Questions 1
(a)Topic-25.1 Acids and bases
(b) Topic-10.1 Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
(c)Topic-25.1 Acids and bases
(d)Topic-25.1 Acids and bases
(a) Disodium phosphate, \((Na^+)_2(HPO_4^{2–})\), reacts with an acid to form monosodium phosphate, \(Na^+(H_2PO_4^–)\).
(i) Identify the ions that are a conjugate acid–base pair in this reaction, using the formulae of the species involved.
(ii) Define buffer solution.
(iii) Write two equations to show how a mixture of \((Na^+)_2(HPO_4^{2–})\) and \(Na^+(H_2PO_4^–)\) can act as a buffer solution.
equation 1 ………………………………………………………………………………………………………….
equation 2 ………………………………………………………………………………………………………….
(iv) Identify one inorganic ion that acts as a buffer in blood.
(b) Compound E is the hydroxide of a Group 2 element. Compound E is a strong alkali. 2.63g of E is dissolved in water to make 250 cm³ of solution F. Solution F has a pH of 13.09 at 298K.
(i) Show that the concentration of hydroxide ions in solution F is 0.123 moldm⁻³.
(ii) Explain why the concentration of compound E in solution F is 0.0615moldm⁻³.
(iii) Use the concentration given in (ii) to identify compound E.
(c) Compound E is much more soluble than magnesium hydroxide. A saturated solution of magnesium hydroxide in water has a concentration of 1.40 × 10⁻⁴ moldm⁻³ at 298K. Calculate the solubility product, \(K_{sp}\), of magnesium hydroxide. Include units.
(d) Explain why compound E is much more soluble than magnesium hydroxide.
▶️Answer/Explanation
Ans
(a)(i) conjugate acid \(H_2PO_4^–\)
conjugate base \(HPO_4^{2–}\)
(ii) allows only small changes in pH / resists changes in pH
when small amounts of acid or alkali/base /H⁺ or OH⁻ are added
(iii) \(Na^+_2HPO_4^{2–} + H^+ \to Na^+H_2PO_4^– + Na^+\)
or \(HPO_4^{2–} + H^+ \to H_2PO_4^–\)
Na⁺H₂PO₄⁻ + NaOH → \(Na^+_2HPO_4^{2–}\) + H₂O
or \(H_2PO_4^– + OH^– \to HPO_4^{2–} + H_2O\)
(iv) \(HCO_3^–\)/ hydrogen carbonate
(b)(i) either \(10^{–13.09}\) or [H+] = 8.1 × \(10^{–14}\) seen [1]
\(8.1 × 10^{–14} × [OH–] = 1 × 10^{–14}\) seen [1]
OR \([OH^–] =\frac{1 × 10^{–14}}{8.1 × 10^{–14}}\) seen
ALLOW alternative method
pOH = 0.91 [1]
\([OH^–] = 10^{–0.91}\) seen
(ii) 2 moles of OH– in one mole of E / X(OH)₂
OR
there are two hydroxide ions in each formula unit of E/X(OH)₂.
(iii) 0.0615 × 0.25 = 0.0154 moles of E
RFM = 2.63 / 0.0154 = 170.8 / 171 OR RAM = 34 – 2.63 / 0.0154 = 137.1 /137
AND barium hydroxide / Ba(OH)₂
c) \(K_{sp} = [Mg^{2+}][OH^–]^2\) OR \(K_{sp} = 1.4 \times 10^{-4} \times (2 \times 1.4 \times 10^{-4})^2\)
\(K_{sp} = 1.10 × 10^{–11} mol^3dm^{-9}\)
(d) M1: \(\Delta H_{latt}\) and \(\Delta H_{hyd}\) are less exothermic (down the group)
OR \(\Delta H_{latt}\)/LE and \(\Delta H_{hyd}\) are more exothermic/more negative for Mg [1]
M2: \(\Delta H_{latt}\) changes more than \(\Delta H_{hyd}\) (down the group)
OR \(\Delta H_{latt}\) changes more / faster for Mg [1]
M3: \(\Delta H_{sol}\) becomes more exothermic (down the group)
OR \(\Delta H_{sol}\) becomes less exothermic / less negative for Mg
Questions 2
(a)Topic-5.1 Enthalpy change, ΔH
(b)Topic-5.1 Enthalpy change, ΔH
(c)Topic-5.1 Enthalpy change, ΔH
(d)Topic-5.1 Enthalpy change, ΔH
(e)Topic-23.4 Gibbs free energy change, ΔG
(f)Topic-23.4 Gibbs free energy change, ΔG
(a) Predict and explain the variation in the enthalpy change of hydration for the ions \(F^–\), \(Cl^–\), \(Br^–\), and \(I^–\).
(b) Fig. 2.1 shows an incomplete energy cycle involving calcium fluoride, CaF₂.
(i) Complete line D. Include state symbols.
(ii) The value of the enthalpy change for process 1 can be calculated using the values of five other enthalpy changes which are not referred to in Fig. 2.1.
process 1: \(Ca(s) + F_2(g) \to Ca^{2+}(g) + 2F^–(g)\)
Identify these five other enthalpy changes, using either names or symbols.
(iii) Define lattice energy, \(\Delta H_{latt}\).
(iv) Complete the expression to give the mathematical relationship between \(\Delta H_{latt}\) of calcium fluoride and the enthalpy changes for processes 1 and 3.
(c) Use data from Table 2.1 to calculate a value for the hydration energy, \(\Delta H_{hyd}\), of fluoride ions, \(F^–(g)\).
(d) Define entropy.
(e) At 298K, the Gibbs free energy change, ΔG, for the solution of compound T is +6.00 kJ mol⁻¹. The enthalpy change of solution, \(\Delta H_{sol}\), of compound T is +30.0 kJ mol⁻¹ at 298K. Calculate the value of the entropy change, ΔS, for the solution of compound T at 298K.
(f) Predict whether compound T becomes more or less soluble as the water is heated from 298K to 360K. Explain your answer.
▶️Answer/Explanation
Ans
(a) becomes less negative/less exothermic (down the group / from F to I)
AND due to increase in (ionic) radius / size
decreased attraction to water
OR weaker ion-dipole force to water
(b)(i) CaF₂(aq) OR Ca²⁺(aq) + 2F⁻(aq)
(iii) energy change / energy released
when one mole of an ionic compound is formed
from its gaseous ion(s)
(iv) \((\Delta H_{latt}\) = ) (change) 3 – (change) 1 [1]
OR
\((\Delta H_{latt}\) = ) \(\Delta H_f(CaF_2(s)) – \Delta H_f(Ca^{2+}(g)) – 2\Delta H_f(F^– (g))\)
(c) expression involves four correct numbers (13, 1395, 1214, 1650)
AND 2 times [1]
1395 – 1650 + 2x = –1214 + 13
\(\Delta H_f(F^–(g)) \times = –473\)
(d) number of possible arrangements of particles / energy in a system
(e) states or clearly uses ΔG = ΔH – TΔS [1]
OR 6000 = 30 000 – (298 \(\times\) ΔS)
ΔS = (+)80.5(4)
(f) becomes more soluble
AND ΔS is positive / TΔS is positive / –TΔS is negative (as T inc) [1]
(so ΔG becomes more negative)
Questions 3
(a)Topic- 26.2 Homogeneous and heterogeneous catalysts
(b)Topic-26.1 Simple rate equations, orders of reaction and rate constants
(c)Topic-26.1 Simple rate equations, orders of reaction and rate constants
(a) A and B react together to give product AB.
\(A + B \to AB\)
When the concentrations of A and B are both 0.0100 moldm⁻³, the rate of formation of AB is 7.62 × 10⁻⁴ moldm⁻³ s⁻¹. When the concentrations of A and B are both 0.0200 moldm⁻³,
the rate of formation of AB is 3.05 × 10⁻³ moldm⁻³ s⁻¹.
(i) Complete the three possible rate equations that are consistent with these data.
rate = …………………………………………………………………………………………………………………
rate = …………………………………………………………………………………………………………………
rate = …………………………………………………………………………………………………………………
(ii) Choose one of the rate equations you have written in (i), and calculate the value of the rate constant, k. Include the units of k.
(iii) Explain why it is not possible to calculate a value for the half-life, \(t_{\frac{1}{2}}\), of this reaction using the value of the rate constant k calculated in (ii) and the equation k = 0.693/\(t_{\frac{1}{2}}\).
(b) Catalysts may be homogeneous or heterogeneous.
(i) Identify two metals that act as heterogeneous catalysts in the removal of \(NO_2\) from the exhaust gases of car engines.
(ii) Iron acts as a heterogeneous catalyst in the Haber process. Describe the mode of action of this iron catalyst.
(iii) Fe²⁺ ions act as a homogeneous catalyst in the reaction between I⁻(aq) and S₂O₈²⁻(aq). Write equations for the two reactions that occur when Fe²⁺(aq) is added to a mixture of I⁻(aq) and S₂O₈²⁻(aq).
(iv) Explain the difference between a homogeneous catalyst and a heterogeneous catalyst.
(c) \(Fe^{2+}\) ions can be oxidised to \(Fe^{3+}\) ions under alkaline conditions by suitable oxidising agents.
(i) Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states.
(ii) The half-equation for the reduction of Fe³⁺ under alkaline conditions, and its \(E^o\) value, are shown.
Four more half-equations for reactions under alkaline conditions, and their \(E^o\) values, are shown.
Select two oxidising agents that can oxidise Fe²⁺ ions to Fe³⁺ ions under alkaline conditions. Write an equation, and give the \(E^0_{cell}\) value, for each of the two reactions that occur
▶️Answer/Explanation
Ans
(a)(i) rate = k[A][B] [1]
rate = \(k[A]^2/ k[A]^2[B]^0\) AND rate = \(k[B]^2/ k[A]^0[B]^2\)
(ii) k = 7.62 / 7.625 / 7.63
\(mol^{–1} dm^3 s^{–1}\)
(iii) reaction is 2nd order
OR reaction is not 1st order
OR reaction does not have constant half-life
(b)(i) Pt, Pd, Rh
(ii) • adsorption (of reactants)
/ bond forms between the surface/catalyst and reactants
• bond weakening (in reactants)
• desorption (of reactants / products)
/ bond breaks between the surface/catalyst and products
(iii) \(S_2O_8^{2–} + 2Fe^{2+} \times 2SO_4^{2–} + 2Fe^{3+}\)
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
(iv) (heterogeneous) catalyst and reactants in different states / phases
AND
(homogeneous) catalyst and reactants in same states / phases
(c)(i) similar energy of the (3)d and (4)s subshells / orbitals
Questions 4
(a)Topic- 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(b)Topic- 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(c)Topic- 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
Transition metal atoms and transition metal ions form complexes by combining with ligands.
(a) Explain why transition elements form complex ions.
(b) \(Co^{2+}\) ions form complex ion G. Each G ion contains two \(Co^{2+}\) ions, both of which are octahedrally coordinated. Each G ion contains one \(O_2\) molecule, which donates one pair of electrons to each \(Co^{2+}\) ion, and one\( NH_2^–\) ion, which donates one pair of electrons to each \(Co^{2+}\) ion. The remaining ligands are NH₃ molecules.
(i) Deduce the formula of complex ion G. Include its overall charge.
formula of G ……………………………………………..
(ii) The d-orbitals of the \(Co^{2+}\) ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each \(Co^{2+}\) ion.
(iii) \(Co^{2+}\) ions form a different complex ion, M. Each M ion contains two \(Co^{2+}\) ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G. Explain why G and M have different colours.
(c) Cadmium forms complex ion X, \([Cd(NH_3)_4]^{2+}\). When a solution containing CN⁻ ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no \(NH_3\) ligands and no \(H_2O\) ligands. Y is in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do not change in this reaction.
(i) Write an ionic equation for this reaction, using the formulae of the complex ions.
(ii) Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl⁻ ions. When NaCl(aq) is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, \(K_{stab}\).
(d) Ethanedioate ions, \(C_2O_4^{2–}\), form complexes with transition element ions. The concentration of \(C_2O_4^{2–}\) ions can be found by reaction with acidified \(Cr_2O_7^{2–}\) ions. \(C_2O_4^{2–}\) ions are protonated and form HOOCCOOH molecules which are oxidised by \(Cr_2O_7^{2–}\). The half-equations are shown.
(i) Construct an equation for the reaction between acidified Cr₂O₇²⁻ and HOOCCOOH
(ii) A 25.0 cm³ sample of a solution of Na₂C₂O₄ reacts with exactly 16.20 cm³ of an acidified solution of 0.0500 moldm⁻³ K₂Cr₂O₇. Calculate the concentration of the solution of Na₂C₂O₄.
▶️Answer/Explanation
Ans
(a) empty (d) orbitals are energetically accessible
OR empty (d) orbitals can form dative bonds with ligands
OR empty (d) orbitals can accept a lone pair from ligands
(b)(i) \([Co_2O_2NH_2(NH_3)_8]^{3+}\) scores [2] must have Co₂.
• \(Co_2O_2NH_2\)
• 8 ammonia so both Co are octahedral (ecf bullet 1)
• correct charge based on ligands present (ecf Co and NH₂)
(ii) 2
3
(iii) ΔE different OR (d–d) energy gap different [1]
different frequency/wavelength/energy from visible light absorbed
Questions 5
(a)Topic- 3.5 Shapes of molecules
(b) Topic-28.2 General characteristic chemical properties of the first set of transition elements, titanium to copper
(c)Topic-28.2 General characteristic chemical properties of the first set of transition elements, titanium to copper
(d)Topic-28.2 General characteristic chemical properties of the first set of transition elements, titanium to copper
The shapes of four different complexes, P, Q, R and S, are shown in Table 5.1. The symbol J represents an atom or ion of a transition element. The symbol L is used to represent a monodentate ligand.
(a) Label one bond angle on each of complexes P, Q, R and S, and identify the size of the angle in degrees.
(b) Identify the shapes of complexes P, Q, R and S.
(c) Two L ligands are exchanged with two different monodentate ligands X and Y in each of complexes P, Q, R and S. Identify all the complexes which form new complexes that show geometrical isomerism.
(d) Three L ligands are exchanged with three different monodentate ligands X, Y and Z in each of complexes P, Q and R. Identify all the complexes which form new complexes that show optical isomerism.
▶️Answer/Explanation
Ans
a) bond angle must go from bond to bond
P 109 OR 109.5°
Q 90° (or 180° if different angle labelled)
R 90° (or 180° if different angle labelled)
S 180°
any two [1] all four [2]
(b) P tetrahedral
Q square planar
R octahedral
S linear
any two [1] all four [2]
(c) Q AND R [1]
(d) P AND R
Questions 6
(a)Topic-21.1 Organic synthesis
(b)Topic-21.1 Organic synthesis
(c)Topic-21.1 Organic synthesis
(d)Topic-21.1 Organic synthesis
(e)Topic-21.1 Organic synthesis
Benzene, \(C_6H_6\), reacts with chloroethane, \(C_2H_5Cl\), in the presence of a suitable catalyst to form ethylbenzene, \(C_6H_5C_2H_5\). In the presence of the catalyst, the ion \(C_2H_5^+\) is formed. This ion reacts with benzene.
(a) Complete the equation for the reaction of \(C_2H_5Cl\), with this catalyst to form \(C_2H_5^+\) as one product.
(b) Ethylbenzene reacts with more \(C_2H_5Cl\), forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene.
(i) Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene.
(ii) Explain why there is very little 1,3-diethylbenzene in the product mixture.
(c) 1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, \(C_6H_4(COOH)_2\).
(i) State the reagent and conditions used for this reaction.
(ii) Complete the overall equation for this reaction. An atom of oxygen from the oxidising agent is represented as [O]. All of the atoms in the two ethyl groups are fully oxidised in this reaction
(iii) Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid.
(d) The proton (1H) NMR spectra of ethylbenzene, \(C_6H_5C_2H_5\), in \(CDCl_3\), and of benzene-1,2-dioic acid, \(C_6H_4(COOH)_2\), in \(CDCl_3\) are shown. They have not been identified.
(i) Explain the use of CDCl₃, instead of CHCl₃, as the solvent when obtaining these spectra.
(ii) Identify the substance shown by the spectrum in Fig. 6.1, and complete Table 6.1.
(iii) Identify the substance shown by the spectrum in Fig. 6.2, and complete Table 6.2.
(iv) When \(D_2O\) is used as a solvent, the spectrum obtained is different from the spectrum in Fig. 6.2. Describe this difference and explain your answer.
(e) Benzene-1,2-dioic acid can be used to produce K.
Suggest the name of this type of reaction.
▶️Answer/Explanation
Ans
(ii) alkyl / ethyl group is 2,4 directing
OR ethyl group is electron donating / positive inductive effect
(c)(i) hot alkaline KMnO₄ / MnO₄⁻
(iii) 4 [1]
(d)(i) CDCl₃ does not cause a peak
OR does not interfere with spectrum / peaks [1]
(ii) • ethylbenzene / C₆H₅C₂H₅
• triplet
• CH₃
• 2H on neighbouring C / next to CH₂
– quartet / quadruplet
– CH₂
– 3H on neighbouring C / next to CH₃ / coupling by CH₃
(iii) • benzene-1,2-dioic acid / \(C_6H_4(COOH)_2\)
• H on the benzene / \(C_6H_4\)
– COOH / carboxylic acid
ALL three [1]
(iv) • COOH peak disappears OR remove peak at 13.1 (ecf (d)(iii) if δ value used)
AND
• as proton / hydrogen exchanges with deuterium / D
BOTH [1]
(e) dehydration / elimination / (auto)condensation
Questions 7
(a)Topic-21.1 Organic synthesis
(b)Topic-21.1 Organic synthesis
(c)Topic-21.1 Organic synthesis
(d)Topic-21.1 Organic synthesis
(e)Topic-21.1 Organic synthesis
(f)Topic-21.1 Organic synthesis
(g)Topic-21.1 Organic synthesis
(h)Topic-21.1 Organic synthesis
(i)Topic-21.1 Organic synthesis
A reaction scheme is shown in Fig. 7.1. The reagents needed for reaction 2 and reaction 3 are stated. Reaction 5 takes place when \(C_2H_5NH_2\) is mixed with compound V. No special conditions are required.
(a) Identify compound U which contains only three elements.
(b) Describe the reagents and conditions for reaction 1.
(c) Identify compound V.
(d) Complete the equation for reaction 3.
\(CH_3COOH + SOCl_2 \to …………..\)
(e) Identify compound W.
(f) Describe the conditions for reaction 4.
(g) Suggest the reagent needed for reaction 6.
(h) Complete Table 7.1 by adding the reaction numbers, 1, 2, 3, 4, 5 and 6, to the right-hand column. Use the reaction numbers given in Fig. 7.1. Each of the numbers 1, 2, 3, 4, 5 and 6 should be used once only.
(i) Compare the basicities of \(C_2H_5NHCOCH_3\), \(C_2H_5NHC_2H_5\), and \(NH_3\). Explain your answer.
▶️Answer/Explanation
Ans
(a) CH₃CN / ethanenitrile [1]
(b) dilute acid / HCl (aq)
AND hot / heat [1]
three components required (acid / aq / heat)
(c) CH₃COCl / ethanoyl chloride
(e) \(C_2H_5Br / C_2H_5Cl\) / bromoethane / chloroethane [1]
(f) heat in ethanol AND
under pressure / in sealed tube [1]
(g) LiAlH₄ / lithium aluminium hydride
Questions 8
(a)Topic-21.1 Organic synthesis
(b)Topic-21.1 Organic synthesis
(c)Topic-21.1 Organic synthesis
(a) An aqueous solution of phenol, \(C_6H_5OH\), is acidic at 298K. Explain why phenol is more acidic than water.
(b) (i) Name the two products formed when phenol reacts with an excess of \(Br_2\) (aq).
……………………………………………………… and ……………………………………………………..
(ii) Draw the structures of the two isomeric organic products, with \(M_r = 139\), that are formed when phenol reacts with HNO₃(aq) at room temperature.
(iii) Write the equation for the reaction between phenol, C₆H₅OH, and sodium metal.
(c) Phenol can be produced from phenylamine in a two-step synthesis.
Describe the reagents and conditions needed in each step.
step one:
reagents ……………………………………………………………………………………………………………………
conditions …………………………………………………………………………………………………………………
step two:
reagents ……………………………………………………………………………………………………………………
conditions …………………………………………………………………………………………………………………
▶️Answer/Explanation
Ans
(a) M1: p orbital / lone pair on the oxygen / O
AND is delocalised into the ring / overlaps with delocalised ring [1]
M2: O-H weakened (in phenol)
OR anion / phenoxide ion / conjugate base
is stabilised/more stable [1]
(b)(i) 2,4,6-tribromophenol (name) AND hydrogen bromide / HBr