9701_w21_qp_41-NehaS
Question
When dilute sulfuric acid is electrolysed, water is split into hydrogen and oxygen.
\(2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)\)
A current of xA is passed through the solution for 14.0 minutes. 462cm³ of hydrogen are produced at the cathode, measured under room conditions.
(a) Calculate the number of hydrogen molecules produced during the electrolysis.
(b) Calculate the total number of electrons transferred to produce this number of hydrogen molecules.
(c) Calculate the quantity of charge, in coulombs, of the total number of electrons calculated in (b).
(d) Calculate the current, x, passed during this experiment.
(e) The standard entropies, \(s^{\theta }\), of three species are given in the table.
(i) Calculate \(\Delta s^{\theta }\) for the reaction \(2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)\).
(ii) \(\Delta H^{\theta }\) for the reaction \(2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)\) is +572kJmol-1.
Calculate \(\Delta G^{\theta }\) for this reaction at 298K.
(iii) Predict the effect of increasing temperature on the spontaneity of this reaction. Explain your answer.
Answer/Explanation
Answer (a) moles of H2 = 462 / 24 000 = 0.01925
molecules of H2 = 0.019 × 6.02 × 1023
= 1.16 × 1022 (1.1 ×1022 / 1.2 × 1022) min 2sf ecf M1
(b) number of electrons = 1.16× 1022 × 2
= 2.32 × 1022 min 2sf ecf 1a
(c) Q = 2.32×1022 × 1.6 × 10–19
= 3.71 × 103 min 2sf ecf 1b
(d) x = 3.71 × 103 / (14 × 60)
= 4.4 (A) min 2sf ecf 1c
(e)(i) ΔS = 262 + 205 – 140 = (+) 327 (JK–1 mol–1)
(e)(ii) Δ G =ΔH – TΔ S OR use of Gibbs equation
Δ G = 572 – (298× 0.327)
= (+)474.6 (kJ mol–1) min 3sf ecf 1e(i)
(e)(iii) becomes more feasible / spontaneous
as T ΔS is more positive / –T ΔS becomes more negative
Question
Solution Y is hydrochloric acid, HCl(aq). Solution Z is aqueous 4-chlorobutanoic acid,
Cl(CH2)3CO2H(aq). The pKa of Cl(CH2)3CO2H(aq) is 4.52. The pH of both solutions is 4.00.
(a) (i) Write an expression for the Ka of Cl(CH2)3CO2H(aq).
(ii) Write a mathematical expression to describe the relationship between Ka and pKa.
(iii) Calculate [H+] in solutions Y and Z.
(iv) Calculate the ratio \(\frac{[HCl]\: dissolved\: in\: solution\: Y}{[Cl(CH_{2})_{3}CO_{2}H]\: dissolved\: in\: solution\: Z}\) .
(b) A buffer solution of pH 5.00 is produced by adding sodium propanoate to 5.00g of propanoic acid
in 100cm³ of distilled water.
Calculate the mass of sodium propanoate that must be used to produce this buffer solution.
The Ka of propanoic acid is 1.35 × 10-5moldm-3.
[Mr : propanoic acid, 74.0; sodium propanoate, 96.0]
(c) Some dilute sulfuric acid is mixed with a small sample of the buffer solution described in (b).
The final pH of the mixture is close to 1.
Explain this observation.
Answer/Explanation
Answer (a)(i)
(a)(ii) pKa =– logKa OR Ka = 10–pKa
(a)(iii) [H+] = 10–4.0 = 1 × 10–4
(a)(iv)
(b)
(c) all of the (sodium) propanoate (ion) has been
protonated / converted to (propanoic) acid / neutralised
H+ is in excess / H+ is 0.1 mol dm–3 (from the H2SO4)
Question
(a) Define the term electron affinity.
(b) Write an equation for the process corresponding to the second ionisation energy of calcium.
Include state symbols.
Some data relating to calcium and oxygen are listed. Select relevant data from this list for your answers to parts (c), (d) and (e).
Show all your working.
(d) (i) Suggest why the first electron affinity of oxygen is negative.
(ii) Suggest why the second electron affinity of oxygen is positive.
(e) Calculate the enthalpy of formation of calcium oxide, CaO(s).
(f) The lattice energy of lithium fluoride, LiF(s), is –1022kJmol-1.
Identify the factor that causes the lattice energy of calcium oxide to be more exothermic than
that of lithium fluoride. Explain why this factor causes the difference in lattice energies.
Answer/Explanation
Answer (a) • enthalpy/energy change
• one mole of electrons gained
• by one mole of atoms
• gaseous (atoms)
(b) Ca+(g) → Ca2+(g) + e–
(c) M1: selecting correct data 951, 844, 142 only
M2: evaluation to give 249 (ΔHatom)
OR 2(951) = BE – 2(142) + 2(844)
M3: evaluation to 498 (2 × 249) ecf M2
951 =ΔHatom –142 + 844
ΔHatom = 249
BE = 498 (kJ mol–1)
(d)(i) attraction between nucleus / protons / nuclear charge
and electron
(d)(ii) repulsion between 1– ion / electrons of O–
and electron
(e) M1: selecting correct data 951, 1933, 3517 only (ignore signs)
M2: evaluation to give –633 (Δ Hf) ecf
ΔHf = 951 + 1933 – 3517 = –633 (kJ mol–1)
(f) ionic charge / charge density (of the ions)
greater (attractive) force between the ions
Question
Separate samples of 0.02mol of calcium carbonate and 0.02mol of barium carbonate are heated until completely decomposed to the metal oxide and carbon dioxide.
(a) State which of these two Group 2 carbonates requires the higher temperature before it begins to decompose. Explain your answer.
(b) After decomposition is complete, the 0.02mol sample of calcium oxide is taken and added to 2.00dm³ of water. A solution is formed with no solid present. Dilute sulfuric acid is then added dropwise until a precipitate is seen.
The same procedure is repeated with the 0.02mol sample of barium oxide, using the same concentration solution of dilute sulfuric acid.
Identify the sample to which most sulfuric acid must be added to cause a precipitate to appear.
Explain your answer. You should refer to the solubilities of the precipitates and relevant energy terms in your answer.
(c) (i) Calculate the mass, in g, of CO2 produced by the decomposition of 0.020 moles of calcium carbonate.
(ii) Calculate the minimum mass, in g, of propane that would, on complete combustion, produce the same mass of CO2 calculated in (c)(i).
Give your answer to three significant figures.
Answer/Explanation
Answer (a) • barium carbonate / Ba / BaCO3
• larger ionic radius OR smaller charge density of cation / M2+
• anion / CO32– / carbonate ion is less distorted / less polarised OR C-O / C=O less weakened
(b) • calcium oxide / calcium hydroxide
• CaSO4 / calcium sulfate is more soluble OR BaSO4 is less soluble
• ΔHlatt and ΔHhyd are less exothermic / more endothermic (for BaSO4) • ΔHhyd is dominant factor / ΔHhyd change is greater OR ΔHlatt changes less
(c)(i) mass of CO2 = 0.02 × 44 = 0.88 g
(c)(ii) (writes correct equation, deduces 3 CO2 per mole)
moles of propane = 0.02 / 3 OR 0.00667 OR 1 / 150
mass of propane = 0.02 / 3 × 44 = 0.293 g ecf M1 × 44 3sf needed
Question
(a) [MnCl 4]2- is a complex ion.
(i) Deduce the oxidation state of manganese in [MnCl 4]2-.
(ii) The [MnCl 4]2- complex does not contain any 180° bond angles.
Draw a three-dimensional diagram to show the shape of the [MnCl 4]2- complex. State one bond angle on your diagram.
(b) A solution of cobalt(II) sulfate contains the complex ion [Co(H2O)6]2+.
A solution containing [Co(H2O)6]2+ is reacted separately with an excess of each of NaOH(aq), NH3(aq) and NaCl(aq).
Write an equation for each of these reactions. State one observation that can be made immediately after the reaction, include the colour and state of the cobalt-containing product.
(i) [Co(H2O)6]2+ and an excess of NaOH(aq)
(ii) [Co(H2O)6]2+ and an excess of NH3(aq)
(iii) [Co(H2O)6]2+ and an excess of NaCl(aq)
(iv) Name the type of reaction that occurs in (b)(iii).
(c) Cobalt forms the complex ion [Co(NH3)2(en)2]2+. The abbreviation en is used for the bidentate ligand 1,2-diaminoethane, H2NCH2CH2NH2. The complex ion shows both geometrical and optical isomerism.
(i) Define the term bidentate ligand.
(ii) Draw three-dimensional diagrams for the two optical isomers of [Co(NH3)2(en)2]2+.
Each en ligand can be represented using 
Answer/Explanation
Answer (a)(i) +2 (a)(ii)
bond angle labelled 109.5°.
(b)(i) [Co(H2O)6]2+ + 2OH– → Co(H2O)4(OH)2 + 2H2O OR [Co(H2O)6]2+ + 2OH– → Co(OH)2 + 6H2O blue precipitate
(b)(ii) [Co(H2O)6]2+ + 6NH3 → [Co(NH3)6]2+ + 6H2O
yellow / brown / straw solution
(b)(iii) [Co(H2O)6]2+ + 4Cl – → [CoCl4]2– + 6H2O
blue solution
(b)(iv) ligand exchange
(c)(i) (a species) that donates two lone pairs to form two dative bonds
to a (transition) metal atom / metal ion
(c)(ii)
Question
An excess of sodium iodide is added to a solution of copper(II) sulfate. Iodine and a white precipitate of copper(I) iodide are formed.
(a) Write an equation for the reaction that occurs.
(b) (i) Explain why the copper(II) sulfate solution is coloured.
(ii) Suggest why the precipitate of copper(I) iodide is white.
(c) Use suitable \(E^{\theta }\) values from the Data Booklet to predict whether iodide ions can reduce Cu2+ to Cu+ under standard conditions. Explain your answer.
(d) An excess of sodium iodide is added to copper(II) sulfate solution. Copper(I) iodide forms as
a precipitate. After precipitation, [Cu+] is much lower than 1.0moldm-3.
Use this information and your answer to (c) to explain how the relevant electrode potentials
change and hence why I– ions can reduce Cu2+ ions.
Answer/Explanation
Answer (a) 2Cu2+ + 4I– → 2CuI + I2 OR 2CuSO4 + 4NaI → 2CuI + I2 + 2Na2SO4
(b)(i) M1: d-d orbital splitting occurs
M2: electron(s) promoted / excited
M3: wavelength / frequency of light is absorbed
M4: colour seen is complementary
OR wavelength / frequency of light not absorbed is seen
(b)(ii) (for Cu+) 3d10 OR 3d subshell full
(c) M1: (Cu2+ / Cu+) \(E^{\theta }\) = (+)0.15 V AND (I2 / I–) \(E^{\theta }\) = (+)0.54 V
M2: No, since (\(E^{\theta }\)cell) negative / –0.39 V
OR No, since (I2 / I–) is more positive than (Cu2+ / Cu+)
OR No, I2 is more easily reduced OR No, I2 stronger oxidant ORA
(d) M1: Cu2+ / Cu+ E becomes more positive as equilibrium shifts to the right
M2: The new E for Cu2+ / Cu+ is more positive than 0.54 /\(E^{\theta }\)(I2 / I– )
Question
The structure of phenylethanoic acid is shown.
(a) Give the number of different peaks in the carbon-13 (13C) NMR spectrum of phenylethanoic acid.
(i) Write an equation for the reaction between chloromethane and AlCl3.
Answer/Explanation
Answer (a) 6
(b) M1: trend phenylethanoic acid > phenol > ethanol
M2: why phenylethanoic acid is the strongest
• negative inductive electron withdrawing effect of C=O
which weakens O-H bond / stabilises anion
M3: why phenol is stronger than ethanol/ weaker than phenylethanoic acid • oxygen lone pair is delocalised into the ring system
which weakens O-H bond / stabilises anion
M4: why ethanol is the weakest
• electron donating alkyl / ethyl group
which strengthens O-H bond / destabilises anion
(c)(i) CH3Cl + AlCl3 → +CH3 + AlCl4–
(c)(ii)
7(d)(i) Br2 + UV light
7(d)(ii)
7(d)(iii) CHECK Q is correct
step 2 – KCN in ethanol + heat
step 3 – HCl(aq) + heat/reflux/boil
(d)(iv)
Question
Phenylamine, C6H5NH2, and ethylamine, C2H5NH2, can be distinguished by adding aqueous bromine.
(a) State what is seen when aqueous bromine is added to phenylamine.
(b) Suggest what is seen when aqueous bromine is added to ethylamine.
(c) Draw the structure of the organic product formed when an excess of aqueous bromine is added to phenylamine.
(d) Name the product you have drawn in (c).
Answer/Explanation
Answer (a) bromine decolourised OR orange / brown to colourless
white precipitate
(b) no change
(c)
(d) 2,4,6-tribromophenylamine ECF 8(c) for a bromophenylamine
Question
Compound T is made by a three-stage synthesis.
(a) In stage 1, phenylethanoic acid reacts with a suitable reagent to form compound R.
Suggest a suitable reagent for stage 1.
(b) In stage 2, compound R reacts with ethylamine to form compound S.
(i) Name the functional group formed in stage 2.
(ii) Identify the other product formed in stage 2.
(c) In stage 3, compound S reacts with a suitable reagent to form compound T.
(i) State the formula of a suitable reagent for stage 3.
(ii) Name the type of reaction that occurs in stage 3.
Draw the structures of the ions responsible for these peaks.
(e) The proton (1H) NMR spectrum of compound T shows hydrogen atoms in different environments. Six of these environments are shown on the structure using letters a, b, c, d, e and f.
Use the letters a, b, c, d, e and f to answer the questions that follow. The questions relate to the proton (1H) NMR spectrum of T.
Proton d does not cause splitting of the peaks for protons c or e under the conditions used.
Each answer may be one, or more than one, of the letters a, b, c, d, e and f.
(i) Identify the proton or protons with a chemical shift (δ) in the range 6.0 to 9.0.
(ii) Identify the proton or protons whose peak will disappear if D2O is added.
(iii) Identify the proton or protons whose peak is a triplet.
(iv) Identify the proton or protons with the lowest chemical shift (δ).
Answer/Explanation
Answer (a) PCl5 OR PCl3 OR SOCl2 (b)(i) amide (b)(ii) HCl / hydrogen chloride OR C2H5NH3Cl / ethylammonium chloride
(c)(i) LiAl H4
(c)(ii) reduction
(d)(i) relative abundance = 2 carbons × 1.1 × 0.62 = 1.36 / 1.4 min 2sf
(d)(ii) CH3CH2 + / C2H5+ C6H5CH2+ (e)(i) a
(e)(iii) d
(e)(iii) b, c, f
(e)(iv) f
Question
Valine (Val) and lysine (Lys) are amino acids. The structures of these amino acids can be found in the Data Booklet.
The isoelectric point of an amino acid is the pH at which it exists as a zwitterion. The isoelectric
point of valine is 6.0. The isoelectric point of lysine is 9.7.
(a) Draw the structure of valine at pH 6.0.
(b) A solution of lysine is produced with pH 9.7. Dilute sulfuric acid is added slowly until the pH of
the solution is 1.0. The sulfuric acid reacts with lysine to produce different organic ions that are
not present in significant concentrations at pH 9.7.
Draw the structures of three of the organic ions that form during the addition of sulfuric acid in
the boxes. Draw the organic ion present at pH 1.0 in box C.
(c) Draw the structure of the dipeptide Val-Lys. The peptide bond should be shown fully displayed.
Answer/Explanation
Answer (a)
(b) A and B any two from: [all net single positive charge]
(c) M1: peptide link correct and displayed unit including C=O M2: everything else correct
