Question 22M.2.hl.TZ1.c(iii)
| Date | May 2022 | Marks available | [Maximum mark: 2] | Reference code | 22M.2.hl.TZ1.c(iii) |
| Level | hl | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate | Question number | c(iii) | 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].]
The standard free energy change, ΔG⦵, for the Haber–Bosch process is –33.0 kJ at 298 K.
Calculate the entropy change for the Haber–Bosch process, in J mol–1 K–1 at 298 K. Use your answer to (b)(i) and section 1 of the data booklet.
[2]
ΔG = «ΔH – TΔS =» –93000 «J» – 298«K» × ΔS = –33000 ✔
ΔS = 〈〈〉〉 = –201 «J mol–1 K–1» ✔
Do not penalize failure to convert kJ to J in both (c)(ii) and (c)(iii).
Award [2] for correct final answer
Award [1 max] for (+) 201 «J mol–1 K–1».
Award [2] for –101 or –100.5 «J mol–1 K–1».
Very good performance; since the unit for S is J mol˗1 K˗1, ΔG and ΔH needed to be converted from kJ to J, but that was not done in some cases.
