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Date November 2016 Marks available 6 Reference code 16N.2.hl.TZ0.1
Level HL Paper 2 Time zone TZ0
Command term Calculate, Comment, and Deduce Question number 1 Adapted from N/A

Question

Ethane-1,2-diol, HOCH2CH2OH, has a wide variety of uses including the removal of ice from aircraft and heat transfer in a solar cell.

(i) Calculate ΔHθ, in kJ, for this similar reaction below using Δ H f θ data from section 12 of the data booklet. Δ H f θ of HOCH2CH2OH(l) is –454.8kJmol-1.

2CO (g) + 3H2 (g)   HOCH2CH2OH (l)

(ii) Deduce why the answers to (a)(iii) and (b)(i) differ.

(iii) ΔSθ for the reaction in (b)(i) is –620.1JK-1. Comment on the decrease in entropy.

(iv) Calculate the value of ΔGθ, in kJ, for this reaction at 298 K using your answer to (b)(i). (If you did not obtain an answer to (b)(i), use –244.0 kJ, but this is not the correct value.)

(v) Comment on the statement that the reaction becomes less spontaneous as temperature is increased.

[6]
b.

Predict the 1HNMR data for ethanedioic acid and ethane-1,2-diol by completing the table.

[2]
f.

Markscheme

i
«ΔH = Σ ΔHf products – ΣΔHf reactants = –454.8 kJ mol-1 – 2(–110.5 kJ mol-1) =» –233.8 «kJ»

 

ii
in (a)(iii) gas is formed and in (b)(i) liquid is formed
OR
products are in different states
OR
conversion of gas to liquid is exothermic
OR
conversion of liquid to gas is endothermic
OR
enthalpy of vapourisation needs to be taken into account

Accept product is «now» a liquid.
Accept answers referring to bond enthalpies being means/averages.

 

iii
«ΔS is negative because five mols of» gases becomes «one mol of» liquid
OR
increase in complexity of product «compared to reactants»
OR
product more ordered «than reactants»

Accept “fewer moles of gas” but not “fewer molecules”.



iv
ΔS ( 620.1 1000 ) «kJ K-1»
ΔG = –233.8 kJ – (298 K  ( 620.1 1000 ) kJ K-1) = –49.0 «kJ»

Award [2] for correct final answer.
Award [1 max] for «+»185 × 103.

If –244.0 kJ used, answer is:
ΔG = –244.0 kJ – (298 K ( 620.1 1000 ) kJ K-1) = –59.2 «kJ»
Award [2] for correct final answer.

 

v
increasing T makes ΔG larger/more positive/less negative
OR
–TΔS will increase

b.

Accept “none/no splitting” for singlet.

f.

Examiners report

[N/A]
b.
[N/A]
f.

Syllabus sections

Additional higher level (AHL) » Topic 15: Energetics/thermochemistry » 15.2 Entropy and spontaneity
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