A-Level Chemistry Practice Test 2

Correct: (C)

Ar = (24 × 78.99 + 25 × 10.00 + 26 × 11.01) / 100 = (1895.76 + 250 + 286.26) / 100 = 24.32.

Correct: (A)

Chromium is an exception to the usual filling pattern: the 3d and 4s energies are so close that one 4s electron moves into the 3d sub-shell, giving the more stable half-filled 3d^5 arrangement.

Correct: (D)

The 3p electron in Al is further from the nucleus and is shielded by the 3s electrons, so it needs less energy to remove it despite the extra proton.

Correct: (B)

Ar = (35 × 75.8 + 37 × 24.2) / 100 = (2653 + 895.4) / 100 = 35.48, which is 35.5 to three significant figures.

Correct: (C)

Melting silicon requires breaking strong covalent bonds throughout the giant structure; melting phosphorus only overcomes weak van der Waals forces between P4 molecules.

Correct: (A)

p = nRT / V = (0.500 × 8.31 × 300) / 0.0200 = 62 325 Pa = 62.3 kPa.

Correct: (D)

Moles of NaOH = 0.200 × 40.0/1000 = 8.00 × 10^-3 mol. H2SO4 + 2NaOH → Na2SO4 + 2H2O, so moles of H2SO4 = 4.00 × 10^-3 mol; concentration = 4.00 × 10^-3 / 0.0250 = 0.160 mol dm^-3.

Correct: (B)

The empirical formula mass is 14; 84 / 14 = 6, so the molecular formula is 6 × CH2 = C6H12.

Correct: (C)

Moles of CaCO3 = 10.0/100 = 0.100 mol, so the theoretical yield of CaO = 0.100 × 56 = 5.60 g. Percentage yield = (4.48 / 5.60) × 100 = 80.0%.

Correct: (A)

Total Mr of reactants = 160 + 3(28) = 244; Mr of desired product = 2 × 56 = 112. Atom economy = (112 / 244) × 100 = 45.9%.

Correct: (D)

Volume = moles / concentration = 0.125 / 0.500 = 0.250 dm^3 = 250 cm^3.

Correct: (B)

In CO2 the carbon has two electron domains (two double bonds) which repel to opposite sides of the atom, giving a linear molecule with a bond angle of 180°.

Correct: (C)

Methanol forms hydrogen bonds between its molecules, which are much stronger than the van der Waals and dipole-dipole forces in the other substances, so far more energy is needed to separate its molecules.

Correct: (A)

Lattice enthalpy of formation is the enthalpy change when 1 mole of an ionic solid is formed from its gaseous ions; equation B is the standard enthalpy of formation and C is the (endothermic) lattice dissociation.

Correct: (D)

The small, highly charged Al^3+ ion has a high charge density and strongly polarises the large Cl^- ion, distorting its electron cloud and giving the bonding considerable covalent character (Fajans' rules).

Correct: (B)

Water is a bent molecule, so the two O-H bond dipoles do not cancel; CO2, CCl4 and BF3 have symmetrical shapes in which the bond dipoles cancel out.

Correct: (C)

In a metal the regular layers of ions can be displaced without breaking the bonding, because the sea of delocalised electrons moves with them; this also explains ductility.

Correct: (A)

q = mcΔT = 200 × 4.18 × 28.8 = 24 077 J = 24.08 kJ; ΔH = −24.08 / 0.0500 = −482 kJ mol^-1.

Correct: (D)

Using a Hess cycle through the combustion products: ΔH = ΔHc(C2H4) + ΔHc(H2) − ΔHc(C2H6) = −1411 + (−286) − (−1560) = −137 kJ mol^-1.

Correct: (B)

Three moles of gas form a liquid: the number of gas molecules decreases and the products are far more ordered, so the entropy decreases.

Correct: (C)

The reaction is feasible when ΔG = ΔH − TΔS becomes negative, i.e. when T > ΔH / ΔS = 50 000 / 200 = 250 K.

Correct: (A)

Bonds broken = 945 + 3(436) = 2253 kJ; bonds formed = 6(391) = 2346 kJ; ΔH = 2253 − 2346 = −93 kJ mol^-1.

Correct: (D)

This is the definition of the standard enthalpy of atomisation of an element (per mole of gaseous atoms formed).

Correct: (B)

3^order = 9, so the order is 2: rate is proportional to [X]^2.

Correct: (C)

k = rate / ([A][B]) = 4.8 × 10^-3 / (0.20 × 0.30) = 4.8 × 10^-3 / 0.060 = 0.080 mol^-1 dm^3 s^-1.

Correct: (A)

ln(k2/k1) = (Ea/R)(1/T1 − 1/T2); 1.386 = (Ea/8.31)(1/300 − 1/320) = (Ea/8.31)(2.083 × 10^-4). So Ea = 1.386 × 8.31 / 2.083 × 10^-4 = 55 300 J mol^-1 = 55 kJ mol^-1.

Correct: (D)

At a higher temperature the Maxwell-Boltzmann distribution shifts so that many more collisions have enough energy to react; the activation energy itself does not change.

Correct: (B)

By definition, the rate of a zero-order reaction is independent of the concentration of the reactant (rate = k).

Correct: (C)

At equilibrium: [ester] = [water] = 0.667; [acid] = [ethanol] = 1.00 − 0.667 = 0.333 mol dm^-3. Kc = (0.667 × 0.667) / (0.333 × 0.333) = 4.0 (no units, as there are equal moles on each side).

Correct: (A)

Kp is the partial pressures of the products (each raised to its stoichiometric power) divided by the partial pressures of the reactants raised to their powers.

Correct: (D)

Raising the temperature shifts the equilibrium in the endothermic (reverse) direction, so there are relatively fewer products and Kc decreases. (A catalyst never changes Kc.)

Correct: (B)

Mole fraction of NH3 = 2.0 / (2.0 + 6.0 + 2.0) = 0.20; partial pressure = 0.20 × 500 = 100 kPa.

Correct: (C)

Kc = [HI]^2 / ([H2][I2]) = (0.80)^2 / (0.10 × 0.10) = 0.64 / 0.010 = 64 (no units).

Correct: (A)

2(Cr) + 7(−2) = −2, so 2(Cr) = +12 and each Cr is +6.

Correct: (D)

EMF = E(more positive) − E(more negative) = +0.80 − (−0.76) = +1.56 V.

Correct: (B)

The strongest oxidising agent is the species most readily reduced, which belongs to the couple with the most positive electrode potential: Ag^+.

Correct: (C)

Q = It = 1.50 × (30.0 × 60) = 2700 C; moles of electrons = 2700 / 96 500 = 0.0280 mol; Ag^+ + e^- → Ag, so moles of Ag = 0.0280; mass = 0.0280 × 108 = 3.02 g.

Correct: (A)

For this reaction EMF = +0.77 − (+0.15) = +0.62 V, which is positive, so Sn^2+ reduces Fe^3+. In each of the other reactions the EMF is negative.

Correct: (D)

pH = −log(3.16 × 10^-3) = 3.00 − log(3.16) = 3.00 − 0.50 = 2.50.

Correct: (B)

pOH = −log(0.0200) = 1.70; pH = 14.00 − 1.70 = 12.30.

Correct: (C)

[H^+] = 10^-3.00 = 1.0 × 10^-3 mol dm^-3. Ka = [H^+]^2 / [HA] = (1.0 × 10^-3)^2 / 0.100 = 1.0 × 10^-5 mol dm^-3.

Correct: (A)

Added OH^- is removed by the large reserve of undissociated CH3COOH: CH3COOH + OH^- → CH3COO^- + H2O, so the [H^+] (and pH) changes only slightly.

Correct: (D)

[OH^-] = Kw / [H^+] = (1.0 × 10^-14) / (1.0 × 10^-3) = 1.0 × 10^-11 mol dm^-3.

Correct: (B)

The small Mg^2+ ion has the highest charge density and polarises the nitrate ion most strongly, weakening its bonds, so magnesium nitrate is the least thermally stable and decomposes most readily.

Correct: (C)

For example, chlorine displaces both bromine and iodine from their halide solutions because it is a stronger oxidising agent. (AgI is in fact a yellow precipitate, iodine is a grey-black solid, and fluorine is made by electrolysis of the molten mixture.)

Correct: (A)

The chloride ions act as ligands and replace the water ligands: [Cu(H2O)6]^2+ + 4Cl^- ⇌ [CuCl4]^2- + 6H2O. The tetrachlorocuprate(II) ion is yellow, and mixed with remaining blue aqua complex the solution looks green.

Correct: (D)

Al(OH)3 is amphoteric: it dissolves in excess NaOH but not in ammonia. Zn(OH)2 would also dissolve in excess ammonia (forming an ammine complex), and Cu^2+ gives a blue precipitate.

Correct: (B)

In the Haber process the solid iron catalyst provides a surface on which the gaseous reactants adsorb; catalyst and reactants are in different phases, so this is heterogeneous catalysis.

Correct: (C)

A termination step removes free radicals by combining two of them to form a stable molecule; steps A and B/D are initiation and propagation steps respectively.

Correct: (A)

Hot concentrated ethanolic KOH causes elimination of HBr to form ethene. (Aqueous KOH would instead give ethanol by substitution.)

Correct: (D)

Electrophilic addition of water follows Markovnikov's rule: the OH group attaches to the carbon with fewer hydrogens (via the more stable secondary carbocation), giving propan-2-ol.

Correct: (B)

A secondary alcohol is oxidised to a ketone, here butanone; it cannot be oxidised further under these conditions. (A primary alcohol would give an aldehyde and then a carboxylic acid.)

Correct: (C)

In a limited supply of oxygen, methane burns incompletely to carbon monoxide (a toxic gas) and water; equation A is complete combustion and equation B produces soot (carbon) instead.

Correct: (A)

A small molecule (water) is removed from a single reactant molecule to form an unsaturated product, which is an elimination (dehydration) reaction.

Correct: (D)

Fehling's solution contains Cu^2+ ions, which are reduced by aldehydes (but not ketones) to brick-red copper(I) oxide, Cu2O.

Correct: (B)

The acid and alcohol undergo esterification (a condensation reaction catalysed by conc. H2SO4): CH3COOH + CH3CH2OH ⇌ CH3COOCH2CH3 + H2O.

Correct: (C)

The AlCl3 catalyst accepts a chloride ion from the acyl chloride, generating the acylium ion CH3C^+=O, a strong electrophile; AlCl3 itself is the catalyst, not the electrophile.

Correct: (A)

The -NH2 group strongly activates the ring, so all three available positions (2, 4 and 6) are brominated, giving a white precipitate of 2,4,6-tribromophenylamine and decolourising the bromine water.

Correct: (D)

Nylon-6,6 forms when the -COOH groups of hexanedioic acid react with the -NH2 groups of 1,6-diaminohexane (hexamethylenediamine), eliminating water and forming amide (peptide) links.

Correct: (B)

There are three types of proton: the CH3 next to the carbonyl, the CH2 next to the oxygen, and the terminal CH3 of the ethyl group, so the spectrum has three peaks (in the ratio 3 : 2 : 3).