Chemistry Paper 2 November 2009

1. A0
2. B1
3. C2
4. D3

1. Arate $= k[NO][Cl_2]$
2. Brate $= k[NO][Cl_2]^2$
3. Crate $= k[NO]^2[Cl_2]^2$
4. Drate $= k[NO]^2[Cl_2]$

1. A0.01
2. B0.1
3. C1.0
4. D10.0

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1. Atemperature of the flask rises sharply, since this reaction releases a very large amount of heat very quickly
2. Bmass of the sealed flask decreases steadily over time, since gas is assumed to escape slowly through the container walls
3. Ctotal colour intensity of the mixture increases, since $NOCl$ is a deep purple gas that intensifies steadily
4. Dtotal pressure (or volume) of the gas mixture decreases, since three moles of gas react to give two moles of gas

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1. Afirst order, since doubling $[Cl_2]$ alone exactly doubles the rate
2. Bsecond order, since doubling $[Cl_2]$ alone should quadruple the rate but does not here
3. Cthird order, matching the overall order of the whole reaction exactly
4. Dzero order, since the rate does not depend on $[Cl_2]$ at all in this reaction

1. Aproviding a solid surface where the gas molecules adsorb, offering an alternative reaction pathway with a lower activation energy
2. Breacting permanently and irreversibly with $Cl_2$ gas to form a new, separate chlorine-containing compound that then slowly decomposes to give the final product
3. Cabsorbing all the heat released, which somehow speeds up how fast the molecules collide with each other
4. Dincreasing the pressure inside the container, which pushes the equilibrium further towards the product side

1. Aoverall order 2; units $mol^{-2}dm^{6}s^{-1}$
2. Boverall order 2; units $mol^{-1}dm^{3}s^{-1}$
3. Coverall order 3; units $mol^{-2}dm^{6}s^{-1}$
4. Doverall order 3; units $mol\,dm^{-3}s^{-1}$

1. Athe cathode, because hydroxide ions are left in solution
2. Bthe anode, because chlorine is discharged there
3. Cthe anode, because sodium ions migrate there
4. Dthe cathode, because chloride ions migrate there

1. A$Cl_2 + 2e^- \rightarrow 2Cl^-$
2. B$2Cl^- + 2H_2O \rightarrow Cl_2 + 2OH^- + H_2$
3. C$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$
4. D$2Cl^- \rightarrow Cl_2 + 2e^-$

1. A7.10 g
2. B0.10 g
3. C3.55 g
4. D4.00 g

Answer this when you sit the paper.

Answer this when you sit the paper.

1. Ahydrogen: $2H^+ + 2e^- \rightarrow H_2$; overall: $2NaCl + 2H_2O \rightarrow 2NaOH + H_2 + Cl_2$
2. Bhydrogen: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$; overall: $2NaCl \rightarrow 2Na + Cl_2$
3. Chydrogen: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$; overall: $2NaCl + 2H_2O \rightarrow 2NaOH + H_2 + Cl_2$
4. Dhydrogen: $2H^+ + 2e^- \rightarrow H_2$; overall: $NaCl + H_2O \rightarrow NaOH + HCl$

1. A$2[Ag^+][CO_3^{2-}]$
2. B$[Ag^+]^2[CO_3^{2-}]$
3. C$[Ag^+][CO_3^{2-}]$
4. D$[Ag^+][CO_3^{2-}]^2$

1. A$2.50 \times 10^{-6}$
2. B$2.15 \times 10^{-4}$
3. C$5.0 \times 10^{-13}$
4. D$7.07 \times 10^{-7}$

1. Aincrease the value of $K_p$
2. Bdecrease the value of $K_p$
3. Cmake $K_p$ equal to zero
4. Dleave the value of $K_p$ unchanged

1. A$2[Ag^+][Br^-]$
2. B$[Ag^+][Br^-]$
3. C$[Ag^+]^2[Br^-]$
4. D$[Ag^+][Br^-]^2$

Answer this when you sit the paper.

1. Asilver bromide, because $7.07\times10^{-7}$ falls within the required $10^{-7}$ to $10^{-6}$ range
2. Bsilver carbonate, because $2.15\times10^{-4}$ is comfortably above the minimum $10^{-7}$ requirement
3. Cneither compound, because both values fall outside the required range entirely
4. Dboth compounds equally, because each provides more than enough silver ions to kill bacteria

1. A$K_p = \dfrac{1}{P_{CO_2}}$
2. B$K_p = \dfrac{P_{Ag_2O}\,P_{CO_2}}{P_{Ag_2CO_3}}$
3. C$K_p = P_{Ag_2O} \times P_{CO_2}$
4. D$K_p = P_{CO_2}$

1. Amakes $K_p$ meaningless, since gases are not normally involved in a solid-solid equilibrium like this one
2. Bdecreases $K_p$, since a rise in temperature is sometimes wrongly thought to always disfavour the forward reaction, regardless of its enthalpy sign
3. Cincreases $K_p$, since raising the temperature shifts an endothermic equilibrium further towards the products, increasing $P_{CO_2}$
4. Dleaves $K_p$ unchanged, since $K_p$ for this reaction depends only on the amount of solid present, not on temperature

Answer this when you sit the paper.

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1. A$CuSO_4 \cdot 5H_2O$
2. B$Cu(OH)_2$
3. CCuO
4. D$[Cu(NH_3)_4(H_2O)_2]^{2+}$

1. A$CuCl_4^{2-}$
2. B$Cu(NH_3)_2^+$
3. C$[Cu(NH_3)_4(H_2O)_2]^{2+}$
4. D$[Cu(H_2O)_6]^{2+}$

1. Amore delocalised electrons and stronger metallic bonding
2. Ba larger atomic radius than calcium
3. Ca lower first ionisation energy than calcium
4. Da giant covalent structure

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1. Aunreacted aqueous copper(II) sulfate, $[Cu(H_2O)_6]^{2+}$, which the ammonia has not yet reached
2. Ba fresh, unreacted batch of dilute aqueous ammonia solution that has simply not yet come into contact with the copper sulfate
3. Cpure water released as a by-product of the precipitation reaction happening in the middle layer of the test tube
4. Ddissolved sodium sulfate formed as a spectator-ion by-product of the reaction taking place higher up

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1. Aexcess $NH_3$ molecules displace water ligands from the copper(II) complex, forming the deeper-blue $[Cu(NH_3)_4(H_2O)_2]^{2+}$ complex ion
2. Bexcess $NH_3$ reacts with the sulfate ions instead, releasing free $Cu^{2+}$ ions that are darker blue than the original complex
3. Cexcess $NH_3$ reduces $Cu^{2+}$ to $Cu^+$, and $Cu^+$ ions are known to give an intensely dark blue colour in solution
4. Dexcess $NH_3$ simply dilutes the precipitate further, and dilute copper(II) solutions are always thought to appear darker blue than concentrated ones

1. Aconducts electricity only when molten or dissolved in water
2. Bforms at least one stable ion with a partially filled d sub-shell
3. Cis found in the s-block of the periodic table, like calcium
4. Dalways has a fixed, single oxidation state in all of its compounds

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1. Athe ammonia gas is prevented from escaping into the surrounding air during the experiment
2. Bthe copper sulfate is kept from dissolving fully, so a visible precipitate always remains at the very end
3. Cthe separate stages of the reaction (unreacted solution, precipitate, complex) can be seen as distinct layers rather than mixed together
4. Dthe reaction proceeds noticeably faster overall, since vigorous shaking would cool the mixture down and slow any chemical change taking place

1. Acondensation
2. Bhydrolysis
3. Coxidation
4. Daddition

1. A$CH \equiv CCH_2CH_3$
2. B$CH_2=CHCH=CH_2$
3. C$CH_3CH=CHCH_3$
4. D$CH_2=CHCH_2CH_3$

1. Aelectrophilic substitution
2. Belectrophilic addition
3. Cnucleophilic substitution
4. Dfree radical substitution

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1. Aeach double-bonded backbone carbon carries two different groups: a hydrogen atom and the continuing polymer chain
2. Bthe polymer chain is extremely long, and cis-trans isomerism only ever occurs in very large molecules
3. Cbutadiene itself is a gas at room temperature, and only gaseous monomers can form geometrical isomers
4. Dthe double bonds are conjugated with the aromatic benzene ring contributed by styrene, and this conjugation is what creates the isomerism

1. Anumber of styrene and butadiene units present in each repeat unit
2. Bspatial arrangement (geometry) around the C=C double bonds, not the order in which atoms are connected
3. Cmolecular formula of the repeat unit, since cis and trans forms are sometimes wrongly assumed to be entirely different compounds
4. Dtype of polymerisation used to make the rubber, from addition to condensation

1. Asimple dehydration; water is removed from the polymer, converting every C=C bond into a $C \equiv C$ triple bond
2. Belectrophilic addition; every C=C double bond gains an $-OH$ group without breaking the polymer chain at all
3. Coxidative cleavage; every C=C double bond along the backbone is broken, cutting the long polymer chain into much shorter fragments
4. Dfree-radical substitution; hydrogen atoms on the aromatic ring are replaced one by one over time, leaving the whole backbone chain fully intact and unbroken

1. Aleaves the double bonds completely untouched; instead substitutes hydrogen atoms on the aromatic ring only
2. Badds H and Br across a C=C double bond without breaking the chain; also does not react with the aromatic ring, which resists addition
3. Cbreaks the polymer chain into short fragments, just like $KMnO_4$; also reacts readily with the aromatic ring
4. Dadds H and Br across a C=C double bond without breaking the chain; but also substitutes bromine onto the aromatic ring