OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 16
15 marks · Medium difficulty · Calculations
Calculate enthalpy changes of hydration, complete an enthalpy cycle, explain hydration and bond enthalpies, and calculate a bond enthalpy using average bond values.
Practise this questionQuestion
Question text
16 This question is about enthalpy changes.
(a) Table 16.1 shows enthalpy changes that can be used to determine the enthalpy change of
hydration of fluoride ions, F–.
Enthalpy change Energy / kJ mol–1
Hydration of Ca2+ –1609
Solution of CaF2 +13
Lattice enthalpy of CaF2 –2630
Table 16.1
(i) Explain what is meant by the term enthalpy change of hydration.
… [2]
(ii) The enthalpy change of hydration of F– can be determined using the enthalpy changes
in Table 16.1 and the incomplete energy cycle below.
On the dotted lines, add the species present, including state symbols.
lattice
enthalpy
[4]
(iii) Calculate the enthalpy change of hydration of fluoride ions, F–.
enthalpy change of hydration = … kJ mol–1 [2]
(iv) Predict how the enthalpy changes of hydration of F– and Cl – would differ.
Explain your answer.
… [2]
(b) Fluorine reacts with steam as shown in the equation below.
2F (g) + 2H O(g) O (g) + 4HF(g) ∆H = −598 kJ mol–1
22 2
Average bond enthalpies are shown in the table.
Bond Average bond enthalpy / kJ mol–1
O–H +464
O=O +498
H–F +568
(i) Explain what is meant by the term average bond enthalpy.
… [2]
(ii) Calculate the bond enthalpy of the F–F bond.
bond enthalpy = … kJ mol–1 [3]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
16 (a) (i) (enthalpy change when) 2 IGNORE ‘energy released’ OR ‘energy required’
1 mole of gaseous ions react
OR 1 mole of hydrated/aqueous ions are formed
gaseous ions dissolve in water
OR gaseous ions form aqueous/hydrated ions
(a) (ii) Ca2+(g) + 2F–(g) 4 Correct species AND state symbols required for each
mark. (mark independently)
On 2nd line, ALLOW Ca2+(g) + 2F–(aq)
Ca2+(aq) + 2F–(g)
(i.e. F– hydrated before Ca2+)
Ca2+(aq) + 2F–(aq)
On 3rd line, ALLOW CaF2(aq)
CaF2(s)
DO NOT ALLOW when first seen but ALLOW ECF for ‘ 2’
missing and for use of the following ions
Fl –
F –
Ca+/3+
(a) (iii) FIRST, CHECK THE ANSWER ON ANSWER LINE 2 IF alternative answer, check to see if there is any ECF
IF answer = –504 (kJ mol–1) award 2 marks credit possible using working below.
IF answer = –1008 (kJ mol–1) award 1 mark
--------------------------------------------------------------------– ‘–‘ sign is needed.
2 ∆ H(F–) COMMON ERRORS for 1 mark:
hyd
= [–2630 + 13] – (–1609) (+)2694: signs all reversed
OR – 2617 + 1609 –2113: sign wrong for –1609
OR –1008 (kJ mol–1) –2126: sign wrong for 2630
–517: sign wrong for 13
– –1008 –1 +504: sign wrong
∆hydH(F ) = 2 = –504 (kJ mol )
IF ALL 3 relevant values from the information at the
start of Q16a(iii) have NOT been used, award zero
marks unless one number has a transcription error,
where 1 mark can be awarded ECF
(a) (iv) Correct comparison of ∆hyd linked to sizes 2 ORA
∆ H (F–) more negative/exothermic (than ∆ H (Cl–)) IGNORE ‘atomic’ before radius when comparing size of
hyd hyd
AND ions
F– has smaller size (than Cl–) IGNORE charge density
Comparison of attraction between ions and water IGNORE electronegativity
F– OR smaller sized ion linked to greater attraction to IGNORE nuclear attraction
H2O DO NOT ALLOW ‘forms stronger hydrogen bonds with
water’ OR ‘forms stronger van der Waals’ forces with
water’
ALLOW ‘forms bonds’ for attraction’
DO NOT ALLOW F- greater attraction to H O if given as
larger ion
Assume ‘F’ / ‘Fluorine’ means ‘ions’ but DO NOT ALLOW
‘F molecules’
(b) (i) Average bond enthalpy 8 2
Breaking of one mole of bonds IGNORE energy required OR energy released IGNORE
heterolytic / homolytic
DO NOT ALLOW bonds formed
DO NOT ALLOW ionic bonds
In gaseous molecules IGNORE species for molecules
(b) (ii) FIRST, CHECK ANSWER ON ANSWER LINE 3 ANNOTATE ANSWER WITH TICKS AND CROSSES
IF answer = (+) 158 award 3 marks
------------------------------------------------------------------- IGNORE sign
IGNORE sign
Bond enthalpy of F–F
(∆H for (O–H) bonds broken =) ALLOW ECF
1856 OR 4 464 (kJ mol–1)
Common errors
(∆H for bonds made =) 2770 (kJ mol–1)
OR 498 AND 2272 (kJ mol–1)
Award 2 marks for;
OR 498 AND 4 568 (kJ mol–1) –158 (Wrong sign)
(±)316 (No ÷ 2)
2770 – 1856 – 598 (+) 622 (use of 2 x 464)
(bond enthalpy) F–F = 2 (+) 457 (omitting – 598)
= (+)158 (kJ mol–1) (+) 756 (use of +598)
Award 1 mark for;
(+) 970 (use of 2 x 464 and +598)
Total 15
How to answer it
Enthalpy Changes & Bond Enthalpies
What this question tests
This question assesses your mastery of energetic cycles (Born-Haber / Enthalpy cycles), precise thermodynamic definitions (hydration and mean bond enthalpy), manipulation of Hess's Law calculations, ionic radius trends affecting hydration, and calculating bond enthalpies from overall reaction enthalpy changes.
Defining Enthalpy Change of Hydration
✅ Correct Answer
The enthalpy change when 1 mole of gaseous ions is dissolved in water to form aqueous/hydrated ions.
💡 Key Knowledge
Key phrasing required by examiners:
- 1 mole of gaseous ions (do not miss "gaseous").
- Forming aqueous or hydrated ions.
❌ Common Errors
Writing "energy released" or "energy required". While hydration is exothermic, definitions must state "enthalpy change", not energy values.
Constructing the Enthalpy Cycle
✅ Correct Answer
Top line: Ca²⁺(g) + 2F⁻(g)
2nd line down: Ca²⁺(aq) + 2F⁻(g) (or hydrate F⁻ first)
3rd line down: Ca²⁺(aq) + 2F⁻(aq) (or CaF₂(aq) )
Bottom line: CaF₂(s)
🧠 Exam Technique
Always include full state symbols (g) , (aq) , and (s) alongside correct ion stoichiometries ( 2F⁻ ) to secure independent layout marks.
Calculating Enthalpy of Hydration of F⁻
📐 Calculation Steps
- State Hess's Law path: Lattice enthalpy + Solution enthalpy = Hydration of Ca²⁺ + 2 × (Hydration of F⁻)
- Substitute values: -2630 + 13 = -1609 + 2(ΔhydH)
- Rearrange: 2(ΔhydH) = -2617 - (-1609) = -1008 kJ mol⁻¹
- Divide by 2: ΔhydH(F⁻) = -1008 / 2 = -504 kJ mol⁻¹
❌ Common Calculation Traps
Forgetting to multiply or divide by the stoichiometric coefficient 2 for fluoride ions yields -1008 kJ mol⁻¹ (1 mark penalty). Missing the negative sign on the final answer will lose accuracy marks.
Comparing F⁻ and Cl⁻ Hydration Enthalpies
✅ Correct Answer
Comparison: Hydration enthalpy of F⁻ is more negative / exothermic than Cl⁻.
Reason: F⁻ has a smaller ionic radius than Cl⁻, leading to stronger electrostatic attraction to the δ⁺ hydrogen atoms of water molecules.
❌ Common Errors
Do NOT refer to "atomic radius" (they are ions, not atoms). Avoid mentioning electronegativity or nuclear attraction; focus strictly on ionic radius and charge density / attraction to water.
Defining Average Bond Enthalpy
✅ Correct Answer
The enthalpy change when one mole of a specified covalent bond is broken, averaged across a range of different compounds.
💡 Key Knowledge
Essential keywords required:
- Breaking (or energy to break)
- 1 mole of bonds
- Gaseous molecules
Calculating F–F Bond Enthalpy
📐 Calculation Steps
Equation: 2F₂(g) + 2H₂O(g) → O₂(g) + 4HF(g) with ΔH = -598 kJ mol⁻¹
- Bonds broken: 2(F–F) + 4(O–H in H₂O) -> 2(x) + 4(464) = 2x + 1856
- Bonds made: 1(O=O) + 4(H–F) -> 1(498) + 4(568) = 498 + 2272 = 2770 kJ mol⁻¹
- Apply formula: ΔH = Σ(Bonds broken) - Σ(Bonds made)
-598 = (2x + 1856) - 2770 - Rearrange for x: 2x = -598 - 1856 + 2770 = +316
- Final value: x = 316 / 2 = +158 kJ mol⁻¹
❌ Common Errors
Forgetting to multiply bond values by stoichiometric coefficients (e.g., missing the 4 for O–H and H–F bonds, or forgetting to divide the final sum by 2 for the two F–F bonds).
Topics
Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 3.2 Physical chemistry · 5.2 Energy
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.