A-Level Chemistry Practice Test 16

Correct: (C)

P(35,35) = 0.75 × 0.75 = 0.5625; P(35,37) = 2 × 0.75 × 0.25 = 0.375; P(37,37) = 0.25 × 0.25 = 0.0625. The ratio 0.5625 : 0.375 : 0.0625 simplifies to 9 : 6 : 1.

Correct: (A)

The first three ionisation energies increase steadily, but there is a very large jump at the fourth, showing that the fourth electron comes from an inner shell; the atom therefore has three outer-shell electrons (this is aluminium).

Correct: (D)

Chromium is an exception to the usual filling pattern: one 4s electron moves into the 3d subshell to give the more stable half-filled arrangement [Ar] 3d^5 4s^1.

Correct: (B)

Time ratio = sqrt(112 / 28) = sqrt(4) = 2, so the heavier ion takes 2.0 units of time.

Correct: (C)

All four ions have the same number of electrons, but Al^3+ has the greatest nuclear charge (13 protons), so its electrons are pulled in most strongly and it has the smallest radius.

Correct: (A)

n = pV / RT = (100000 × 1.20 × 10^-4) / (8.31 × 290) = 12 / 2410 = 4.98 × 10^-3 mol. Mr = 0.29 / 4.98 × 10^-3 = 58 g mol^-1 (the gas is butane).

Correct: (D)

Moles of CO2 = 0.88 / 44 = 0.020, so mass of C = 0.020 × 12 = 0.24 g. Moles of H2O = 0.36 / 18 = 0.020, so moles of H = 0.040 and mass of H = 0.04 g. Mass of O = 0.60 – 0.24 – 0.04 = 0.32 g, so moles of O = 0.32 / 16 = 0.020. Ratio C:H:O = 0.020 : 0.040 : 0.020 = 1 : 2 : 1, giving CH2O.

Correct: (B)

Moles of NaOH = 0.0500 × 0.0200 = 1.00 × 10^-3. Ethanoic acid reacts with NaOH in a 1:1 ratio, so concentration = 1.00 × 10^-3 / 0.0250 = 0.0400 mol dm^-3.

Correct: (C)

Atom economy = (88 / (60 + 46)) × 100 = (88 / 106) × 100 = 83%.

Correct: (A)

Each formula unit of CaCl2 contains two Cl^- ions, so moles of Cl^- = 2 × 0.050 = 0.100 mol. Number of ions = 0.100 × 6.02 × 10^23 = 6.02 × 10^22.

Correct: (D)

Moles of Mg = 6.0 / 24 = 0.25 mol, which would need 0.50 mol of HCl. Only 0.40 mol of HCl is present, so the acid is the limiting reactant and moles of H2 = 0.40 / 2 = 0.20 mol. Volume = 0.20 × 24 = 4.8 dm^3.

Correct: (B)

Lone pair-lone pair and lone pair-bond pair repulsions are stronger than bond pair-bond pair repulsion (VSEPR theory), so the two bonding pairs are pushed closer together.

Correct: (C)

From the Born-Haber cycle: -594 = 159 + 520 + 79 – 328 + ΔHlatt = 430 + ΔHlatt, so ΔHlatt = -594 – 430 = -1024 kJ mol^-1.

Correct: (A)

Larger molecules have more electrons and larger temporary dipoles, so the induced dipole-dipole (van der Waals) attractions between molecules are stronger and more energy is needed to separate them.

Correct: (D)

In ice each water molecule is hydrogen-bonded in a rigid, open tetrahedral network containing large spaces; this makes ice less dense than liquid water, where the molecules are packed more closely.

Correct: (B)

The molecule is linear and symmetrical (O=C=O), so the dipole of one C=O bond is exactly cancelled by the dipole of the other.

Correct: (C)

Lattice enthalpy increases with the charge on the ions and decreases with their size. MgO has 2+ and 2- ions that are smaller than Na^+ and Cl^-, so the attraction is much stronger.

Correct: (A)

q = mcΔT = 100 × 4.18 × 6.8 = 2842 J. Moles of water formed = 1.00 × 0.0500 = 0.0500 mol. ΔH = -2842 / 0.0500 = -56 840 J mol^-1 ≈ -57 kJ mol^-1.

Correct: (D)

ΔHf = [(-394) + 2(-286)] – (-890) = -966 + 890 = -76 kJ mol^-1.

Correct: (B)

ΔS = S(products) – S(reactants) = (2 × 70) – [(2 × 131) + 205] = 140 – 467 = -327 J K^-1 mol^-1. The change is negative because three moles of gas form a liquid.

Correct: (C)

ΔG = ΔH – TΔS becomes negative when T > ΔH / ΔS = 100000 / 200 = 500 K.

Correct: (A)

By definition, standard enthalpy of formation refers to the formation of one mole of a compound from its constituent elements in their standard states at 298 K and 100 kPa.

Correct: (D)

Bonds broken = 945 + (3 × 436) = 2253 kJ. Bonds formed = 6 × 391 = 2346 kJ (two NH3 molecules each contain three N-H bonds). ΔH = 2253 – 2346 = -93 kJ mol^-1.

Correct: (B)

The reaction is second order in A and first order in B, so the overall order is 2 + 1 = 3.

Correct: (C)

For a first-order reaction, k = ln 2 / t0.5 = 0.693 / 20 = 0.0347 s^-1. The half-life of a first-order reaction is independent of concentration.

Correct: (A)

From the Arrhenius equation, gradient = -Ea / R, so Ea = 6500 × 8.31 = 54 015 J mol^-1 ≈ 54 kJ mol^-1.

Correct: (D)

At higher temperature the distribution spreads out (the area stays the same because the number of molecules is unchanged), the peak shifts right and flattens, and the fraction of molecules exceeding Ea increases, so the rate increases.

Correct: (B)

In a zero-order reaction the rate is constant and independent of concentration, so concentration falls by equal amounts in equal times, giving a straight line of slope -k.

Correct: (C)

Kc = [SO3]^2 / ([SO2]^2 × [O2]) = (0.80)^2 / [(0.20)^2 × 0.10] = 0.64 / 0.004 = 160 mol^-1 dm^3.

Correct: (A)

Kp (like Kc) depends only on temperature. Changing pressure or concentration shifts the position of equilibrium, but the value of Kp stays the same at a given temperature.

Correct: (D)

Kc = ([ester] × [water]) / ([acid] × [ethanol]). All species are in the same volume, so amounts can be used directly: Kc = (0.67 × 0.67) / (0.33 × 0.33) = 0.449 / 0.109 = 4.1 (no units).

Correct: (B)

There are 2 moles of gas on each side, so increasing the pressure favours neither side; the position of equilibrium (and Kc) is unchanged, although equilibrium is reached faster.

Correct: (C)

Increasing the temperature favours the endothermic direction, so more products are present at equilibrium and the equilibrium constant increases.

Correct: (A)

Let the oxidation state of Mn be x: x + 4(-2) = -1, so x = +7.

Correct: (D)

EMF = E(more positive electrode) – E(more negative electrode) = (+0.80) – (-0.44) = +1.24 V.

Correct: (B)

The strongest reducing agent is the species in the couple with the most negative electrode potential. Zinc most readily loses electrons to form Zn^2+, so Zn is the strongest reducing agent.

Correct: (C)

Eθcell = (+1.09) – (+0.77) = +0.32 V. A positive Eθcell shows the reaction is feasible: Br2 oxidises Fe^2+ to Fe^3+.

Correct: (A)

Al^3+ + 3e^- → Al, so three moles of electrons produce one mole of aluminium: 0.30 / 3 = 0.10 mol.

Correct: (D)

Moles of H^+ added = 0.100 × 0.0500 = 5.00 × 10^-3; moles of OH^- = 0.100 × 0.0250 = 2.50 × 10^-3. Excess H^+ = 2.50 × 10^-3 mol in a total volume of 75.0 cm^3, so [H^+] = 2.50 × 10^-3 / 0.0750 = 0.0333 mol dm^-3. pH = -log(0.0333) = 1.48.

Correct: (B)

[H^+] ≈ sqrt(Ka × c) = sqrt(1.7 × 10^-5 × 0.050) = sqrt(8.5 × 10^-7) = 9.2 × 10^-4 mol dm^-3. pH = -log(9.2 × 10^-4) = 3.03.

Correct: (C)

The ethanoate ion hydrolyses: Kb = Kw / Ka = 1.0 × 10^-14 / 1.8 × 10^-5 = 5.6 × 10^-10. [OH^-] = sqrt(5.6 × 10^-10 × 0.100) = 7.5 × 10^-6 mol dm^-3, so pOH = 5.13 and pH = 14.00 – 5.13 = 8.87.

Correct: (A)

The equivalence point of a strong acid-weak base titration is acidic (pH about 3-7), so an indicator that changes colour in this range, such as methyl orange, is needed.

Correct: (D)

Water is neutral, so [H^+] = sqrt(Kw) = sqrt(5.48 × 10^-14) = 2.34 × 10^-7 mol dm^-3. pH = -log(2.34 × 10^-7) = 6.63. The value is below 7 because the ionisation of water is endothermic, so Kw increases with temperature.

Correct: (B)

The thermal stability of Group 2 carbonates increases down the group, because the smaller cations higher up the group have a greater charge density and polarise (distort) the carbonate ion more, weakening it. MgCO3 therefore decomposes most easily.

Correct: (C)

Silver chloride is a white precipitate that dissolves in dilute ammonia. Silver bromide is cream and dissolves only in concentrated ammonia, while silver iodide is yellow and insoluble in ammonia.

Correct: (A)

Platinum(II) is a d^8 metal ion forming four coordinate bonds, and its complexes such as cisplatin are square planar. The cis isomer is the anti-cancer drug; geometry matters for its activity.

Correct: (D)

Fe^2+(aq) forms green iron(II) hydroxide, Fe(OH)2, which is rapidly oxidised by air at the surface to brown iron(III) hydroxide.

Correct: (B)

In the Contact process, V2O5 catalyses the oxidation of sulfur dioxide to sulfur trioxide: 2SO2 + O2 ⇌ 2SO3. (The Haber process uses an iron catalyst.)

Correct: (C)

In a propagation step a radical reacts with a molecule to form a product and a new radical, keeping the chain going. Option A is initiation and option B is termination.

Correct: (A)

In ethanolic solution the hydroxide ion acts as a base and removes a hydrogen from the carbon next to the C-X bond, so the halogenoalkane eliminates HX to form an alkene. (Aqueous KOH gives substitution to the alcohol instead.)

Correct: (D)

Cold, dilute manganate(VII) is a mild oxidising agent that adds two -OH groups across the double bond (di-hydroxylation), and the purple MnO4^- colour disappears.

Correct: (B)

2-Methylpropan-2-ol is a tertiary alcohol; it has no hydrogen on the carbon bearing the -OH group, so it cannot be oxidised by acidified dichromate and the orange colour remains.

Correct: (C)

Elimination of water can give but-1-ene or but-2-ene, but but-2-ene is the major product because it is the more substituted (and therefore more stable) alkene, as predicted by Saytzeff's rule.

Correct: (A)

In a termination step two radicals combine to form a stable molecule with no new radical, ending the chain; here two methyl radicals combine to form ethane.

Correct: (D)

NaBH4 reduces ketones to secondary alcohols; the C=O bond of propanone is reduced to give propan-2-ol. (NaBH4 does not reduce C=C bonds or carboxylic acids.)

Correct: (B)

The acyl chloride CH3COCl, with the AlCl3 catalyst (which generates the CH3CO^+ electrophile), substitutes a COCH3 group onto the benzene ring. CH3Cl/AlCl3 would give alkylation instead.

Correct: (C)

Carboxylic acids react with carbonates to give a salt, water and carbon dioxide gas (the effervescence). Phenol is too weak an acid to react with sodium carbonate.

Correct: (A)

Terylene is formed from a diol and a dicarboxylic acid with a molecule of water eliminated for each ester link formed. The others are addition polymers formed from a single alkene monomer with no by-product.

Correct: (D)

Primary aromatic amines form diazonium salts at low temperature (0-10 °C); the diazonium ion is an electrophile that couples with compounds such as phenol to form brightly coloured azo dyes.

Correct: (B)

The molecule is symmetrical, so all six hydrogen atoms are in the same environment and give a single singlet peak.