Chemistry mock paper 2.

Answer: Cathode $\mathrm{Al^{3+} + 3e^- \rightarrow Al}$; anode $\mathrm{C + 2O^{2-} \rightarrow CO_2 + 4e^-}$

Cathode: $\mathrm{Al^{3+} + 3e^- \rightarrow Al}$ [1].

Anode: $\mathrm{C + 2O^{2-} \rightarrow CO_2 + 4e^-}$ [1]. Also accept $\mathrm{2O^{2-} \rightarrow O_2 + 4e^-}$ followed by $\mathrm{C + O_2 \rightarrow CO_2}$. Electrons on the wrong side lose the mark.

Answer: $1.07\times10^{10}$ C

$n(\mathrm{Al}) = 1.000\times10^{6}/26.98 = 3.706\times10^{4}$ mol [1].

$Q = 3nF = 3\times3.706\times10^{4}\times96\,485 = 1.073\times10^{10}$ C [1].

Forgetting the factor 3 ($3.58\times10^{9}$ C) loses the second mark.

Answer: 10.6 h

Useful current $= 0.940\times3.00\times10^{5} = 2.82\times10^{5}$ A [1].

$t = 1.073\times10^{10}/2.82\times10^{5} = 3.80\times10^{4}$ s $= 10.6$ h [1].

Ignoring the efficiency gives 9.93 h and loses the first mark. Allow error carried forward from (b).

Answer: $1.33\times10^{4}$ kW h (13.3 MW h)

Total charge passed $= 1.073\times10^{10}/0.940 = 1.141\times10^{10}$ C, so energy $= QV = 1.141\times10^{10}\times4.20 = 4.79\times10^{10}$ J [1].

$4.79\times10^{10}/3.60\times10^{6} = 1.33\times10^{4}$ kW h [1].

Using the charge from (b) without dividing by 0.940 gives $1.25\times10^{4}$ kW h and loses the first mark. Also accept $Pt$: $3.00\times10^{5}\times4.20\times3.80\times10^{4}$ s.

Answer: 334 kg of carbon; 667 m³ of CO₂

4 mol of electrons per mol of C, so $n(\mathrm{C}) = 3\times3.706\times10^{4}/4 = 2.780\times10^{4}$ mol, mass $= 2.780\times10^{4}\times12.01 = 3.34\times10^{5}$ g $= 334$ kg [1].

$V(\mathrm{CO_2}) = 2.780\times10^{4}\times24.0 = 6.67\times10^{5}$ dm³ $= 667$ m³ [1].

Equivalently: 3 mol of C for every 4 mol of Al, from $\mathrm{2Al_2O_3 + 3C \rightarrow 4Al + 3CO_2}$.

Answer: Hot anodes also burn in air; some carbon monoxide is formed instead of CO₂ (also: the wasted charge oxidises carbon; anode stubs are discarded)

Any two, [1] each:

• the hot carbon anodes react with oxygen of the air above the bath (they burn);
• some CO forms instead of CO₂, which uses two carbon atoms for every four electrons instead of one;
• the 6% of wasted charge also oxidises carbon (for example aluminium dissolved in the melt is reoxidised by CO₂, and the charge has to be passed again);
• the ends (butts) of used anodes cannot be consumed and are thrown away or recycled.

\"Impurities\" alone scores 0.

Answer: $A_r \approx 207$; X is lead, Pb

In XCl₂, 70.90 g of Cl is 25.5% of the mass, so $M(\mathrm{XCl_2}) = 70.90/0.255 = 278$ g mol^-1 [1].

$A_r(\mathrm{X}) = 278 - 70.9 = 207$, so X is lead [1].

Answer: A Pb(NO₃)₂, B PbI₂, C NO₂, D PbO; $\mathrm{2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2}$

A lead(II) nitrate, Pb(NO₃)₂, and B lead(II) iodide, PbI₂ [1].

C nitrogen dioxide, NO₂, and D lead(II) oxide, PbO [1].

$\mathrm{2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2}$ [1]. Allow error carried forward from a wrong X if the chemistry is consistent.

Answer: Pb₃O₄ $=$ 2PbO$\cdot$PbO₂: two Pb(II) for each Pb(IV)

Pb $90.67/207.2 = 0.4376$, O $9.33/16.00 = 0.583$; ratio $1 : 1.333 = 3 : 4$, so Pb₃O₄ [1].

Average oxidation state $+8/3$ is not a whole number; it is 2PbO$\cdot$PbO₂, with two Pb(+2) for every one Pb(+4) [1].

Check: Pb₃O₄ has $64.00/685.6 = 9.33\%$ O.

Answer: $\mathrm{Pb_3O_4 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + PbO_2 + 2H_2O}$

$\mathrm{Pb_3O_4 + 4HNO_3 \rightarrow 2Pb(NO_3)_2 + PbO_2 + 2H_2O}$ [1]. F is lead(IV) oxide, PbO₂. The acid reacts only with the basic PbO part; this is an acid-base reaction, not a redox one.

Answer: $\mathrm{PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O}$; down group 14 the +2 state becomes more stable than +4 (inert pair effect)

$\mathrm{PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O}$ [1].

Down group 14 the two outer s electrons (6s² for lead) are held more tightly and are less often used in bonding (the inert pair effect), so +2 is the more stable state for lead and Pb(IV) is easily reduced to Pb(II); for C and Si the +4 state is the stable one [1].

\"Lead is a metal\" alone scores 0.

Answer: 0.62 g dm^-3

$K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^-}]^2 = s(2s)^2 = 4s^3$, so $s = (9.8\times10^{-9}/4)^{1/3} = 1.35\times10^{-3}$ mol dm^-3 [1].

$M(\mathrm{PbI_2}) = 461.0$ g mol^-1, so solubility $= 1.35\times10^{-3}\times461.0 = 0.62$ g dm^-3 [1].

Using $K_{sp} = s^3$ (gives $2.1\times10^{-3}$ mol dm^-3, 0.99 g dm^-3) or $s\cdot2s$ loses the first mark.

Answer: Empirical formula C₄H₅ ($M = 53$); $106/53 = 2$, so C₈H₁₀

C $90.5/12.01 = 7.54$, H $9.5/1.008 = 9.42$; ratio $1 : 1.25 = 4 : 5$, empirical formula C₄H₅ [1].

$M(\mathrm{C_4H_5}) = 53.1$; $106/53.1 = 2$, so C₈H₁₀ [1].

Answer: 4; a benzene ring (one ring and three C=C) that does not add bromine

$(2\times8 + 2 - 10)/2 = 4$; a benzene ring accounts for exactly 4 (one ring and three double bonds), and benzene rings do not decolourise bromine water, unlike alkenes [1]. Both the value and the reason are needed.

Answer: W ethylbenzene; X 1,3-dimethylbenzene; Y 1,2-dimethylbenzene; Z 1,4-dimethylbenzene

W gives benzoic acid, C₆H₅COOH, so it has one side chain of two carbons: ethylbenzene, C₆H₅CH₂CH₃ [1].

X, Y, Z give benzenedicarboxylic acids, so they are the three dimethylbenzenes. Only the 1,2-acid (COOH groups next to each other) can close a five-membered anhydride ring, so Y is 1,2-dimethylbenzene [1].

Z gives one ring monobromination product, so all four ring hydrogens are equivalent: Z is 1,4-dimethylbenzene, and X is 1,3-dimethylbenzene [1].

Old names (o-, m-, p-xylene) are accepted.

Answer: X: 3; Y: 2

X (1,3-dimethylbenzene) has ring hydrogens at C2, C4/C6 (equivalent) and C5: 3 products [1].

Y (1,2-dimethylbenzene) has ring hydrogens at C3/C6 and C4/C5: 2 products [1].

Allow error carried forward from wrong structures in (c) if the counting is right for the structures given.

Answer: 2 signals, ratio 3 : 2 (6H : 4H), both singlets

Two environments: the six CH₃ hydrogens and the four ring hydrogens, area ratio $6 : 4 = 3 : 2$ [1].

Both signals are singlets: the CH₃ hydrogens have no H on the neighbouring carbon, and the four ring hydrogens are all equivalent, so they do not split each other [1].

Answer: (1-Bromoethyl)benzene; the benzylic radical is stabilised by the ring; chiral, but formed as a racemic mixture, so no rotation

(1-Bromoethyl)benzene, C₆H₅CHBrCH₃; removing H from CH₂ gives a radical next to the ring, which is stabilised by delocalisation of the unpaired electron into the ring [1].

The product is chiral, but the planar radical is attacked equally from both faces, so a racemic (50 : 50) mixture forms and there is no overall rotation [1].

\"Yes, it is chiral\" alone loses the second mark.

Answer: $C = 10.06$ kJ K^-1

Energy released by the benzoic acid $= 0.8155\times26.434 = 21.56$ kJ [1].

$C = 21.56/2.142 = 10.06$ kJ K^-1 [1].

Using $mc\Delta T$ with $c = 4.18$ J g^-1 K^-1 cannot work: the mass of water is not given, and the steel vessel also absorbs energy. That is the point of calibrating.

Answer: $\Delta_c U = -5151$ kJ mol^-1

Energy released $= 10.06\times2.402 = 24.16$ kJ and $n = 0.6012/128.16 = 4.691\times10^{-3}$ mol [1].

$\Delta_c U = -24.16/4.691\times10^{-3} = -5151$ kJ mol^-1, with the negative sign [1]. Accept $-5149$ to $-5155$ (the unrounded heat capacity gives $-5153$). Allow error carried forward from (a).

A positive value loses the second mark: the reaction releases energy.

Answer: $\mathrm{C_{10}H_8(s) + 12O_2(g) \rightarrow 10CO_2(g) + 4H_2O(l)}$; $\Delta_c H = -5156$ kJ mol^-1

$\mathrm{C_{10}H_8(s) + 12O_2(g) \rightarrow 10CO_2(g) + 4H_2O(l)}$, so $\Delta n_{\mathrm{gas}} = 10 - 12 = -2$ [1].

$\Delta_c H = \Delta_c U + \Delta n_{\mathrm{gas}}RT = -5151 + (-2)(8.314\times10^{-3})(298) = -5151 - 5.0 = -5156$ kJ mol^-1 [1]. Accept $-5154$ to $-5160$, or error carried forward from (b).

Common errors: counting the water as gas ($\Delta n_{\mathrm{gas}} = +2$, giving $-5146$), or using $R = 8.314$ without changing J to kJ.

Answer: $\Delta_f H^\circ$(s) $= +78$ kJ mol^-1; $\Delta_f H^\circ$(g) $= +151$ kJ mol^-1

$\Delta_f H^\circ(\mathrm{s}) = 10(-393.5) + 4(-285.8) - (-5156) = -5078.2 + 5156 = +78$ kJ mol^-1 [1]. Accept $+74$ to $+82$, or error carried forward from (c): it is a small difference of two large numbers.

$\Delta_f H^\circ(\mathrm{g}) = 78 + 72.6 = +151$ kJ mol^-1 [1].

Getting the sign of the combustion term wrong (giving about $-10\,234$) loses the first mark.

Answer: 5 C=C, 6 C-C, 8 C-H; estimate $+468$ kJ mol^-1; stabilisation about 317 kJ mol^-1

5 C=C, 6 C-C and 8 C-H bonds (11 carbon-carbon bonds in all) [1].

Atomising the elements: $10(717) + 4(436) = 8914$ kJ; forming the bonds releases $5(612) + 6(347) + 8(413) = 8446$ kJ; estimate $\Delta_f H(\mathrm{g}) = 8914 - 8446 = +468$ kJ mol^-1 [1].

Stabilisation energy $= 468 - 151 = 317$ kJ mol^-1: real naphthalene is this much more stable than the Kekulé structure [1]. Accept 310 to 325, or error carried forward from (d).

Common errors: comparing with the solid (giving about 390), although bond enthalpies only apply to gases; breaking 8 H-H bonds instead of 4 (the 8 H atoms come from 4 H₂).

Answer: Less per ring (317 < 2 x 172 = 344), so less stabilisation is lost when it reacts

$317 \lt 2\times172 = 344$ kJ mol^-1, so naphthalene is less stabilised per ring; attacking one ring still leaves a complete benzene ring behind, so less delocalisation energy is lost and reaction is easier [1]. Both the comparison and the reason are needed for the mark.

Answer: $\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}$

$\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$ and $\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$ [1].

$\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}$ [1].

Answer: First permanent pale pink colour

The manganate(VII) is decolourised while Fe²⁺ remains; the end point is the first permanent pale pink colour from one drop of excess MnO₄^- [1]. \"Turns purple\" is accepted only if \"pale pink\" or \"first permanent\" is also given.

Answer: MnO₄^- would also oxidise Cl^- to Cl₂, so the titre would be too large

Manganate(VII) is a strong enough oxidising agent to oxidise chloride ions to chlorine, so extra manganate(VII) would be used and the titre would be too high [1].

Answer: 63.1 mg per tablet, 97.1% of the label value

$n(\mathrm{MnO_4^-}) = 22.60\times10^{-3}\times0.00500 = 1.130\times10^{-4}$ mol, so $n(\mathrm{Fe}) = 5.650\times10^{-4}$ mol in 25.00 cm³ [1].

In 250.0 cm³ (five tablets): $5.650\times10^{-3}$ mol; per tablet $1.130\times10^{-3}$ mol $\times55.85 = 63.1$ mg, which is $63.1/65.0 = 97.1\%$ of the label value [1].

Forgetting to divide by five tablets (316 mg) or to multiply by ten aliquots (6.31 mg) loses the second mark.

Answer: 92.0%

Titration 1 measures Fe²⁺ only; titration 2 measures Fe²⁺ plus the Fe³⁺ reduced by zinc, that is all the iron [1].

$\%\,\mathrm{Fe^{2+}} = 20.80/22.60\times100 = 92.0\%$ [1].

Giving 8.0% (the iron(III) fraction) loses the second mark unless the iron(II) value is also stated.

Answer: $\mathrm{Zn + 2Fe^{3+} \rightarrow Zn^{2+} + 2Fe^{2+}}$; leftover zinc would reduce MnO₄^- (or keep re-forming Fe²⁺), making the titre too high

$\mathrm{Zn + 2Fe^{3+} \rightarrow Zn^{2+} + 2Fe^{2+}}$ [1].

Excess zinc would itself react with the manganate(VII) (directly, or by reducing the Fe³⁺ formed back to Fe²⁺), so the titre would be too large and there would be no sharp end point [1].

Note that Zn²⁺ left in solution does not react with manganate(VII).

Answer: $A \approx 0.25$

$n(\mathrm{Fe^{3+}})$ in 25.00 cm³ $= (22.60 - 20.80)\times10^{-3}\times0.00500\times5 = 4.50\times10^{-5}$ mol, so $[\mathrm{Fe^{3+}}] = 1.80\times10^{-3}$ mol dm^-3 in the tablet solution [1].

After dilution by 50: $3.60\times10^{-5}$ mol dm^-3, so $A = \varepsilon cl = 7.0\times10^{3}\times3.60\times10^{-5}\times1.00 = 0.25$ [1].

Forgetting the dilution ($A = 12.6$) loses the second mark; allow error carried forward from (e).