9701_m23_qp_42
A paper of Chemistry, 9701
Questions:
6
Year:
2023
Paper:
4
Variant:
2
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The most common zinc mineral contains zinc(sulfide, ZnS. Complete the electrons in boxes diagram in to show the electronic configuration of a zinc(ion. [Ar] 3d 4s 4p Complete to show the Born–Haber diagram for the ionic solid ZnS. Include state symbols of relevant species. Zn2++ S+ 2e– ZnSZn2++ S–+ e– Zn2++ S2–IE2 IE1 ΔHf ΔHlatt ΔHat(S) EA2 EA1 ΔHat(Zn) Describe the trend in the first electron affinity of the Group 16 elements S to Te. Explain your answer. Explain why the lattice energy, ∆Hlatt, of ZnO is more exothermic than that of ZnS. Zinc metal can be obtained in a two-step process as shown. step 1 2ZnS+ 3O22ZnO+ 2SO2step 2 ZnO+ CZn+ COThe reactions are carried out at 800 °C. Predict the sign of the entropy change, ∆S o , of the reaction in step 1. Explain your answer. Use the data in Table 1.1 to calculate ∆S o of the reaction shown in step 2. Table 1.1 chemical ZnOCZnCOS o / J K–1 mol–1 43.7 5.7 50.8 197.7 ∆S o = J K–1 mol–1 An equation for the direct reduction of ZnS by carbon is shown. 2ZnS+ C2Zn+ CS2∆H o = +733 kJ mol–1 ∆S o = +218 J K–1 mol–1 This reaction is not feasible at 800 °C. Calculate ∆G o for this reaction at 800 °C. ∆G o = kJ mol–1 Zn(NO3)2 undergoes thermal decomposition when heated. The reaction is similar to the thermal decomposition of Group 2 nitrates. Construct an equation for the thermal decomposition of Zn(NO3)2. The radii of some Group 2 cations and Zn2+ are shown in Table 1.2. Table 1.2 cation Mg2+ Ca2+ Sr2+ Ba2+ Zn2+ radius / pm State and explain the trend in thermal stability of the Group 2 nitrates down the group. Use Table 1.2 to suggest which Group 2 nitrates are less thermally stable than zinc nitrate.
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  →  2.3. Formulas
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Hypophosphorous acid is an inorganic acid. The conjugate base of hypophosphorous acid is H2PO2 –. Give the formula of hypophosphorous acid. H2PO2 – is a strong reducing agent. It can be used to reduce metal cations without the need for electrolysis. equation 1 HPO3 2– + 2H2O + 2e– H2PO2 – + 3OH– E o = –1.57 V In an experiment, an alkaline HPO3 2–/H2PO2 – half-cell is constructed with [H2PO2 –] = 0.050 mol dm–3. All other ions are at their standard concentration. Predict how the value of E of this half-cell differs from its E o value. Explain your answer. The Cr3+/Cr half-cell has a standard electrode potential of –0.74 V. An electrochemical cell consists of an alkaline HPO3 2–/H2PO2 – half-cell and a Cr3+/Cr half-cell. Calculate the standard cell potential, Ecell. Ecell = V o o Complete the diagram in to show how the standard electrode potential of the Cr3+/Cr half-cell can be measured relative to that of the standard hydrogen electrode. Identify the chemicals, conditions and relevant pieces of apparatus. Label to show: • which is the positive electrode • the direction of electron flow in the external circuit. H2PO2 – reduces Ni2+ to Ni in alkaline conditions. Use equation 1 to construct the ionic equation for this reaction. equation 1 HPO3 2– + 2H2O + 2e– H2PO2 – + 3OH– H2PO2 –reacts with OH–. H2PO2 –+ OH–HPO3 2–+ H2Table 2.1 shows the results of a series of experiments used to investigate the rate of this reaction. Table 2.1 experiment [H2PO2 –] / mol dm–3 [OH–] / mol dm–3 volume of H2 produced in 60 s / cm3 0.40 2.00 6.4 0.80 2.00 12.8 1.20 1.00 4.8 The volume of H2 was measured under room conditions. Use the molar volume of gas, Vm, and the data from experiment 1 to calculate the rate of reaction in mol dm–3 s–1. rate of reaction = mol dm–3 s–1 The rate equation was found to be: rate = k [H2PO2 –] [OH–]2 Show that the data in Table 2.1 is consistent with the rate equation. State the units of the rate constant, k, for the reaction. The experiment is repeated using a large excess of OH–. Under these conditions, the rate equation is: rate = k1 [H2PO2 –] k1 = 8.25 × 10–5 s–1 Calculate the value of the half-life, t1 2, of the reaction. t1 2 = s Describe how an increase in temperature affects the value of the rate constant, k1. A student suggests that the reaction between H2PO2 –and OH–might happen more quickly in the presence of a heterogeneous catalyst. Describe the mode of action of a heterogeneous catalyst.
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
11. Group 17  →  11.4. The reactions of chlorine
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Vanadium is a transition element in Period 4 of the Periodic Table. Define transition element. Vanadium shows typical chemical properties of transition elements, including variable oxidation states. State two other typical chemical properties of transition elements. Explain why transition elements have variable oxidation states. VO2 + can be reduced to V2+ by C2O4 2– in acidic conditions. equation 2 2VO2 + + 3C2O4 2– + 8H+ 2V2+ + 6CO2 + 4H2O In a titration, 25.00 cm3 of 0.0300 mol dm–3 VO2 +is added to 10 cm3 of dilute sulfuric acid. A solution of 0.0400 mol dm–3 C2O4 2–is then added from a burette until the end-point is reached. The titration is repeated and concordant results obtained, as shown in Table 3.1. Table 3.1 volume of C2O4 2–added / cm3 28.15 28.10 Show that these results are consistent with the stoichiometry of equation 2. An excess of C2O4 2– reacts with VO2 + to form a mixture of two octahedral complex ions. The complex ions are stereoisomers of each other. Each complex ion contains a V2+ cation and three C2O4 2– ions. Complete the diagram to show the three-dimensional structure of one of the complex ions. Include the charge of the complex ion. Use O O to represent a C2O4 2– ion. V V2+can be oxidised by H2O2. Table 3.2 gives some relevant data. Table 3.2 half-equation E o / V H2O2+ 2H++ 2e– 2H2O+1.77 VO2 ++ 2H++ e– VO2++ H2O+1.00 VO2++ 2H++ e– V3++ H2O+0.34 V3++ e– V2+–0.26 Identify the vanadium species that forms when an excess of H2O2reacts with V2+under standard conditions. Explain your answer with reference to the data in Table 3.2. Concentrated acidified H2O2 can react with V2+ to form red VO2 3+ ions. VO2 3+ contains vanadium combined with the peroxide anion, O2 2–. Deduce the oxidation state of vanadium in VO2 3+.
9. The Periodic Table: chemical periodicity  →  9.2. Periodicity of chemical properties of the elements in Period 3
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
10. Group 2  →  10.1. Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
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Ethylamine and phenylamine are primary amines. NH2 phenylamine ethylamine CH3CH2NH2 These two compounds are synthesised by different methods. Several methods can be used to form ethylamine. Ethylamine forms when ethanamide, CH3CONH2, is reduced by LiAl H4. Write an equation for this reaction. Use [H] to represent one atom of hydrogen from the reducing agent. Ethylamine is a product of the reaction of bromoethane with ammonia. Name the mechanism of this reaction and state the conditions used. mechanism conditions The reaction in also forms secondary and tertiary amines. Suggest the identity of a secondary or tertiary amine formed by the reaction in . Ethylamine is a weak base. State the relative basicities of ammonia, ethylamine and phenylamine. Explain your answer. least basic most basic Pure phenylamine, C6H5NH2, can be prepared from benzene in two steps. Draw the structure of the intermediate compound. Suggest reagents and conditions for each step. shows some reactions of phenylamine. NH2 N2Cl OH N N phenylamine reaction 1 reaction 2 excess Br2room temperature reaction 4 W reaction 3 X Y benzenediazonium chloride Draw the structure of W, the organic product of reaction 1. State the reagents used in reaction 2. Benzenediazonium chloride, C6H5N2Cl, and X react together in reaction 4 to form Y, an azo compound. Name X, the organic product of reaction 3. State the necessary conditions for reaction 4 to occur. Suggest a use for Y. Methylamine, CH3NH2, is another primary amine. CH3NH2 can act as a monodentate ligand. Define monodentate ligand. Cu2+reacts with CH3NH2 to form [Cu(CH3NH2)2(H2O)4]2+. Draw three-dimensional diagrams to show the two geometrical isomers of [Cu(CH3NH2)2(H2O)4]2+. Cu Cu State the coordination number of copper in [Cu(CH3NH2)2(H2O)4]2+. Cd2+ions form tetrahedral complexes with CH3NH2, OH– and Cl – ions, as shown in equilibria 1, 2 and 3. equilibrium 1 Cd2++ 4CH3NH2[Cd(CH3NH2)4]2+Kstab = 3.3 × 106 equilibrium 2 Cd2++ 4OH–Cd(OH)4 2–Kstab = 5.0 × 108 equilibrium 3 Cd2++ 4Cl –CdCl 4 2–Kstab = 6.3 × 102 Give the units of Kstab for equilibrium 1. Write an expression for Kstab for equilibrium 3. Kstab = A solution of Cl –is added to Cd2+and allowed to reach equilibrium. The equilibrium concentrations are given. [Cd2+] = 0.043 mol dm–3 [Cl –] = 0.072 mol dm–3 Use your expression in to calculate the concentration of CdCl 4 2–in the equilibrium mixture. [CdCl 4 2–] = mol dm–3 When CH3NH2is added to Cd2+, a mixture of [Cd(CH3NH2)4]2+and [Cd(OH)4]2–forms. Suggest how the [Cd(OH)4]2–is formed. Cd2+exists as a complex ion, [Cd(H2O)6]2+. Identify the most stable and the least stable of the complexes in Table 4.1 by placing one tick (3) in each column. Explain your answer. Table 4.1 complex most stable least stable [Cd(H2O)6]2+[Cd(OH)4]2–[Cd(CH3NH2)4]2+[CdCl4]2–explanation
19. Nitrogen compounds  →  19.1. Primary amines
15. Halogen compounds  →  15.1. Halogenoalkanes
12. Nitrogen and sulfur  →  12.1. Nitrogen and sulfur
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Tulobuterol is used in some medicines. Cl OH H N tulobuterol Tulobuterol contains a benzene ring in its structure. Describe and explain the shape of benzene. In your answer, include: • the bond angle between carbon atoms • the hybridisation of the carbon atoms • how orbital overlap forms v and r bonds between the carbon atoms. In a synthesis of tulobuterol, the first step involves the formation of chlorobenzene. Benzene reacts with Cl2 in the presence of an Al Cl3 catalyst. Cl step 1 Cl2 and AlCl3 Write an equation to show how Cl2 reacts with Al Cl3 to generate an electrophile. Complete the mechanism in for the reaction of benzene with the electrophile generated in . Include all relevant curly arrows and charges. Draw the structure of the intermediate. Cl intermediate The second step of the synthesis involves the reaction of chlorobenzene with Cl COCH2Cl, also in the presence of an Al Cl 3 catalyst, forming compound Q. Cl Cl Cl Cl O O Cl Q step 2 and AlCl3 Name the mechanism of the reaction in step 2. Draw the structure of an isomer of Q that forms as an organic by-product of the reaction in step 2. The reactants used in step 2 contain acyl chloride, alkyl chloride and aryl chloride functional groups. State and explain the relative ease of hydrolysis of acyl chlorides, alkyl chlorides and aryl chlorides. , , easiest to hydrolyse hardest to hydrolyse Tulobuterol is produced from Q as shown in . Cl O Cl Q Z step 3 step 4 Cl OH H N tulobuterol Suggest reagents and conditions for steps 3 and 4. Draw the structure of Z in the box. step 3 step 4 Z The synthesis produces two enantiomers of tulobuterol. Define enantiomers. Suggest one disadvantage of producing two enantiomers in this synthesis. Suggest a method of adapting the synthesis to produce a single enantiomer. Cl OH H N tulobuterol Predict the number of peaks that would be seen in the carbon-13 NMR spectrum of tulobuterol. The proton (1H) NMR spectrum of tulobuterol dissolved in D2O shows peaks in four different types of proton environment. The peak for the —CH2N— environment is a doublet in the chemical shift range d = 2.0–3.0 ppm. Give details for each of the other three peaks in the proton NMR spectrum of tulobuterol, to include: • chemical shift • environment of the proton • splitting pattern • number of 1H atoms responsible. Table 5.1 gives information about typical chemical shift values.
21. Organic synthesis  →  21.1. Organic synthesis
14. Hydrocarbons  →  14.2. Alkenes
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A student uses thin-layer chromatography (TLC) to analyse a mixture containing different metal cations. The student repeats the experiment using different solvents. shows the chromatogram obtained by the student using water as a solvent. solvent front cm baseline M Suggest a compound that could be used as the stationary phase in this experiment. Table 6.1 shows the Rf values for different metal cations when separated by TLC using water as a solvent. Table 6.1 cation Rf value Cd2+0.40 Co2+0.77 Cu2+0.32 Fe3+0.12 Hg2+0.23 Ni2+0.75 Suggest the identity of the cation that causes the spot at M in . Explain your answer. The student repeats the experiment using butan-1-ol as a solvent. The metal cations do not travel as far up the TLC plate in this experiment. Suggest why the metal cations do not move as far up the TLC plate with butan-1-ol as a solvent. The student sprays the TLC plate in with KSCN. The colour of some of the spots changes, as some of the metal cations undergo a ligand exchange reaction. Identify the ligands involved in the ligand exchange reaction. exchanges with In a third experiment, the pH of the mixture of metal ions is kept constant using a buffer solution. The student prepares the buffer solution by mixing 20.0 cm3 of 0.150 mol dm–3 KOHand 50.0 cm3 of 0.100 mol dm–3 C8H5O4K. C8H5O4K is a weak carboxylic acid that has pKa = 5.40. OH O–K+ O O C8H5O4K Complete the equation for the reaction of C8H5O4Kwith KOH. C8H5O4K + Calculate the pH of the buffer solution. Show all your working. pH =
7. Equilibria  →  7.2. Brønsted–Lowry theory of acids and bases
11. Group 17  →  11.3. Some reactions of the halide ions
11. Group 17  →  11.4. The reactions of chlorine
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