A-Level Chemistry Practice Test 5

Correct: (C)

Ar = (0.200 × 10) + (0.800 × 11) = 2.00 + 8.80 = 10.8

Correct: (A)

There is a large jump between the 2nd and 3rd ionisation energies, so the 3rd electron is removed from an inner shell; X has 2 outer electrons and is in Group 2.

Correct: (D)

Fe is [Ar] 4s^2 3d^6; the 4s electrons are lost first, then one 3d electron, giving [Ar] 3d^5.

Correct: (B)

Across the period each atom has one more proton but the electrons are added to the same shell, so the effective nuclear charge increases and the atomic radius decreases.

Correct: (A)

A chlorine atom has 17 protons and 17 electrons; the Cl^- ion has gained one extra electron, giving 18 electrons.

Correct: (B)

V = nRT/p = (0.200 × 8.31 × 300)/100000 = 4.99 × 10^-3 m^3 = 4.99 dm^3

Correct: (D)

Moles of NaOH = 0.0250 × 0.100 = 2.50 × 10^-3; since H2SO4 reacts with 2NaOH, moles of H2SO4 = 1.25 × 10^-3; concentration = 1.25 × 10^-3/0.0200 = 0.0625 mol dm^-3

Correct: (C)

Percentage yield = (6.30/8.40) × 100 = 75.0%

Correct: (A)

Moles: C = 40.0/12.0 = 3.33; H = 6.67/1.00 = 6.67; O = 53.3/16.0 = 3.33; the ratio is 1 : 2 : 1, so the empirical formula is CH2O.

Correct: (D)

Atom economy = 44.0/(16.0 + 64.0) × 100 = 44.0/80.0 × 100 = 55.0%

Correct: (B)

Moles of NaCl = 5.85/58.5 = 0.100; concentration = 0.100/0.250 = 0.400 mol dm^-3

Correct: (A)

NH3 has three bonding pairs and one lone pair; the lone pair repels more strongly, giving a trigonal pyramidal shape with a bond angle of 107°.

Correct: (C)

Born-Haber cycle: −411 = 107 + 121 + 496 + (−349) + lattice enthalpy, so lattice enthalpy = −411 − 375 = −786 kJ mol^-1

Correct: (B)

Fluorine is small and highly electronegative, so HF molecules are held together by strong hydrogen bonds which need extra energy to break; boiling does not break the covalent bonds inside the molecules.

Correct: (D)

CH3Cl has polar bonds arranged asymmetrically, so the bond dipoles do not cancel; the dipoles cancel in CO2, CCl4 and BF3 because of their symmetrical shapes.

Correct: (A)

When a force is applied the layers of ions slide into new positions, but the sea of delocalised electrons still attracts the ions, so the metal changes shape without breaking.

Correct: (C)

Lattice enthalpy is more exothermic for small ions with large charges; Mg^2+ and O^2- are both small and doubly charged, so MgO has the most exothermic lattice enthalpy.

Correct: (D)

q = mcΔT = 50.0 × 4.18 × 12.5 = 2613 J = 2.61 kJ; ΔH = −2.61/0.0500 = −52.3 kJ mol^-1

Correct: (B)

ΔH = [−394 + 2(−286)] − [−75] = −966 + 75 = −891 kJ mol^-1

Correct: (A)

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

Correct: (C)

At the limit of feasibility ΔG = ΔH − TΔS = 0, so T = ΔH/ΔS = 100000/200 = 500 K

Correct: (B)

Energy absorbed breaking bonds = 436 + 242 = 678; energy released forming bonds = 2 × 431 = 862; ΔH = 678 − 862 = −184 kJ mol^-1

Correct: (D)

ΔG = ΔH − TΔS becomes negative when TΔS exceeds ΔH, i.e. when T is above 50000/200 = 250 K.

Correct: (C)

The reaction is first order with respect to A and second order with respect to B, so rate = k[A][B]^2

Correct: (A)

k = rate/concentration^2, so the units are (mol dm^-3 s^-1)/(mol dm^-3)^2 = mol^-1 dm^3 s^-1

Correct: (D)

ln(k2/k1) = −(50000/8.31) × (1/310 − 1/300) = +0.647, so k2/k1 = e^0.647 = 1.9

Correct: (B)

Doubling [X] doubles the rate (first order in X) and doubling [Y] quadruples the rate (second order in Y); k = 0.020/(0.10 × 0.10^2) = 20 mol^-2 dm^6 s^-1

Correct: (C)

The rate is determined by the slow (rate-determining) step, which involves two molecules of NO2, so rate = k[NO2]^2

Correct: (A)

0.20 mol dm^-3 of N2O4 reacts to form 0.40 mol dm^-3 of NO2; Kc = [NO2]^2/[N2O4] = (0.40)^2/0.80 = 0.20 mol dm^-3

Correct: (D)

Kp is the partial pressures of the products, each raised to its stoichiometric power, divided by those of the reactants: pNH3^2/(pN2 × pH2^3)

Correct: (B)

Equilibrium constants only change with temperature, so Kp is unchanged; the higher pressure shifts the equilibrium towards the side with fewer gas molecules, increasing the yield of ammonia.

Correct: (C)

Kc = [SO3]^2/([SO2]^2[O2]); the units are (mol dm^-3)^2/(mol dm^-3)^3 = dm^3 mol^-1

Correct: (A)

Partial pressures: NH3 = 0.20 × 200 = 40 kPa, N2 = 40 kPa, H2 = 120 kPa; Kp = 40^2/(40 × 120^3) = 2.31 × 10^-5 kPa^-2

Correct: (D)

2(+1) + 2(x) + 7(−2) = 0, so 2x = +12 and x = +6

Correct: (B)

EMF = E° of the positive electrode − E° of the negative electrode = 0.34 − (−0.76) = +1.10 V

Correct: (C)

The most positive electrode potential corresponds to the species that is most readily reduced, so Cl2 is the strongest oxidising agent.

Correct: (A)

Q = It = 2.00 × 1800 = 3600 C; moles of electrons = 3600/96500 = 0.0373; moles of Cu = 0.0373/2 = 0.0187; mass = 0.0187 × 63.5 = 1.18 g

Correct: (D)

Standard electrode potentials are measured at 298 K, 100 kPa and 1.00 mol dm^-3 with effectively no current flowing (a high-resistance voltmeter is used).

Correct: (B)

Hydrochloric acid is fully ionised, so [H^+] = 0.0200; pH = −log(0.0200) = 1.70

Correct: (C)

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

Correct: (A)

pH = pKa + log([salt]/[acid]); the concentrations are equal so the log term is zero; pKa = −log(1.8 × 10^-5) = 4.74, so pH = 4.74

Correct: (D)

[H^+] = Kw/[OH^-] = 1.00 × 10^-14/0.0100 = 1.00 × 10^-12; pH = 12.00

Correct: (B)

At half-neutralisation [HA] = [A^-], so pH = pKa = −log(1.6 × 10^-5) = 4.80

Correct: (C)

The solubility of the Group 2 sulfates decreases down the group, so BaSO4 is the least soluble; the hydroxides show the opposite trend.

Correct: (A)

Cl2 + 2NaOH → NaCl + NaClO + H2O; chlorine is both oxidised (to ClO^-) and reduced (to Cl^-) in a disproportionation reaction.

Correct: (D)

Zn^2+ has a full 3d subshell (3d^10), so no d-d electron transitions can occur and its aqueous ion is colourless.

Correct: (B)

Fe^2+ + 2OH^- → Fe(OH)2, which forms a green precipitate (it slowly turns brown at the surface as it is oxidised by air).

Correct: (C)

Chloride ligands replace the water ligands to form [CuCl4]^2-, which is yellow; mixed with some remaining blue complex the solution often appears green.

Correct: (A)

Ultraviolet light causes homolytic fission of Cl2 to form two chlorine radicals; steps C and D are propagation steps and there is no ionic step.

Correct: (D)

The C-I bond is the weakest carbon-halogen bond, so it breaks most easily and 1-iodobutane reacts fastest; the C-F bond is the strongest.

Correct: (B)

Electrophilic addition follows Markovnikov's rule: the hydrogen adds to the carbon that already has more hydrogen atoms, via the more stable secondary carbocation, giving 2-bromopropane.

Correct: (C)

Under reflux with excess oxidising agent, the primary alcohol is fully oxidised to the carboxylic acid, propanoic acid; propanal is obtained only if it is distilled off as it forms.

Correct: (A)

In but-2-ene, each carbon of the C=C double bond is attached to two different groups (H and CH3), so E/Z isomerism is possible; in the others at least one double-bond carbon has two identical groups.

Correct: (D)

Hot ethanolic potassium hydroxide removes HBr from bromoethane to give ethene; aqueous KOH would instead give ethanol by substitution.

Correct: (B)

Tollens' reagent oxidises aldehydes but not ketones; the Ag^+ ions are reduced to metallic silver by propanal.

Correct: (C)

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

Correct: (A)

The two acids react together to generate the nitronium ion, NO2^+, which is the electrophile that attacks the benzene ring.

Correct: (D)

The ethyl group pushes electron density onto the nitrogen, making its lone pair more available to accept a proton; in phenylamine the lone pair is delocalised into the ring and in ethanamide into the carbonyl group, so both are much weaker bases.

Correct: (B)

The CH3 protons, the CH2 protons and the OH proton are in three different environments, so there are three signals in the ^1H NMR spectrum (ignoring splitting).

Correct: (C)

In the solid state, a proton is transferred from the carboxylic acid group to the amine group, forming a zwitterion; this ionic character explains the high melting points of amino acids.