutshell

At AS you wrote K_c expressions and used le Chatelier's principle to predict shifts. This subtopic makes equilibrium quantitative.

You will work out how much of each substance is present at equilibrium, write K_c and K_p for homogeneous and heterogeneous equilibria, find their units, and explain why only temperature changes the value of an equilibrium constant.

Assumed knowledge: Chemical equilibrium: dynamic equilibrium, le Chatelier's principle and writing K_c; Amount of substance: moles, concentration and reacting ratios.

Core content

Equilibrium: mole fraction and partial pressure

K_p is written in terms of the pressures of the individual gases in a mixture. To get those pressures you need two terms.

The mole fraction of a gas is its share of the total moles of gas:

$x_A = \frac{\text{moles of A}}{\text{total moles of gas in the mixture}}$

The partial pressure of a gas is its share of the total pressure: the pressure it would exert if it alone occupied the whole volume of the mixture.

$p(A) = x_A \times P$

where P is the total pressure.

Two checks that catch most slips:

• The mole fractions of all the gases add up to 1.
• The partial pressures of all the gases add up to the total pressure.

Use whatever pressure unit the question gives (Pa, kPa, MPa or atm) and keep it the same all the way through.

At a total pressure of 250 kPa, each gas's slice of the pressure is the same as its slice of the moles: 50 kPa, 25 kPa and 175 kPa. Worked example 1 does this calculation in full.

Equilibrium: calculating quantities present at equilibrium

Questions rarely give you equilibrium amounts directly. Usually you get the initial amounts and one piece of equilibrium data (an amount of one substance, or a percentage that reacted). You work out the rest from the balanced equation.

Set the working out as a table with three rows:

Then convert to what the expression needs:

• For K_c: equilibrium concentration = equilibrium moles ÷ volume in dm³.
• For K_p: mole fraction = equilibrium moles ÷ total equilibrium moles of gas, then partial pressure = mole fraction × total pressure.

Still don't get it? · working out the moles at equilibrium

Think of a sandwich shop. One sandwich needs 2 slices of bread and 1 slice of cheese. You start with 10 slices of bread and 8 slices of cheese. At closing time, 3 sandwiches have been made.

You do not need to count what is left. You can work it out:

1. 3 sandwiches used 3 × 2 = 6 slices of bread and 3 × 1 = 3 slices of cheese.
2. Bread left = 10 - 6 = 4. Cheese left = 8 - 3 = 5.
3. Sandwiches = 0 + 3 = 3.

The recipe is the balanced equation. The "3 sandwiches made" is the one piece of equilibrium data the question gives you.

Now the chemistry. For N₂ + 3H₂ ⇌ 2NH₃, if 0.20 mol of N₂ reacts:

1. The equation says 1 N₂ reacts with 3 H₂, so 3 × 0.20 = 0.60 mol of H₂ is used.
2. The equation says 1 N₂ makes 2 NH₃, so 2 × 0.20 = 0.40 mol of NH₃ forms.
3. Subtract the used amounts from the reactants. Add the formed amount to the product.

Exam version: equilibrium moles of a reactant = initial moles - moles reacted; equilibrium moles of a product = initial moles + moles formed, with the moles reacted and formed always in the ratio of the balancing numbers. Only equilibrium amounts go into K_c or K_p.

Practical: determining quantities present at equilibrium

OCR expects you to know a technique for finding the amounts in an equilibrium mixture, so that K_c can be calculated. (This is for K_c only: OCR does not ask for a practical method for K_p.)

The standard example is the esterification equilibrium, where every species is a liquid:

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

ethanoic acid + ethanol ⇌ ethyl ethanoate + water

The idea: the only acid that can be titrated is the unreacted ethanoic acid (plus the acid catalyst). So a titration with standard sodium hydroxide tells you how much ethanoic acid is left at equilibrium.

Method

1. Measure known volumes of ethanoic acid and ethanol from burettes into a boiling tube. Add a known volume of dilute hydrochloric acid as the catalyst.
2. Work out the initial moles of each liquid: mass = density × volume, then moles = mass ÷ M.
3. Stopper the tube (a closed system, and it stops the volatile ester and ethanol escaping). Leave it for about a week at a constant temperature, so it reaches equilibrium.
4. Separately, titrate the same volume of the HCl catalyst on its own with the sodium hydroxide. This tells you how much of the titre is due to the catalyst.
5. Transfer the equilibrium mixture, with washings, into a 250.0 cm³ volumetric flask and make up to the mark with distilled water. Invert to mix.
6. Pipette 25.0 cm³ portions into a conical flask, add phenolphthalein and titrate with standard NaOH(aq) until the first permanent pale pink. Repeat until you have concordant titres.

Why each step matters

• Constant temperature: the value of K_c depends on temperature, so the temperature must not change.
• Leave for a week: the reaction is slow. To check equilibrium has been reached, set up identical mixtures, leave them for different times, and confirm they give the same value of K_c.
• Subtract the catalyst: the NaOH neutralises the HCl catalyst as well as the ethanoic acid. Only the rest of the titre is ethanoic acid.
• Water in the catalyst: dilute HCl is mostly water, so the initial amount of water is not zero. Include it in the "initial" row.
• Titrate promptly: excess hydroxide ions slowly react with the ester, so the pink end point can fade. Take the first permanent pale pink.

Worked example 6 turns a set of results from this practical into a value of K_c.

Equilibrium: K_c and K_p for homogeneous equilibria

A homogeneous equilibrium has all species in the same phase (all gases, or all in solution, or all liquids).

For a homogeneous equilibrium aA + bB ⇌ cC + dD:

$K_c = \frac{[\mathrm{C}]^c\,[\mathrm{D}]^d}{[\mathrm{A}]^a\,[\mathrm{B}]^b}$

If the species are gases, you can also write K_p, using partial pressures instead of concentrations:

$K_p = \frac{p(\mathrm{C})^c\,p(\mathrm{D})^d}{p(\mathrm{A})^a\,p(\mathrm{B})^b}$

• Products on top, reactants on the bottom.
• Balancing numbers become powers.
• Square brackets [ ] mean equilibrium concentration. In K_p use p( ) for partial pressure, never square brackets.
• K_p contains gases only.

In the esterification equilibrium from the practical, all four species are liquids in one phase. That makes it homogeneous, so water is included in K_c:

$K_c = \frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}$

Equilibrium: K_c and K_p for heterogeneous equilibria

A heterogeneous equilibrium has species in different phases (for example a solid with gases).

The rule: the concentration of a solid or a liquid is constant, so it is left out of heterogeneous K_c and K_p expressions.

Read the state symbols before you write anything. In the second row, H₂O is a gas, so it stays in.

Equilibrium: units of K_c and K_p

At A level you must give units. Their units depend on the expression, so work them out every time.

Method: substitute the unit for every term, then cancel.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g):

$K_c \text{ units} = \frac{(\mathrm{mol\ dm^{-3}})^2}{(\mathrm{mol\ dm^{-3}})(\mathrm{mol\ dm^{-3}})^3} = \frac{1}{(\mathrm{mol\ dm^{-3}})^2} = \mathrm{dm^{6}\ mol^{-2}}$

$K_p \text{ units} = \frac{\mathrm{kPa}^2}{\mathrm{kPa} \times \mathrm{kPa}^3} = \mathrm{kPa^{-2}}$

A quick check: count the total power on top minus the total power on the bottom. That number is the power of the unit.

If the equation is written the other way round, K becomes 1/K and the units are inverted. Always tie a K value to its equation.

Equilibrium: calculating K_c, K_p and related quantities

Every K calculation follows the same order:

1. Write the expression from the balanced equation (leave out solids and liquids if heterogeneous).
2. Find the equilibrium concentrations or partial pressures (use the initial / change / equilibrium table if needed).
3. Substitute, apply the powers, and calculate.
4. Work out the units and give the answer to the right number of significant figures.

A "related quantity" question gives you K and asks for one missing equilibrium concentration or partial pressure. Rearrange the expression for the unknown, then substitute. OCR will not ask you to solve a quadratic equation.

Worked examples 1 to 6 cover each calculation type.

Equilibrium: how temperature changes K

Temperature is the only factor that changes the value of K_c or K_p. Whether K goes up or down depends on the sign of ΔH for the forward reaction:

Two examples:

• N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δ_rH^⦵ = -92 kJ mol^-1. Exothermic, so raising the temperature decreases K_p and less ammonia is present at equilibrium.
• N₂O₄(g) ⇌ 2NO₂(g), Δ_rH^⦵ = +57 kJ mol^-1. Endothermic, so raising the temperature increases K_p and more NO₂ is present.

This matches le Chatelier's principle: heating favours the endothermic direction. The difference at A level is that you can now say why: the equilibrium constant itself has changed.

Equilibrium: what does not change K

The value of K is constant (at a fixed temperature) when you change:

• Concentration of a reactant or product.
• Pressure.
• Adding a catalyst.

Concentration and pressure can shift the position of equilibrium, but K keeps the same value.

A catalyst changes neither K nor the position. It increases the rates of the forward and reverse reactions by the same amount, so equilibrium is simply reached faster.

Equilibrium: how K controls the position of equilibrium

This is the A level explanation behind le Chatelier's principle. The equilibrium constant is fixed, so the concentrations (or partial pressures) have to change until their ratio equals K again.

Changing concentration (K_c unchanged)

1. Adding a reactant makes the bottom of the K_c expression bigger.
2. The ratio of concentrations is now smaller than K_c: the system is no longer at equilibrium.
3. Reactants turn into products, increasing the top and decreasing the bottom, until the ratio equals K_c again.
4. So the position shifts right.

Here is that happening for a simple equilibrium A ⇌ B with K_c = 2.0, when extra A is added at 40 s:

The ratio drops when A is added, then climbs back to exactly 2.0. The new mixture contains more B than before, but K_c has not changed.

Changing pressure (K_p unchanged)

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), write each partial pressure as x × P:

$K_p = \frac{(x_{\mathrm{SO_3}}P)^2}{(x_{\mathrm{SO_2}}P)^2\,(x_{\mathrm{O_2}}P)} = \frac{x_{\mathrm{SO_3}}^2}{x_{\mathrm{SO_2}}^2\,x_{\mathrm{O_2}}} \times \frac{1}{P}$

1. There is one more P on the bottom than on the top (3 moles of gas on the left, 2 on the right).
2. Increasing P makes the 1/P term smaller.
3. K_p stays the same, so the mole fraction part must get bigger: x(SO₃) increases.
4. So the position shifts right, to the side with fewer moles of gas.

Changing temperature (K does change)

1. The temperature change changes the value of K.
2. The existing ratio of concentrations no longer equals the new K.
3. The position shifts until the ratio equals the new K.
4. For an exothermic forward reaction, a higher temperature gives a smaller K, so the position shifts left.

Still don't get it? · how K controls the position of equilibrium

Think of a recipe for squash: always 1 part cordial to 4 parts water. That ratio is fixed, like K.

Step by step:

1. You have a jug mixed perfectly at 1 : 4.
2. Someone pours in extra cordial. The ratio is now wrong: too much cordial.
3. To get back to 1 : 4, you have to add water until the ratio is right again.
4. At the end the jug holds more of everything, but the ratio is back to exactly 1 : 4. The recipe never changed.

In an equilibrium, the "recipe" is K, and the system itself "adds water" by turning reactants into products (or the reverse) until the ratio fits.

Changing the temperature is different. That is like someone rewriting the recipe to 1 : 5. Now the jug is wrong even though nobody poured anything in, and the mixture has to change to match the new recipe.

Exam version:

• Concentration or pressure: K does not change. The change makes the concentration (or partial pressure) ratio no longer equal to K, so the position of equilibrium shifts until the ratio equals K again.
• Temperature: the value of K changes (decreases for an exothermic forward reaction when temperature rises), and the position shifts to give the new value of K.

Equilibrium: other equilibrium constants

Everything above applies to any equilibrium constant, not just K_c and K_p. You will meet two more in Acids, bases and buffers:

• K_a, the acid dissociation constant: HA(aq) ⇌ H⁺(aq) + A^-(aq), $K_a = \dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$, units mol dm^-3.
• K_w, the ionic product of water: $K_w = [\mathrm{H^+}][\mathrm{OH^-}]$, units mol² dm^-6. Its value at 298 K is given on the data sheet.

The same rules carry over:

• Write the expression from the equation, products over reactants.
• Work out the units by cancelling.
• Only temperature changes the value. For example, the ionisation of water is endothermic, so K_w increases as temperature rises.

Worked examples

1. Mole fractions and partial pressures

An equilibrium mixture contains 0.500 mol SO₂, 0.250 mol O₂ and 1.75 mol SO₃. The total pressure is 250 kPa. Calculate the partial pressure of each gas.

Step 1: total moles of gas.

$n_{\text{total}} = 0.500 + 0.250 + 1.75 = 2.50\ \mathrm{mol}$

Step 2: mole fractions.

$x_{\mathrm{SO_2}} = \frac{0.500\ \mathrm{mol}}{2.50\ \mathrm{mol}} = 0.200$

$x_{\mathrm{O_2}} = \frac{0.250\ \mathrm{mol}}{2.50\ \mathrm{mol}} = 0.100$

$x_{\mathrm{SO_3}} = \frac{1.75\ \mathrm{mol}}{2.50\ \mathrm{mol}} = 0.700$

Check: 0.200 + 0.100 + 0.700 = 1.000.

Step 3: partial pressures.

$p(\mathrm{SO_2}) = 0.200 \times 250\ \mathrm{kPa} = 50.0\ \mathrm{kPa}$

$p(\mathrm{O_2}) = 0.100 \times 250\ \mathrm{kPa} = 25.0\ \mathrm{kPa}$

$p(\mathrm{SO_3}) = 0.700 \times 250\ \mathrm{kPa} = 175\ \mathrm{kPa}$

Check: 50.0 + 25.0 + 175 = 250 kPa, the total pressure.

Step 4 (follow-on): K_p for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

$K_p = \frac{p(\mathrm{SO_3})^2}{p(\mathrm{SO_2})^2\,p(\mathrm{O_2})} = \frac{(175\ \mathrm{kPa})^2}{(50.0\ \mathrm{kPa})^2 \times 25.0\ \mathrm{kPa}}$

$K_p = \frac{30625\ \mathrm{kPa^2}}{62500\ \mathrm{kPa^3}} = 0.490\ \mathrm{kPa^{-1}}$

Where marks go:

• Dividing by the moles of one gas, or by the moles of reactants only, instead of the total moles of gas.
• Forgetting to square p(SO₂) gives 24.5: wrong.
• Leaving off the unit kPa^-1 loses the units mark.

2. K_c from initial amounts (including a volume conversion)

0.400 mol of PCl₅ is heated in a sealed 500 cm³ container:

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

At equilibrium, 0.150 mol of Cl₂ is present. Calculate K_c, with units.

Step 1: initial / change / equilibrium table (moles). The ratio PCl₅ : PCl₃ : Cl₂ is 1 : 1 : 1.

Step 2: convert the volume and find concentrations.

$V = \frac{500\ \mathrm{cm^3}}{1000} = 0.500\ \mathrm{dm^3}$

$[\mathrm{PCl_5}] = \frac{0.250\ \mathrm{mol}}{0.500\ \mathrm{dm^3}} = 0.500\ \mathrm{mol\ dm^{-3}}$

$[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = \frac{0.150\ \mathrm{mol}}{0.500\ \mathrm{dm^3}} = 0.300\ \mathrm{mol\ dm^{-3}}$

Step 3: substitute.

$K_c = \frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} = \frac{(0.300\ \mathrm{mol\ dm^{-3}})(0.300\ \mathrm{mol\ dm^{-3}})}{0.500\ \mathrm{mol\ dm^{-3}}}$

$K_c = 0.180\ \mathrm{mol\ dm^{-3}}$

Units: (mol dm^-3 × mol dm^-3) ÷ mol dm^-3 = mol dm^-3. Answer to 3 s.f., matching the data.

Where marks go:

• Putting the initial 0.400 mol into the expression instead of the equilibrium 0.250 mol.
• Using moles instead of concentrations gives 0.0900: wrong. You can only skip the volume when the powers on top and bottom are equal (K has no units). Here they are not.
• Dividing by 500 instead of 0.500 (forgetting cm³ to dm³) gives 1.80 × 10^-4: wrong.

3. K_p from initial amounts and a percentage reacted

1.00 mol of N₂ and 3.00 mol of H₂ are mixed and allowed to reach equilibrium:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

At equilibrium, 20.0% of the N₂ has reacted and the total pressure is 2.00 × 10⁴ kPa. Calculate K_p, with units.

Step 1: moles of N₂ that reacted.

$0.200 \times 1.00\ \mathrm{mol} = 0.200\ \mathrm{mol}$

Step 2: initial / change / equilibrium table. Ratio N₂ : H₂ : NH₃ = 1 : 3 : 2.

Step 3: total equilibrium moles and mole fractions.

$n_{\text{total}} = 0.800 + 2.40 + 0.400 = 3.60\ \mathrm{mol}$

$x_{\mathrm{N_2}} = \frac{0.800}{3.60} = 0.2222 \qquad x_{\mathrm{H_2}} = \frac{2.40}{3.60} = 0.6667 \qquad x_{\mathrm{NH_3}} = \frac{0.400}{3.60} = 0.1111$

Step 4: partial pressures. Keep the unrounded values in your calculator.

$p(\mathrm{N_2}) = 0.2222 \times 2.00 \times 10^{4}\ \mathrm{kPa} = 4444\ \mathrm{kPa}$

$p(\mathrm{H_2}) = 0.6667 \times 2.00 \times 10^{4}\ \mathrm{kPa} = 13333\ \mathrm{kPa}$

$p(\mathrm{NH_3}) = 0.1111 \times 2.00 \times 10^{4}\ \mathrm{kPa} = 2222\ \mathrm{kPa}$

Step 5: substitute.

$K_p = \frac{p(\mathrm{NH_3})^2}{p(\mathrm{N_2})\,p(\mathrm{H_2})^3} = \frac{(2222\ \mathrm{kPa})^2}{(4444\ \mathrm{kPa}) \times (13333\ \mathrm{kPa})^3}$

$K_p = 4.69 \times 10^{-10}\ \mathrm{kPa^{-2}}$

Units: kPa² ÷ (kPa × kPa³) = kPa^-2. 3 s.f., matching the data.

Where marks go:

• Using 1 : 1 : 1 instead of the balancing numbers: the H₂ used is 3 × 0.200, and the NH₃ formed is 2 × 0.200.
• Dividing by the initial total moles (4.00) instead of the equilibrium total (3.60) gives 5.79 × 10^-10: wrong.
• Rounding the partial pressures to 3 s.f. before cubing gives 4.72 × 10^-10: the cube magnifies rounding errors, so carry unrounded values through.

4. A related quantity: an unknown concentration from K_c

For H₂(g) + I₂(g) ⇌ 2HI(g), K_c = 48.0 at a certain temperature. An equilibrium mixture contains [H₂] = [I₂] = 0.0100 mol dm^-3. Calculate [HI].

Step 1: write the expression. Top minus bottom = 2 - 2 = 0, so K_c has no units.

$K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$

Step 2: rearrange for the unknown.

$[\mathrm{HI}]^2 = K_c \times [\mathrm{H_2}][\mathrm{I_2}]$

$[\mathrm{HI}] = \sqrt{K_c \times [\mathrm{H_2}][\mathrm{I_2}]}$

Step 3: substitute.

$[\mathrm{HI}] = \sqrt{48.0 \times 0.0100\ \mathrm{mol\ dm^{-3}} \times 0.0100\ \mathrm{mol\ dm^{-3}}} = \sqrt{4.80 \times 10^{-3}\ \mathrm{mol^2\ dm^{-6}}}$

$[\mathrm{HI}] = 0.0693\ \mathrm{mol\ dm^{-3}}$

Where marks go:

• Stopping at [HI]² = 4.80 × 10^-3 and forgetting the square root.
• Rearranging upside down (dividing K_c by the concentrations).

5. Heterogeneous expressions and their units

Steam is passed over hot carbon:

C(s) + H₂O(g) ⇌ CO(g) + H₂(g)

Write K_c and K_p, and give the units of each (pressure in kPa).

Step 1: decide what goes in. C is a solid, so it is left out. H₂O is a gas here, so it stays in.

Step 2: write the expressions.

$K_c = \frac{[\mathrm{CO}][\mathrm{H_2}]}{[\mathrm{H_2O}]} \qquad K_p = \frac{p(\mathrm{CO})\,p(\mathrm{H_2})}{p(\mathrm{H_2O})}$

Step 3: units. Top minus bottom = 2 - 1 = +1.

$K_c: \frac{(\mathrm{mol\ dm^{-3}})(\mathrm{mol\ dm^{-3}})}{\mathrm{mol\ dm^{-3}}} = \mathrm{mol\ dm^{-3}} \qquad K_p: \frac{\mathrm{kPa} \times \mathrm{kPa}}{\mathrm{kPa}} = \mathrm{kPa}$

Where marks go:

• Including [C] on the bottom: solids are always left out.
• Leaving out H₂O because "water is a liquid". Check the state symbol: here it is (g).

6. K_c from titration results (the esterification practical)

A student mixes 0.100 mol of ethanoic acid and 0.100 mol of ethanol with dilute HCl catalyst. The catalyst contains 0.00200 mol of H⁺ and 0.0550 mol of water. After a week the mixture is made up to 250.0 cm³. A 25.0 cm³ portion needs a mean titre of 21.50 cm³ of 0.200 mol dm^-3 NaOH. Calculate K_c.

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

Step 1: moles of NaOH in the titre.

$n(\mathrm{NaOH}) = 0.200\ \mathrm{mol\ dm^{-3}} \times \frac{21.50}{1000}\ \mathrm{dm^3} = 4.30 \times 10^{-3}\ \mathrm{mol}$

Step 2: scale up to the whole 250.0 cm³ (ten 25.0 cm³ portions). NaOH reacts 1 : 1 with H⁺.

$n(\mathrm{H^+})_{\text{total}} = 4.30 \times 10^{-3} \times 10 = 0.0430\ \mathrm{mol}$

Step 3: subtract the catalyst.

$n(\mathrm{CH_3COOH})_{\text{eq}} = 0.0430 - 0.00200 = 0.0410\ \mathrm{mol}$

Step 4: initial / change / equilibrium table. All ratios are 1 : 1 : 1 : 1; acid reacted = 0.100 - 0.0410 = 0.0590 mol.

Step 5: substitute. Top minus bottom = 2 - 2 = 0, so the volume cancels and the equilibrium moles can go straight in. K_c has no units.

$K_c = \frac{n(\mathrm{ester}) \times n(\mathrm{H_2O})}{n(\mathrm{acid}) \times n(\mathrm{alcohol})} = \frac{0.0590 \times 0.1140}{0.0410 \times 0.0410}$

$K_c = \frac{6.726 \times 10^{-3}}{1.681 \times 10^{-3}} = 4.00 \ \text{(3 s.f., no units)}$

Where marks go:

• Not subtracting the catalyst (using 0.0430 mol as the acid) gives 3.45: wrong.
• Ignoring the water in the catalyst (initial water = 0) gives 2.07: wrong.
• Forgetting to multiply by 10 for the 250.0 cm³ flask gives only 0.00430 mol of H⁺, which is impossible once the catalyst is subtracted.

Common exam mistakes

• Including solids or liquids in a heterogeneous expression, for example writing [CaCO₃] or p(C). Their concentrations are constant, so they are omitted. But in a homogeneous liquid mixture (esterification), water is included.
• Using square brackets in K_p. [ ] means concentration. Partial pressures are written p(NH₃).
• Putting initial amounts into K. Only equilibrium concentrations or partial pressures go in.
• Ignoring the balancing numbers when working out amounts reacted. If 0.20 mol of N₂ reacts, 0.60 mol of H₂ reacts and 0.40 mol of NH₃ forms.
• Using moles instead of concentrations when K_c has units. Divide by the volume in dm³ (convert from cm³ first).
• Dividing by the wrong total for mole fractions: use the total equilibrium moles of gas, not the initial total and not one substance.
• Wrong or missing units. Work them out by cancelling every time: for N₂ + 3H₂ ⇌ 2NH₃, K_c is in dm⁶ mol^-2, not mol² dm^-6. If the powers cancel, write no units.
• Rounding too early, especially before cubing or squaring. Keep full calculator values until the final answer.
• Saying K changes when concentration or pressure changes. Only temperature changes K. Concentration and pressure can shift the position, but K is constant.
• Saying a catalyst increases K or the yield. It changes neither; it increases the forward and reverse rates by the same amount.
• Getting the temperature effect on K the wrong way round. For an exothermic forward reaction, increasing temperature decreases K. Check the sign of ΔH first.
• Explaining with le Chatelier when the question asks about K. In a "use K" answer, say the ratio no longer equals K, so the position shifts until it equals K again; for temperature, say K itself changes.
• In the esterification practical, forgetting to subtract the acid catalyst from the titre, or treating the initial water as zero.

Key definitions

• Mole fraction of gas A: $x_A = \dfrac{\text{moles of A}}{\text{total moles of gas in the mixture}}$. The mole fractions in a mixture add up to 1.
• Partial pressure of gas A: the pressure that A would exert if it alone occupied the whole volume of the mixture; $p(A) = x_A \times P$, where P is the total pressure. The partial pressures add up to the total pressure.
• Homogeneous equilibrium: an equilibrium in which all species are in the same phase.
• Heterogeneous equilibrium: an equilibrium in which the species are in different phases.
• K_c for aA + bB ⇌ cC + dD: $K_c = \dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$, where [ ] is the equilibrium concentration in mol dm^-3.
• K_p for aA(g) + bB(g) ⇌ cC(g) + dD(g): $K_p = \dfrac{p(\mathrm{C})^c\,p(\mathrm{D})^d}{p(\mathrm{A})^a\,p(\mathrm{B})^b}$, where p is the equilibrium partial pressure; gases only.
• Heterogeneous K_c and K_p: solid and liquid concentrations are constant and are omitted from the expression.
• Effect of conditions on K: the value of an equilibrium constant changes only with temperature; it is unchanged by changes in concentration, pressure or the presence of a catalyst.

Specification

• I can use the terms mole fraction and partial pressure, and calculate the partial pressure of each gas in a mixture. (5.1.2 a)
• I can calculate the amounts, concentrations or partial pressures present at equilibrium from initial amounts and given equilibrium data. (5.1.2 b)
• I can describe a technique to determine the quantities present at equilibrium, such as titrating an esterification mixture, and explain each step. (5.1.2 c)
• I can write K_c and K_p expressions for homogeneous and heterogeneous equilibria, leaving out solids and liquids in heterogeneous expressions. (5.1.2 d)
• I can calculate K_c and K_p, or a related quantity such as a missing equilibrium concentration, and determine their units. (5.1.2 e)
• I can state the qualitative effect of increasing or decreasing temperature on K for exothermic and endothermic reactions. (5.1.2 f i)
• I can state that K is unchanged by changes in concentration or pressure, or by adding a catalyst. (5.1.2 f ii)
• I can explain how the equilibrium constant controls the position of equilibrium when concentration, pressure or temperature changes. (5.1.2 g)
• I can apply these principles to other equilibrium constants, such as K_a and K_w. (5.1.2 h)

Related notes

• Chemical equilibrium: dynamic equilibrium, le Chatelier's principle and K_c at AS
• Amount of substance: moles, concentrations and reacting ratios
• Acids: acid-base titration technique
• How fast?: rate equations and the effect of temperature on the rate constant
• Acids, bases and buffers: K_a and K_w, more equilibrium constants
• Enthalpy and entropy: feasibility and free energy
• Carboxylic acids and esters: esterification