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GCSE · Chemistry

C6b Reversible Reactions and Equilibrium

AQA 8464 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Industrial production of ammonia: the Haber process

Original context passage written for Revision Library.

Ammonia (NH3) is manufactured on an industrial scale by the Haber process. Nitrogen is obtained from the air and hydrogen is obtained from natural gas. The two gases are purified, mixed in the correct ratio, and passed over an iron catalyst inside a reaction vessel held at a temperature of about 450 degrees C and a pressure of about 200 atmospheres. The forward reaction, N2(g) + 3H2(g) <=> 2NH3(g), is exothermic. Not all of the nitrogen and hydrogen react, since the reaction reaches dynamic equilibrium rather than going to completion. As the mixture leaves the reactor it is cooled so that the ammonia liquefies and can be separated. The unreacted nitrogen and hydrogen are recycled back into the reactor.

Some chemical reactions are reversible.

(a)Define the term 'reversible reaction'.(1)

(b)State what the symbol ≤> represents when written in a chemical equation.(1)

(Total for Question 1 is 2 marks)

Cobalt chloride paper is used in the laboratory as a test for water. Anhydrous cobalt chloride, CoCl2, is blue. When water is added, it forms hydrated cobalt chloride, CoCl2.6H2O, which is pink. This is a reversible reaction:
CoCl2( ) + 6H2O( ) ≤> CoCl2.6H2O( )

(a)Complete the equation above by writing the correct state symbol in each of the three brackets.(1)

(b)State the colour change observed when water is added to anhydrous cobalt chloride.(1)

(c)State whether the forward reaction (hydration, left to right as written) is exothermic or endothermic.(1)

(Total for Question 2 is 3 marks)

When solid ammonium chloride is heated strongly at one end of a horizontal test tube, a white solid reforms further along the cooler part of the tube. This is a reversible reaction:
NH4Cl(s) ≤> NH3(g) + HCl(g)

(a)Name the type of change that happens to the ammonium chloride when it is heated.(1)

(b)State whether the forward reaction (heating the solid) is exothermic or endothermic.(1)

(c)Explain why a white solid reforms further along the cooler part of the tube.(1)

(Total for Question 3 is 3 marks)

Blue hydrated copper sulfate crystals, CuSO4.5H2O, can be heated to form white anhydrous copper sulfate powder, CuSO4. This is a reversible reaction:
CuSO4.5H2O(s) ≤> CuSO4(s) + 5H2O( )

(a)Complete the state symbol for the water in the equation above, given that it leaves the crystals as steam.(1)

(b)State the colour change seen when hydrated copper sulfate crystals are heated.(1)

(Total for Question 4 is 2 marks)

Higher tier only. Reversible reactions can reach dynamic equilibrium if they take place in a closed system.

(a)State what is meant by a 'closed system'.(1)

(b)State two features of a reaction mixture that show it has reached dynamic equilibrium.(2)

(Total for Question 5 is 3 marks)

A student investigates the reversible reaction between hydrated and anhydrous copper sulfate.
Method:
1. Weigh an empty boiling tube.
2. Add a sample of blue hydrated copper sulfate crystals (CuSO4.5H2O) and reweigh.
3. Heat the tube gently and continuously until no further colour change occurs, then allow it to cool and reweigh.
4. Add water dropwise to the cooled white solid.
CuSO4.5H2O(s) ≤> CuSO4(s) + 5H2O(g)

(a)Suggest one piece of safety equipment the student should use during this practical, other than a lab coat.(1)

(b)Explain how the student knows that the dehydration reaction is complete.(1)

(c)Predict and explain what the student observes when water is added dropwise to the cooled white solid.(2)

(d)The student repeats the heating and weighing three times and records the following masses of water lost: 0.42 g, 0.40 g, 0.85 g. Identify the anomalous result and suggest one possible reason for it.(2)

(e)Calculate the mean mass of water lost, excluding the anomalous result.(1)

(f)Suggest one improvement to the method that would make the results more accurate.(1)

(g)Suggest why the student allows the tube to cool before adding water in step 4.(1)

(h)State one control variable that should be kept the same across all three repeats to ensure the results are comparable, and explain why.(2)

(i)Suggest one way the student could adapt the method to show how the mass of the tube changes over time as the crystals are heated, rather than measuring only the total change at the end.(1)

(Total for Question 6 is 12 marks)

Use the information above about the Haber process. N2(g) + 3H2(g) ≤> 2NH3(g) (relative formula masses: N2 = 28, H2 = 2, NH3 = 17)

(a)Calculate the maximum mass, in kg, of ammonia that could be produced from 28 kg of nitrogen reacting with an excess of hydrogen.(3)

(b)In the actual industrial process, only 21.25 kg of ammonia is obtained from this reaction. Calculate the percentage yield.(2)

(Total for Question 7 is 5 marks)

Higher tier only. N2(g) + 3H2(g) ≤> 2NH3(g) (forward reaction exothermic). Extra nitrogen is added to the equilibrium mixture at constant temperature and pressure.

(a)Predict the effect on the position of equilibrium of adding extra nitrogen.(1)

(b)Use Le Chatelier's principle to explain your answer to part (a).(2)

(c)In industry, ammonia is continually removed from the equilibrium mixture by cooling and liquefying it. State and explain the effect this has on the overall yield of ammonia.(2)

(Total for Question 8 is 5 marks)

Higher tier only. N2(g) + 3H2(g) ≤> 2NH3(g) (forward reaction exothermic). The temperature of the reaction vessel is increased, with the pressure held constant.

(a)Predict the effect on the position of equilibrium.(1)

(b)Use Le Chatelier's principle to explain your answer.(2)

(c)State the effect of this temperature increase on the percentage yield of ammonia at equilibrium.(1)

(Total for Question 9 is 4 marks)

Higher tier only. N2(g) + 3H2(g) ≤> 2NH3(g)

(a)State the total number of moles of gas on the reactant side and on the product side of this equation.(1)

(b)Predict and explain the effect of increasing the pressure on the position of equilibrium.(2)

(c)State the effect of increasing the pressure on the percentage yield of ammonia.(1)

(Total for Question 10 is 4 marks)

Higher tier only. Sulfur dioxide reacts with oxygen in another industrial equilibrium reaction, used to manufacture sulfuric acid:
2SO2(g) + O2(g) ≤> 2SO3(g) (forward reaction exothermic)

(a)State the total number of moles of gas on each side of the equation.(1)

(b)Predict and explain the effect of increasing the pressure on the yield of sulfur trioxide (SO3) at equilibrium.(2)

(c)The temperature of the reaction vessel is increased instead of the pressure. Predict and explain the effect on the yield of sulfur trioxide.(2)

(Total for Question 11 is 5 marks)

*Higher tier only. Ammonia is manufactured industrially using the Haber process at a temperature of about 450 degrees C, a pressure of about 200 atmospheres, and an iron catalyst. These conditions do not give the highest possible equilibrium yield of ammonia.
Explain why these particular conditions are used.

(Total for Question 12 is 6 marks)

Higher tier only. A student uses a colorimeter to monitor a reversible reaction, X ≤> Y, as it approaches equilibrium in a closed flask. The graph below shows how the concentrations of reactant X and product Y change with time.

(a)State the time at which the reaction mixture first reaches equilibrium.(1)

(b)Explain how the graph shows that the mixture has reached dynamic equilibrium, rather than the reaction simply stopping.(2)

(c)Explain why the concentrations of X and Y are not equal to each other at equilibrium, even though both are constant.(1)

(Total for Question 13 is 4 marks)

Higher tier only. A chemical company operates a reactor at equilibrium, producing ammonia by the Haber process. The engineers first decrease the temperature of the reactor, and then also increase the pressure, while keeping the amounts of nitrogen and hydrogen the same.

(a)State and explain the effect of decreasing the temperature alone on the yield of ammonia and on the rate at which equilibrium is reached.(2)

(b)State and explain the additional effect of then increasing the pressure.(2)

(c)Despite these yield increases, industrial chemists do not use the lowest possible temperature and the highest possible pressure. Explain why.(2)

(Total for Question 14 is 6 marks)

Higher tier only. A student makes the following claim: 'Adding an iron catalyst to the Haber process increases the equilibrium yield of ammonia, because it speeds up the forward reaction more than the reverse reaction.'
Evaluate this claim.

(Total for Question 15 is 3 marks)

The dehydration of hydrated copper sulfate (heating it to drive off water) is endothermic and requires 120 kJ of energy to be transferred to the crystals per mole of water lost.

(a)State the energy change (exothermic or endothermic) for the reverse reaction, when water is added back to anhydrous copper sulfate, and give the amount of energy transferred.(1)

(b)Describe how the reaction profile (energy level diagram) for the reverse (hydration) reaction compares with the profile for the forward (dehydration) reaction.(1)

(c)Name the scientific law that explains why the size of the energy change must be the same in both directions.(1)

(Total for Question 16 is 3 marks)

For each statement about reversible reactions and equilibrium, state whether it is true or false.

(a)'A closed system means that no matter can enter or leave the reaction mixture.'(1)

(b)'At dynamic equilibrium, the forward and reverse reactions have both stopped.'(1)

(c)'At equilibrium, the concentrations of reactants and products are always exactly equal to each other.'(1)

(d)'The symbol ≤> in an equation shows that a reaction is reversible.'(1)

(Total for Question 17 is 4 marks)

Higher tier only. A sealed container initially holds 4.0 mol of nitrogen and 9.0 mol of hydrogen. The mixture is left until equilibrium is reached:
N2(g) + 3H2(g) ≤> 2NH3(g)
At equilibrium, 1.5 mol of ammonia has been formed.

(a)Use the mole ratio in the equation to calculate the number of moles of nitrogen and of hydrogen that have reacted.(2)

(b)Calculate the number of moles of nitrogen and of hydrogen present at equilibrium.(2)

(c)Even after a very long time, not all of the nitrogen and hydrogen are converted into ammonia. Suggest why.(1)

(Total for Question 18 is 5 marks)

Mark scheme · C6b Reversible Reactions and Equilibrium

Question 1

(a) B1 a reaction in which the products can react together (under different conditions) to reform the original reactants oe
(a) Answer: A reversible reaction is one in which the products of the reaction can react to reform the original reactants.
(b) B1 shows that the reaction is reversible / can proceed in both the forward and reverse directions oe
(b) Answer: It shows that the reaction is reversible: it can proceed in both the forward and the reverse direction.

Question 2

(a) B1 all three correct: CoCl2(s) + 6H2O(l) ≤> CoCl2.6H2O(s) cao
(a) Answer: CoCl2(s) + 6H2O(l) ≤> CoCl2.6H2O(s)
(b) B1 changes from blue to pink oe
(b) Answer: The solid changes colour from blue to pink.
(c) B1 exothermic
(c) Answer: Exothermic.

Question 3

(a) B1 thermal decomposition (thermal dissociation) oe
(a) Answer: Thermal decomposition (thermal dissociation).
(b) B1 endothermic
(b) Answer: Endothermic.
(c) B1 the ammonia and hydrogen chloride gases cool as they move along the tube and recombine (the reverse reaction occurs) to reform solid ammonium chloride oe
(c) Answer: The ammonia and hydrogen chloride gases cool as they move along the tube and recombine, reforming solid ammonium chloride.

Question 4

(a) B1 (g)
(a) Answer: (g)
(b) B1 changes from blue to white oe
(b) Answer: The crystals change colour from blue to white.

Question 5

(a) B1 a system in which no reactants or products (no matter) can enter or leave oe
(a) Answer: A closed system is one in which no reactants or products can enter or leave.
(b) B1 the rate of the forward reaction equals the rate of the reverse reaction
(b) B1 the concentrations of the reactants and the products remain constant (but are not necessarily equal to each other) oe
(b) Answer: At dynamic equilibrium, the forward and reverse reactions occur at the same rate, and the concentrations of reactants and products remain constant (though not necessarily equal to each other).

Question 6

(a) B1 eye protection (safety goggles); or a heatproof mat; or tongs/a tube holder to handle the hot tube oe
(a) Answer: Eye protection (safety goggles) should be worn, since a solid is being heated.
(b) B1 there is no further change in colour (the solid is uniformly white), or no further loss in mass on reheating oe
(b) Answer: There is no further change in colour (the solid is uniformly white) and no further loss in mass on reheating.
(c) B1 the solid turns blue again (as CuSO4.5H2O reforms)
(c) B1 the tube becomes warm/hot, because the reverse reaction (hydration) is exothermic oe
(c) Answer: The solid turns blue again as CuSO4.5H2O reforms, and the tube becomes warm because the reverse (hydration) reaction is exothermic.
(d) B1 0.85 g is anomalous (it does not agree with/is much greater than the other two results)
(d) B1 a plausible reason, e.g. the crystals were not fully dehydrated (heated to constant mass) in the other repeats, or a weighing/measurement error occurred in this repeat oe
(d) Answer: 0.85 g is anomalous, since it is far higher than the other two results; a possible reason is a weighing error, or that this sample was not heated for as long as the others before being reweighed.
(e) B1 0.41 g cao (allow ft from candidate's identification of the anomaly in part d)
(e) Answer: 0.41 g
(f) B1 heat and reweigh repeatedly until the mass is constant (heat to constant mass) to ensure all the water has been driven off; or use a balance that measures to more decimal places oe
(f) Answer: Heat, cool and reweigh the tube repeatedly until the mass is constant (heating to constant mass), to make sure all the water of crystallisation has been driven off.
(g) B1 to avoid the water boiling/spitting violently, or causing a burn from the hot glass/tube (safety) oe
(g) Answer: Adding water to the hot tube could cause the water to boil and spit violently, risking a burn, so the tube is allowed to cool first.
(h) B1 a valid control variable, e.g. the mass of hydrated copper sulfate crystals used each time, or the heating time/method used each time oe
(h) B1 so that any difference in the mass of water lost is caused only by genuine variation in the reaction, rather than by differences in the starting mass or heating conditions oe
(h) Answer: The mass of hydrated crystals used should be kept the same in each repeat, so that any difference in the mass of water lost is a genuine result and not simply caused by starting with a different mass of crystals.
(i) B1 reweigh the tube and its contents at regular time intervals during heating (e.g. every minute) so that a mass-time graph can be plotted oe
(i) Answer: Reweigh the tube and its contents at regular time intervals during heating (for example, every minute), so the mass loss over time can be plotted on a graph.

Question 7

(a) M1 moles of N2 = 28000 / 28 = 1000 mol oe (or equivalent working in kg: 28/28 = 1000 kmol)
(a) M1 moles of NH3 = 1000 x 2 = 2000 mol, using the mole ratio N2 : NH3 = 1 : 2
(a) A1 mass of NH3 = 2000 x 17 = 34000 g = 34 kg cao
(a) Answer: 34 kg
(b) M1 percentage yield = (actual yield / theoretical yield) x 100 = (21.25 / 34) x 100 (allow ft from candidate's answer to part a)
(b) A1 62.5% cao (ft consistent with their part a answer)
(b) Answer: 62.5%

Question 8

(a) B1 the position of equilibrium shifts to the right (towards the products/ammonia)
(a) Answer: The position of equilibrium shifts to the right, towards the products (ammonia).
(b) B1 the system acts to oppose (counteract) the increase in nitrogen concentration oe
(b) B1 by reacting more of the nitrogen (and hydrogen) to form more ammonia, using up the extra nitrogen that was added oe
(b) Answer: The system opposes the increase in nitrogen concentration by reacting more nitrogen and hydrogen together to form more ammonia, using up the extra nitrogen added.
(c) B1 the yield of ammonia increases
(c) B1 because removing ammonia shifts the position of equilibrium to the right (forward) to replace the ammonia that has been removed oe
(c) Answer: The yield of ammonia increases, because removing ammonia from the mixture shifts the equilibrium position to the right to replace it, forming more ammonia.

Question 9

(a) B1 the position of equilibrium shifts to the left (towards the reactants, N2 and H2)
(a) Answer: The position of equilibrium shifts to the left, towards the reactants.
(b) B1 the system acts to oppose the increase in temperature oe
(b) B1 by favouring the endothermic (reverse) reaction, which absorbs (takes in) the extra heat energy oe
(b) Answer: The system opposes the increase in temperature by favouring the endothermic reverse reaction, which absorbs the extra heat energy.
(c) B1 the percentage yield decreases
(c) Answer: The percentage yield of ammonia decreases.

Question 10

(a) B1 reactant side = 4 mol of gas (1 + 3); product side = 2 mol of gas, both stated correctly
(a) Answer: Reactant side: 4 mol of gas (1 mol N2 + 3 mol H2). Product side: 2 mol of gas (2 mol NH3).
(b) B1 the position of equilibrium shifts to the right/towards the products (ammonia)
(b) B1 because this is the side with fewer moles of gas, so shifting this way reduces the pressure, opposing the increase oe
(b) Answer: The equilibrium shifts to the right, towards ammonia, because that is the side with fewer moles of gas, which reduces the pressure and opposes the increase.
(c) B1 the percentage yield increases
(c) Answer: The percentage yield of ammonia increases.

Question 11

(a) B1 reactant side = 3 mol (2 + 1); product side = 2 mol, both stated correctly
(a) Answer: Reactant side: 3 mol of gas (2 mol SO2 + 1 mol O2). Product side: 2 mol of gas (2 mol SO3).
(b) B1 the yield of SO3 increases (equilibrium shifts to the right)
(b) B1 because the product side has fewer moles of gas, so increasing pressure shifts equilibrium towards that side, opposing the increase in pressure oe
(b) Answer: The yield of SO3 increases, because the product side has fewer moles of gas, so the equilibrium shifts towards the products to oppose the increase in pressure.
(c) B1 the yield of SO3 decreases (equilibrium shifts to the left)
(c) B1 because the forward reaction is exothermic, so increasing temperature favours the endothermic reverse reaction, opposing the increase in temperature oe
(c) Answer: The yield of SO3 decreases, because the forward reaction is exothermic, so raising the temperature favours the endothermic reverse reaction instead.

Question 12

Level 1 (1-2): Simple, isolated statement(s) are made about temperature, pressure or the catalyst, with little or no link to rate, yield or cost.
Level 2 (3-4): Some explanation is given of how temperature and/or pressure affect rate and yield, with limited or no reference to the catalyst, or to the idea of an overall economic compromise.
Level 3 (5-6): A clear, logically linked explanation is given, covering temperature, pressure and the catalyst, correctly relating each to rate and/or yield, and clearly explaining why a compromise is chosen for economic reasons.
Indicative content:
- The forward reaction is exothermic, so a lower temperature would give a higher equilibrium yield of ammonia.
- However, a lower temperature gives a slower rate of reaction, so equilibrium would take too long to reach to be economical.
- 450 degrees C is a compromise: it gives a fast enough rate while still achieving a reasonably good yield.
- Increasing pressure shifts the equilibrium towards the side with fewer gas molecules (the ammonia side), increasing the yield.
- However, very high pressures need stronger, more expensive reaction vessels and pipework, cost more energy to generate, and increase safety risk (of leaks or explosion).
- 200 atmospheres is a compromise between a higher yield and the cost/safety of achieving very high pressure.
- The iron catalyst increases the rate of both the forward and reverse reactions equally, so it does not change the position of equilibrium or the yield of ammonia.
- Using a catalyst means a lower temperature can be used than would otherwise be needed for a reasonable rate, saving energy costs without a large loss of yield.
- Overall, the conditions chosen balance the cost of achieving them against the value of the ammonia produced and the rate of production, in order to maximise profit.

Question 13

(a) B1 40 s
(a) Answer: 40 s.
(b) B1 both concentrations become constant (the lines are horizontal) from t = 40 s onwards oe
(b) B1 at dynamic equilibrium the forward and reverse reactions are still happening, at equal rates, rather than the reaction having stopped completely oe
(b) Answer: Both concentration lines become horizontal (constant) from t = 40 s onwards. This shows dynamic equilibrium, where the forward and reverse reactions continue to occur at equal rates, rather than the reaction having stopped completely.
(c) B1 dynamic equilibrium only requires the forward and reverse rates to be equal, not the concentrations of reactant and product, which depend on the position of equilibrium for this reaction oe
(c) Answer: Dynamic equilibrium only requires the forward and reverse rates to be equal; the actual concentrations of X and Y depend on the position of equilibrium for this particular reaction, so they need not be equal.

Question 14

(a) B1 the yield of ammonia increases, because the forward reaction is exothermic and the equilibrium shifts to oppose the decrease in temperature by favouring the exothermic (forward) reaction oe
(a) B1 the rate of reaching (a new) equilibrium decreases (is slower), because a lower temperature means a lower rate of reaction oe
(a) Answer: The yield of ammonia increases, since the forward reaction is exothermic and the equilibrium shifts to oppose the temperature decrease by favouring the forward reaction. However, the rate of reaching a new equilibrium decreases, because lower temperature means a slower rate of reaction.
(b) B1 the yield of ammonia increases further
(b) B1 because increasing pressure shifts the equilibrium further towards the side with fewer gas molecules (ammonia), opposing the increase in pressure oe
(b) Answer: The yield of ammonia increases further, because increasing the pressure shifts the equilibrium even more towards the ammonia side (fewer gas molecules), opposing the pressure increase.
(c) B1 using the lowest possible temperature would make the rate of reaction far too slow to be economical, even though the equilibrium yield would be higher oe
(c) B1 using extremely high pressure would require very strong, expensive equipment and higher running (energy/safety) costs, which would outweigh the benefit of the extra yield oe
(c) Answer: The lowest possible temperature would give too slow a rate of reaction to be economical, and extremely high pressure would require very strong, expensive equipment and much higher running and safety costs, outweighing the benefit of the extra yield.

Question 15

B1 a catalyst increases the rate of both the forward and the reverse reactions by the same factor (equally) oe
B1 so it does not change the position of equilibrium, and therefore does not change the equilibrium yield of ammonia oe
B1 the claim is incorrect: the catalyst only allows equilibrium to be reached more quickly (in less time); it does not increase how much ammonia is present at equilibrium oe
Answer: The claim is incorrect. An iron catalyst speeds up the forward and reverse reactions equally, so it does not shift the position of equilibrium or change the equilibrium yield of ammonia; it only allows equilibrium to be reached faster.

Question 16

(a) B1 exothermic; 120 kJ (per mole of water) is released/transferred to the surroundings oe
(a) Answer: The reverse reaction is exothermic, releasing 120 kJ per mole of water added.
(b) B1 it is the same diagram reversed (inverted): the reactants and products swap places, so the energy level that was the products' level becomes the reactants' level, and vice versa oe
(b) Answer: The reverse reaction profile is the forward profile reversed: the reactants and products swap energy levels, so what was the lower (product) level becomes the higher (reactant) level.
(c) B1 the law of conservation of energy
(c) Answer: The law of conservation of energy.

Question 17

(a) B1 True
(a) Answer: True.
(b) B1 False (they continue to occur, at equal rates) oe
(b) Answer: False: the forward and reverse reactions continue to occur, at equal rates.
(c) B1 False
(c) Answer: False.
(d) B1 True
(d) Answer: True.

Question 18

(a) M1 moles N2 reacted = 1.5 / 2 = 0.75 mol, using the ratio N2 : NH3 = 1 : 2
(a) A1 moles H2 reacted = 0.75 x 3 = 2.25 mol cao, using the ratio H2 : NH3 = 3 : 2
(a) Answer: N2 reacted = 0.75 mol; H2 reacted = 2.25 mol
(b) M1 moles N2 at equilibrium = 4.0 - 0.75 = 3.25 mol (allow ft from candidate's answer to part a)
(b) A1 moles H2 at equilibrium = 9.0 - 2.25 = 6.75 mol cao (allow ft from candidate's answer to part a)
(b) Answer: N2 at equilibrium = 3.25 mol; H2 at equilibrium = 6.75 mol
(c) B1 the reaction reaches dynamic equilibrium in this closed system, at which point the forward and reverse reactions occur at equal rates, so the reaction does not go to completion and some N2 and H2 remain oe
(c) Answer: The reaction reaches dynamic equilibrium in the closed container: the forward and reverse reactions continue at equal rates, so the reaction never goes to completion and some nitrogen and hydrogen always remain.

Reversible Reactions and Equilibrium - Worksheets, Questions and Revision