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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	Formulate	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.

ΔH^⦵_f PCl₃(g) = −306.4 kJ mol⁻¹

ΔH^⦵_f PCl₅(g) = −398.9 kJ mol⁻¹

[1]

c(i).

Calculate the entropy change, ΔS, in J K⁻¹mol⁻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⁻¹K⁻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, K_c, 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^–1mol^–1 – (311.7 J K^–1mol^–1 + 223.0 J K^–1mol^–1)=» –170.2 «J K^–1mol^–1» ✔

c(ii).

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

«ΔG = –92.5 kJ mol^–1 – (298 K × –0.1702 kJ mol^–1K^–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 = $- \frac{8.31 J {mol}^{- 1} K^{- 1}}{1000}$ × 298 K × lnK»
OR
«ΔG = –41800 J mol^–1 = –8.31 J mol^–1K^–1 × 298 K × lnK»

«lnK = =» 16.9 ✔

«K = e^16.9 =» 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).

«K_c =» $\frac{[{PCl}_{5}]}{[{PCl}_{3}] [{Cl}_{2}]}$ ✔

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 1: Stoichiometric relationships » 1.1 Introduction to the particulate nature of matter and chemical change

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