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Date	May 2019	Marks available	1	Reference code	19M.2.sl.TZ1.3
Level	SL	Paper	2	Time zone	TZ1
Command term	Suggest	Question number	3	Adapted from	N/A

Question

This question is about compounds of sodium.

Sodium peroxide is used in diving apparatus to produce oxygen from carbon dioxide.

2Na₂O₂ (s) + 2CO₂ (g) → 2Na₂CO₃ (s) + O₂ (g)

Describe the structure and bonding in solid sodium oxide.

[2]

a(i).

Write equations for the separate reactions of solid sodium oxide and solid phosphorus(V) oxide with excess water and differentiate between the solutions formed.

Sodium oxide, Na₂O:

Phosphorus(V) oxide, P₄O₁₀:

Differentiation:

[3]

a(ii).

Sodium peroxide, Na₂O₂, is formed by the reaction of sodium oxide with oxygen.

2Na₂O (s) + O₂ (g) → 2Na₂O₂ (s)

Calculate the percentage yield of sodium peroxide if 5.00 g of sodium oxide produces 5.50 g of sodium peroxide.

[2]

Determine the enthalpy change, ΔH, in kJ, for this reaction using data from the table and section 12 of the data booklet.

[3]

c(i).

Outline why bond enthalpy values are not valid in calculations such as that in (c)(i).

[1]

c(ii).

The reaction of sodium peroxide with excess water produces hydrogen peroxide and one other sodium compound. Suggest the formula of this compound.

[1]

State the oxidation number of carbon in sodium carbonate, Na₂CO₃.

[1]

Markscheme

«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions» [✔]

electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Na⁺ and O²⁻ ions [✔]

Note: Do not accept “ionic” without description.

a(i).

Sodium oxide:
Na₂O(s) + H₂O(l) → 2NaOH (aq) [✔]

Phosphorus(V) oxide:
P₄O₁₀ (s) + 6H₂O(l) → 4H₃PO₄ (aq) [✔]

Differentiation:
NaOH / product of Na₂O is alkaline/basic/pH > 7 AND H₃PO₄ / product of P₄O₁₀ is acidic/pH < 7 [✔]

a(ii).

n(Na₂O₂) theoretical yield «= $\frac{{5.00{\text{ g}}}}{{61.98{\text{ mo}}{{\text{l}}^{ - 1}}}}$ » = 0.0807/8.07 × 10⁻² «mol»
OR
mass Na₂O₂ theoretical yield «= $\frac{{5.00{\text{ g}}}}{{61.98{\text{ mo}}{{\text{l}}^{ - 1}}}}$ × 77.98 gmol⁻¹» = 6.291 «g» [✔]

% yield «= $\frac{{5.50{\text{ g}}}}{{6.291{\text{ g}}}}$ × 100» OR « $\frac{{0.0705}}{{0.0807}}$ x 100» = 87.4 «%» [✔]

Note: Award [2] for correct final answer.

ΣΔH_f products = 2 × (−1130.7) / −2261.4 «kJ» [✔]

ΣΔH_f reactants = 2 × (−510.9) + 2 × (−393.5) / −1808.8 «kJ» [✔]

ΔH = «ΣΔH_f products − ΣΔH_f reactants = −2261.4 −(−1808.8) =» −452.6 «kJ» [✔]

Note: Award [3] for correct final answer.

Award [2 max] for “+452.6 «kJ»”.

c(i).

only valid for covalent bonds
OR
only valid in gaseous state [✔]

c(ii).

NaOH [✔]

Note: Accept correct equation showing NaOH as a product.

IV [✔]

Examiners report

Disappointingly many students did not realise that sodium oxide was held by ionic bonds, many said it was covalent or metallic bonding. The ones that knew it was ionic failed to describe it adequately to earn the 2 marks.

a(i).

Very few students could correctly write out the two equations and so often were unable to realise it was acid/base behaviour that would differentiate the oxides.

a(ii).

Many candidates were able to correctly calculate the % yield but some weaker candidates just used 5.0/5.5 to find %.

The calculation of the enthalpy change using enthalpies of formation was generally answered well but common mistakes were students forgetting to multiply by 2 or adding extra terms for oxygen.

c(i).

Most students didn’t gain a mark and “values are average” was the most common incorrect answer. The fact this was an ionic compound did not register with them. Some students did gain a mark for stating that the substances were not in a gaseous state.

c(ii).

Some students correctly identified sodium hydroxide as the correct product, but hydrogen, oxygen and sodium oxide were common answers.

Oxidation number of +4 was often correctly identified.

Syllabus sections

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

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