DP Chemistry (last assessment 2024)

Question 19M.2.hl.TZ1.3

Date	May 2019	Marks available	[Maximum mark: 22]	Reference code	19M.2.hl.TZ1.3
Level	hl	Paper	2	Time zone	TZ1
Command term	Calculate, Compare, Describe, Determine, Draw, Justify, Outline, State, Suggest, Write	Question number	3	Adapted from	N/A

[Maximum mark: 22]

19M.2.hl.TZ1.3

This question is about sodium and its compounds.

(a)

Plot the relative values of the first four ionization energies of sodium.

[1]

Markscheme

[✔]

Notes: Accept curve showing general trend.
Award mark only if the energy difference between the first two points is larger than that between points 2/3 and 3/4.

Examiners report

Generally well done with a correct plot of ionization energies.

(b)

Outline why the alkali metals (group 1) have similar chemical properties.

[1]

Markscheme

same number of electrons in outer shell

all are s¹ [✔]

Examiners report

The majority answered correctly stating same number of valence electrons as the reason. Some candidates stated same size or similar ionization energy but the majority scored well.

(c)

Describe the structure and bonding in solid sodium oxide.

[2]

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.

Examiners report

Many candidates lost one or two marks for missing “electrostatic forces” between “oppositely charged ions”, or “lattice”. Some candidates’ answers referred to covalent bonds and shapes of molecules.

The Born-Haber cycle for sodium oxide is shown (not to scale).

(d(i))

Calculate values for the following changes using section 8 of the data booklet.

ΔH_atomisation (Na) = 107 kJ mol⁻¹
ΔH_atomisation (O) = 249 kJ mol⁻¹

$\frac{1}{2}$ O₂(g) → O^2-(g):

Na (s) → Na⁺ (g):

[2]

Markscheme

$\frac{1}{2}$ O₂(g) → O^2-(g)

«ΔH_atomisation (O) + 1st EA + 2nd EA = 249 k Jmol⁻¹ − 141 kJmol⁻¹ + 753 kJmol⁻¹ =» «+»861 «kJmol⁻¹» [✔]

Na (s) → Na⁺ (g)

«ΔH_atomisation (Na) + 1st IE = 107 kJmol⁻¹ + 496 kJmol⁻¹ =» «+»603 «kJmol⁻¹» [✔]

Examiners report

Good performance with typical error being in the calculation for the first equation, ½O2 (g) → O₂⁻ (g), where the value for the first electron affinity of oxygen was left out.

(d(ii))

The standard enthalpy of formation of sodium oxide is −414 kJ mol⁻¹. Determine the lattice enthalpy of sodium oxide, in kJ mol⁻¹, using section 8 of the data booklet and your answers to (d)(i).

(If you did not get answers to (d)(i), use +850 kJ mol⁻¹ and +600 kJ mol⁻¹ respectively, but these are not the correct answers.)

[2]

Markscheme

lattice enthalpy = 861 «kJ mol⁻¹» + 2 × 603 «kJ mol⁻¹» −(−414 «kJ mol⁻¹») [✔]

«= +» 2481 «kJ mol⁻¹» [✔]

Note: Award [2] for correct final answer.

If given values are used:
M1: lattice enthalpy = 850 «kJ mol⁻¹» +
2 × 600 «kJ mol⁻¹» −(−414 «kJ mol⁻¹»)
M2: «= +» 2464 «kJ mol⁻¹»

Examiners report

Many candidates earned some credit for ECF based on (d)(i).

(d(iii))

Justify why K₂O has a lower lattice enthalpy (absolute value) than Na₂O.

[1]

Markscheme

K⁺ ion is larger than Na⁺

smaller attractive force because of greater distance between ion «centres» [✔]

Examiners report

Average performance with answers using atomic size rather than ionic size or making reference to electronegativities of K and Na.

(e)

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]

Markscheme

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 [✔]

Examiners report

An average of 1.1 out of 3 earned here. Many candidates could write a balanced equation for the reaction of sodium oxide with water but not phosphorus(V) oxide. Mediocre performance in identifying the acid/base nature of the solutions formed.

(f)

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 5.00g of sodium oxide produces 5.50g of sodium peroxide.

[2]

Markscheme

n(Na₂O₂) theoretical yield «= $\frac{{5.00{\text{g}}}}{{61.98{\text{g mo}}{{\text{l}}^{ - 1}}}}$ » = 0.0807/8.07 × 10⁻² «mol»

mass of Na₂O₂ theoretical yield «= $\frac{{5.00{\text{g}}}}{{61.98{\text{g mo}}{{\text{l}}^{ - 1}}}}$ × 77.98 gmol⁻¹» = 6.291 «g» [✔]

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

Note: Award [2] for correct final answer.

Examiners report

The majority earned one or two marks in finding a % yield.

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

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

(g(i))

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

[3]

Markscheme

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

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

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

Note: Award [3] for correct final answer.

Award [2 max] for “+ 452.6 «kJ»”.

Examiners report

The average was 2.2 out 3 for this question on enthalpy of formation. Enthalpy calculations were generally well done.

(g(ii))

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

[1]

Markscheme

only valid for covalent bonds

only valid in gaseous state [✔]

Examiners report

The majority of candidates referred to “bond enthalpy values are average”, rather than not valid for solids or only used for gases.

(h)

An allotrope of molecular oxygen is ozone. Compare, giving a reason, the bond enthalpies of the O to O bonds in O₂ and O₃.

[1]

Markscheme

bond in O₃ has lower enthalpy AND bond order is 1.5 «not 2» [✔]

Note: Accept “bond in ozone is longer”.

Examiners report

Some candidates recognized that ozone had a resonance structure but then only compared bond length between ozone and oxygen rather than bond enthalpy.

(i)

Outline why a real gas differs from ideal behaviour at low temperature and high pressure.

[1]

Markscheme

Any one of:

finite volume of particles «requires adjustment to volume of gas» [✔]

short-range attractive forces «overcomes low kinetic energy» [✔]

Examiners report

Few candidates could distinguish the cause for difference in behaviour between real and ideal gases at low temperature or high pressure. Many answers were based on increase in number of collisions or faster rate or movement of gas particles.

(j)

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

[1]

Markscheme

NaOH [✔]

Examiners report

Na₂O was a common formula in many candidates’ answers for the product of the reaction of sodium peroxide with water.

(k)

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

[1]

Markscheme

IV [✔]