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Date	November 2016	Marks available	1	Reference code	16N.2.sl.TZ0.1
Level	SL	Paper	2	Time zone	TZ0
Command term	Suggest	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.

Ethane-1,2-diol can be formed according to the following reaction.

2CO (g) + 3H₂(g) $\rightleftharpoons$ HOCH₂CH₂OH (g)

(i) Deduce the equilibrium constant expression, K_c, for this reaction.

(ii) State how increasing the pressure of the reaction mixture at constant temperature will affect the position of equilibrium and the value of K_c.

Position of equilibrium:

K_c:

(iii) Calculate the enthalpy change, ΔH^θ, in kJ, for this reaction using section 11 of the data booklet. The bond enthalpy of the carbon–oxygen bond in CO (g) is 1077kJmol^-1.

(iv) The enthalpy change, ΔH^θ, for the following similar reaction is –233.8 kJ.

2CO(g) + 3H₂(g) $\rightleftharpoons$ HOCH₂CH₂OH (l)

Deduce why this value differs from your answer to (a)(iii).

[7]

Determine the average oxidation state of carbon in ethene and in ethane-1,2-diol.

Ethene:

Ethane-1,2-diol:

[2]

Explain why the boiling point of ethane-1,2-diol is significantly greater than that of ethene.

[2]

Ethane-1,2-diol can be oxidized first to ethanedioic acid, (COOH)₂, and then to carbon dioxide and water. Suggest the reagents to oxidize ethane-1,2-diol.

[1]

Markscheme

(i)
$\ll {K_{\text{C}}} = \gg \frac{{\left[ {{\text{HOC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}} \right]}}{{{{\left[ {{\text{CO}}} \right]}^{\text{2}}} \times {{\left[ {{{\text{H}}_{\text{2}}}} \right]}^{\text{3}}}}}$

(ii)
Position of equilibrium: moves to right OR favours product
K_c: no change OR is a constant at constant temperature

(iii)
Bonds broken: 2C≡O + 3(H-H) / 2(1077kJmol^-1) + 3(436kJmol^-1) / 3462 «kJ»

Bonds formed: 2(C-O) + 2(O-H) + 4(C-H) + (C-C) / 2(358kJmol^-1) + 2(463kJmol^-1) + 4(414kJmol^-1) + 346kJmol^-1 / 3644 «kJ»

«Enthalpy change = bonds broken - bonds formed = 3462 kJ - 3644 kJ =» -182 «kJ»

Award [3] for correct final answer.
Award [2 max] for «+»182 «kJ».

(iv)
in (a)(iii) gas is formed and in (a)(iv) 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.

Ethene: –2

Ethane-1,2-diol: –1

Do not accept 2–, 1– respectively.

ethane-1,2-diol can hydrogen bond to other molecules «and ethene cannot»

ethane-1,2-diol has «significantly» greater van der Waals forces

Accept converse arguments.
Award [0] if answer implies covalent bonds are broken

hydrogen bonding is «significantly» stronger than other intermolecular forces

acidified «potassium» dichromate«(VI)»/H⁺ AND K₂Cr₂O₇/H⁺ AND Cr₂O₇^2-