9701_w24_qp_43
A paper of Chemistry, 9701
Questions:
8
Year:
2024
Paper:
4
Variant:
3
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Disodium phosphate, (Na+)2(HPO4 2–), reacts with an acid to form monosodium phosphate, Na+(H2PO4 –). Identify the ions that are a conjugate acid–base pair in this reaction, using the formulae of the species involved. conjugate acid conjugate base Define buffer solution. Write two equations to show how a mixture of (Na+)2(HPO4 2–) and Na+(H2PO4 –) can act as a buffer solution. equation 1 equation 2 Identify one inorganic ion that acts as a buffer in blood. Compound E is the hydroxide of a Group 2 element. Compound E is a strong alkali. 2.63 g of E is dissolved in water to make 250 cm3 of solution F. Solution F has a pH of 13.09 at 298 K. Show that the concentration of hydroxide ions in solution F is 0.123 mol dm–3. Explain why the concentration of compound E in solution F is 0.0615 mol dm–3. Use the concentration given in to identify compound E. compound E Compound E is much more soluble than magnesium hydroxide. A saturated solution of magnesium hydroxide in water has a concentration of 1.40 × 10–4 mol dm–3 at 298 K. Calculate the solubility product, Ksp, of magnesium hydroxide. Include units. Ksp = units Explain why compound E is much more soluble than magnesium hydroxide.
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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Predict and explain the variation in enthalpy change of hydration for the ions F–, Cl –, Br – and I–. shows an incomplete energy cycle involving calcium fluoride, CaF2. process 1: enthalpy change of formation of Ca2+plus twice the enthalpy change of formation of F–, ΔHf Ca2++ 2ΔHf F–process 3: enthalpy change of formation of calcium fluoride, ΔHf CaF2process 2: enthalpy change of hydration of calcium ions plus twice the enthalpy change of hydration of fluoride ions, ΔHhyd Ca2++ 2ΔHhyd F–line A: Ca+ F2line B: Ca2++ 2F–line C: CaF2line D: process 4: enthalpy change of solution of calcium fluoride, ΔHsol CaF2Complete line D. Include state symbols. The value of the enthalpy change for process 1 can be calculated using the values of five other enthalpy changes which are not referred to in . process 1: Ca+ F2Ca2++ 2F–Identify these five other enthalpy changes, using either names or symbols. Define lattice energy, ΔHlatt. Complete the expression to give the mathematical relationship between ΔHlatt of calcium fluoride and the enthalpy changes for processes 1 and 3. ΔHlatt = Use data from Table 2.1 to calculate a value for the hydration energy, ΔHhyd, of fluoride ions, F–. Table 2.1 value / kJ mol–1 enthalpy change of solution of calcium fluoride, CaF2+13 overall enthalpy change of process 1 in +1395 enthalpy change of formation of calcium fluoride –1214 enthalpy change of hydration of Ca2+–1650 ΔHhyd F–= kJ mol–1 Define entropy. At 298 K, the Gibbs free energy change, ΔG, for the solution of compound T is +6.00 kJ mol–1. The enthalpy change of solution, ΔHsol, of compound T is +30.0 kJ mol–1 at 298 K. Calculate the value of the entropy change, ΔS, for the solution of compound T at 298 K. ΔS = J K–1 mol–1 Predict whether compound T becomes more or less soluble as the water is heated from 298 K to 360 K. Explain your answer.
5. Chemical energetics  →  5.1. Enthalpy change, \(\Delta H\)
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A and B react together to give product AB. A + B AB When the concentrations of A and B are both 0.0100 mol dm–3, the rate of formation of AB is 7.62 × 10–4 mol dm–3 s–1. When the concentrations of A and B are both 0.0200 mol dm–3, the rate of formation of AB is 3.05 × 10–3 mol dm–3 s–1. Complete the three possible rate equations that are consistent with these data. rate = rate = rate = Choose one of the rate equations you have written in , and calculate the value of the rate constant, k. Include the units of k. k = units Explain why it is not possible to calculate a value for the half-life, t1 , of this reaction using the value of the rate constant k calculated in and the equation k = 0.693 / t1 . Catalysts may be homogeneous or heterogeneous. Identify two metals that act as heterogeneous catalysts in the removal of NO2 from the exhaust gases of car engines. and Iron acts as a heterogeneous catalyst in the Haber process. Describe the mode of action of this iron catalyst. Fe2+ ions act as a homogeneous catalyst in the reaction between I–and S2O8 2–. Write equations for the two reactions that occur when Fe2+is added to a mixture of I–and S2O8 2–. equation 1 S2O8 2– + equation 2 Explain the difference between a homogeneous catalyst and a heterogeneous catalyst. Fe2+ ions can be oxidised to Fe3+ ions under alkaline conditions by suitable oxidising agents. Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states. The half-equation for the reduction of Fe3+ under alkaline conditions, and its E o value, are shown. Fe(OH)3 + e– Fe(OH)2 + OH– E o = – 0.56 V Four more half-equations for reactions under alkaline conditions, and their E o values, are shown. Al (OH)4 – + 3e– Al + 4OH– E o = –2.35 V Cl O– + H2O + 2e– Cl – + 2OH– E o = +0.89 V O2 + 2H2O + 4e– 4OH– E o = +0.40 V Zn(OH)4 2– + 2e– Zn + 4OH– E o = –1.22 V Select two oxidising agents that can oxidise Fe2+ ions to Fe3+ ions under alkaline conditions. Write an equation, and give the E cell value, for each of the two reactions that occur. oxidising agent 1: equation: E cell = V oxidising agent 2: equation: E cell = V o o o
8. Reaction kinetics  →  8.1. Rate of reaction
8. Reaction kinetics  →  8.3. Homogeneous and heterogeneous catalysts
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Transition metal atoms and transition metal ions form complexes by combining with ligands. Explain why transition elements form complex ions. Co2+ ions form complex ion G. Each G ion contains two Co2+ ions, both of which are octahedrally coordinated. Each G ion contains one O2 molecule, which donates one pair of electrons to each Co2+ ion, and one NH2 – ion, which donates one pair of electrons to each Co2+ ion. The remaining ligands are NH3 molecules. Deduce the formula of complex ion G. Include its overall charge. formula of G The d-orbitals of the Co2+ ions present in complex ion G are split. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level in each Co2+ ion. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level Co2+ ions form a different complex ion, M. Each M ion contains two Co2+ ions, both of which are octahedrally coordinated, but the ligands are different from the ligands in G. Explain why G and M have different colours. Cadmium forms complex ion X, [Cd(NH3)4]2+. When a solution containing CN– ions is added to an aqueous solution of X, a ligand exchange reaction takes place, forming complex ion Y. Y contains no NH3 ligands and no H2O ligands. Y is in a much higher concentration in the mixture than X. The oxidation state and coordination number of cadmium do not change in this reaction. Write an ionic equation for this reaction, using the formulae of the complex ions. Cadmium forms complex ion Z in the same oxidation state and with the same coordination number as in X. All the ligands in Z are Cl – ions. When NaCl is added to a solution of X, very little Z forms. Write the three cadmium complexes, X, Y and Z, in order of increasing stability constant, Kstab. smallest value of Kstab largest value of Kstab Ethanedioate ions, C2O4 2–, form complexes with transition element ions. The concentration of C2O4 2– ions can be found by reaction with acidified Cr2O7 2– ions. C2O4 2– ions are protonated and form HOOCCOOH molecules which are oxidised by Cr2O7 2–. The half-equations are shown. Cr2O7 2– + 14H+ + 6e– 2Cr3+ + 7H2O 2CO2 + 2H+ + 2e– HOOCCOOH Construct an equation for the reaction between acidified Cr2O7 2– and HOOCCOOH. A 25.0 cm3 sample of a solution of Na2C2O4 reacts with exactly 16.20 cm3 of an acidified solution of 0.0500 mol dm–3 K2Cr2O7. Calculate the concentration of the solution of Na2C2O4. [Na2C2O4] = mol dm–3
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
11. Group 17  →  11.3. Some reactions of the halide ions
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The shapes of four different complexes, P, Q, R and S, are shown in Table 5.1. The symbol J represents an atom or ion of a transition element. The symbol L is used to represent a monodentate ligand. Table 5.1 L P Q R S L L J L J L L L L J L L L J L L L L L Label one bond angle on each of complexes P, Q, R and S, and identify the size of the angle in degrees. Identify the shapes of complexes P, Q, R and S. P Q R S Two L ligands are exchanged with two different monodentate ligands X and Y in each of complexes P, Q, R and S. Identify all the complexes which form new complexes that show geometrical isomerism. Three L ligands are exchanged with three different monodentate ligands X, Y and Z in each of complexes P, Q and R. Identify all the complexes which form new complexes that show optical isomerism.
13. An introduction to AS Level organic chemistry  →  13.3. Shapes of organic molecules; σ and π bonds
13. An introduction to AS Level organic chemistry  →  13.4. Isomerism: structural isomerism and stereoisomerism
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Benzene, C6H6, reacts with chloroethane, C2H5Cl, in the presence of a suitable catalyst to form ethylbenzene, C6H5C2H5. In the presence of the catalyst, the ion C2H5 + is formed. This ion reacts with benzene. Complete the equation for the reaction of C2H5Cl with this catalyst to form C2H5 + as one product. C2H5Cl + C2H5 + + Ethylbenzene reacts with more C2H5Cl, forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene. Draw the structures of 1,2-diethylbenzene and 1,4-diethylbenzene. 1,2-diethylbenzene 1,4-diethylbenzene Explain why there is very little 1,3-diethylbenzene in the product mixture. 1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, C6H4(COOH)2. COOH COOH benzene-1,2-dioic acid State the reagent and conditions used for this reaction. Complete the overall equation for this reaction. An atom of oxygen from the oxidising agent is represented as [O]. All of the atoms in the two ethyl groups are fully oxidised in this reaction. + [O] C6H4(COOH)2 + + (1,2-diethylbenzene) Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioic acid. The proton (1H) NMR spectra of ethylbenzene, C6H5C2H5, in CDCl 3 and of benzene-1,2-dioic acid, C6H4(COOH)2, in CDCl 3 are shown. They have not been identified. δ / ppm δ / ppm Explain the use of CDCl 3, instead of CHCl 3, as the solvent when obtaining these spectra. Identify the substance shown by the spectrum in , and complete Table 6.1. substance Table 6.1 peak at δ = 1.2 peak at δ = 2.6 name of splitting pattern group responsible for peak explanation of splitting pattern Identify the substance shown by the spectrum in , and complete Table 6.2. substance Table 6.2 peak at δ = 7.8 peak at δ = 13.1 group responsible for peak When D2O is used as a solvent, the spectrum obtained is different from the spectrum in . Describe this difference and explain your answer. Benzene-1,2-dioic acid can be used to produce K. COOH COOH C C O O K O benzene-1,2-dioic acid heat Suggest the name of this type of reaction.
14. Hydrocarbons  →  14.1. Alkanes
14. Hydrocarbons  →  14.2. Alkenes
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A reaction scheme is shown in . The reagents needed for reaction 2 and reaction 3 are stated. Reaction 5 takes place when C2H5NH2 is mixed with compound V. No special conditions are required. compound U compound W compound V C2H5NH2 C2H5NHC2H5 C2H5NHCOCH3 CH3COOH reaction 1 reaction 6 reaction 2 reagent = LiAl H4 reaction 4 reaction 5 reaction 3 reagent = SOCl2 Identify compound U which contains only three elements. Describe the reagents and conditions for reaction 1. Identify compound V. Complete the equation for reaction 3. CH3COOH + SOCl2 Identify compound W. Describe the conditions for reaction 4. Suggest the reagent needed for reaction 6. Complete Table 7.1 by adding the reaction numbers, 1, 2, 3, 4, 5 and 6, to the right-hand column. Use the reaction numbers given in . Each of the numbers 1, 2, 3, 4, 5 and 6 should be used once only. Table 7.1 type of reaction reaction numberhydrolysis addition reduction substitution Compare the basicities of C2H5NHCOCH3, C2H5NHC2H5 and NH3. Explain your answer. most basic least basic
15. Halogen compounds  →  15.1. Halogenoalkanes
17. Carbonyl compounds  →  17.1. Aldehydes and ketones
18. Carboxylic acids and derivatives  →  18.1. Carboxylic acids
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An aqueous solution of phenol, C6H5OH, is acidic at 298 K. Explain why phenol is more acidic than water. Name the two products formed when phenol reacts with an excess of Br2. and Draw the structures of the two isomeric organic products, with Mr = 139, that are formed when phenol reacts with HNO3at room temperature. Write the equation for the reaction between phenol, C6H5OH, and sodium metal. Phenol can be produced from phenylamine in a two-step synthesis. phenylamine intermediate compound phenol Describe the reagents and conditions needed in each step. step one: reagents conditions step two: reagents conditions step one step two
21. Organic synthesis  →  21.1. Organic synthesis
14. Hydrocarbons  →  14.2. Alkenes
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