Questions 1
(a)Topic -5.1 Enthalpy change, ΔH
(b)Topic- 5.1 Enthalpy change, ΔH
(c)Topic-5.1 Enthalpy change, ΔH
(d)Topic-23.3 Entropy change, ΔS
(e)Topic-21.1 Organic synthesis
Potassium iodide, KI, is used as a reagent in both inorganic and organic chemistry.
(a) KI forms an ionic lattice that is soluble in water.
(i) Define enthalpy change of solution, \(ΔH_{sol}\).
(ii) KI(s) has a high solubility in water although its enthalpy change of solution is endothermic. Explain how this high solubility is possible.
(b) Table 1.1 gives some data about the halide ions, \(Cl^–, Br^–\) and \(I^–\), and their potassium salts.
(i) Explain the trend in the enthalpy change of hydration of the halide ions.
(ii) The \(ΔH_{sol}\) values of these potassium halides are almost constant. Use the \(ΔH_{hyd}\) and \(ΔH_{latt}\) data in Table 1.1 to suggest why.
(iii) The enthalpy change of solution of KI(s) is +21.0 kJ mol⁻¹. Use this information and the data in Table 1.1 to calculate the enthalpy change of hydration of the potassium ion, K⁺(g).
(iv) Solid PbI₂ forms when KI(aq) is mixed with Pb²⁺(aq) ions. The solubility product, \(K_{sp}\), of \(PbI_2\) is 7.1 × 10⁻⁹ mol³ dm⁻⁹ at 25°C. Calculate the solubility, in moldm⁻³, of PbI₂(s).
(v) The ionic radius of \(Pb^{2+}\) is 0.120 nm compared to 0.133 nm for \(K^+\). Suggest how the \(ΔH^o_{latt}\) of PbI₂(s) differs from \(ΔH^o_{latt}\) of KI(s). Explain your answer.
(c) KI slowly oxidises in air, forming I₂.
Table 1.2 shows some data relevant to this question.
(i) Calculate the standard entropy change, \(ΔS^o\) , of reaction 1.
(ii) Use your answer to (c)(i) to show that reaction 1 is spontaneous at 298K.
(iii) The Group 1 carbonates are much more thermally stable than the Group 2 carbonates. State and explain the trend in the thermal stability of the Group 2 carbonates.
(d) A student electrolyses a solution of KI(aq) for 8 minutes using a direct current. The half-equation for the reaction that occurs at the anode is given.
\(2I^–(aq) \to I_2(aq) + 2e^–\)
(i) Write a half-equation for the reaction that occurs at the cathode. Include state symbols.
(ii) After the electrolysis, the \(I_2(aq)\) produced requires 21.35 cm³ of 0.100 moldm–³ Na₂S₂O₃(aq) to react completely.
\(I_2(aq) + 2Na_2S_2O_3(aq) \to 2NaI(aq) + Na_2S_4O_6(aq)\)
Calculate the average current used in 8 minutes during the electrolysis
(e) KI is used as a source of I⁻ ions in organic synthesis. One example of this is shown in the synthetic route in Fig. 1.1.
(i) Identify the reagents required for steps 1 and 2.
(ii) Step 3 occurs in two stages.
stage I NaNO₂ and HCl undergo an acid–base reaction to produce HNO₂.
stage II HNO₂ reacts with C, C₆H₅NH₂, to produce D, C₆H₅N₂⁺. Complete the equations for stage I and for stage II.
stage I NaNO₂ + HCl …………………………………………………………………………………….
stage II ……………………………………………………………………………………………………………..
(iii) The \(I^–\) from KI reacts with D in step 4. The mechanism is shown in Fig. 1.1. Suggest the name for this mechanism.
▶️Answer/Explanation
Ans:
(a)(i) (enthalpy change when) one mole of a substance / solute
AND dissolves in water / turns into an aqueous solution
(to form a solution of infinite dilution)
(ii) there is a (large) increase in entropy OR ΔS is positive OR TΔS is positive
so ΔG is negative / TΔS outweighs ΔH.
(b)(i) anionic charge density decreases (down the group / \(Cl^–\) to \(I^–\))
(so hydration enthalpies become less negative / less exothermic because)
less attraction of ion to water / the dipole–ion force weakens
(ii) the difference between \(\Delta H_{latt}\) and \(\Delta H_{hyd}\) remains roughly constant
OR \(\Delta H_{latt}\) and \(\Delta H_{hyd}\) become less exothermic by a similar amount
(iii) \(\Delta H_{hyd}(K^+(g)) = –629 + 21.0 –(–293) = –315 kJ mol^{–1}\)
(iv) solubility of \(PbI_2 = 7.1 \times 10^{–9} = x·(2x)^2\)
∴ x = ∛(¼ × 7.1 × 10⁻⁹)
solubility of \(PbI_2 = 1.21 \times 10^{–3} mol dm^{–3}\) min 2sf
(v) • \(Pb^{2+}\) has a greater charge
• \(Pb^{2+}\) is smaller (than \(K^+\)) OR \((Pb^{2+})\) smaller ionic radius
• greater attraction between \(Pb^{2+}\) and \(I^–\)
OR ionic bond between \(Pb^{2+}\) and \(I^–\) is stronger
OR lattice energy is more exothermic / more negative
(c)(i) ΔS = 2(155.5) + 2(116.1) – 4(106.3) – 2(213.6) – 205.2 1
ΔS = –514.4 (\(J K^{–1} mol^{–1})\) min 3sf ECF
(ii) ΔG = ΔH – TΔS AND use of 298 K for T OR clear working to represent this
ΔG = –203.4 – 298(–514.4 / 1000)
ΔG = –50.1 OR –50.2 (\(kJ mol^{–1})\) OR ΔG = –50 108.8 J min 3sf ECF from (c)(i)
(negative so spontaneous)
(iii) increases (in thermal stability down the group)
AND (cat)ionic radius / ion size increases (down the group)
less polarisation of anion / C—O bond / distortion of carbonate ion / CO₃²⁻
OR C–O is less weakened / stronger (down the group)
(d)(i) 2H⁺(aq) + 2e⁻ → H₂(g) state symbols required
(ii) M1 moles of \(S_2O_3^{2–} =\frac{21.35}{1000} × 0.100= 2.135 \times 10^{-3}\)
M2 calc of Q = 2.135 × 10⁻³ × 0.5 × 2 × 96500 = 206.02 / 205.68 ECF
M3 = answer I = 206 / (8 × 60) = 0.428 OR 0.429 A min 2sf ECF
(e)(i) reaction 1 concentrated HNO3 AND concentrated H2SO4 (concentrated seen once)
reaction 2 Sn AND (concentrated) HCl
(ii) step 1 NaNO₂ + HCl → HNO₂ + NaCl
step 2 C₆H₅NH₂ + HNO₂ + H⁺ → C₆H₅N₂⁺ + 2H₂O
(iii) nucleophilic (aromatic) substitution
Questions 2
(a)Topic – 3.6 Intermolecular forces, electronegativity and bond properties
(b)Topic -25.1 Acids and bases
(c)Topic – 23.3 Entropy change, ΔS
(d)Topic – 24.2 Standard electrode potentials, standard cell potentials and the Nernst equation
Water is an amphoteric compound that also acts as a good solvent of polar and ionic compounds.
(a) Equation 1 shows water acting as a Brønsted–Lowry acid.
equation 1 \(H_2O + NO_2^– \rightleftharpoons HNO_2 + OH^–\)
(i) Identify the two conjugate acid–base pairs in equation 1
(ii) Water also behaves as a Brønsted–Lowry acid when it dissolves CH₃NH₂. Explain the ability of CH₃NH₂ to act as a base.
(iii) Write an equation to show water acting as a base with CH₃COOH.
(b) The ionic product of water, \(K_w\), measures the extent to which water dissociates.
\(H_2O(l) \rightleftharpoons H^+(aq) + OH^–(aq)\)
Fig. 2.1 shows how \(K_w\) varies with temperature
(i) Write an expression for \(K_w\).
(ii) Use information from Fig. 2.1 to deduce whether the dissociation of water is an exothermic or an endothermic process. Explain your answer
(iii) An aqueous solution has pH = 7.00 at 30°C. Use information from Fig. 2.1 to explain why this solution can be considered to be alkaline at 30°C.
(c) The three physical states of \(H_2O\) have different standard entropies, \(S^o\) , associated with them. Table 2.1 shows these \(S^o\) values
(i) Explain the difference in the \(S^o\) values of \(H_2O(s)\) and \(H_2O(l)\).
(ii) Explain why the increase in \(S^o\) is much greater when H2O boils than when it melts.
(iii) The energy changes for \(H_2O(s) → H_2O(l)\) are shown.
ΔG = 0.00 kJ mol⁻¹
ΔH = +6.03 kJ mol⁻¹
Use these data to show that the melting point of \(H_2O(s)\) is 0°C.
(d) Metal–air batteries are electrochemical cells that generate electrical energy from the reaction of metal anodes with air. The standard electrode potentials for the zinc–air battery are shown.
(i) Calculate the standard cell potential, \(E^o_{cell}\), of the zinc–air battery.
(ii) The zinc–air battery usually operates at pH 11 and 298 K. The overall cell potential is dependent on \([OH^–]\). The Nernst equation shows how the electrode potential at the cathode changes with \([OH^–]\).Calculate the electrode potential, E, at pH 11.
▶️Answer/Explanation
Ans:
(a)(i) ● conjugate base of acid I = \(OH^–\)
● acid II = HNO₂
● conjugate base of acid II = \(NO_2^–\)
(ii) lone pair on the N can be donated to a proton/\(H^+\).
OR lone pair on the N can accept / gain a proton /\(H^+\)
OR lone pair on the N can form a dative bond to a proton/\(H^+\)
(iii) \(H_2O + CH_3COOH \to H_3O^+ + CH_3COO^–\)
2(b)(i) \((K_w) = [H^+][OH^–]\) ALLOW \((K_w) = K_a \times K_b\)
(ii) endothermic
AND equilibrium position moves right OR as water dissociates more
OR \(K_w\) increases with temperature
(iii) M1 (\(K_w\) increases with temperature so) pH of neutral solution decreases
OR (from graph) \(K_w = 1.50 \times 10^{–14} = [H^+]\)
∴ neutral pH = –½ log (1.50 × 10⁻¹⁴) = 6.91
OR \([H^+]\) = 10⁻⁷
∴ [OH⁻] = 1.50 × 10⁻¹⁴ / 10⁻⁷ = 1.50 × 10⁻⁷
M2 pH 7 is therefore above neutral pH / is alkaline
OR \([OH^–] > [H^+]\) (so alkaline)
(c)(i) \(H_2O(l)\) particles / molecules has more randomness / disorder
OR
\(H_2O(l)\) has more ways to arrange particles / energy (than in solid)
(ii) \(H_2O(g)\) particles / molecules has much more randomness / disorder
OR
\(H_2O\)(g) has many more ways to arrange particles / energy (than in liquid)
(iii) +6030 / (70.1 – 48.0) = 272.85 / 272.9 / 273 K (which is 0 °C)
(d)(i) (+)1.62 V
(ii) \([OH^–] = 10^{–14}/10^{–11} = 1 \times 10^{–3}\)
E = +0.40 – ½ × 0.059 × \(log (10^{–3})^2\) = +0.40 – 0.059(–3) = +0.577 (V) min 2sf
Questions 3
(a)Topic – 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(b)Topic -28.3 Colour of complexes
(c)Topic – 7.1 Chemical equilibria: reversible reactions, dynamic equilibrium
Iron is a transition metal in Group 8 of the Periodic Table.
(a) (i) Explain why iron has variable oxidation states.
(ii) Complete the shorthand electronic configurations of Fe and Fe³⁺.
(b) An aqueous solution of \(Fe(NO_3)_3\) contains the complex \([Fe(H_2O)_6]^{3+}\). When solutions of KSCN(aq) and \([Fe(H_2O)_6]^{3+}\)(aq) are mixed, a colour change is observed. The red complex \([Fe(H_2O)_5SCN]^{2+}\) forms.
(i) Define complex.
(ii) State the coordination number of Fe in \([Fe(H_2O)_6]^{3+}\).
(iii) The H—O—H bond angle in water is 104.5°. Suggest the H—O—H bond angle in \([Fe(H_2O)_6]^{3+}\). Explain your answer.
(iv) Explain why iron complexes are coloured.
(v) Aqueous solutions of complexes \([Fe(H_2O)_6]^{3+}\) and \([Fe(H_2O)_5SCN]^{2+}\) are different colours. Explain why these complexes are different colours.
(c) Table 3.1 gives values for the stability constants, \(K_{stab}\), of different complexes of iron.
(i)\([Fe(H_2O)_5(H_2PO_4)]^{2+}\) can form when H₃PO₄ reacts with \([Fe(H_2O)_6]^{3+}\). Write an equation for this reaction.
(ii) Write an expression for \(K_{stab}\) of \([Fe(H_2O)_5SCN]^{2+}\) and give its units.
(iii) Use the stability constant data in Table 3.1 to calculate the value of the equilibrium constant, \(K_c\), for the following equilibrium.
\([Fe(H_2O)_5(H_2PO_4)]^{2+} + SCN^– \rightleftharpoons [Fe(H_2O)_5SCN]^{2+} + H_2PO_4^–\)
▶️Answer/Explanation
Ans:
(a)(i) energies / energy levels of the 3d and the 4s (sub-shells / orbitals) are similar
OR the difference between the 3d and the 4s is small
(ii) Fe =\( [Ar] 3d^6 4s^2 \) AND
Fe³⁺ = [Ar] 3d⁵
(b)(i) (a molecule or ion formed by a central) metal atom / ion
surrounded / bonded by one or more ligands
(ii) 6 / six
(iii) bond angle 106–108°
AND lone pair from O is donated (in a bond)
OR one more of O’s electron pairs is now a bond/bonding pair (so repels less)
(iv) M1 (degenerate) d orbitals split (into two energy levels)
OR (degenerate) d orbitals become non-degenerate
M2 as an electron moves up / to a higher energy level
M3 absorption of light energy / energy in the visible region
AND colour seen is complementary (to colour absorbed)
(v) (d–d) energy gap / \(\Delta E / \Delta _{oct}\) is different
different frequency / wavelength (of light) absorbed / transmitted / reflected
(c)(i) \([Fe(H_2O)_6]^{3+} + H_3PO_4 \to [Fe(H_2O)_5(H_2PO_4)]^{2+} + H_3O^+\)
(ii)\(K_{stab} =\frac{[[Fe(H_2O)_5 (SCN)]^{2+} ]}{[[Fe(H_2O)_6 ]^{3+} ] [SCN^-]}\)
\(mol^{–1} dm^3 u/c\)
(iii) \(1.30 \times 10^2/ 59.0 = 2.2(03)\) min 2sf
Questions 4
(a)Topic – 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(b)Topic – 28.2 General characteristic chemical properties of the first set of transition elements, titanium to copper
(c)Topic – 35.1 Condensation polymerisation
Ruthenium and osmium are transition metals below iron in Group 8 of the Periodic Table.
(a) Two different complex ions, X and Y, can form when anhydrous RuCl₃ reacts with water under certain conditions. X and Y have octahedral geometry. Aqueous samples of X and Y react separately with an excess of AgNO₃ (aq). Different amounts of AgCl are precipitated:
• 1 mole of complex ion X produces 2 moles of AgCl
• 1 mole of complex ion Y produces 1 mole of AgCl.
(i) Complete Table 4.1 to suggest formulae for X and Y
(ii) Both complexes react with an excess of bipyridine, bipy, to form a mixture of two stereoisomers of [Ru(bipy)₃]³⁺.
Bipyridine is a bidentate ligand. Draw three-dimensional diagrams of the two stereoisomers of [Ru(bipy)₃]³⁺. Use to represent the bipy ligand in your structures.
(b) Fig. 4.1 shows another ruthenium complex.
This complex contains the neutral ligand pyrazine.
(i) Suggest how pyrazine is able to bond to two separate ruthenium ions.
(ii) Pyrazine is an aromatic compound. The bonding and structure of pyrazine is similar to that of benzene. Describe and explain the shape of pyrazine. In your answer, include:
• the hybridisation of the nitrogen and carbon atoms
• how orbital overlap forms π bonds between the atoms in the ring.
(iii) Predict the number of peaks seen in the carbon–13 NMR spectrum of pyrazine. Explain your answer.
(iv) The overall charge of the ruthenium complex in Fig. 4.1 is 5+.
Deduce the possible oxidation states of the two ruthenium ions in the complex.
(c) Osmium tetroxide, OsO₄, reacts with alkenes in a similar manner to cold dilute acidified MnO₄⁻. Fig. 4.2 shows a proposed synthesis of a condensation polymer G.
(i) Suggest a reagent for step 1.
(ii) Draw the structure of exactly one repeat unit of the condensation polymer G. The ester linkage should be shown fully displayed.
▶️Answer/Explanation
Ans:
(b)(i) (pyrazine has) a lone pair on each N (atoms) / two lone pairs (on the N’s)
AND which can be donated / form a coordinate / dative bond (with Ru)
(ii) • shape is (hexagonal ring) planar / (trigonal) planar / 120°
• carbons and nitrogens are \(sp^2\) hybridised
• a p orbital (from each atom) overlaps sideways/laterally
(with each other above and below the ring forming π bonds)
(iii) 1/one
all the carbon atoms are equivalent / in same environment
OR pyrazine is a symmetrical molecule u/c
4(b)(iv) +2 / 2+ AND +3 / 3+ OR +4 / 4+ AND +1 / 1 + 1
4(c)(i) SOCl₂ OR PCl₃ OR PCl₅
Questions 5
(a)Topic -21.1 Organic synthesis
(b)Topic -21.1 Organic synthesis
Compound Q can be synthesised from chlorobenzene in seven steps, using the route shown in Fig. 5.1.
(a) (i) Write an equation for the formation of the electrophile for step 1.
(ii) Complete the mechanism in Fig. 5.2 for step 1, the alkylation of chlorobenzene. Include all relevant curly arrows and charges. Draw the structure of the intermediate.
(iii) Step 2 is an oxidation reaction. Construct an equation for the reaction in step 2. Use [O] to represent an atom of oxygen from an oxidising agent.
(iv) Suggest reagents for the conversion of K to M in steps 3 and 4.
(v) Identify the type of reaction that occurs in step 5.
(vi) Step 7 takes place when P is heated with a weak base such as K₂CO₃(aq).
Suggest why a strong base such as NaOH(aq) is not used for this reaction.
(vii) Q is optically active. Explain the meaning of optically active.
(viii) Give two reasons why it might be desirable to synthesise a single optical isomer of Q for use as a drug.
(b) Q is commonly used in conjunction with aspirin.
Aspirin is a weak Brønsted–Lowry acid.
(i) The \(pK_a\) of aspirin is 3.49. 75mg of aspirin dissolves in water to form 100 cm³ of an aqueous solution. Calculate the pH of this solution.
[\(M_r\): aspirin, 180.0]
(ii) Aspirin undergoes acid hydrolysis in the stomach. Give the structures of the organic products of this acid hydrolysis.
▶️Answer/Explanation
Ans:
(a)(i) AlCl₃ + CH₃Cl → AlCl₄⁻ + CH₃⁺
(iii) \(C_6H_4(Cl)CH_3 + 2[O] → C_6H_4(Cl)CHO + H_2O\)
(iv) step 3 HCN AND KCN (cat) OR KCN AND \(H_2SO_4\)/HCl
step 4 H₂SO₄ (aq) OR HCl (aq)
(v) addition–elimination / condensation
(vi) (a strong base) it would hydrolyse the ester
(vii) (a substance able to) rotate the plane of plane-polarised light
(viii) any two of:
• reduced / different biological activity of ‘other’ enantiomer ORA
• avoids need to separate the optical isomers to form the pure active isomer
• lower dosage required OR (drug is) more potent
• higher yield (of biologically-active molecule)
• no / less (harmful) side effects OR other isomer can have side effects
(b)(i) M1 [HA] = \((75 × 10^–3/ 180) ÷ 0.100 = 4.17 × 10^–3\) OR \(1 / 240 (mol dm^{–3})\)
M2 \([H^+] = (K_a \times [HA])½= (10^{–3.49} \times 4.17 \times 10^{–3})½ = 1.16 \times 10^{–3}(mol dm^{–3})\)
M3 pH = –log \([H^+]\) = 2.93 to 2.94 min 2sf
Questions 6
(a)Topic – 34.4 Amino acids
(b)Topic -21.1 Organic synthesis
(c)Topic -21.1 Organic synthesis
(d) Topic- 37.3 Carbon-13 NMR spectroscopy
Amino acids are molecules that contain \(—NH_2\) and —COOH functional groups. Glycine, \(H_2NCH_2COOH\), is the simplest stable amino acid.
(a) The isoelectric point of glycine is 6.2.
(i) Define isoelectric point.
(ii) Draw the structure of glycine at pH 4.
(b) Fig. 6.1 shows two syntheses starting with glycine.
(i) State the essential conditions for reaction 1.
(ii) Identify the reagent used in reaction 2.
(iii) Draw the structure of the organic product U that forms when hippuric acid reacts with an excess of \(LiAlH_4\) in reaction 3.
(iv) A molecule of phenylalanine, R, can react with a molecule of glycine to form two dipeptides, S and T. S and T are structural isomers.
Draw the structures of these dipeptides. The peptide bond formed should be shown fully displayed.
(c) A student proposes a synthesis of hippuric acid by the reaction of benzamide, C₆H₅CONH₂, and chloroethanoic acid, ClCH₂COOH. The reaction does not work well because benzamide is a very weak base.
(i) Explain why amides are weaker bases than amines.
(ii) The \(pK_a\) of chloroethanoic acid is 2.86 whereas the \(pK_a\) of ethanoic acid is 4.76. Explain the difference between these two \(pK_a\) values.
(d) Compound V is another amino acid. The proton (1H) NMR spectrum of V shows hydrogen atoms in five different environments, a, b, c, d and e, as shown in Fig. 6.2.
(i) Complete Table 6.2 for the proton \((^1H)\) NMR spectrum of V taken in CDCl₃. Table 6.1 gives some relevant data.
(ii) Complete Table 6.3 by placing a tick (✓) to indicate any protons whose peaks are still present in the proton \((^1H)\) NMR spectrum of V taken in \(D_2O\).
▶️Answer/Explanation
Ans:
(a)(i) pH at which a molecule has
no overall charge / is neutral/ is a Zwitterion / charges cancel out
(ii)
(b)(i) ethanol AND heat in a sealed tube
OR ethanol AND high pressure
(ii) \(C_6H_5COCl\) / benzoyl chloride / benzoyl anhydride
(c)(i) M1 nitrogen lone pair in amides is delocalised with C=O
M2 lone pair less available for donation / to accept \(H^+\)
OR less electron density on N/NH₂, so less able to accept \(H^+\)
(ii) chloroethanoic acid is a stronger acid (than ethanoic acid)
because electron-withdrawing (–I / inductive) effect of Cl substituent
AND weakens O—H / carboxylate anion stabilised