Question 22M.2.hl.TZ1.c(iv)
Date | May 2022 | Marks available | [Maximum mark: 1] | Reference code | 22M.2.hl.TZ1.c(iv) |
Level | hl | Paper | 2 | Time zone | TZ1 |
Command term | Outline | Question number | c(iv) | 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.
Outline, with reference to the reaction equation, why this sign for the entropy change is expected.
[1]
«forward reaction involves» decrease in number of moles «of gas» ✔
Average performance for sign of the entropy change expected for the reaction. Some answers were based on ΔG value rather than in terms of decrease in number of moles of gas or had no idea how to address the question.
