Question 23M.2.HL.TZ2.2a
| Date | May 2023 | Marks available | [Maximum mark: 5] | Reference code | 23M.2.HL.TZ2.2a |
| Level | HL | Paper | 2 | Time zone | TZ2 |
| Command term | Calculate, Outline | Question number | a | Adapted from | N/A |
Nitrogen (IV) oxide exists in equilibrium with dinitrogen tetroxide, N2O4 (g), which is colourless.
2NO2 (g) ⇌ N2O4 (g)
At 100 °C Kc for this reaction is 0.0665. Outline what this indicates about the extent of this reaction.
[1]
reaction hardly proceeds
OR
reverse reaction/formation of NO2 is favoured
OR
«concentration of» reactants greater than «concentration of» products «at equilibrium» ✓
Accept equilibrium lies to the left.
Calculate the Gibbs free energy change, ΔG⦵, for this equilibrium at 100 °C. Use sections 1 and 2 of the data booklet.
[1]
ΔG⦵ = «−RTlnK = −8.31 × 373 × ln(0.0665) =»
8.40 «kJ mol−1» ✓
Calculate the value of Kc at 100 °C for the equilibrium:
N2O4 (g) ⇌ 2NO2 (g)
[1]
«Kc = =» 15.0 ✓
Calculate the standard enthalpy change, in kJ mol−1, for the reaction:
N2O4 (g) → 2NO2 (g)
| ΔH⦵f (kJ mol−1) | |
| NO2 |
33.18 |
| N2O4 |
9.16 |
[1]
« ΔH⦵ = 2(33.18) − 9.16 =» «+» 57.20 «kJ mol−1» ✓
Calculate the standard entropy change, in J mol−1, for the reaction:
N2O4 (g) → 2NO2 (g)
| S⦵ (J mol−1) | |
| NO2 |
240.06 |
| N2O4 |
304.29 |
[1]
« ΔS⦵ = 2(240.06) − 304.29 =»
«+»175.83 «J K−1 mol−1» ✓