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).

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

Ethene:

Ethane-1,2-diol:

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

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.

(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»

OR

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-

OR

«acidified potassium» manganate(VII)/ «H⁺» KMnO₄ /«H⁺» MnO₄^-

Accept Accept H₂SO₄ or H₃PO₄ for H⁺.
Accept “permanganate” for “manganate(VII)”.

The following reaction was allowed to reach equilibrium at 761 K.

H₂ (g) + I₂ (g) $\rightleftharpoons$ 2HI (g)               ΔH^θ < 0

Outline the effect, if any, of each of the following changes on the position of equilibrium, giving a reason in each case.

Identify two different amphiprotic species in the above reactions.

State what is meant by the term conjugate base.

State the conjugate base of the hydroxide ion, OH^–.

A student working in the laboratory classified HNO₃, H₂SO₄, H₃PO₄ and HClO₄ as acids based on their pH. He hypothesized that “all acids contain oxygen and hydrogen”.

Evaluate his hypothesis.

Award [1 max] if both effects are correct.

Reason for increasing volume:

Accept “concentration of all reagents reduced by an equal amount so cancels out in K_c expression”.

Accept “affects both forward and backward rates equally”.

HCO₃^– AND H₂O

species that has one less proton/H⁺ ion «than its conjugate acid»

OR

species that forms its conjugate acid by accepting a proton

OR

species that is formed when an acid donates a proton

Do not accept “differs by one proton/H⁺ from conjugate acid”.

oxide ion/O^2–

insufficient data to make generalization

OR

need to consider a «much» larger number of acids

OR

hypothesis will continue to be tested with new acids to see if it can stand the test of time

«hypothesis is false as» other acids/HCl/HBr/HCN/transition metal ion/BF₃ do not contain oxygen

OR

other acids/HCl/HBr/HCN/transition metal ion/BF₃ falsify hypothesis

correct inductive reasoning «based on limited sample»

«hypothesis not valid as» it contradicts current/accepted theories/Brønsted-Lowry/Lewis theory

[Max 2 Marks]

Sketch the Lewis (electron dot) structure of the P₄ molecule, containing only single bonds.

Write an equation for the reaction of white phosphorus (P₄) with chlorine gas to form phosphorus trichloride (PCl₃).

Deduce the electron domain and molecular geometry using VSEPR theory, and estimate the Cl–P–Cl bond angle in PCl₃.

Explain the polarity of PCl₃.

Calculate the standard enthalpy change (ΔH^⦵) for the forward reaction in kJ mol⁻¹.

ΔH^⦵_f PCl₃(g) = −306.4 kJ mol⁻¹

ΔH^⦵_f PCl₅(g) = −398.9 kJ mol⁻¹

State the equilibrium constant expression, K_c, for this reaction.

State, with a reason, the effect of an increase in temperature on the position of this equilibrium.

Accept any diagram with each P joined to the other three.

Accept any combination of dots, crosses and lines.

P₄(s) + 6Cl₂ (g) → 4PCl₃ (l) ✔

Electron domain geometry: tetrahedral ✔

Molecular geometry: trigonal pyramidal ✔

Bond angle: 100«°» ✔

Accept any value or range within the range 91−108«°» for M3.

polar AND unsymmetrical distribution of charge
OR
polar AND dipoles do not cancel
OR
«polar as» dipoles «add to» give a «partial» positive «charge» at P and a «partial» negative «charge» at the opposite/Cl side of the molecule ✔

Accept “polar AND unsymmetrical molecule”.

«−398.9 kJ mol⁻¹ − (−306.4 kJ mol⁻¹) =» −92.5 «kJ mol⁻¹» ✔

«K_c =» $\frac{\left[{\mathrm{PCl}}_5\right]}{\left[{\mathrm{PCl}}_3\right]\left[{\mathrm{Cl}}_2\right]}$ ✔

«shifts» left/towards reactants AND «forward reaction is» exothermic/ΔH is negative ✔

Distinguish between the terms reaction quotient, Q, and equilibrium constant, K_c.

The equilibrium constant, K_c, is 0.282 at temperature T.

Deduce, showing your work, the direction of the initial reaction.

Q: non-equilibrium concentrations AND K_c: equilibrium concentrations

OR

Q: «measured» at any time AND K_c: «measured» at equilibrium

[1 mark]

«Q = $\frac{{[\text{S}{\text{O}}_3]}^2}{{[\text{S}{\text{O}}_2]}^2[{\text{O}}_2]}=\frac{{1.00}^2}{{1.00}^2\times 2.00}$» = 0.500

reverse reaction favoured/reaction proceeds to the left AND Q > K_c/0.500 > 0.282

Do not award M2 without M1.

[2 marks]

Deduce the equilibrium constant expression, K_c, for the decomposition of PCl₅(g).

Deduce, giving a reason, the factor responsible for establishing the new equilibrium after 14 minutes.

Deduce the Lewis (electron dot) structure and molecular geometry of PCl₃.

«K_c» = $\frac{\left[\text{PC}{\text{l}}_3\right]\left[\text{C}{\text{l}}_2\right]}{\left[\text{PC}{\text{l}}_5\right]}$

[1 mark]

decrease in temperature

endothermic «reaction» AND «equilibrium» shifts to the left/reactants
OR
endothermic «reaction» AND K_c decreases
OR
endothermic «reaction» AND concentration of PCl₅ increased/concentration of PCl₃ and Cl₂ decreased
OR
«equilibrium» shifts in exothermic direction

Do not accept “temperature change”.

Accept “ΔH positive” in place of “endothermic”.

Accept “products” instead of “PCl₃ and Cl₂”.

[2 marks]

Lewis structure:

Molecular geometry:

trigonal/triangular pyramidal

Penalize missing lone pairs once only between this question and 4(b).

Accept any combination of lines, dots or crosses to represent electrons.

Do not apply ECF.

[2 marks]

Estimate the H−N−H bond angle in methanamine using VSEPR theory.

Ammonia reacts reversibly with water.

NH₃(g) + H₂O(l) $\rightleftharpoons$ NH₄⁺(aq) + OH⁻(aq)

Explain the effect of adding H⁺(aq) ions on the position of the equilibrium.

Hydrazine reacts with water in a similar way to ammonia. Deduce an equation for the reaction of hydrazine with water.

Outline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.

Hydrazine has been used as a rocket fuel. The propulsion reaction occurs in several stages but the overall reaction is:

N₂H₄(l) → N₂(g) + 2H₂(g)

Suggest why this fuel is suitable for use at high altitudes.

Determine the enthalpy change of reaction, ΔH, in kJ, when 1.00 mol of gaseous hydrazine decomposes to its elements. Use bond enthalpy values in section 11 of the data booklet.

N₂H₄(g) → N₂(g) + 2H₂(g)

The standard enthalpy of formation of N₂H₄(l) is +50.6 kJ$$mol⁻¹. Calculate the enthalpy of vaporization, ΔH_vap, of hydrazine in kJ$$mol⁻¹.

N₂H₄(l) → N₂H₄(g)

(If you did not get an answer to (f), use −85 kJ but this is not the correct answer.)

Calculate, showing your working, the mass of hydrazine needed to remove all the dissolved oxygen from 1000 dm³ of the sample.

Calculate the volume, in dm³, of nitrogen formed under SATP conditions. (The volume of 1 mol of gas = 24.8 dm³ at SATP.)

107^°

Accept 100° to < 109.5°.

Literature value = 105.8°

[1 mark]

removes/reacts with OH⁻

moves to the right/products «to replace OH⁻ ions»

Accept ionic equation for M1.

[2 marks]

N₂H₄(aq) + H₂O(l) $\rightleftharpoons$ N₂H₅⁺(aq) + OH^–(aq)

Accept N₂H₄(aq) + 2H₂O(l) $\rightleftharpoons$ N₂H₆²⁺(aq) + 2OH^–(aq).

Equilibrium sign must be present.

[1 mark]

bubbles
OR
gas
OR
magnesium disappears

2NH₄⁺(aq) + Mg(s) → Mg²⁺(aq) + 2NH₃(aq) + H₂(g)

Do not accept “hydrogen” without reference to observed changes.

Accept "smell of ammonia".

Accept 2H⁺(aq) + Mg(s) → Mg²⁺(aq) + H₂(g)

Equation must be ionic.

[2 mark]

no oxygen required

[1 mark]

bonds broken:
E(N–N) + 4E(N–H)
OR
158 «kJ$$mol^–1» + 4 x 391 «kJ$$mol^–1» / 1722 «kJ»

bonds formed:
E(N≡N) + 2E(H–H)
OR
945 «kJ$$mol^–1» + 2 x 436 «kJ$$mol^–1» / 1817 «kJ»

«ΔH = bonds broken – bonds formed = 1722 – 1817 =» –95 «kJ»

Award [3] for correct final answer.

Award [2 max] for +95 «kJ».

[3 marks]

OR
ΔH_vap= −50.6 kJ$$mol⁻¹ − (−95 kJ$$mol⁻¹)

«ΔH_vap =» +44 «kJ$$mol⁻¹»

Award [2] for correct final answer.

Award [1 max] for −44 «kJ$$mol⁻¹».

Award [2] for:
ΔH_vap − = 50.6 kJ$$mol⁻¹ − (−85 kJ$$mol⁻¹) + = 34 «kJ$$mol⁻¹».

Award [1 max] for −34 «kJ$$mol⁻¹».

[2 marks]

total mass of oxygen «= 8.0 x 10^–3 g$$dm^–3 x 1000 dm³» = 8.0 «g»

n(O₂) «$=\frac{8.0\text{ g}}{32.00\text{ g}\text{mo}{\text{l}}^{-1}}=$» 0.25 «mol»

OR
n(N₂H₄) = n(O₂)
«mass of hydrazine = 0.25 mol x 32.06 g$$mol^–1 =» 8.0 «g»

Award [3] for correct final answer.

[3 marks]

«n(N₂H₄) = n(O₂) $=\frac{8.0\text{ g}}{32.00\text{ g}\text{mo}{\text{l}}^{-1}}=$» 0.25 «mol»

«volume of nitrogen = 0.25 mol x 24.8 dm³$$mol^–1» = 6.2 «dm³»

Award [1] for correct final answer.

[1 mark]

Calculate the percentage by mass of nitrogen in urea to two decimal places using section 6 of the data booklet.

Suggest how the percentage of nitrogen affects the cost of transport of fertilizers giving a reason.

The structural formula of urea is shown.

Predict the electron domain and molecular geometries at the nitrogen and carbon atoms, applying the VSEPR theory.

Urea can be made by reacting potassium cyanate, KNCO, with ammonium chloride, NH₄Cl.

                                      KNCO(aq) + NH₄Cl(aq) → (H₂N)₂CO(aq) + KCl(aq)

Determine the maximum mass of urea that could be formed from 50.0 cm³ of 0.100 mol dm⁻³ potassium cyanate solution.

Urea can also be made by the direct combination of ammonia and carbon dioxide gases.

                                   2NH₃(g) + CO₂(g) $\rightleftharpoons$ (H₂N)₂CO(g) + H₂O(g)     ΔH < 0

Predict, with a reason, the effect on the equilibrium constant, K_c, when the temperature is increased.

Suggest one reason why urea is a solid and ammonia a gas at room temperature.

Sketch two different hydrogen bonding interactions between ammonia and water.

The combustion of urea produces water, carbon dioxide and nitrogen.

Formulate a balanced equation for the reaction.

The mass spectrum of urea is shown below.

Identify the species responsible for the peaks at m/z = 60 and 44.

The IR spectrum of urea is shown below.

Identify the bonds causing the absorptions at 3450 cm⁻¹ and 1700 cm⁻¹ using section 26 of the data booklet.

Predict the number of signals in the ¹H NMR spectrum of urea.

molar mass of urea «= 4 × 1.01 + 2 × 14.01 + 12.01 + 16.00» = 60.07 «g mol^–1»

«% nitrogen = $\frac{2\times 14.01}{60.07}$ × 100 =» 46.65 «%»

Award [2] for correct final answer.

Award [1 max] for final answer not to two decimal places.

[2 marks]

«cost» increases AND lower N% «means higher cost of transportation per unit of nitrogen»

OR

«cost» increases AND inefficient/too much/about half mass not nitrogen

Accept other reasonable explanations.

Do not accept answers referring to safety/explosions.

[1 mark]

Note: Urea’s structure is more complex than that predicted from VSEPR theory.

[3 marks]

n(KNCO) «= 0.0500 dm³ × 0.100 mol dm^–3» = 5.00 × 10^–3 «mol»

«mass of urea = 5.00 × 10^–3 mol × 60.07 g mol^–1» = 0.300 «g»

Award [2] for correct final answer.

[2 marks]

«K_c» decreases AND reaction is exothermic

OR

«Kc» decreases AND ΔH is negative

OR

«K_c» decreases AND reverse/endothermic reaction is favoured

[1 mark]

Any one of:

urea has greater molar mass

urea has greater electron density/greater London/dispersion

urea has more hydrogen bonding

urea is more polar/has greater dipole moment

Accept “urea has larger size/greater van der Waals forces”.

Do not accept “urea has greater intermolecular forces/IMF”.

[1 mark]

Award [1] for each correct interaction.

If lone pairs are shown on N or O, then the lone pair on N or one of the lone pairs on O MUST be involved in the H-bond.

Penalize solid line to represent H-bonding only once.

[2 marks]

2(H₂N)₂CO(s) + 3O₂(g) → 4H₂O(l) + 2CO₂(g) + 2N₂(g)

correct coefficients on LHS

correct coefficients on RHS

Accept (H₂N)₂CO(s) + $\frac{3}{2}$O₂(g) → 2H₂O(l) + CO₂(g) + N₂(g).

Accept any correct ratio.

[2 marks]

60: CON₂H₄⁺

44: CONH₂⁺

Accept “molecular ion”.

[2 marks]

3450 cm^–1: N–H

1700 cm^–1: C=O

Do not accept “O–H” for 3450 cm^–1.

[2 marks]

1

[1 mark]

Predict the electron domain and molecular geometries around the oxygen atom of molecule A using VSEPR.

The IR spectrum of one of the compounds is shown:

COBLENTZ SOCIETY. Collection © 2018 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.

Deduce, giving a reason, the compound producing this spectrum.

Compound A and B are isomers. Draw two other structural isomers with the formula ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$.

The equilibrium constant, $K_{\mathrm{c}}$, for the conversion of A to B is $1.0\times {10}^8$ in water at $298 \mathrm{K}$.

Deduce, giving a reason, which compound, A or B, is present in greater concentration when equilibrium is reached.

Electron domain geometry: tetrahedral ✔

Molecular geometry: bent/V-shaped ✔

B AND $\mathrm{C}=\mathrm{O}$ absorption/$1750 «{\mathrm{cm}}^{-1}»$ 
OR
B AND absence of $\mathrm{O}–\mathrm{H} /3200-3600 «{\mathrm{cm}}^{-1} \mathrm{absorption}»$ ✔

Accept any value between $\mathit{1700}\mathit{-}\mathit{1750}\mathit{ }c{m}^{\mathit{-}\mathit{1}}$.

Accept any two $C_3{H}_{6}O$ isomers except for propanone and propen-2-ol:

✔✔

Penalize missing hydrogens in displayed structural formulas once only.

$\mathrm{B}$ AND $K_{\mathrm{c}}$ is greater than $1$/large ✔

Half of the candidates answered correctly. The rest of the candidates often answered the question in terms of the carbon atom indicating that they did not read the question carefully.

About 50% of the candidates answered correctly. Quite a few, however, gave compounds other than A or B, indicating not reading the question properly or being confused by the skeletal formulas given in the question.

Nearly half of the candidates gave two correct isomers. Propanal was often given as one of the isomers. Some candidates repeated the compounds given in the question and a few gave structures with 5 bonds on a carbon atom.

Half of the candidates answered correctly. A common mistake was K > 0 instead of K > 1.

Explain why Si has a smaller atomic radius than Al.

Explain the decrease in radius from Na to Na⁺.

State the condensed electron configurations for Cr and Cr3⁺.

Describe metallic bonding and how it contributes to electrical conductivity.

Deduce the Lewis (electron dot) structure and molecular geometry of sulfur dichloride, SCl₂.

Suggest, giving reasons, the relative volatilities of SCl₂ and H₂O.

Consider the following equilibrium reaction:

2SO₂(g) + O₂(g) $\rightleftharpoons$ 2SO₃(g)

State and explain how the equilibrium would be affected by increasing the volume of the reaction container at a constant temperature.

nuclear charge/number of protons/Z/Z_eff increases «causing a stronger pull on the outer electrons» ✓

same number of shells/«outer» energy level/shielding ✓

Na⁺ has one less energy level/shell
OR
Na⁺ has 2 energy levels/shells AND Na has 3 ✓

less shielding «in Na⁺ so valence electrons attracted more strongly to nucleus»
OR
effective nuclear charge/Z_eff greater «in Na⁺ so valence electrons attracted more strongly to nucleus» ✓

Accept “more protons than electrons «in Na⁺»” OR “less electron-electron repulsion «in Na⁺»” for M2.

Cr:
[Ar] 4s¹3d⁵ ✓

Cr³⁺:
[Ar] 3d³ ✓

Accept “[Ar] 3d⁵4s¹”.

Accept “[Ar] 3d³4s⁰”.

Award [1 max] for two correct full electron configurations “1s²2s²2p⁶3s²3p⁶4s¹3d⁵ AND 1s²2s²2p⁶3s²3p⁶3d³”.

Award [1 max] for 4s¹3d⁵ AND 3d³.

electrostatic attraction ✓

between «a lattice of» cations/positive «metal» ions AND «a sea of» delocalized electrons ✓

mobile electrons responsible for conductivity
OR
electrons move when a voltage/potential difference/electric field is applied ✓

Do not accept “nuclei” for “cations/positive ions” in M2.

Accept “mobile/free” for “delocalized” electrons in M2.

Accept “electrons move when connected to a cell/battery/power supply” OR “electrons move when connected in a circuit” for M3.

H₂O forms hydrogen bonding «while SCl₂ does not» ✓

SCl₂ «much» stronger London/dispersion/«instantaneous» induced dipole-induced dipole forces ✓

Alternative 1:
H₂O less volatile AND hydrogen bonding stronger «than dipole–dipole and dispersion forces» ✓

Alternative 2:
SCl₂ less volatile AND effect of dispersion forces «could be» greater than hydrogen bonding ✓\

Ignore reference to Van der Waals.

Accept “SCl₂ has «much» larger molar mass/electron density” for M2.

pressure decrease «due to larger volume» ✓

reactant side has more moles/molecules «of gas» ✓

reaction shifts left/towards reactants ✓

Award M3 only if M1 OR M2 is awarded.

The diagram shows the Maxwell-Boltzmann curve for the uncatalyzed reaction.

Draw a distribution curve at a lower temperature (T₂) and show on the diagram how the addition of a catalyst enables the reaction to take place more rapidly than at T₁.

The hydrogen peroxide could cause further oxidation of the methanol. Suggest a possible oxidation product.

Determine the overall equation for the production of methanol.

8.00 g of methane is completely converted to methanol. Calculate, to three significant figures, the final volume of hydrogen at STP, in dm³. Use sections 2 and 6 of the data booklet.

Determine the enthalpy change, ΔH, in kJ. Use section 11 of the data booklet.

Bond enthalpy of CO = 1077 kJ mol⁻¹.

State the expression for K_c for this stage of the reaction.

State and explain the effect of increasing temperature on the value of K_c.

curve higher AND to left of T₁ ✔

new/catalysed E_a marked AND to the left of E_a of curve T₁ ✔

Do not penalize curve missing a label, not passing exactly through the origin, or crossing x-axis after E_a.
Do not award M1 if curve drawn shows significantly more/less molecules/greater/smaller area under curve than curve 1.
Accept E_a drawn to T₁ instead of curve drawn as long as to left of marked E_a.

methanoic acid/HCOOH/CHOOH
OR
methanal/HCHO ✔

Accept “carbon dioxide/CO₂”.

CH₄(g) + H₂O(g) $\rightleftharpoons$ CH₃OH(l) + H₂(g) ✔

Accept arrow instead of equilibrium sign.

amount of methane = « $\frac{8.00 \mathrm{g}}{16.05 \mathrm{g} {\mathrm{mol}}^{-1}}$ = » 0.498 «mol» ✔

amount of hydrogen = amount of methane / 0.498 «mol» ✔

volume of hydrogen = «0.498 mol × 22.7 dm³mol⁻¹ = » 11.3 «dm³» ✔

Award [3] for final correct answer.
Award [2 max] for 11.4 «dm³ due to rounding of mass to 16/moles to 0.5. »

Σbonds broken = 4 × 414 «kJ» + 2 × 463 «kJ» / 2582 «kJ» ✔

Σbonds formed = 1077 «kJ» + 3 × 436 «kJ» / 2385 «kJ» ✔

ΔH «= Σbonds broken − Σbonds formed =( 2582 kJ − 2385 kJ)» = «+»197«kJ» ✔

Award [3] for final correct answer.
Award [2 Max] for final answer of −197 «kJ»

$K_{\mathit{c}}=\frac{\left[\mathrm{CO}\right]{\left[{\mathrm{H}}_2\right]}^3}{\left[{\mathrm{CH}}_4\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}$ ✔

K_c increases AND «forward» reaction endothermic ✔