AQA A-Level Chemistry Paper 3, 2023: Question 5

10 marks · Medium difficulty

Enthalpy of Vaporisation and Polarity This question involves calculating the molar enthalpy of vaporisation ( Δ 𝐻 vap ΔH vap ) of water, explaining the polarity of ethanol, ethylamine, and dimethyl ether based on their molecular shapes and electronegativities, and analysing the trend in their boiling points by discussing intermolecular forces.

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Question text

05 The molar enthalpy of vaporisation (∆Hvap) of a liquid is the enthalpy change when one

mole of liquid is converted to vapour at the boiling point of the liquid.

A student does an experiment to determine ∆Hvap for water.

The student:

• places a large beaker on a balance

• pours 500 cm3 of water into the beaker

• uses a 2.4 kW heater to raise the temperature of the water to 100 °C

• records the mass of the beaker and hot water

• uses the 2.4 kW heater to boil the water for 100 s

• records the mass of the beaker and remaining water.

The loss in mass is 103 g

05.1 Calculate ∆Hvap for water.

[1 kW = 1 kJ s–1]

[3 marks]

15 –1

∆Hvap kJmol

Table 2 shows some data about three compounds that all contain the same number

of electrons.

Table 2

Compound CH3CH2OH CH3CH2NH2 CH3OCH3

Boiling point / K 352 290 248

05.2 All three compounds in Table 2 are polar.

Ethanol is the most polar and ethylamine is the least polar.

Explain why all three molecules are polar and why ethylamine is the least polar.

In your answer refer to the shapes around, and relative electronegativities of, the most

electronegative atoms.

[4 marks]

05.3 Explain the trend in the boiling points of the three compounds.

Refer to the intermolecular forces in all three compounds in your answer.

[3 marks]

Mark scheme

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Question Answers Additional comments/Guidelines Mark

5.1

M1 energy transferred = 2.4 x 100 (= 240kJ / 240000 J) M1 IGNORE sign

M2 amount of water vaporised = (= 5.72mol)

ALLOW ecf in M3 (if q = mcΔT used for incorrect

M1 in kJ 3

M3 H = = (+)41.9 / 42.0 (kJmol–1) M1 then a value in kJ could score ecf in M3)

vap

ALLOW 41.9 to 42.11 to two or more sig figs

M3 NOT if negative

Correct answer = 3 marks

5.2 ALLOW ‘different electronegativities’ PLUS

M1 O AND N more electronegative than C and/or H (so all have

diagrams labelled ẟ+ and ẟ-

polar bonds)

M2 CH3CH2OH and CH3OCH3 both v-shaped/non-linear/bent

ALLOW angular for v-shaped in M2

AND CH3CH2NH2 (trigonal) pyramidal

Diagrams from M2 do not require lone pairs

M3 shapes are not symmetrical (so molecules are polar)

ALLOW M3 if diagrams in M2 show asymmetry

Correct diagrams of the three shapes gives M2

M4 O more electronegative than N (so ethylamine is least polar)

and M3

– A-LEVEL CHEMISTRY – –

5.3

M1 hydrogen bonding in CH3CH2OH and CH3CH2NH2 AND IGNORE van der Waals’ / temporary/induced

(permanent) dipole-dipole forces in CH3OCH3 dipole-dipole forces

M1 NOT any reference to breaking covalent

bonds

M2 hydrogen bonding stronger (than other (intermolecular) forces)

24 3

M3 hydrogen bonding stronger in CH3CH2OH than in CH3CH2NH2 M3 ALLOW reference to O being more/most

electronegative (than N) OR ethanol has greater

dipole moment / more polar than ethylamine

If none of M1, M2 or M3 have been awarded:

ALLOW one mark for an indication that higher

boiling point = stronger intermolecular forces but

NOT if reference to breaking covalent bonds

How to answer it

Step-by-Step Guide: Molar Enthalpy of Vaporisation

Question (a): Calculate ΔHvap for Water

Step 1: Calculate Energy Transferred

Energy transferred = power × time

Calculation:

Power = 2.4 kW = 2400 J/s
Time = 100 s
Energy transferred = 2400 × 100 = 240,000 J or 240 kJ

Step 2: Calculate Moles of Water Vaporised

Moles of water = mass ÷ molar mass

Calculation:

Mass of water vaporised = 103 g
Molar mass of water = 18 g mol^-1
Moles of water = 103 ÷ 18 = 5.72 mol

Step 3: Calculate ΔHvap

ΔHvap = Energy transferred ÷ moles

Calculation:

ΔHvap = 240 kJ ÷ 5.72 mol = 41.9 kJ mol^-1

Teacher Tip: Always keep track of significant figures. Use 2 or 3 sf unless otherwise instructed.

Question (b): Polarity of the Molecules

Step 1: Explain Why All Molecules Are Polar

All three molecules are polar because they contain bonds between atoms with different electronegativities (e.g., O-H, N-H, or C-O).

Ethanol: V-shaped due to lone pairs on oxygen, leading to a dipole.
Ethylamine: Trigonal pyramidal around the nitrogen atom, with a net dipole moment.
Dimethyl ether: V-shaped due to oxygen, creating a polar molecule.

Step 2: Explain Why Ethylamine Is Least Polar

Ethylamine is the least polar because nitrogen is less electronegative than oxygen, resulting in a weaker dipole moment compared to ethanol or dimethyl ether.

Teacher Tip: When describing polarity, always reference the molecular shape and relative electronegativities of atoms.

Question (c): Trend in Boiling Points

Step 1: Identify the Intermolecular Forces

Ethanol: Hydrogen bonding (strongest).
Ethylamine: Hydrogen bonding (weaker than ethanol).
Dimethyl ether: Permanent dipole-dipole interactions (weaker than hydrogen bonding).

Step 2: Explain the Trend

The boiling points follow the strength of intermolecular forces:

Ethanol (352 K): Strong hydrogen bonds due to O-H groups.
Ethylamine (290 K): Weaker hydrogen bonds because N-H bonds are less polar than O-H bonds.
Dimethyl ether (248 K): No hydrogen bonding, only weaker dipole-dipole forces.

Teacher Tip: Always link boiling points to the type and strength of intermolecular forces. Avoid discussing covalent bond strength, as this is irrelevant here.

Topics

Physical Chemistry · 3.1.2 Amount of Substance · 3.1.3 Bonding

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.