9701_w23_qp_42
A paper of Chemistry, 9701
Questions:
9
Year:
2023
Paper:
4
Variant:
2
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Propanone, CH3COCH3 , reacts with iodine, I2 , in the presence of an acid catalyst. CH3COCH3 + I2 CH3COCH2I + H+ + I– The rate equation for this reaction is shown. rate = k[CH3COCH3][H+] Complete Table 1.1 to describe the order of the reaction. Table 1.1 order of the reaction with respect to [CH3COCH3 ] order of the reaction with respect to [I2] order of the reaction with respect to [H+] overall order of the reaction An experiment is performed using a large excess of CH3COCH3 and a large excess of H+. The initial concentration of I2 is 1.00 × 10–5 mol dm–3. The initial rate of decrease in the I2 concentration is 2.27 × 10–7 mol dm–3 s–1. Use the axes to draw a graph of [I2 ] against time for the first 10 seconds of the reaction. 1.00 × 10–5 0.90 × 10–5 0.80 × 10–5 0.70 × 10–5 0.60 × 10–5 0.50 × 10–5 0.40 × 10–5 0.30 × 10–5 0.20 × 10–5 0.10 × 10–5 [I2] / mol dm–3 time / s State whether it is possible to calculate the numerical value of the rate constant, k, for this reaction from your graph. Explain your answer. The experiment is repeated at a different temperature. The initial concentrations of H+ ions, I2 and CH3COCH3 are all 0.200 mol dm–3. The value of k at this temperature is 2.31 × 10–5 mol–1 dm3 s–1. Calculate the initial rate of this reaction. rate = mol dm–3 s–1 The experiment is repeated using an excess of H+. The new rate equation is shown. rate = k1[CH3COCH3] The value of k1 is 1.1 × 10–3 s–1. Calculate the value of the half-life, t1 2 . t1 2 = s Use your answer to to draw a graph of [CH3COCH3] against time for this reaction. The initial value of [CH3COCH3 ] on your graph should be 0.200 mol dm–3. The final value of [CH3COCH3 ] on your graph should be 0.0250 mol dm–3. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200X [CH3COCH3] / mol dm–3 time / s A four-step mechanism is suggested for the overall reaction. CH3COCH3 + I2 CH3COCH2I + H+ + I– rate = k [CH3COCH3][H+ ] Part of this mechanism is shown. step 1: CH3COCH3 + H+ CH3C+(OH)CH3 step 2: CH3C+(OH)CH3 CH3C(OH)=CH2 + H+ step 3: step 4: CH3C+(OH)CH2I CH3COCH2I + H+ Write an equation for step 3. Suggest the slowest step of the mechanism. Explain your answer. Identify one conjugate acid-conjugate base pair in the mechanism. conjugate acid conjugate base
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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Benzoic acid, C6H5COOH, is a weak acid. The Ka of benzoic acid is 6.31 × 10–5 mol dm–3 at 298 K. A 1.00 dm3 buffer solution is made at 298 K containing 1.00 g of C6H5COOH and a slightly greater mass of sodium benzoate, C6H5COO–Na+. This buffer solution has a pH of 4.15. Define buffer solution. Write equations to show how this solution acts as a buffer solution when the named substances are added to it: dilute aqueous sodium hydroxide dilute aqueous nitric acid. Calculate the H+ concentration and the C6H5COOH concentration in the buffer solution described. Use the expression for the Ka of C6H5COOH to calculate the concentration of C6H5COO–Na+ in the buffer solution. Show your working and give each answer to a minimum of three significant figures. [H+] = mol dm–3 [C6H5COOH] = mol dm–3 [C6H5COO–Na+] = mol dm–3 A 10.0 cm3 sample of the buffer solution is mixed with 10.0 cm3 of 1.00 mol dm–3 KOH. Both solutions are at 298 K. A reaction is allowed to occur without stirring. Two observations are recorded: ● the temperature, after the reaction is complete, is fractionally above 298 K ● the pH, after the reaction, is greater than 13. Explain these two observations. Magnesium benzoate, Mg(C6H5COO)2, has a solubility in water of less than 1.00 g dm–3 at 298 K. Ksp = [Mg2+][C6H5COO–]2 = 1.76 × 10–7 at 298 K Calculate the solubility of Mg(C6H5COO)2 in water at 298 K. Give your answer in g dm–3. Show your working. [Mr: Mg(C6H5COO)2 , 266.3] solubility = g dm–3 An excess of Mg(C6H5COO)2 is added to a sample of 0.50 mol dm–3 MgSO4 at 298 K. State whether the equilibrium concentration of Mg(C6H5COO)2 is higher than, the same as, or lower than your answer to . Explain your answer. The concentration is the concentration in . explanation
7. Equilibria  →  7.2. Brønsted–Lowry theory of acids and bases
7. Equilibria  →  7.1. Chemical equilibria: reversible reactions, dynamic equilibrium
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Some electrode potentials are shown in Table 3.1. Table 3.1 electrode reaction E o– / V V2+ + 2e– V –1.20 V3+ + e– V2+ –0.26 VO2+ + 2H+ + e– V3+ + H2O +0.34 VO2 + + 2H+ + e– VO2+ + H2O +1.00 Fe2+ + 2e– Fe –0.44 Fe3+ + 3e– Fe –0.04 Fe3+ + e– Fe2+ +0.77 2H+ + 2e– H2 0.00 Cl O– + H2O + 2e– Cl – + 2OH– +0.89 Complete the diagram to show a standard hydrogen electrode. Label your diagram. Identify all substances. You do not need to state standard conditions. An electrochemical cell is set up using an Fe3+ / Fe2+ electrode and a standard hydrogen electrode. Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit. positive electrode Electrons flow from the electrode to the electrode. The vanadium-containing species in the electrode reactions given in Table 3.1 are V, V2+, V3+, VO2+ and VO2 +. Identify one vanadium-containing species that does not react with Fe2+ ions under standard conditions. Use data from Table 3.1 to explain your answer. Identify all the vanadium-containing species that will react with Fe2+ ions under standard conditions. Write an equation for one of the possible reactions identified in . Another electrochemical cell is set up using an Fe3+ / Fe2+ electrode and an alkaline Cl O– / Cl – electrode. The concentration of Fe3+ is 1000 times greater than the concentration of Fe2+ in the Fe3+ / Fe2+ electrode. All other conditions are standard. Use the Nernst equation to calculate the E value of the Fe3+ / Fe2+ electrode. Show your working. E = V Write an equation for the reaction that occurs in the cell, under these conditions. Another electrochemical cell is set up using an Fe2+ / Fe electrode and an alkaline Cl O– / Cl – electrode under standard conditions. Calculate the value of ΔG o– for the cell. ΔGo – = kJ mol–1 A solution of iron(sulfate, FeSO4is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode. A current of 0.640 A is passed for 17.0 minutes. The mass of the cathode increases by 0.185 g. Use these results to calculate an experimental value for the Avogadro constant, L. Show your working. L = mol–1 Iron(chloride, FeCl2, is oxidised by chlorine to form iron(chloride, FeCl 3, under standard conditions. 2FeCl 2+ Cl 22FeCl 3ΔH o– = –128 kJ mol–1 Table 3.2 species S o– / J K–1 mol–1 Cl 2223 FeCl 2120 FeCl 3142 Use Table 3.2 and other data to calculate the Gibbs free energy change, ΔG o–, for this reaction. Show your working. ΔG o– = kJ mol–1 Predict whether this reaction becomes more or less feasible at a higher temperature. Explain your answer. The reaction becomes feasible. explanation
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
1. Atomic structure  →  1.3. Electrons, energy levels and atomic orbitals
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The structure of the polydentate ligand, EDTA4–, is shown in . NCH2CH2N CCH2 O O O CCH2 O CH2C O O O CH2C O The stability constants, at 298 K, of five octahedral complexes are given in Table 4.1. Table 4.1 complex Kstab [Cu(EDTA)]2– 6.31 × 1019 [Cr(EDTA)]2– 1.00 × 1013 [Cr(EDTA)]– 1.00 × 1024 [Fe(EDTA)]2– 2.00 × 1014 [Fe(EDTA)] – 1.26 × 1025 Define stability constant. Calculate the oxidation states of Cu in [Cu(EDTA)]2– and Cr in [Cr(EDTA)]–. Cu Cr Deduce the number of lone pairs donated by each EDTA4– ligand in a single [Fe(EDTA)]2– complex ion. Identify the most stable complex in Table 4.1. Explain your choice. In a solution at equilibrium at 298 K, [[Cu(H2O)6]2+] = 3.00 × 10–10 mol dm–3 and [EDTA4–] = 5.00 × 10–12 mol dm–3. Use the expression for Kstab to calculate the concentration of [Cu(EDTA)]2– in this solution. Show your working. [[Cu(EDTA)]2–] = mol dm–3 A solution of [Cu(EDTA)]2– ions is pale blue while a solution of [Cu(NH3)4(H2O)2]2+ ions is deep blue. Explain this difference in colour.
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
11. Group 17  →  11.3. Some reactions of the halide ions
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Some of the ionic compounds of Group 2 elements undergo thermal decomposition. Thermal decomposition of solid anhydrous magnesium ethanedioate, MgC2O4 , occurs above 650 °C. The products are magnesium oxide and a mixture of two different gases, one of which gives a white precipitate with saturated calcium hydroxide solution. Complete the equation for the thermal decomposition of MgC2O4. MgC2O4 Suggest which of MgC2O4 or CaC2O4 undergoes thermal decomposition at a lower temperature. Explain your answer. The ethanedioate ion is oxidised by acidified KMnO4. 5C2O4 2– + 2MnO4 – + 16H+ 10CO2 + 2Mn2+ + 8H2O An experiment is performed to find the solubility of MgC2O4 in water. A 40.0 cm3 sample of saturated aqueous MgC2O4 requires 27.05 cm3 of 0.00200 mol dm–3 acidified KMnO4 to oxidise all the C2O4 2– ions. Calculate the solubility, in mol dm–3, of MgC2O4 in water. Show your working. solubility = mol dm–3
10. Group 2  →  10.1. Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
11. Group 17  →  11.4. The reactions of chlorine
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Phosphine, :PH3, and carbon monoxide, :CO, are monodentate ligands found in some transition element complexes. Define monodentate ligand. Define transition element complex. Explain why transition elements form complexes. The formulae of six complexes are given in Table 6.1. The abbreviation en is used for 1,2-diaminoethane. The abbreviation dien is used for the tridentate ligand H2NCH2CH2NHCH2CH2NH2. The dien ligand forms three bonds to the gold ion in [Au(H2O)2Cl ]2+ and AuCl3. These three bonds all lie in the same plane. The CO ligand coordinates through the carbon atom in [Rh(CO)2Cl 2]+. Table 6.1 formula isomerism shown geometry [Rh2Cl 2]+ yes [Rh(CO)2Cl 2]+ yes [Au(H2O)2Cl ]2+ AuCl 3 no octahedral Ni(PH3)2Cl 2 no [Ni(H2O)2(NH3)4]2+ yes Complete Table 6.1 to state the geometry of the first three complexes. Each complex is either square planar, tetrahedral or octahedral. Use complexes [Au(H2O)2Cl ]2+ and AuCl 3 to write an equation showing ligand exchange. Draw the three-dimensional structure of AuCl 3 in the box. The dien ligand can be drawn as . N N N Au Draw the three-dimensional structure of Ni(PH3)2Cl 2 in the box. Ni One of the complexes, [Rh2Cl 2]+ or [Rh(CO)2Cl 2]+, can exist in three isomeric forms. Identify this complex and the types of isomerism shown. Draw the three-dimensional structures of the two isomers of [Ni(H2O)2(NH3)4]2+ in the boxes and identify the type of isomerism shown. Ni Ni type of isomerism shown
3. Chemical bonding  →  3.4. Covalent bonding and coordinate (dative covalent) bonding
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
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Benzene can be used to make benzoic acid in the two-step process shown in . benzene methylbenzene CH3 step 1 step 2 benzoic acid COOH Give the reagents and conditions for step 1 and step 2. step 1 step 2 Methylbenzene and benzoic acid each have five different peaks in the carbon (13C) NMR spectrum. Table 7.1 hybridisation of the carbon atom environment of carbon atom example chemical shift range / ppm sp3 alkyl CH3–, –CH2–, –CHCC=Cpeakin the chemical shift range of and ● has peakin the chemical shift range of . The carbon (13C) NMR spectrum of benzoic acid: ● has peakin the chemical shift range of and ● has peakin the chemical shift range of . When treated with Cl 2 under suitable conditions, methylbenzene forms compound J. When treated with Cl 2 under different conditions with different reagents, methylbenzene forms compound K. Suggest and draw structures of compounds J and K in the boxes. The molecular formula of each compound is given. methylbenzene J C7H7Cl K C7H7Cl CH3 State the reagents and conditions required to form each product. to form compound J to form compound K When treated with a chlorine-containing reagent under suitable conditions, benzoic acid forms compound L. When treated with a different chlorine-containing reagent under different conditions, benzoic acid forms compound M. Suggest and draw structures of compounds L and M in the boxes. The molecular formula of each product is given. benzoic acid L C7H5ClO2 M C7H5ClO COOH State the reagents and conditions to form compound M from benzoic acid.
13. An introduction to AS Level organic chemistry  →  13.1. Formulas, functional groups and the naming of organic compounds
21. Organic synthesis  →  21.1. Organic synthesis
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Lactic acid, CH3CH(OH)COOH, is the only monomer needed to form the polymer polylactic acid, PLA. Draw a short length of the PLA polymer chain, including a minimum of two monomer residues. The methyl groups may be written as –CH3 but all other bonds should be shown fully displayed. Label one repeat unit of polylactic acid on your diagram. Give the name of the type of polymerisation involved in the formation of PLA and the name of the functional group that forms between the monomers. type of polymerisation functional group Predict whether PLA is readily biodegradable. Explain your answer. The proton (1H) NMR spectrum of CH3CH(OH)COOH in CDCl 3 is shown in . The proton NMR chemical shift ranges are shown in Table 8.1. Lactic acid δ / ppm 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 Use and Table 8.1 to complete Table 8.2. Table 8.2 proton environment chemical shift (δ) name of splitting pattern –COOH CH –OH –CH3 Name the substance responsible for the peak at δ = 0.0. Explain why CDCl 3 is a better solvent than CHCl 3 for use in proton NMR. An impure sample of CH3CH(OH)COOH contains pentan-3-one as the only contaminant. The mixture is analysed using gas/liquid chromatography. The pentan-3-one is found to have a longer retention time than the lactic acid. Explain what is meant by retention time. Suggest suitable substances, or types of substances, that could be used as the mobile and stationary phases. mobile phase stationary phase Describe how the percentage composition of the mixture can be determined from the gas / liquid chromatogram.
20. Polymerisation  →  20.1. Addition polymerisation
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State the reactants and conditions for two different types of reactions that both produce diethylamine, CH3CH2NHCH2CH3. reaction one reaction two Describe the relative basicities of diethylamine, phenylamine and ammonia in aqueous solution. Explain your answer in terms of structure. least basic most basic explanation Phenylamine reacts with HNO2at 4° C to form compound P. Compound P reacts with phenol under alkaline conditions at 4° C. The product of this reaction is acidified, forming azo compound Q. Draw the structure of compound Q. Circle the azo group on your structure. State one use of an azo compound such as Q. compound Q: An azo compound can be used . CH3CH2NHCH2CH3 reacts with ethanoyl chloride, CH3COCl, to give the amide N,N-diethylethanamide, CH3CON(C2H5)2. An incomplete description of the mechanism of this reaction is shown in . box one reactants N H C2H5 C2H5 Cl C O H3C N H C2H5 C2H5 Cl C O H3C box two intermediate box three products and O C N H3C C2H5 C2H5 Complete the mechanism in . You should include: ● all relevant dipoles (δ+ and δ–) and full electric charges (+ and –) on the species in box one and in box two ● all relevant lone pairs on the species in box one and in box two ● all relevant curly arrows to show the movement of electron pairs in box one and in box two ● the formula of the second product in box three. Name this mechanism.
13. An introduction to AS Level organic chemistry  →  13.2. Characteristic organic reactions
14. Hydrocarbons  →  14.2. Alkenes
18. Carboxylic acids and derivatives  →  18.1. Carboxylic acids
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