A-Level Chemistry Practice Test 15

Correct: (B)

(69 × 0.60) + (71 × 0.40) = 41.4 + 28.4 = 69.8

Correct: (D)

First ionisation energy generally increases across Period 3 as the nuclear charge increases and the atoms get smaller; phosphorus (1012 kJ mol^-1) is higher than magnesium, aluminium and silicon.

Correct: (C)

Iron is [Ar] 3d^6 4s^2. When ions form, the 4s electrons are removed first, then one 3d electron, giving [Ar] 3d^5.

Correct: (A)

The very large increase between the second and third ionisation energies shows that the third electron is removed from an inner shell, so the atom has only two outer-shell electrons.

Correct: (B)

Across the period the nuclear charge increases while the electrons are added to the same shell, so atoms get smaller and ionisation energy generally rises (with small dips at aluminium and sulfur).

Correct: (D)

n = pV / RT = (100000 × 5.00 × 10^-4) / (8.31 × 300) = 50 / 2493 = 0.020 mol

Correct: (C)

Percentage yield = (9.40 / 12.5) × 100 = 75.2%

Correct: (A)

Moles of NaOH = 0.100 × 0.0250 = 2.50 × 10^-3. H2SO4 reacts with NaOH in a 1:2 ratio, so moles of H2SO4 = 1.25 × 10^-3. Concentration = 1.25 × 10^-3 / 0.0200 = 0.0625 mol dm^-3.

Correct: (B)

C: 40.0/12 = 3.33; H: 6.70/1 = 6.70; O: 53.3/16 = 3.33. Ratio 1:2:1, so the empirical formula is CH2O.

Correct: (D)

Atom economy = (Mr of desired product / total Mr of all products) × 100 = (56 / 100) × 100 = 56%

Correct: (C)

Moles of CO2 = 4.40 / 44 = 0.100 mol. Number of molecules = 0.100 × 6.02 × 10^23 = 6.02 × 10^22.

Correct: (A)

From the Born-Haber cycle: -411 = 107 + 121 + 496 – 349 + ΔHlatt, so ΔHlatt = -411 – 375 = -786 kJ mol^-1.

Correct: (B)

Nitrogen has three bonding pairs and one lone pair. The lone pair repels more strongly than the bonding pairs, pushing the bonds together and reducing the angle from 109.5° to 107°.

Correct: (D)

Hydrogen bonding occurs when hydrogen is bonded to a small, highly electronegative atom such as nitrogen, oxygen or fluorine. Only NH3 qualifies here.

Correct: (C)

Each carbon atom in graphite forms three covalent bonds, leaving one electron per atom delocalised; these electrons can move and carry charge.

Correct: (A)

The small, highly charged Al^3+ ion has the greatest polarising power and distorts the electron cloud of the large Cl^- ion most, giving AlCl3 the most covalent character (Fajans' rules).

Correct: (B)

The doubly charged ions attract each other strongly in all directions in the giant ionic lattice, so a very large amount of thermal energy is needed to melt it.

Correct: (D)

q = mcΔT = 100 × 4.18 × 12.5 = 5225 J = 5.225 kJ. ΔH = -5.225 / 0.0100 = -523 kJ mol^-1 (energy released per mole burned).

Correct: (C)

ΔHf = [2(-394) + 3(-286)] – (-1367) = (-788 – 858) + 1367 = -279 kJ mol^-1.

Correct: (A)

Bonds broken = 436 + (0.5 × 498) = 685 kJ. Bonds formed = 2 × 464 = 928 kJ. ΔH = 685 – 928 = -243 kJ mol^-1.

Correct: (B)

A gas is produced from a solid, so the number of particles and the degree of disorder increase, giving a positive ΔS.

Correct: (D)

ΔG = ΔH – TΔS = 0 when T = ΔH / ΔS = 100000 / 200 = 500 K. The reaction is feasible below this temperature.

Correct: (C)

ΔHsolution = ΔHdissociation + ΔHhydration = 701 + (-322 – 364) = 701 – 686 = +15 kJ mol^-1.

Correct: (A)

k = rate / [A]^2 = (2.0 × 10^-4) / (0.10)^2 = (2.0 × 10^-4) / 0.010 = 0.020 mol^-1 dm^3 s^-1.

Correct: (B)

Rate is proportional to [X]^2, since 2^2 = 4; the reaction is second order with respect to X.

Correct: (D)

Gradient = -Ea / R, so Ea = 1.20 × 10^4 × 8.31 = 99 720 J mol^-1 ≈ 100 kJ mol^-1.

Correct: (C)

The rate equation is determined by the slow (rate-determining) step, which involves two molecules of NO2, so rate = k[NO2]^2. CO appears only in the fast step and does not affect the rate.

Correct: (A)

A catalyst offers a different route with a lower activation energy; it does not change the energies of the particles or the position of equilibrium.

Correct: (B)

Formation of 0.600 mol dm^-3 of C consumes 0.300 mol dm^-3 of A and of B, so [A] = [B] = 0.700 mol dm^-3. Kc = (0.600)^2 / (0.700 × 0.700) = 0.36 / 0.49 = 0.73 (no units, as top and bottom powers cancel).

Correct: (D)

Kp = (pNO2)^2 / pN2O4 = (40)^2 / 60 = 1600 / 60 = 26.7 kPa.

Correct: (C)

Raising the temperature favours the endothermic (reverse) direction, so the equilibrium shifts left and the equilibrium constant Kc becomes smaller.

Correct: (A)

A catalyst speeds up the forward and reverse reactions equally, so the position of equilibrium and the value of Kc are unchanged; equilibrium is simply reached faster.

Correct: (B)

Let x mol of H2 react: Kc = (2x)^2 / (0.50 – x)^2 = 64. Taking the square root: 2x / (0.50 – x) = 8, so 2x = 4.0 – 8x and x = 0.40. HI formed = 2x = 0.80 mol.

Correct: (D)

Let the oxidation state of Cr be x: 2x + 7(-2) = -2, so 2x = +12 and x = +6.

Correct: (C)

EMF = E(reduction side) – E(oxidation side) = (+0.34) – (-0.76) = +1.10 V.

Correct: (A)

The strongest oxidising agent is the species in the couple with the most positive electrode potential; Br2 (+1.09 V) has the greatest tendency to gain electrons and be reduced.

Correct: (B)

Eθcell = (+0.77) – (+0.54) = +0.23 V. A positive Eθcell means the reaction is feasible, so Fe^3+ oxidises I^- to I2 and is itself reduced to Fe^2+.

Correct: (D)

Charge = It = 0.50 × 3860 = 1930 C, which is 1930 / 96500 = 0.0200 mol of electrons. Cu^2+ + 2e^- → Cu, so moles of Cu = 0.0100 and mass = 0.0100 × 63.5 = 0.64 g.

Correct: (C)

[OH^-] = 0.020 mol dm^-3, so pOH = -log(0.020) = 1.70 and pH = 14.00 – 1.70 = 12.30.

Correct: (A)

[H^+] ≈ sqrt(Ka × c) = sqrt(1.8 × 10^-5 × 0.100) = sqrt(1.8 × 10^-6) = 1.34 × 10^-3 mol dm^-3. pH = -log(1.34 × 10^-3) = 2.87.

Correct: (B)

Pure water is neutral, so [H^+] = sqrt(Kw) = sqrt(2.92 × 10^-14) = 1.71 × 10^-7 mol dm^-3. pH = -log(1.71 × 10^-7) = 6.77. (Neutral pH is below 7 because Kw is larger at higher temperature.)

Correct: (D)

pH = pKa + log([salt] / [acid]) = 4.76 + log(2) = 4.76 + 0.30 = 5.06.

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)

The solubility of Group 2 sulfates decreases down the group; barium sulfate is the least soluble and is effectively insoluble.

Correct: (B)

The oxidation state of chlorine changes from 0 to -1 in NaCl (reduction) and from 0 to +1 in NaClO (oxidation), so the chlorine disproportionates. The NaClO formed is used as bleach.

Correct: (D)

Ammonia first precipitates pale blue Cu(OH)2, which dissolves in excess ammonia as four NH3 ligands replace water ligands, forming the deep blue tetraamminecopper(II) complex ion.

Correct: (C)

Fe^3+(aq) forms insoluble iron(III) hydroxide, which appears as an orange-brown precipitate.

Correct: (A)

Zn^2+ has a completely filled 3d subshell (3d^10), so no d-d electron transitions are possible and its compounds absorb no visible light.

Correct: (B)

In initiation, UV light provides the energy for homolytic fission of the Cl-Cl bond, forming two chlorine free radicals.

Correct: (D)

2-Bromo-2-methylpropane is a tertiary halogenoalkane and hydrolyses by an SN1 mechanism through a stable tertiary carbocation, so it reacts much faster than primary halogenoalkanes.

Correct: (C)

The electrophile H^+ adds to the double bond to give the more stable secondary carbocation (Markovnikov addition), which then reacts with Br^- to form 2-bromopropane as the major product.

Correct: (A)

Distillation removes the aldehyde before it can be oxidised further, so the primary alcohol is oxidised only to propanal. (Under reflux, propanoic acid would form.)

Correct: (B)

Concentrated sulfuric acid acts as a dehydrating agent; the alcohol eliminates water to form the alkene but-1-ene.

Correct: (D)

With insufficient oxygen, incomplete combustion occurs and the toxic gas carbon monoxide is produced instead of carbon dioxide.

Correct: (C)

Tollens' reagent is a mild oxidising agent; aldehydes reduce the Ag^+ ions to a silver mirror, whereas ketones cannot be oxidised further and give no reaction.

Correct: (A)

Alkaline hydrolysis of an ester gives the sodium salt of the carboxylic acid and the alcohol: CH3COOC2H5 + NaOH → CH3COONa + C2H5OH.

Correct: (B)

The nitrating mixture of concentrated nitric and sulfuric acids generates the NO2^+ electrophile; the temperature is kept near 50 °C to avoid multiple nitration.

Correct: (D)

The alkyl group in ethylamine pushes electrons towards the nitrogen, making its lone pair more available to accept a proton than in ammonia. In phenylamine the lone pair is delocalised into the benzene ring, making it a much weaker base.

Correct: (C)

In alkaline solution the -NH3^+ group of the zwitterion loses a proton to OH^-, so the amino acid exists mainly as the anion H2N-CH2-COO^-.

Correct: (A)

The three CH3 groups are in identical environments and the single CH carbon is in a different environment, so there are two different carbon environments and two peaks.