9701_s23_qp_42
A paper of Chemistry, 9701
Questions:
9
Year:
2023
Paper:
4
Variant:
2
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1
Group 2 carbonates decompose when heated to form the metal oxide and carbon dioxide. Suggest a mechanism for the decomposition of the carbonate ion by adding two curly arrows in . O O– –O C CO2 O2– + Describe the variation in the thermal stability of Group 2 carbonates. Explain your answer. Define lattice energy. The lattice energy of the Group 2 carbonates, ΔH o latt(MCO3), becomes less exothermic down the group. The lattice energy of the Group 2 oxides, ΔH o latt(MO), also becomes less exothermic down the group. ΔH o latt(MCO3) and ΔH o latt(MO) change by different amounts going down the group. Suggest how the standard enthalpy change of the decomposition reaction for Group 2 carbonates changes down the group. Explain your reasoning in terms of the relative sizes of the anions and the relative changes in lattice energy down the group. Potassium sulfite, K2SO3, is used as a food additive. The concentration of sulfite ions, SO3 2–, can be determined by titration using aqueous acidified manganate(ions, MnO4 –. • A 250 cm3 solution contains 3.40 g of impure K2SO3. • 25.0 cm3 of this solution requires 22.40 cm3 of 0.0250 mol dm–3 acidified MnO4 – to reach the end-point. All the SO3 2– ions are oxidised. None of the other species in the impure K2SO3 are oxidised. The reaction occurs as shown by the two half-equations. H2O + SO3 2– SO4 2– + 2H+ + 2e– MnO4 – + 8H+ + 5e– Mn2+ + 4H2O Give the ionic equation for the reaction between SO3 2– and acidified MnO4 –. Calculate the percentage purity of the sample of K2SO3. Show your working. percentage purity of K2SO3 = Potassium disulfite, K2S2O5, is another food additive. The disulfite ion, S2O5 2–, has the displayed formula shown in . O O O O– O– S S α Deduce the geometry around the S(α) atom in S2O5 2–. geometry around S(α)
10. Group 2  →  10.1. Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
2. Atoms, molecules and stoichiometry  →  2.4. Reacting masses and volumes (of solutions and gases)
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State two typical chemical properties of a transition element. Aqueous solutions of cobalt(salts contain the complex ion [Co(H2O)6]2+. Define complex ion. Samples of [Co(H2O)6]2+ are reacted separately with an excess of aqueous ammonia, with an excess of concentrated HCl and with an excess of aqueous sodium hydroxide, as shown in . excess NH3excess NaOHexcess concentrated HCl [Co(H2O)6]2+A C B Complete Table 2.1 about the reactions shown by [Co(H2O)6]2+. Table 2.1 reagent added to [Co(H2O)6]2+formula of cobalt species formed colour and state of cobalt species formed type of reaction an excess of NH3A = an excess of concentrated HCl B = an excess of NaOHC = The ethanedioate ion, C2O4 2–, can act as a bidentate ligand. Explain what is meant by a bidentate ligand. The complex [Co(H2O)2(C2O4)BrCl ]− exists as stereoisomers. Complete the three-dimensional diagrams in to show four stereoisomers of [Co(H2O)2(C2O4)BrCl ]−. The C2O4 2– ligand is represented using O O . O Co O O Co O O Co O O Co O State the oxidation state of cobalt in this complex and a type of stereoisomerism shown. oxidation state of cobalt type of stereoisomerism
17. Carbonyl compounds  →  17.1. Aldehydes and ketones
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
11. Group 17  →  11.3. Some reactions of the halide ions
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Complete Table 3.1 by placing one tick (3) in each row to indicate the sign of each type of energy change under standard conditions. Table 3.1 energy change always positive always negative can be either negative or positive bond energy enthalpy change of atomisation enthalpy change of formation Define standard enthalpy change of atomisation. Table 3.2 shows some energy changes. Table 3.2 energy change value / kJ mol–1 standard enthalpy change of atomisation of silver +285 first ionisation energy of silver +731 second ionisation energy of silver +2074 bond energy of O=O +496 bond energy of O–O +150 first electron affinity of oxygen –141 second electron affinity of oxygen +798 first ionisation energy of oxygen +1314 standard enthalpy change of formation of silver oxide, Ag2O–31 Calculate the lattice energy, ΔH o latt, of Ag2Ousing relevant data from Table 3.2. It may be helpful to draw a labelled energy cycle. Show your working. ΔH o latt of Ag2O= kJ mol–1 Suggest the trend in the magnitude of the lattice energies of the silver compounds Ag2S, Ag2O and Ag2Se. Explain your answer. least exothermic most exothermic Silver sulfite, Ag2SO3, is sparingly soluble in water. Give an expression for the solubility product, Ksp, of Ag2SO3. Ksp = Calculate the equilibrium concentration of Ag+ in a saturated solution of Ag2SO3 at 298 K. [Ksp: Ag2SO3, 1.50 × 10–14 mol3 dm–9 at 298 K] [Ag+] = mol dm–3 The standard enthalpy change of solution, ΔH o sol, of AgNO3in water is +22.6 kJ mol–1. Suggest how the feasibility of dissolving AgNO3in water changes with temperature. Explain your answer.
5. Chemical energetics  →  5.1. Enthalpy change, \(\Delta H\)
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In aqueous solution, iron(ions react with iodide ions, as shown. 2Fe3+ + 2I– 2Fe2+ + I2 A series of experiments is carried out using different concentrations of Fe3+ and I–, as shown in Table 4.1. Table 4.1 experiment [Fe3+] / mol dm–3 [I–] / mol dm–3 initial rate / mol dm–3 s–1 0.0400 0.0200 2.64 × 10–4 0.1200 0.0200 7.92 × 10–4 0.0800 0.0400 2.11 × 10–3 Explain what is meant by overall order of reaction. Use the data in Table 4.1 to deduce the order of reaction with respect to Fe3+ and with respect to I–. Explain your reasoning. Use your answer to to construct the rate equation for this reaction. rate = Use your answer to and the data from experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. k = units Describe qualitatively the effect of an increase in temperature on the rate constant and on the rate of this reaction. In aqueous solution, iodide ions react with acidified hydrogen peroxide, as shown. 2I– + H2O2 + 2H+ I2 + 2H2O The initial rate of reaction is found to be first order with respect to I–, first order with respect to H2O2 and zero order with respect to H+. shows a possible four-step mechanism for this reaction. step 1 H2O2 + I– IO– + H2O step 2 H+ + IO– HIO step 3 HIO + I– I2 + OH– step 4 OH– + H+ H2O Suggest which of the steps, 1, 2, 3 or 4, in this mechanism is the rate-determining step. Explain your answer. Identify a step in that involves a redox reaction. Explain your answer in terms of oxidation numbers. Suggest the role of HIO in this mechanism. Explain your reasoning.
8. Reaction kinetics  →  8.1. Rate of reaction
8. Reaction kinetics  →  8.2. Effect of temperature on reaction rates and the concept of activation energy
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Methylbenzene can undergo different reactions, as shown in . reaction 1 H2 / Pt Br2 / UV light Br2 / FeBr3 + reaction 2 reaction 3 Draw structures in for the possible organic products of the three reactions shown. Complete Table 5.1. Table 5.1 type of reaction mechanism reaction 1 reaction 2 When methylbenzene reacts with an electrophile, a substitution reaction occurs. No addition reaction takes place under these conditions. Explain why no addition reaction takes place. The reaction of methylbenzene with thionyl bromide, SOBr2, in the presence of an iron(bromide catalyst, FeBr3, is shown in . FeBr3 + SOBr2 + HBr S O Br The mechanism of this reaction is similar to that of the bromination of benzene. The first step of the mechanism generates the SOBr+ electrophile, as shown. SOBr2 + FeBr3 SOBr+ + FeBr4 – The reaction of methylbenzene with SOBr+ ions is shown in . Complete the mechanism in . Include all relevant curly arrows and charges. Draw the structure of the organic intermediate. S+ O Br intermediate + S O Br The reaction shown in produces a small amount of a by-product, P, with the molecular formula C14H14OS. Suggest a structure for by-product P. Acyl bromides, RCOBr, can be synthesised by the reaction of a carboxylic acid and SOBr2. This is a similar reaction to the synthesis of acyl chlorides using SOCl 2. Give an equation for the reaction between ethanoic acid and SOBr2. Suggest the relative ease of hydrolysis of acyl bromides, RCOBr, acyl chlorides, RCOCl, and alkyl chlorides, RCl. Explain your answer. > > easiest to hydrolyse hardest to hydrolyse
14. Hydrocarbons  →  14.2. Alkenes
15. Halogen compounds  →  15.1. Halogenoalkanes
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Perindopril is a drug used to treat heart disease. O N N O O O OH H perindopril State the number of chiral carbon atoms present in one molecule of perindopril. Suggest one benefit and one disadvantage of producing a drug such as perindopril as a single pure optical isomer. benefit disadvantage Name all the functional groups in perindopril. A sample of perindopril is hydrolysed with hot aqueous acid. Draw the structures of the three organic products of the complete acid hydrolysis of perindopril.
13. An introduction to AS Level organic chemistry  →  13.4. Isomerism: structural isomerism and stereoisomerism
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Explain why phenol is brominated much more easily than benzene is brominated. Iodine monobromide, I–Br, reacts with benzene in the presence of an Al Br3 catalyst. Predict whether the organic product will be bromobenzene or iodobenzene. Explain your answer. shows some reactions of phenol. OH OH OH N N O2N Q R reaction 1 reaction 2 excess Br2NaGive an equation for the reaction of phenol with Na. Draw the structure of the organic product, R, formed when phenol reacts with an excess of Br2. State the reagents and conditions for reaction 1 and reaction 2 in . reaction 1 reaction 2
15. Halogen compounds  →  15.1. Halogenoalkanes
8. Reaction kinetics  →  8.3. Homogeneous and heterogeneous catalysts
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Describe the difference in reactivity between chloroethane and chlorobenzene with OH–. Explain your answer. Compound T, C5H9O2Cl, is a useful synthetic intermediate. shows some reactions of T. T Z, C6H13NO2 O O Cl O O NC step 1 step 2 W X excess NaOHOH heat + Y + excess HCl heat Give the systematic name for T. Draw the structures of W, X, Y and Z in . State the reagents and conditions for steps 1 and 2 in . step 1 step 2 The proton (1H) NMR spectrum of compound T, C5H9O2Cl, in CDCl 3 is shown in . T O O chemical shift δ / ppm δ 3.9 δ 3.7 δ 2.8 δ 1.2 Cl Table 8.1 environment of proton example chemical shift range δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0 alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 alkyl amine R–NH– 1.0–5.0 aryl amine Ar–NH2 3.0–6.0 amide RCONHR 5.0–12.0 Suggest why CDCl 3 is used as a solvent instead of CHCl 3 for the proton (1H) NMR spectrum. Complete Table 8.2 for the proton (1H) NMR spectrum of T. Table 8.2 chemical shift δ / ppm environment of proton splitting pattern number of 1H atoms responsible for the peak 1.2 2.8 3.7 3.9 Explain the splitting pattern of the peak at δ 3.9 ppm.
15. Halogen compounds  →  15.1. Halogenoalkanes
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Define standard cell potential, E o cell. An electrochemical cell is set up to measure E o cell of a cell consisting of an Fe3+ / Fe2+ half-cell and a Cl 2 / Cl – half-cell. Draw a labelled diagram of this electrochemical cell. Include all necessary substances. It is not necessary to state conditions used. The cell reaction for the electrochemical cell in is shown. Cl 2 + 2Fe2+ 2Fe3+ + 2Cl – E o cell = +0.59 V Calculate ΔG o , in kJ mol–1, for this cell reaction. ΔG o = kJ mol–1
1. Atomic structure  →  1.3. Electrons, energy levels and atomic orbitals
7. Equilibria  →  7.1. Chemical equilibria: reversible reactions, dynamic equilibrium
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