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Date November 2021 Marks available 1 Reference code 21N.2.hl.TZ0.3
Level HL Paper 2 Time zone TZ0
Command term Calculate Question number 3 Adapted from N/A

Question

White phosphorus is an allotrope of phosphorus and exists as P4.

An equilibrium exists between PCl3 and PCl5.

PCl3 (g) + Cl2 (g) PCl5 (g)

Sketch the Lewis (electron dot) structure of the P4 molecule, containing only single bonds.

 

[1]
a(i).

Write an equation for the reaction of white phosphorus (P4) with chlorine gas to form phosphorus trichloride (PCl3).

[1]
a(ii).

Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl3.

[3]
b(i).

Outline the reason why PCl5 is a non-polar molecule, while PCl4F is polar.

[3]
b(ii).

Calculate the standard enthalpy change (ΔH) for the forward reaction in kJ mol−1.

ΔHf PCl3 (g) = −306.4 kJ mol−1

ΔHf PCl5 (g) = −398.9 kJ mol−1

[1]
c(i).

Calculate the entropy change, ΔS, in J K−1 mol−1, for this reaction.

 

Chemistry 2e, Chpt. 21 Nuclear Chemistry, Appendix G: Standard Thermodynamic Properties for Selected Substances https://openstax.org/books/chemistry-2e/pages/g-standard-thermodynamic-properties-for- selectedsubstances# page_667adccf-f900-4d86-a13d-409c014086ea © 1999-2021, Rice University. Except where otherwise noted, textbooks on this site are licensed under a Creative Commons Attribution 4.0 International License. (CC BY 4.0) https://creativecommons.org/licenses/by/4.0/.

[1]
c(ii).

Calculate the Gibbs free energy change (ΔG), in kJ mol−1, for this reaction at 25 °C. Use section 1 of the data booklet.

If you did not obtain an answer in c(i) or c(ii) use −87.6 kJ mol−1 and −150.5 J mol−1 K−1 respectively, but these are not the correct answers.

[2]
c(iii).

Determine the equilibrium constant, K, for this reaction at 25 °C, referring to section 1 of the data booklet.

If you did not obtain an answer in (c)(iii), use ΔG = –43.5 kJ mol−1, but this is not the correct answer.

[2]
c(iv).

State the equilibrium constant expression, Kc, for this reaction.

[1]
c(v).

State, with a reason, the effect of an increase in temperature on the position of this equilibrium.

[1]
c(vi).

Markscheme

Accept any diagram with each P joined to the other three.
Accept any combination of dots, crosses and lines.

a(i).

P4 (s) + 6Cl2 (g) → 4PCl3 (l) ✔

a(ii).

Electron domain geometry: tetrahedral ✔

Molecular geometry: trigonal pyramidal ✔

Bond angle: 100«°» ✔


Accept any value or range within the range 91−108«°» for M3.

b(i).

PCl5 is non-polar:

symmetrical
OR
dipoles cancel ✔

 

PCl4F is polar:

P–Cl has a different bond polarity than P–F ✔

non-symmetrical «dipoles»
OR
dipoles do not cancel ✔


Accept F more electronegative than/different electronegativity to Cl for M2.

b(ii).

«−398.9 kJ mol−1 − (−306.4 kJ mol−1) =» −92.5 «kJ mol−1» ✔

c(i).

«ΔS = 364.5 J K–1 mol–1 – (311.7 J K–1 mol–1 + 223.0 J K–1 mol–1)=» –170.2 «J K–1 mol–1» ✔

c(ii).

«ΔS =» –0.1702 «kJ mol–1 K–1»
OR
298 «K» ✔

«ΔG = –92.5 kJ mol–1 – (298 K × –0.1702 kJ mol–1 K–1) =» –41.8 «kJ mol–1» ✔

 

Award [2] for correct final answer.

If –87.6 and -150.5 are used then –42.8.

c(iii).

«ΔG = –41.8 kJ mol–1 = -8.31Jmol-1K-11000 × 298 K × lnK»
OR
«ΔG = –41800 J mol–1 = –8.31 J mol–1 K–1 × 298 K × lnK»

«lnK = =» 16.9 ✔

«K = e16.9 =» 2.19 × 107

 

Award [2] for correct final answer.

Accept range of 1.80 × 106–2.60 × 107.

If –43.5 is used then 4.25 × 107.

c(iv).

«Kc =» PCl5PCl3Cl2

c(v).

«shifts» left/towards reactants AND «forward reaction is» exothermic/ΔH is negative ✔

c(vi).

Examiners report

[N/A]
a(i).
[N/A]
a(ii).
[N/A]
b(i).
[N/A]
b(ii).
[N/A]
c(i).
[N/A]
c(ii).
[N/A]
c(iii).
[N/A]
c(iv).
[N/A]
c(v).
[N/A]
c(vi).

Syllabus sections

Core » Topic 5: Energetics/thermochemistry » 5.2 Hess’s Law
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