Chemistry Paper 2 June 2015

1. A$pVT = nR$
2. B$pV = nRT$
3. C$pV = mRT$
4. D$p/V = nRT$

1. Ahave negligible volume and no attractions
2. Boccupy a finite volume and attract one another
3. Ccollide elastically with the walls of the vessel
4. Dmove in straight lines between collisions

1. Ahigh pressure and low temperature
2. Bhigh pressure and high temperature
3. Clow pressure and low temperature
4. Dlow pressure and high temperature

1. Athe molecules collide perfectly elastically with each other and with the container walls only when the gas is compressed into a very small, fixed volume
2. Bthe gas constant R itself changes value, cancelling out any deviation from the ideal gas equation
3. Cthe molecules move fast enough to escape through the container walls, leaving behind a perfect vacuum
4. Dintermolecular attractions and the molecules' own volume become negligible compared with the container's volume and the molecules' kinetic energy

Answer this when you sit the paper.

1. A$3.0\times10^3$
2. B$6.0\times10^3$
3. C60
4. D600

1. Aequals the sum of the bond energies of the reactants
2. Bis always negative
3. Cis independent of the route taken
4. Ddepends on the amount of catalyst used

1. A$-1696.6$
2. B$-136.8$
3. C$+136.8$
4. D$+1696.6$

1. Athe temperature and the pressure
2. Bthe ionic charges and the ionic radii
3. Cthe melting point and the solubility
4. Dthe electronegativity and the mass of the ions

1. Aone mole of a substance is completely burnt in excess oxygen, all reactants and products in their standard states
2. Bone mole of bonds is broken in the gaseous state
3. Cone mole of a solute dissolves completely in a large excess of water
4. Done mole of a compound is formed from its constituent elements, each in their standard states, at the standard pressure

1. A$\Delta H_c(C_2H_6) + \Delta H_c(C_2H_4) + \Delta H_c(H_2)$
2. B$\Delta H_c(C_2H_6) - \Delta H_c(C_2H_4) - \Delta H_c(H_2)$
3. C$\Delta H_c(C_2H_4) - \Delta H_c(H_2) - \Delta H_c(C_2H_6)$
4. D$\Delta H_c(C_2H_4) + \Delta H_c(H_2) - \Delta H_c(C_2H_6)$

1. Adecreases, because a higher ionic charge actually weakens the bond holding the lattice together
2. Bstays the same, since charge has no effect on lattice energy
3. Cdepends only on the size of the ions, not their charge
4. Dincreases, because the electrostatic attraction between the ions strengthens

1. Astays exactly constant, because lattice energy in this simple ionic model depends only on the ionic charges present
2. Bincreases only if the cation is larger than the anion
3. Cdecreases, because the ions cannot approach as closely, weakening the electrostatic attraction between them
4. Dincreases, because larger ions hold more charge on their surface

1. Avery slowly, giving magnesium hydroxide and hydrogen
2. Bexplosively, giving magnesium hydride
3. Cnot at all, even on boiling
4. Dvigorously, giving magnesium oxide and hydrogen

1. Aas a fuel in rockets
2. Bas a bleaching agent
3. Cas a fertiliser
4. Das a refractory lining for furnaces

1. Ais more easily reduced than $Mg^{2+}$
2. Bforms a covalent carbonate
3. Cis larger, so it polarises the carbonate ion less
4. Dcarries a higher charge than $Mg^{2+}$

Answer this when you sit the paper.

1. A$K_{sp} = [Ca^{2+}][SO_4^{2-}]$
2. B$K_{sp} = [Ca^{2+}]^2[SO_4^{2-}]$
3. C$K_{sp} = [Ca^{2+}]/[SO_4^{2-}]$
4. D$K_{sp} = [Ca^{2+}] + [SO_4^{2-}]$

Answer this when you sit the paper.

1. Adoes not occur, because the ionic product is smaller than $K_{sp}$
2. Bcannot be predicted without knowing the temperature of mixing
3. Coccurs, because the ionic product, $2.5\times10^{-5}$, exceeds $K_{sp}$, $2\times10^{-5}$
4. Ddoes not occur, because $CaSO_4$ is completely soluble in water

1. Athe lattice energy of the sulphate increases sharply while its hydration enthalpy stays almost completely constant
2. Bthe hydration enthalpy of the cation falls faster than the lattice energy of the sulphate falls, making dissolving less exothermic
3. Cthe sulphate ion itself becomes more strongly hydrated, trapping the calcium ions in solid form
4. Dcalcium sulphate becomes covalent, so it can no longer dissolve as separate ions

1. Adecreases steadily
2. Bstays approximately constant
3. Cincreases then sharply decreases
4. Dincreases steadily as the group is descended, mirroring the same steady increase seen in the solubility trend of the Group 2 hydroxides

1. Astays approximately constant
2. Bincreases then decreases
3. Cincreases steadily
4. Ddecreases steadily

1. Aan alkene which decolourises bromine
2. Ban ester which is volatile and has a pleasant smell
3. Can acid which reacts with the skin
4. Dan alcohol which evaporates slowly

1. Aethanoic acid and 3-methylbutan-1-ol
2. Bpropanoic acid and butan-1-ol
3. Cmethanol and pentanoic acid
4. Dethanol and 3-methylbutanoic acid

1. Athe bromine is decolourised and a white precipitate of 2,4,6-tribromophenol forms
2. Ba yellow precipitate of triiodomethane
3. Cno visible change; no reaction occurs
4. Dan orange precipitate; a phenylhydrazone

1. Afree radical substitution
2. Belectrophilic substitution onto the aromatic ring
3. Celectrophilic addition
4. Dnucleophilic substitution

1. Abromine dissolves better in phenol than in benzene, increasing the local bromine concentration
2. Bphenol's aromatic ring is noticeably smaller in diameter than benzene's, so a bromine molecule fits into it more easily
3. Cthe $-OH$ group donates electron density into the ring by resonance, strongly activating it towards attack by bromine
4. Dphenol is a stronger acid than benzene, so it reacts with bromine as an acid-base reaction

1. Aaddition polymerisation of an alkene
2. Bnucleophilic addition to a carbonyl group
3. Cesterification, the reaction between an alcohol and a carboxylic acid
4. Dsaponification of the ester using hot, concentrated aqueous sodium hydroxide solution

1. Aaldehyde and ester
2. Bcarboxylic acid and ketone
3. Cphenol and amine
4. Dalcohol and alkene

1. Aa carboxylic acid
2. Ban alkene
3. Ca ketone
4. Dan ester

1. Aa condensation polymer
2. Ban addition polymer
3. Ca polyamide
4. Da polyester

1. Aelectrophilic substitution
2. Bnucleophilic addition
3. Cfree radical substitution
4. Delectrophilic addition

1. Ais completely insoluble in the aqueous solvents used to carry out both of these particular tests
2. Bcontains no carbonyl (aldehyde or ketone) group, which both reagents require
3. Ccontains too many double bonds for either reagent to attack selectively
4. Dhas already reacted fully with the alcohol group present

Answer this when you sit the paper.

1. Aa diol, with an $-OH$ group added to each of the two carbons of the double bond
2. Ba single alcohol group at one end of the chain only
3. Can alkane, with both double bonds of the isoprene unit fully reduced by the cold dilute oxidant
4. Da carboxylic acid at each double bond

1. Aoxidatively cleaving each double bond, converting the fragments into carbonyl or carboxylic acid groups
2. Breducing them to single bonds, converting isoprene into an alkane
3. Cadding two hydroxyl groups across each double bond in turn, without breaking any carbon-carbon bonds at all
4. Dpolymerising the isoprene units into a longer chain