9701_w22_qp_42
A paper of Chemistry, 9701
Questions:
9
Year:
2022
Paper:
4
Variant:
2
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Calciumchloride, CaCl 2, is an ionic solid. The values of some energy changes are shown in Table1.1. Table 1.1 energy change value / kJ mol–1 lattice energy, , CaCl 2–2237 standard enthalpy change of atomisation of calcium +193 first ionisation energy of calcium +590 second ionisation energy of calcium +1150 standard enthalpy change of atomisation of chlorine +121 first electron affinity of chlorine –364 Define lattice energy. Use the data in Table1.1 to calculate the standard enthalpy change of formation, , of calciumchloride. It may be helpful to draw an energy cycle. Show all your working.  (CaCl 2) = kJ mol–1 Three possible values for the first electron affinity of bromine are shown in Table1.2. One of them is correct. Place a tick by the correct value. Explain your choice. Table 1.2 possible values place one tick (✓) in this column –342 kJ mol–1 –364 kJ mol–1 –386 kJ mol–1 explanation The enthalpy change of hydration of the chloride ion can be calculated using the lattice energy of calciumchloride and the data shown in Table1.3. Table 1.3 energy change value / kJ mol–1 standard enthalpy change of solution of CaCl 2–83 standard enthalpy change of hydration of Ca2+–1650 Define the following terms. enthalpy change of solution enthalpy change of hydration  Calculate the standard enthalpy change of hydration of the chloride ion, Cl –. It may be helpful to draw an energy cycle. Show all your working.  (Cl –) = kJ mol–1 Calciumfluoride, CaF2, can be synthesised directly from its elements. The value of (CaF2) is –1214 kJ mol–1. Predict the sign of the entropy change, ∆S o, for this synthesis. Explain your answer. The sign of the entropy change is . explanation  Use the value of (CaF2) given in and your answer to to predict how the feasibility for this synthesis will change with increasing temperature. 
5. Chemical energetics  →  5.1. Enthalpy change, \(\Delta H\)
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Nitrogenmonoxide, NO, reacts with ozone, O3. NO+ O3→ NO2+ O2This reaction is first order with respect to both NO and O3. At 298 K, the rate constant k = 11 500 mol–1 dm3 s–1. Complete the rate equation for this reaction. rate =  A reaction is carried out in which the initial concentrations of NO and O3 are both 1.20×10–6 mol dm–3. Calculate the initial rate of the reaction. State its units.  rate of reaction = units = The reaction described in is monitored over a period of time. Predict whether or not the graph of [NO] against time, under these conditions, shows that the reaction has a constant half-life. Explain your answer. prediction explanation Nitrousoxide, N2O, decomposes into its elements. 2N2O→ 2N2+ O2At a high temperature, a small amount of platinum wire is added to a large amount of nitrousoxide. The reaction follows zero order kinetics. The platinum wire behaves as a catalyst. Sketch a graph, on the axes below, of reaction rate against time for the catalysed decomposition of N2O under these conditions. reaction rate time  Sketch a graph, on the axes below, of [N2O] against time for this reaction. [N2O] time  Platinum behaves as a heterogeneous catalyst in this reaction. Describe the mode of action of a heterogeneous catalyst. Suggest a reason why this reaction has zero order kinetics when the amount of nitrousoxide is large and the amount of platinum is small. 
8. Reaction kinetics  →  8.1. Rate of reaction
12. Nitrogen and sulfur  →  12.1. Nitrogen and sulfur
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Data should be selected from Table3.1 in order to answer some parts of this question. Table 3.1 electrode reaction E o / V Mg2+ + 2e– Mg –2.38 Mn2+ + 2e– Mn –1.18 Mn3+ + e– Mn2+ +1.49 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 An electrochemical cell can be constructed from a Mg2+ / Mg half-cell and a MnO4 – / Mn2+ half‑cell. The standard cell potential of this cell can be calculated using the standard electrode potentials of the two half-cells. Define standard electrode potential. Include details of the standard conditions used. Complete the diagram below to show an electrochemical cell constructed from a Mg2+ / Mg half‑cell and a MnO4 – / Mn2+ half-cell. Label your diagram.  Use a positive (+) sign and a negative (–) sign to identify the polarity of each of the two electrodes in your diagram. Use an arrow and the symbol ‘e’ to show the direction of electron flow in the external circuit. Calculate the standard cell potential, , of this cell.  = V Construct an equation for the cell reaction. Predict how the cell reaction will change, if at all, when the solution in the Mg2+ / Mg half‑cell is diluted by the addition of a large volume of water. Explain your answer. A molten magnesium salt is electrolysed for 15.0minutes by a constant current. 4.75×1022 magnesium atoms are produced at the cathode. Calculate the value of the current used.  current = A 
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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The value of the solubility product, Ksp, of iron(hydroxide, Fe(OH)3, is given by the following expression. Ksp = [Fe3+][OH–]3 = 2.0 × 10–39 mol4 dm–12 Calculate the solubility of Fe(OH)3 in water.  solubility = mol dm–3 Calculate the solubility of Fe(OH)3 in 0.010 mol dm–3 bariumhydroxide, Ba(OH)2.  solubility = mol dm–3 Fe(OH)3 is less soluble in Ba(OH)2than it is in pure water. Name this effect. The numerical value of the Ka of HBrO is 2.00 × 10–9. X is a solution of HBrO which contains 4.00 × 10–3 mol of HBrO in 100 cm3 of solution. In this solution the following equilibrium is established in which there are two conjugate acid-base pairs. HBrO + H2O BrO– + H3O+ Define conjugate acid-base pair. Identify the two conjugate acid-base pairs shown in the equation above. pair one acid base pair two acid base  Calculate the pH of solution X. Show all your working.  pH = A solution containing 2.00 × 10–3 mol of NaOH is added to solution X. A buffer solution is formed. Calculate the pH of this buffer solution.  pH = 
7. Equilibria  →  7.2. Brønsted–Lowry theory of acids and bases
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Copper is a transition element. It forms a wide variety of compounds. Define transition element. An aqueous solution of copper(sulfate, CuSO4, contains [Cu(H2O)6]2+ complex ions. If an excess of concentrated hydrochloricacid is added to this solution a ligand exchange reaction occurs and [CuCl 4]2– complex ions are formed. Complete Table5.1 to state the geometry, the coordination number of copper, and one bond angle in each of the two complex ions. Table 5.1 complex ion geometry coordination number bond angle [Cu(H2O)6]2+ [CuCl 4]2–  In an isolated Cu2+ ion the d-orbitals are all degenerate. In a complex ion such as [Cu(H2O)6]2+ the d-orbitals are non-degenerate. Define degenerate and non-degenerate in this context. degenerate non-degenerate  Explain why the solutions of the two complex ions in Table5.1 are different colours. Cu2+ forms a complex ion containing water molecules and ethanedioate ions, C2O4 2–, as ligands. The formula of the complex is [Cu(C2O4)2(H2O)2]2–. The ethanedioate ion is a bidentate ligand. Explain what is meant by bidentate. There are three stereoisomers with the formula [Cu(C2O4)2(H2O)2]2–. Complete the three-dimensional diagrams to show these three stereoisomers. stereoisomer 1 Cu stereoisomer 2 Cu stereoisomer 3 Cu  Use your answer to to answer this question. Stereoisomers 1, 2 and 3 show two different types of isomerism. Name these two types of isomerism. For each type of isomerism identify the pair of stereoisomers that demonstrate this isomerism. type of isomerism pair of stereoisomers and and  A solution contains 3.70 g of Na2[Cu(C2O4)2(H2O)2] dissolved in 100 cm3 of solution. A 25.0 cm3 sample of this solution is warmed and then oxidised by 0.0100 mol dm–3 acidified potassiummanganate(. The equation for the redox reaction is shown. 5C2O4 2– + 2MnO4 – + 16H+ → 10CO2 + 2Mn2+ + 8H2O Calculate the minimum volume of 0.0100 mol dm–3 acidified potassiummanganate(needed to oxidise all of the ethanedioate ions, C2O4 2–, in the 25.0 cm3 sample. Show all your working. [Mr: Na2[Cu(C2O4)2(H2O)2], 321.5]  minimum volume = cm3 Copper(nitrate, Cu(NO3)2, and bariumnitrate, Ba(NO3)2, both decompose when heated. Copper(nitrate decomposes at a lower temperature than barium nitrate. Suggest a reason for this difference. Explain your answer. 
13. An introduction to AS Level organic chemistry  →  13.4. Isomerism: structural isomerism and stereoisomerism
11. Group 17  →  11.3. Some reactions of the halide ions
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An aqueous solution of cobalt(chloride is a pink colour due to the presence of [Co(H2O)6]2+ complex ions. Complete Table6.1 to describe what is observed when the named reagent is added to an aqueous solution of cobalt(chloride. Table 6.1 reagent colour of cobalt-containing product state of cobalt-containing product NaOHan excess of conc. HCl  Write an equation for the reaction between [Co(H2O)6]2+ ions and NaOH. Write an equation for the reaction between [Co(H2O)6]2+ ions and an excess of conc.HCl. Define stability constant. Write an expression for the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion. Kstab =  Give the units of the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion.  units = The numerical value of the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion is 7.7×104. In an aqueous solution the concentration of the [Co(NH3)6]2+ complex ion is 0.0740 mol dm–3 and the concentration of NH3 is 0.480 mol dm–3 at equilibrium. Calculate the concentration of [Co(H2O)6]2+ in this solution.  concentration = mol dm–3 
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
11. Group 17  →  11.3. Some reactions of the halide ions
7. Equilibria  →  7.1. Chemical equilibria: reversible reactions, dynamic equilibrium
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The structural and displayed formulae of three aromatic compounds, A, B and C, are shown in Fig.7.1. C6H5CH2CH2COOH A HO O OH CH3C6H4OH B C6H5CH2CHCl CO2H C HO O Cl Compare the relative acidities of A, B and C. > > most acidic least acidic Explain your answer. Methylbenzene, C6H5CH3, can be made from benzene by an electrophilic substitution reaction. Identify a compound that reacts with benzene to form methylbenzene. Identify the catalyst used. compound catalyst  The first step in the reaction is the generation of the CH3 + electrophile. Write an equation for the reaction that generates this electrophile. Describe the mechanism for the reaction between benzene and the CH3 + electrophile. Include all relevant curly arrows and charges. CH3 +  Identify a suitable reagent to oxidise methylbenzene to benzoic acid, C6H5COOH. Write an equation for this reaction using [O] to represent one atom of oxygen from the oxidising agent. reagent equation  Methylbenzene and benzoicacid are both nitrated with a mixture of concentrated nitricacid and sulfuricacid to give mononitrated products. The structural formulae of these products are CH3C6H4NO2 and HOOCC6H4NO2 respectively. Draw the structures of these two products.  A reaction scheme is shown in Fig.7.2. CH3C6H4NO2 CH3C6H4NH2 reaction 1 reaction 2 CH3C6H4OH CH3C6H4N2 +Cl – dye molecule Y reaction 3 reaction 4 Describe the reagents and conditions to produce CH3C6H4N2 +Cl – from CH3C6H4NH2 in reaction2. reagents conditions  Describe how CH3C6H4OH can be produced from CH3C6H4N2 +Cl – in reaction3. Draw the structure of the dye moleculeY formed when CH3C6H4N2 +Cl – and CH3C6H4OH react together in reaction4. Describe the conditions for this reaction. structure conditions  
14. Hydrocarbons  →  14.2. Alkenes
14. Hydrocarbons  →  14.1. Alkanes
18. Carboxylic acids and derivatives  →  18.1. Carboxylic acids
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When answering this question it should be assumed that together all the hydrogen atoms in a benzene ring result in a single unsplit peak at δ = 7.2 in a proton (1H) NMR spectrum. The structures of five isomeric ketones, P, Q, R, S and T are given. P C6H5COCH(CH3)2 S C6H5CH2CH2COCH3 Q C6H5COCH2CH2CH3 T C6H5CH(CH3)COCH3 R C6H5CH2COCH2CH3 Identify all the chiral carbon atoms on the structures above. Label each chiral carbon atom with an asterisk (*). The proton (1H) NMR spectrum of one of the five isomers, P, Q, R, S or T, is shown in Fig.8.1. / ppm Identify which of the compounds P, Q, R, S or T gives this spectrum. Draw the displayed formula of the compound you have identified. Identify the protons responsible for the peaks at δ = 3.7, δ = 2.5 and δ = 1.0 on the structure you have drawn.  Name the splitting pattern of the peak at δ = 3.7. Explain why it has this splitting pattern. Choose from the letters P, Q, R, S and T to identify: the two compounds that each have a doublet peak in the proton (1H) NMR spectrum the compound with only three peaks in its proton (1H) NMR spectrum. Suggest a suitable solvent that should be used for obtaining the spectrum shown in Fig.8.1. The proton (1H) NMR spectrum of compoundT is compared in the presence of D2O and in the absence of D2O. Describe any difference between the two spectra. Explain your answer. Complete Table8.1 below to give the number of peaks in the carbon-13 NMR spectrum of each compound. Table 8.1 compound number of peaks compound number of peaks  
22. Analytical techniques  →  22.2. Mass spectrometry
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Tyrosine and lysine, shown in Fig.9.1, are naturally occurring amino acids found in proteins. HO OH O NH2 tyrosine OH O NH2 H2N lysine The isoelectric point of lysine is 9.47. Draw the structure of lysine at the stated pH in the boxes below. lysine at pH 7.00 lysine at pH 9.47 lysine at pH 12.00  When ethanoicacid is treated with PCl 5 product D is formed. When D is added to tyrosine two different isomeric products, E and F, are formed. E has an ester linkage, F does not. Draw the structures of D, E and F in the boxes below. D E F  Complete Table9.1 by drawing the structure of the organic product formed when tyrosine reacts with each named reagent. Ignore the directing effect of the –CH2CHNH2COOH substituent. Table 9.1 reagent structure an excess of Br2an excess of NaOHan excess of HNO3 A mixture of tyrosine and lysine can be separated by thin-layer chromatography. Under certain conditions the Rf value of lysine is 0.14 and the Rf value of tyrosine is 0.45. Explain what is meant by Rf value. Suggest an explanation for the difference in Rf values. 
17. Carbonyl compounds  →  17.1. Aldehydes and ketones
13. An introduction to AS Level organic chemistry  →  13.1. Formulas, functional groups and the naming of organic compounds
13. An introduction to AS Level organic chemistry  →  13.4. Isomerism: structural isomerism and stereoisomerism
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