Question 22M.2.hl.TZ1.b(i)
| Date | May 2022 | Marks available | [Maximum mark: 3] | Reference code | 22M.2.hl.TZ1.b(i) |
| Level | hl | Paper | 2 | Time zone | TZ1 |
| Command term | Determine | Question number | b(i) | Adapted from | N/A |
Ammonia is produced by the Haber–Bosch process which involves the equilibrium:
N2 (g) + 3 H2 (g) 2 NH3 (g)
The percentage of ammonia at equilibrium under various conditions is shown:
[The Haber Bosch Process [graph] Available at: https://commons.wikimedia.org/wiki/File:Ammonia_yield.png
[Accessed: 16/07/2022].]
One factor affecting the position of equilibrium is the enthalpy change of the reaction.
Determine the enthalpy change, ΔH, for the Haber–Bosch process, in kJ. Use Section 11 of the data booklet.
[3]
bonds broken: N≡N + 3(H-H) /«1 mol×»945 «kJ mol–1» + 3«mol»×436 «kJ mol–1» / 945 «kJ» + 1308 «kJ» / 2253 «kJ» ✔
bonds formed: 6(N-H) / 6«mol»×391 «kJ mol–1» / 2346 «kJ» ✔
ΔH = «2253 kJ - 2346 kJ = » -93 «kJ» ✔
Award [2 max] for (+)93 «kJ».
Good performance; often the bond energy for single N–N bond instead of using it for the triple bond and not taking into consideration the coefficient of three in calculation of bond enthalpies of ammonia. Also, instead of using BE of bonds broken minus those that were formed, the operation was often reversed. Students should be encouraged to draw the Lewis structures in the equations first to determine the bonds being broken and formed.
