Question 19M.2.sl.TZ1.3
| Date | May 2019 | Marks available | [Maximum mark: 13] | Reference code | 19M.2.sl.TZ1.3 |
| Level | sl | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate, Describe, Determine, Outline, State, Suggest, Write | Question number | 3 | Adapted from | N/A |
This question is about compounds of sodium.
Describe the structure and bonding in solid sodium oxide.
[2]
«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.
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.
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]
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 [✔]
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.
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]
n(Na2O2) theoretical yield «= » = 0.0807/8.07 × 10−2 «mol»
OR
mass Na2O2 theoretical yield «= × 77.98 gmol−1» = 6.291 «g» [✔]
% yield «= × 100» OR « x 100» = 87.4 «%» [✔]
Note: Award [2] for correct final answer.
Many candidates were able to correctly calculate the % yield but some weaker candidates just used 5.0/5.5 to find %.
Sodium peroxide is used in diving apparatus to produce oxygen from carbon dioxide.
2Na2O2 (s) + 2CO2 (g) → 2Na2CO3 (s) + O2 (g)
Determine the enthalpy change, ΔH, in kJ, for this reaction using data from the table and section 12 of the data booklet.
[3]
ΣΔHf products = 2 × (−1130.7) / −2261.4 «kJ» [✔]
ΣΔHf reactants = 2 × (−510.9) + 2 × (−393.5) / −1808.8 «kJ» [✔]
ΔH = «ΣΔHf products − ΣΔHf reactants = −2261.4 −(−1808.8) =» −452.6 «kJ» [✔]
Note: Award [3] for correct final answer.
Award [2 max] for “+452.6 «kJ»”.
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.
Outline why bond enthalpy values are not valid in calculations such as that in (c)(i).
[1]
only valid for covalent bonds
OR
only valid in gaseous state [✔]
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.
The reaction of sodium peroxide with excess water produces hydrogen peroxide and one other sodium compound. Suggest the formula of this compound.
[1]
NaOH [✔]
Note: Accept correct equation showing NaOH as a product.
Some students correctly identified sodium hydroxide as the correct product, but hydrogen, oxygen and sodium oxide were common answers.
State the oxidation number of carbon in sodium carbonate, Na2CO3.
[1]
IV [✔]
Oxidation number of +4 was often correctly identified.