Questions on dynamic equilibrium, Le Chatelier's principle and industrial compromise

Question 1

Explain why the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹, is run at about 450 °C and 200 atm over an iron catalyst with the unreacted gases recycled, with reference to equilibrium yield, rate and cost.

Accepted Answer

Indicative content

the forward reaction is exothermic, so a low temperature would shift the position of equilibrium to the right and give a higher equilibrium yield of ammonia
at a low temperature too few collisions have energy equal to or greater than the activation energy, even on the catalyst, so the rate and the output per hour would be far too low
about 450 °C is therefore a compromise between equilibrium yield and rate, accepting a lower yield to make ammonia quickly
four moles of gas become two, so a high pressure shifts the position to the right, towards fewer moles of gas, and also raises the rate by raising the concentrations
a very high pressure needs much more energy for compression and thicker, more expensive pipes and vessels, with a greater hazard from leaks, so about 200 atm balances yield against cost and safety
the iron catalyst raises the forward and reverse rates equally without changing the yield, and recycling the unreacted nitrogen and hydrogen makes about 98% of them into ammonia overall although each pass converts only about 15%

Levels

Level 3 (5–6 marks) all three conditions are explained, each as a compromise: the temperature between equilibrium yield and rate for the exothermic reaction, the pressure between yield and the cost and hazard of compression, and the catalyst with the recycling that raises the overall conversion
Level 2 (3–4 marks) temperature and pressure are explained as compromises through Le Chatelier's principle; the catalyst or the recycling is not explained, or one compromise is argued from one side only
Level 1 (1–2 marks) a low temperature or a high pressure is said to favour ammonia; the catalyst is commonly said to raise the yield, and nothing is weighed against cost or rate

Question 2

Explain, using Le Chatelier's principle, why raising the pressure increases the equilibrium yield of SO₃ in 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) but has no effect on the equilibrium yield of HI in H₂(g) + I₂(g) ⇌ 2HI(g).

Accepted Answer

B1 if a system at equilibrium is subjected to a change in conditions, the position of equilibrium shifts in the direction that opposes the change B1 for SO₃ there are three moles of gas on the left and two on the right, so the position shifts right, to fewer moles of gas, which lowers the pressure; allow 'fewer gas molecules' B1 for HI there are two moles of gas on each side of the equation B1 a shift in either direction would not change the number of gas molecules, so it could not oppose the change in pressure and the yield of HI is unchanged

Question 3

Explain what is seen when the plunger of a sealed syringe containing an equilibrium mixture of N₂O₄ and NO₂ is pulled out quickly and held. N₂O₄(g) ⇌ 2NO₂(g); N₂O₄ is colourless and NO₂ is brown.

Accepted Answer

B1 the brown colour pales at once, because the same molecules now occupy a larger volume, so the concentration of NO₂ falls B1 the pressure has decreased, so the position of equilibrium shifts to the right, the side with more moles of gas B1 N₂O₄ molecules dissociate into NO₂, which raises the pressure, so the brown colour then deepens again B1 the final colour is still paler than at the start, because the shift only partly opposes the change

Question 4

Explain why adding a catalyst to a reaction at equilibrium does not change the position of equilibrium, and explain how a catalyst active at a lower temperature than iron could reduce the energy used by a Haber process plant.

Accepted Answer

B1 the catalyst lowers the activation energy of the forward and the reverse reaction by the same amount B1 both rates increase by the same factor and stay equal, so the concentrations do not change; equilibrium is only reached sooner B1 the converter could run at a lower temperature, where the exothermic forward reaction gives a higher equilibrium yield on each pass, at the same rate B1 less unreacted gas would be recycled and recompressed, or a lower pressure could be used, so less energy is spent on compression

Question 5

Evaluate a proposal to raise the pressure in a methanol plant from its usual 50 to 100 atm to 300 atm. The reaction is CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH = −91 kJ mol⁻¹, run at about 250 °C.

Accepted Answer

B1 three moles of gas become one, so a higher pressure shifts the position to the right and raises the equilibrium yield B1 a higher pressure also increases the concentration of the gases and so the rate B1 against this, compression needs much more energy and the plant needs thicker, more expensive vessels and pipes, with a greater hazard from leaks B1 a judgement: the proposal is justified only if the value of the extra methanol exceeds the extra energy and plant cost, and with recycling the gain is likely to be small; allow either conclusion if it follows from the points made

Question 6

Compare the conditions chosen for the Haber process with those for the Contact process step 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), giving a reason for each difference.

Accepted Answer

B1 both run at about 450 °C as a compromise between yield and rate for an exothermic forward reaction B1 the Haber process uses about 200 atm, because its equilibrium yield at low pressure is too small B1 the Contact process uses only 1 to 2 atm, because the conversion over several catalyst beds is already above 99%, so compression would not pay for itself B1 the Haber process uses an iron catalyst and recycles its unreacted gas; the Contact process uses vanadium(V) oxide, excess air and cooling between the beds

Question 7

Deduce, from the percentages of ammonia at equilibrium given in the table, the sign of ΔH for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), and explain the effect of pressure shown by the data.

Accepted Answer

B1 at every pressure the percentage of ammonia falls as the temperature rises B1 heating shifts the position to the left, which is the endothermic direction, so the forward reaction is exothermic and ΔH is negative B1 at every temperature the percentage of ammonia rises as the pressure rises B1 four moles of gas become two, so a higher pressure shifts the position to the right, the side with fewer moles of gas

Question 8

Give the temperature, the pressure and the catalyst used in the industrial hydration of ethene to ethanol.

Accepted Answer

B1 a temperature of about 300 °C; allow 250 to 350 °C B1 a pressure of about 60 to 70 atm; allow 6 to 7 MPa B1 a catalyst of phosphoric acid on a silica support; accept phosphoric(V) acid or H₃PO₄

Question 9

Describe how the rates of the forward and reverse reactions change from the moment hydrogen and iodine are sealed in a hot flask until equilibrium is reached.

Accepted Answer

B1 at the start the forward rate is at its highest, because hydrogen and iodine are at their highest concentrations, and the reverse rate is zero B1 as hydrogen and iodine are used up the forward rate falls, and as hydrogen iodide builds up the reverse rate rises B1 equilibrium is reached when the two rates become equal, after which the concentrations no longer change

Question 10

Explain why calcium carbonate heated in a sealed container reaches equilibrium, CaCO₃(s) ⇌ CaO(s) + CO₂(g), but calcium carbonate heated in an open lime kiln decomposes completely.

Accepted Answer

B1 in the sealed container the carbon dioxide cannot escape, so it builds up and the reverse reaction with calcium oxide takes place B1 when the reverse rate equals the forward rate a dynamic equilibrium is set up; this needs a closed system B1 in the open kiln the carbon dioxide is carried away, so the reverse reaction cannot occur and the forward reaction continues until all the calcium carbonate has decomposed

Question 11

Suggest the effect on the equilibrium yield of hydrogen of raising the pressure and of raising the temperature for CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), ΔH = −41 kJ mol⁻¹, and suggest one reason for using a catalyst.

Accepted Answer

B1 raising the pressure has no effect on the yield, because there are two moles of gas on each side B1 raising the temperature lowers the yield, because the forward reaction is exothermic and the position shifts in the endothermic direction, to the left B1 a catalyst allows a lower temperature, which favours the yield, while keeping the rate high enough

Question 12

Suggest the sign of ΔH for the forward reaction in the equilibrium pink [Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ blue [CoCl₄]²⁻(aq) + 6H₂O(l), given that the mixture turns blue when heated, and suggest the colour seen when concentrated hydrochloric acid is added and when water is added.

Accepted Answer

B1 ΔH is positive: heating shifts the position to the right, which is the endothermic direction B1 adding concentrated hydrochloric acid raises the chloride concentration, shifting the position right, so the solution turns blue B1 adding water dilutes the solution and lowers the chloride concentration, so the position shifts to the left and the solution turns pink

Question 13

Calculate the mass of ammonia formed on one pass through a Haber converter, using the feed and conversion data given in the table.

Accepted Answer

M1 nitrogen converted = 0.150 × 600 = 90.0 mol M1 ammonia formed = 2 × 90.0 = 180 mol, from the 1 : 2 ratio in N₂ + 3H₂ → 2NH₃ A1 mass = 180 × 17.0 = 3060 g, which is 3.06 kg to three significant figures; error carried forward from M2

Question 14

Calculate the amounts of nitrogen and hydrogen in 1.00 × 10⁴ mol of a Haber equilibrium mixture that contains 27.0% ammonia by moles, the rest being nitrogen and hydrogen in the ratio 1 : 3.

Accepted Answer

M1 ammonia = 0.270 × 1.00 × 10⁴ = 2700 mol, so nitrogen and hydrogen together = 7300 mol M1 nitrogen is one quarter of this and hydrogen three quarters, from the 1 : 3 ratio A1 nitrogen = 1830 mol and hydrogen = 5480 mol, to three significant figures; accept 1825 and 5475 mol

Question 15

Outline how ammonia is separated from the gas leaving a Haber converter, and why the conversion of only about 15% per pass is acceptable.

Accepted Answer

B1 the gas mixture is cooled, and ammonia condenses to a liquid while nitrogen and hydrogen remain as gases B1 the liquid ammonia is run off and the unreacted nitrogen and hydrogen are recycled to the converter B1 over many passes about 98% of the reactants are converted, so little is wasted despite the low yield per pass

Question 16

Explain how the colours seen when tubes of a purple solution of the equilibrium [Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l) are put in hot water and in iced water show that the forward reaction is endothermic. [Co(H₂O)₆]²⁺ is pink and [CoCl₄]²⁻ is blue.

Accepted Answer

B1 in hot water the solution turns blue, so the position has shifted to the right B1 in iced water the solution turns pink, so the position has shifted to the left B1 a rise in temperature shifts the position in the endothermic direction, so the forward reaction, favoured by heating, is endothermic

Question 17

Describe how three boiling tubes of a purple cobalt(II) chloride solution could be used to show the effect of temperature on the position of an equilibrium.

Accepted Answer

B1 one tube is placed in hot water at about 80 °C and one in iced water B1 the third tube is left at room temperature as a control, against which the colours of the other two are compared B1 the hot tube turns blue and the cold tube pink, and both return to purple at room temperature, showing a reversible shift in the position of equilibrium

Question 18

State two features of a reversible reaction that has reached dynamic equilibrium.

Accepted Answer

B1 the forward and reverse reactions are both still taking place, at equal rates B1 the concentrations of the reactants and products remain constant; reject 'equal concentrations'

Temperature / °C	100 atm	200 atm	400 atm
300	52%	63%	76%
400	25%	39%	55%
500	11%	18%	32%

Quantity	Value
Nitrogen fed in	600 mol
Hydrogen fed in	1800 mol
Nitrogen converted on one pass	15.0%
Mr of NH₃	17.0