Chemistry › Practise › Questions › The equilibrium constant Kc
Questions on the equilibrium constant Kc
Writing Kc for homogeneous equilibria and working out its units, calculating Kc from equilibrium amounts and from initial amounts with an ICE table, using Kc to find an equilibrium amount, Kc for a reversed equation, heterogeneous equilibria and the solvent left out of Kc, quenching a sample for titration, the reaction quotient Q, the effect of concentration, pressure, catalysts and temperature on Kc, and the partition coefficient Kpc for a solute between two solvents.
21 original questions · 80 marks · the equilibrium constant Kc notes · Chemical equilibria · Flashcards for this unit
To print: the written questions (PDF, 78 KB) · the mark schemes (PDF, 83 KB)
How to use this page
- Every question here is written for this library rather than taken from a past paper. Write your answer out before opening a mark scheme: the schemes award marks point by point, and the points are much easier to see when you have something of your own to compare against.
- The player marks nothing for you. It shows one question, waits, then shows the scheme so you can mark yourself, and brings a question back sooner when it went badly.
- Each line of a scheme is one mark: B1 is a point that stands on its own, M1 a method mark, and A1 the answer or an accuracy point. A six-mark question marked by level lists what a full answer covers, then says which level an answer reaches.
- Each question’s code is the one printed in the workbook. Type inklearning.co.uk/q/ and a code to jump to its question.
- A question with board names beside its code is set by those boards only; one with none is on every specification here.
- The PDF links above print the same questions, with the answers in a separate book.
- As a rough guide, allow about 1.4 minutes a mark; each board’s papers run at their own pace.
Question codes are permanent, so the numbers on this page can skip or run out of order; nothing is missing.
Explain how each of these changes affects the position of equilibrium and the value of Kc for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −197 kJ mol⁻¹: adding more oxygen, raising the pressure, adding a catalyst, and raising the temperature.
Mark scheme
Indicative content
- adding oxygen raises [O₂], so the ratio [SO₃]² ÷ ([SO₂]²[O₂]) falls below Kc and the forward reaction runs faster than the reverse
- the position shifts to the right until the ratio equals Kc again, so more SO₃ forms while Kc itself is unchanged
- raising the pressure raises every concentration by the same factor, and with three moles of gas on the left and two on the right the ratio falls below Kc
- the position shifts right, towards fewer moles of gas, but Kc is unchanged because only a change of temperature changes Kc
- a catalyst increases the forward and reverse rates equally, so neither the position nor Kc changes and equilibrium is only reached sooner
- raising the temperature decreases Kc because the forward reaction is exothermic, so the ratio is now above the new Kc and the position shifts left, lowering the yield of SO₃
Levels
- Level 3 (5–6 marks) all four changes are explained, each with its effect on the position and on Kc stated separately and correctly, and the reasoning uses the ratio compared with Kc rather than Le Chatelier's principle alone; the answer says that only temperature changes Kc and explains why in terms of the exothermic forward reaction
- Level 2 (3–4 marks) three of the four changes are right for both position and Kc, or all four are covered and Kc goes wrong for one of them
- Level 1 (1–2 marks) the shift in position is given for one or two changes, with Kc said to rise when oxygen is added or the pressure raised
Show that two 20.0 cm³ portions of an organic solvent extract more of a solute from 50.0 cm³ of water than one 40.0 cm³ portion does. The water holds 1.00 g of the solute, and Kpc (organic ÷ aqueous) is 9.00.
Mark scheme
- M1 one 40.0 cm³ portion: (m ÷ 40.0) ÷ ((1.00 − m) ÷ 50.0) = 9.00, so 50.0m = 360(1.00 − m)
- A1 m = 360 ÷ 410 = 0.878 g extracted by one portion
- M1 first 20.0 cm³ portion: 50.0m = 180(1.00 − m), so m = 0.783 g, leaving 0.217 g in the water
- M1 second 20.0 cm³ portion takes the same fraction, 180 ÷ 230, of the 0.217 g left, which is 0.170 g
- A1 two portions extract 0.953 g in total, more than the 0.878 g from one portion; error carried forward from M2
Give the expression for Kc, and its units, for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) and for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
Mark scheme
- B1 Kc = [SO₃]² ÷ ([SO₂]²[O₂]); square brackets are required
- B1 units: (mol dm⁻³)² ÷ (mol dm⁻³)³ = mol⁻¹ dm³; accept dm³ mol⁻¹
- B1 Kc = [NH₃]² ÷ ([N₂][H₂]³)
- B1 units: (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = mol⁻² dm⁶; accept dm⁶ mol⁻²
Calculate Kc for CH₃COOH(l) + CH₃CH₂OH(l) ⇌ CH₃COOCH₂CH₃(l) + H₂O(l) from this experiment: 1.00 mol of ethanoic acid and 1.50 mol of ethanol are mixed with a catalyst and left to reach equilibrium, and 0.230 mol of ethanoic acid remains.
Mark scheme
- M1 acid reacted = 1.00 − 0.230 = 0.770 mol, so 0.770 mol each of ester and water form
- M1 ethanol at equilibrium = 1.50 − 0.770 = 0.730 mol
- M1 the volume cancels, so Kc = (0.770 × 0.770) ÷ (0.230 × 0.730)
- A1 Kc = 3.53, with no units; error carried forward from M2
Calculate Kc for H₂(g) + I₂(g) ⇌ 2HI(g) from the experiment described in the table, in which hydrogen and iodine are heated in a sealed flask until equilibrium is reached.
Quantity Value H₂ at the start 0.300 mol I₂ at the start 0.200 mol Volume of the flask 1.50 dm³ HI at equilibrium 0.320 mol Mark scheme
- M1 HI formed = 0.320 mol, so H₂ and I₂ each fall by 0.320 ÷ 2 = 0.160 mol
- M1 at equilibrium H₂ = 0.140 mol and I₂ = 0.0400 mol
- M1 Kc = [HI]² ÷ ([H₂][I₂]) and the volume cancels, so Kc = 0.320² ÷ (0.140 × 0.0400)
- A1 Kc = 18.3, with no units; ignore any unit given if the working shows the volume cancelling
Calculate Kc for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) from this experiment: 0.250 mol of PCl₅ is heated in a sealed 5.00 dm³ vessel, and at equilibrium the vessel contains 0.0950 mol of Cl₂.
Mark scheme
- M1 at equilibrium PCl₃ = 0.0950 mol, Cl₂ = 0.0950 mol and PCl₅ = 0.250 − 0.0950 = 0.155 mol
- M1 concentrations, dividing by 5.00 dm³: [PCl₃] = [Cl₂] = 0.0190 mol dm⁻³ and [PCl₅] = 0.0310 mol dm⁻³
- M1 Kc = [PCl₃][Cl₂] ÷ [PCl₅] = 0.0190² ÷ 0.0310
- A1 Kc = 0.0116 mol dm⁻³, to three significant figures; an answer of 0.0582 from amounts in moles scores M1 only
Calculate the amount of HI at equilibrium when 0.500 mol of H₂ and 0.500 mol of I₂ are sealed in a flask at a temperature at which Kc for H₂(g) + I₂(g) ⇌ 2HI(g) is 49.0.
Mark scheme
- M1 if y mol of H₂ reacts, equilibrium amounts are H₂ (0.500 − y), I₂ (0.500 − y) and HI 2y mol
- M1 the volume cancels: (2y)² ÷ (0.500 − y)² = 49.0
- M1 taking square roots, 2y ÷ (0.500 − y) = 7.00, so 9.00y = 3.50 and y = 0.389
- A1 amount of HI = 2y = 0.778 mol, to three significant figures
Explain why the volume of the flask is not needed to calculate Kc for the esterification of ethanoic acid from equilibrium amounts but is needed for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), and why doubling the volume of an equilibrium mixture of H₂, I₂ and HI leaves the amount of HI unchanged.
Mark scheme
- B1 each concentration is the amount divided by the same volume V, and for the ester there are two terms on top and two on the bottom, so V cancels
- B1 for PCl₅ there are two terms on top and one below, so one factor of V remains and the amounts must be divided by the volume
- B1 doubling the volume halves every concentration, and with two powers on top and two on the bottom for HI the ratio is unchanged
- B1 the ratio still equals Kc, so there is no shift and the amount of HI stays the same
Evaluate a student's statement that, since Kc for H₂(g) + I₂(g) ⇌ 2HI(g) is 46.0 at 763 K, Kc for 2HI(g) ⇌ H₂(g) + I₂(g) is also 46.0 and Kc for HI(g) ⇌ ½H₂(g) + ½I₂(g) is 23.0 at 763 K.
Mark scheme
- B1 Kc is defined for the equation as written, even though the equilibrium mixture is the same
- B1 reversing the equation inverts the expression, so Kc for 2HI(g) ⇌ H₂(g) + I₂(g) is 1 ÷ 46.0 = 0.0217
- B1 halving every coefficient takes the square root of Kc, not half of it, so Kc for the third equation is √0.0217 = 0.147
- B1 both of the student's values are wrong
Explain how Kc is written for Fe²⁺(aq) + Ag⁺(aq) ⇌ Fe³⁺(aq) + Ag(s), with its units, and why the solid silver is filtered off before the silver ions in the equilibrium mixture are titrated with thiocyanate.
Mark scheme
- B1 Kc = [Fe³⁺] ÷ ([Fe²⁺][Ag⁺]), with units of mol⁻¹ dm³
- B1 silver is left out because the concentration of a solid is fixed by its density and does not change, so it is folded into the constant
- B1 the titration removes silver ions, so with silver present the reverse reaction of Ag with Fe³⁺ would replace them during the titration
- B1 the titre would then be larger than the amount of Ag⁺ at equilibrium; without the solid the reverse reaction cannot happen
Calculate Kc at 298 K for Fe²⁺(aq) + Ag⁺(aq) ⇌ Fe³⁺(aq) + Ag(s) from the experiment described in the table, in which equal volumes of the two solutions are mixed and left to reach equilibrium.
Quantity Value Concentration of iron(II) nitrate before mixing 0.200 mol dm⁻³ Concentration of silver nitrate before mixing 0.200 mol dm⁻³ [Ag⁺] at equilibrium 0.0797 mol dm⁻³ Mark scheme
- M1 after mixing, [Fe²⁺] = [Ag⁺] = 0.100 mol dm⁻³ at the start
- M1 [Ag⁺] falls by 0.0203 mol dm⁻³, so at equilibrium [Fe²⁺] = 0.0797 and [Fe³⁺] = 0.0203 mol dm⁻³
- M1 Kc = [Fe³⁺] ÷ ([Fe²⁺][Ag⁺]) = 0.0203 ÷ 0.0797², with silver left out as a solid
- A1 Kc = 3.2 mol⁻¹ dm³, to two significant figures, the subtraction 0.100 − 0.0797 leaving only two; accept 3.20
Explain why water is left out of Kc for the chromate–dichromate equilibrium 2CrO₄²⁻(aq) + 2H⁺(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l) in dilute solution but is included in Kc for the esterification of ethanoic acid with ethanol, and give Kc for the chromate equilibrium with its units.
Mark scheme
- B1 in dilute solution water is the solvent in vast excess, so its concentration is effectively constant and is left out
- B1 in the ester mixture water is one of four liquids in similar amounts, and its concentration changes significantly as the reaction proceeds, so it is included
- B1 Kc = [Cr₂O₇²⁻] ÷ ([CrO₄²⁻]²[H⁺]²)
- B1 units: mol dm⁻³ ÷ (mol dm⁻³)⁴ = mol⁻³ dm⁹
Calculate Kc for CO(g) + 2H₂(g) ⇌ CH₃OH(g), with its units, from the equilibrium amounts in a 0.500 dm³ vessel given in the table.
Substance Amount at equilibrium / mol CO 0.0500 H₂ 0.0800 CH₃OH 0.0200 Mark scheme
- M1 concentrations, dividing by 0.500 dm³: [CO] = 0.100, [H₂] = 0.160 and [CH₃OH] = 0.0400 mol dm⁻³
- M1 Kc = [CH₃OH] ÷ ([CO][H₂]²) = 0.0400 ÷ (0.100 × 0.160²)
- M1 units: (mol dm⁻³) ÷ (mol dm⁻³)³ = mol⁻² dm⁶
- A1 Kc = 15.6 mol⁻² dm⁶, to three significant figures; an answer of 62.5 from amounts in moles scores the units mark only
Calculate the partition coefficient of ammonia between trichloromethane and water from the titration results given in the table. Each layer was titrated with hydrochloric acid after the mixture had been shaken and left to reach equilibrium.
Layer Volume titrated / cm³ HCl concentration / mol dm⁻³ Titre / cm³ Aqueous 10.0 0.200 24.00 Trichloromethane 25.0 0.0500 4.80 Mark scheme
- M1 aqueous layer: 0.200 × 24.00 ÷ 1000 = 4.80 × 10⁻³ mol in 10.0 cm³, so [NH₃]aqueous = 0.480 mol dm⁻³
- M1 organic layer: 0.0500 × 4.80 ÷ 1000 = 2.40 × 10⁻⁴ mol in 25.0 cm³, so [NH₃]organic = 9.60 × 10⁻³ mol dm⁻³
- M1 Kpc = [NH₃]organic ÷ [NH₃]aqueous = 9.60 × 10⁻³ ÷ 0.480
- A1 Kpc = 0.0200, with no units; accept 2.00 × 10⁻²
Suggest what each of these values of Kc at 298 K tells you about the equilibrium mixture: 1.0 × 10¹² for reaction A, 4.0 for reaction B and 1.0 × 10⁻¹² for reaction C.
Mark scheme
- B1 A: the position lies far to the right and the mixture is almost all products, so the reaction goes virtually to completion
- B1 B: the mixture contains substantial amounts of both reactants and products
- B1 C: the position lies far to the left and almost none of the reactants has reacted
Calculate Kc for N₂O₄(g) ⇌ 2NO₂(g) from an equilibrium mixture containing N₂O₄ at 0.0450 mol dm⁻³ and NO₂ at 0.0300 mol dm⁻³.
Mark scheme
- M1 Kc = [NO₂]² ÷ [N₂O₄]
- M1 Kc = 0.0300² ÷ 0.0450 = 0.000900 ÷ 0.0450
- A1 Kc = 0.0200 mol dm⁻³, to three significant figures; accept 0.02 mol dm⁻³
Calculate the reaction quotient Q for the esterification mixture whose amounts are given in the table, and deduce which way the mixture will move. Kc for the esterification is 4.0.
Substance Amount / mol Ethanoic acid 0.250 Ethanol 0.400 Ethyl ethanoate 0.300 Water 0.300 Mark scheme
- M1 Q = (0.300 × 0.300) ÷ (0.250 × 0.400), the volume cancelling
- B1 Q is less than Kc, so the forward reaction runs faster and more ester forms until Q equals 4.0
- A1 Q = 0.900, with no units
Explain why adding more ethanol to an equilibrium mixture of ethanoic acid, ethanol, ethyl ethanoate and water increases the amount of ester but does not change Kc.
Mark scheme
- B1 Kc depends only on temperature, which has not changed
- B1 the added ethanol makes the denominator larger, so the ratio of concentrations falls below Kc
- B1 the forward reaction runs faster than the reverse until the ratio equals Kc again, which forms more ester
Suggest the reason for each of these steps in finding the amount of ethanoic acid in an esterification mixture at equilibrium: a sample is run into ice-cold water and titrated with sodium hydroxide, and a second titration is done on the sulfuric acid catalyst alone.
Mark scheme
- B1 cooling and diluting slows the reaction almost to a stop; accept 'quenches the reaction'
- B1 so the position of equilibrium does not shift while the alkali removes ethanoic acid during the titration
- B1 the alkali also neutralises the sulfuric acid, so the titre for the catalyst alone is subtracted to leave the titre for the ethanoic acid
Deduce the sign of ΔH for the forward reaction of a gas-phase equilibrium from the values of Kc given in the table, and state the effect of raising the temperature on the equilibrium yield.
Temperature / K Kc 500 1.2 × 10⁻³ 600 8.5 × 10⁻³ 700 4.6 × 10⁻² Mark scheme
- B1 Kc increases as the temperature rises
- B1 so the forward reaction is endothermic and ΔH is positive
- B1 raising the temperature shifts the position to the right and increases the equilibrium yield of products
Show that Kc = 4.0 for CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l), given that 1.00 mol of ethanoic acid and 1.00 mol of ethanol, left with an acid catalyst until equilibrium is reached, give 0.667 mol of ethyl ethanoate.
Mark scheme
- M1 at equilibrium: ester 0.667 mol, water 0.667 mol, acid 1.00 − 0.667 = 0.333 mol and ethanol 0.333 mol, since the equation is 1 : 1 : 1 : 1
- M1 Kc = [CH₃COOC₂H₅][H₂O] ÷ ([CH₃COOH][C₂H₅OH]) = (0.667 × 0.667) ÷ (0.333 × 0.333); the volume cancels because there are two terms above the line and two below
- A1 Kc = 0.445 ÷ 0.111 = 4.0, with no units; allow 4.01