9701_s24_qp_42
A paper of Chemistry, 9701
Questions:
9
Year:
2024
Paper:
4
Variant:
2
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1
Describe the trend in the solubility of the sulfates of magnesium, calcium and strontium. Explain your answer. > > most soluble least soluble Define lattice energy, ΔH latt. State and explain the main factors that affect the magnitude of lattice energies. Table 1.1 shows some energy changes. Table 1.1 energy change value / kJ mol–1 standard enthalpy change of atomisation of potassium +89 first ionisation energy of potassium +419 second ionisation energy of potassium +3070 standard enthalpy change of atomisation of sulfur +279 S–S bond energy +265 first ionisation energy of sulfur +1000 second ionisation energy of sulfur +2260 first electron affinity of sulfur –200 second electron affinity of sulfur +640 standard enthalpy change of formation of potassium sulfide K S381 Born–Haber cycles can be used to determine the lattice energies of ionic compounds. Complete the Born–Haber cycle in for potassium sulfide, K2S. Include state symbols for all of the species. 2K++ S+ 2e– 2K+ SK2SCalculate the lattice energy, ΔH o latt, of K2Susing relevant data from Table 1.1. Show your working. ΔH o latt of K2S= kJ mol–1 ,  ,
9. The Periodic Table: chemical periodicity  →  9.2. Periodicity of chemical properties of the elements in Period 3
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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2
Lithium nitrate, LiNO3, decomposes on heating in a similar way to Group 2 nitrates to give the metal oxide, a brown gas and oxygen. Write an equation for the decomposition of LiNO3. The other Group 1 nitrates, MNO3, decompose on heating to form the metal nitrite, MNO2, and oxygen. The thermal stability of these nitrates increases down the group. Suggest why the thermal stability of MNO3 increases down the group. Acidified manganate(ions, MnO4 –, can be used to analyse solutions containing nitrite ions, NO2 –, by titration. X is a solution of NaNO2. 250.0 cm3 of X is added to 50.0 cm3 of 0.125 mol dm–3 acidified MnO4 –. The MnO4 –ions are in excess; all the NO2 – ions are oxidised in the reaction. The unreacted MnO4 –required 22.50 cm3 of 0.0400 mol dm–3 Fe2+to reach the end‑point. The relevant half‑equations are shown. NO2 – + H2O NO3 – + 2H+ + 2e– MnO4 – + 8H+ + 5e– Mn2+ + 4H2O Fe2+ Fe3+ + e– Calculate the concentration, in mol dm–3, of NaNO2 in X. Table 2.1 shows electrode potentials for some electrode reactions involving manganese compounds. Table 2.1 electrode reaction E o / V Mn2+ + 2e– Mn –1.18 MnO2 + 4H+ + 2e– Mn2+ + 2H2O +1.23 MnO4 – + e– MnO4 2– +0.56 MnO4 – + 4H+ + 3e– MnO2 + 2H2O +1.67 MnO4 – + 8H+ + 5e– Mn2+ + 4H2O +1.52 MnO4 – + 2H2O + 3e– MnO2 + 4OH– +0.59 MnO4 2– + 2H2O + 2e– MnO2 + 4OH– +0.60 MnO4 2– + 4H+ + 2e– MnO2 + 2H2O +1.70 Aqueous manganate(ions, MnO4 2–, are unstable in acidic conditions and undergo a disproportionation reaction. The E o cell for this reaction is +1.14 V. Construct an overall ionic equation for this disproportionation reaction. Suggest and explain how the Ecell value of the disproportionation reaction changes with an increase in pH. ,  ,
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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Carbon disulfide, CS2, is flammable and reacts readily with oxygen, as shown in reaction 1. reaction 1 CS2+ 3O2CO2+ 2SO2Table 3.1 shows the standard enthalpy of formation, ΔH o f , and the standard entropy, S o, for some substances. Table 3.1 CS2O2CO2SO2ΔH o f / kJ mol–1 116.7 0.0 –393.5 –296.8 S o/ J K–1 mol–1 237.8 205.2 213.8 248.2 Calculate the standard Gibbs free energy change, ΔG o, in kJ mol–1, for reaction 1 at 25 °C. ΔG o = kJ mol–1 Carbon disulfide reacts with chlorine to form tetrachloromethane, as shown in reaction 2. reaction 2 CS2 + 3Cl2 CCl 4 + S2Cl2 ΔH o = –261.6 kJ mol–1 ΔS o = –365.5 J K–1 mol–1 Calculate the maximum temperature, in K, for reaction 2 to be feasible. temperature = K
5. Chemical energetics  →  5.1. Enthalpy change, \(\Delta H\)
8. Reaction kinetics  →  8.2. Effect of temperature on reaction rates and the concept of activation energy
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4
Explain why transition elements have variable oxidation states. Sketch the shape of a 3dz2 orbital in . z y x Samples of [Cu(H2O)6]2+are reacted separately with an excess of solution A and with an excess of solution B. The reaction of [Cu(H2O)6]2+with solution A is a precipitation reaction. The reaction of [Cu(H2O)6]2+with solution B is a ligand substitution reaction. Suggest a possible identity for solution A and for solution B. Give relevant observations and the formula of the copper‑containing product for each reaction. solution A observations formula of the copper‑containing product solution B observations formula of the copper‑containing product Solutions containing the [Ag(NH3)2]+ complex are colourless. Explain why this complex is colourless. Two bidentate ligands are shown in . en dpys S N N NH2 H2N Explain what is meant by a bidentate ligand. ,   , Ruthenium(ions, Ru3+, form an octahedral complex, [Ru2Cl2]+, with the ligands dpys and chloride ions. This complex shows the same kind of stereoisomerism as [Ru(NH3)4Cl2]+ but also shows a different type of stereoisomerism. Complete the three‑dimensional diagrams in to show the three different stereoisomers of [Ru2Cl2]+. The dpys ligand can be represented using N N. isomer 1 Ru isomer 2 Ru isomer 3 Ru State the different types of stereoisomerism shown by [Ru2Cl2]+. Deduce which stereoisomers in are non-polar. Explain your answer. ,   ,
11. Group 17  →  11.3. Some reactions of the halide ions
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
9. The Periodic Table: chemical periodicity  →  9.2. Periodicity of chemical properties of the elements in Period 3
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5
Nitrosyl chloride, NOCl, can be formed by the reaction between nitrogen monoxide and chlorine, as shown. 2NO + Cl 2 2NOCl The initial rate of this reaction is investigated, starting with different concentrations of NO and Cl 2. The results obtained are shown in Table 5.1. Table 5.1 experiment [NO] / mol dm–3 [Cl 2] / mol dm–3 initial rate / mol dm–3 min–1 0.0250 0.0150 3.68 × 10–2 0.0750 0.0150 3.32 × 10–1 0.0500 0.0600 5.89 × 10–1 Use the data in Table 5.1 to deduce the rate equation for this reaction. Explain your reasoning. Use your rate equation from and the data from experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. k = units NO2Cl is another compound containing nitrogen, oxygen and chlorine. In sunlight, NO2Cl can undergo homolytic fission to release chlorine radicals which can catalyse the conversion of ozone, O3, into oxygen. Complete the mechanism for this process. initiation (homolytic fission) NO2Cl + propagation step 1 + O3 + propagation step 2 + + Ozone reacts with nitrogen dioxide, as shown. O3 + 2NO2 N2O5 + O2 The rate of reaction is first order with respect to O3 and first order with respect to NO2. Suggest equations for a two‑step mechanism for this reaction. step 1 step 2 ,  ,
11. Group 17  →  11.4. The reactions of chlorine
12. Nitrogen and sulfur  →  12.1. Nitrogen and sulfur
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Aqueous solutions of methanoic acid, HCOOH, and propanoic acid, CH3CH2COOH, are mixed together. An equilibrium is set up between two conjugate acid–base pairs. Define conjugate acid–base pair. The pKa of HCOOH is 3.75 and of CH3CH2COOH is 4.87. Complete the equation for the Brønsted–Lowry equilibrium between the stronger of these two acids and water. + H2O + Write an expression for the acid dissociation constant, Ka, for butanoic acid, CH3CH2CH2COOH. Ka = The pKa of CH3CH2CH2COOH is 4.82. A solution of CH3CH2CH2COOHhas a pH of 3.25. Calculate the concentration, in mol dm–3, of CH3CH2CH2COOH in this solution. concentration of CH3CH2CH2COOH = mol dm–3 Define buffer solution. A buffer solution containing a mixture of CH3COOH and CH3COONa is prepared as follows. A solution of 600 cm3 of CH3COOH is mixed with 400 cm3 of 0.125 mol dm–3 CH3COONa. The buffer solution has pH 5.70. The Ka of CH3COOH is 1.78 × 10–5 mol dm–3. Calculate the initial concentration, in mol dm–3, of CH3COOH used. concentration of CH3COOH = mol dm–3 A fuel cell is an electrochemical cell that can be used to generate electrical energy by using oxygen to oxidise a fuel. Methanoic acid, HCOOH, is being investigated as a fuel in fuel cells. When the cell operates, HCOOH is oxidised to carbon dioxide. The half‑equation for the reaction at the cathode is: O2 + 4H+ + 4e– 2H2O. In this fuel cell, the overall cell reaction is the same as that for the complete combustion of HCOOH. Deduce the half‑equation for the reaction at the anode. Calculate the volume, in cm3, of oxygen used when a current of 3.75 A is delivered by the cell for 40.0 minutes. Assume the cell operates at room conditions. volume of oxygen = cm3 , ,
7. Equilibria  →  7.2. Brønsted–Lowry theory of acids and bases
18. Carboxylic acids and derivatives  →  18.1. Carboxylic acids
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Methyl red can be synthesised as shown in . COOH step 1 P Q NO2 step 2 step 3 methyl red R S COOH N N N Give the systematic name of P. P can be synthesised as shown in . CH3 P NO2 COOH NO2 Suggest reagents and conditions for this reaction. A student attempts to synthesise P by an alternative route, as shown in . Compound T is the major product in this reaction rather than P. COOH T conc. HNO3 conc. H2SO4 COOH NO2 Explain why T is the major product in this reaction. S reacts in a similar way to phenol in step 3. Draw the structures of Q, R and S in the boxes in . Suggest reagents and conditions for steps 1 and 2 in . step 1 step 2 [T t l 9] , ,
21. Organic synthesis  →  21.1. Organic synthesis
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State the relative basicities of phenylamine, C6H5NH2, benzylamine, C6H5CH2NH2, and ammonia, NH3, in aqueous solution. Explain your answer. > > most basic least basic An excess of Br2is added to separate samples of C6H5NH2 and benzene, C6H6. C6H5NH2 reacts readily with Br2to form organic product M. State the expected observations for this reaction. Draw the structure of M. observations structure of M C6H6 does not react with Br2. Suggest why Br2reacts with C6H5NH2 but not with C6H6. Explain why benzamide, C6H5CONH2, is a much weaker base than ammonia, NH3. C6H5CONH2 is formed by reacting benzoyl chloride, C6H5COCl, with NH3. Complete the mechanism in for the reaction of C6H5COCl with NH3. Include all relevant lone pairs of electrons, curly arrows, charges and dipoles. Draw the structure of the organic intermediate. O NH3 organic intermediate Cl C C O NH2 HCl + , , Phenylalanine, C6H5CH2CH(NH2)COOH, is an amino acid with an isoelectric point of 5.5. State what is meant by isoelectric point. Draw the structure of C6H5CH2CH(NH2)COOH at pH 10. C6H5CH2CH(NH2)COOH and alanine, CH3CH(NH2)COOH, react to form a dipeptide containing both amino acid residues. Draw the structure of this dipeptide. The peptide functional group formed should be displayed. ,  ,
12. Nitrogen and sulfur  →  12.1. Nitrogen and sulfur
21. Organic synthesis  →  21.1. Organic synthesis
15. Halogen compounds  →  15.1. Halogenoalkanes
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Explain why trichloroethanoic acid, CCl 3COOH, is more acidic than ethanoic acid, CH3COOH. Acyl chlorides are formed by reacting carboxylic acids with thionyl chloride, SOCl 2. Ethanedioyl chloride, (COCl )2, can be prepared by reacting ethanedioic acid, (COOH)2, with an excess of SOCl 2. Write an equation for this reaction. Samples of (COCl )2 are reacted separately with an excess of warm acidified KMnO4and with H2NCH2CH2NH2. The carbon‑containing product from the reaction with H2NCH2CH2NH2 has the molecular formula C4H6N2O2. Complete the boxes in to suggest the structure of the carbon‑containing product in each reaction. with warm acidified KMnO4with H2NCH2CH2NH2 A polyester can be synthesised from the reaction of (COCl )2 with ethane‑1,2‑diol, HOCH2CH2OH. Draw two repeat units of the polymer formed. Any functional groups should be displayed. Compound H, C6H10O3, reacts with alkaline I2to form yellow precipitate J but does not react with Na2CO3. The proton (1H) NMR spectrum of H in CDCl 3 is shown in . 4.0 3.0 2.0 chemical shift δ / ppm 1.0 Table 9.1 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 , , Identify yellow precipitate J. Complete Table 9.2 for the proton (1H) NMR spectrum of H, C6H10O3. Table 9.2 chemical shift δ / ppm splitting pattern number of 1H atoms responsible for the peak number of protons on adjacent carbon atoms 1.15 2.25 3.60 3.95 Suggest a structure for H, C6H10O3. , ,
17. Carbonyl compounds  →  17.1. Aldehydes and ketones
14. Hydrocarbons  →  14.2. Alkenes
18. Carboxylic acids and derivatives  →  18.1. Carboxylic acids
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