9701_w23_qp_41
A paper of Chemistry, 9701
Questions:
9
Year:
2023
Paper:
4
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1
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Fluorine reacts with chlorine dioxide, Cl O2, as shown. F2+ 2Cl O22FCl O2The rate of the reaction is first order with respect to the concentration of F2 and first order with respect to the concentration of Cl O2. No catalyst is involved. Suggest a two-step mechanism for this reaction. step 1 step 2 Identify the rate-determining step in this mechanism. Explain your answer. When the rate of the reaction is measured in mol dm−3 s−1 the numerical value of the rate constant, k, is 1.22 under certain conditions. Complete the rate equation for this reaction, stating the overall order of the reaction. rate = overall order of reaction = Use your rate equation in to calculate the rate of the reaction when the concentrations of F2 and Cl O2 are both 2.00 × 10−3 mol dm−3. rate = mol dm−3 s−1 Under different conditions, and in the presence of a large excess of ClO2, the rate equation is as shown. rate = k1[F2] The half-life, t1 2, of the concentration of F2 is 4.00 s under these conditions. Calculate the numerical value of k1, giving its units. Give your answer to three significant figures. k1 = units An experiment is performed under these conditions in which the starting concentration of F2 is 0.00200 mol dm–3. Draw a graph on the grid in to show how the concentration of F2 changes over the first 12 s of the reaction. 0.0002 0.0004 0.0006 0.0008 0.0010 0.0012 0.0014 0.0016 0.0018 0.0020 [F2] / mol dm–3 time / s Use your graph in to find the rate of the reaction when the concentration of F2 is 0.00100 mol dm–3. Show your working on the graph. rate = mol dm–3 s−1
8. Reaction kinetics  →  8.1. Rate of reaction
11. Group 17  →  11.4. The reactions of chlorine
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Define Kw mathematically by completing the expression. Kw = Two solutions, V and W, are described. • V is HCl . • W is NaOH. • The concentration of HCl in V is the same as the concentration of NaOH in W. • The pH values of V and W differ by exactly 11.00 at 298 K. Calculate the concentration of HCl in V. concentration of HCl in V = mol dm−3 Equal volumes of the two solutions V and W are mixed, giving solution X. Name solution X and state its pH. solution X pH A 1 cm3 sample of 1.0 mol dm−3 HNO3 is added to 100 cm3 of solution X, forming mixture Y. A 1 cm3 sample of 1.0 mol dm−3 KOH is added to 100 cm3 of solution X, forming mixture Z. Estimate the pH of mixtures Y and Z. No calculations are required. mixture Y mixture Z CH3CH2COOH, CH3CCl2COOH and H2SO4 are all acidic. Suggest the trend in the relative acid strength of these three compounds. Explain your answer. strongest acid weakest acid explanation When concentrated H2SO4 is added to water a series of acid-base reactions occurs. There are three conjugate acid-base pairs that can be identified during this series of reactions. Write the formulae of these three conjugate acid-base pairs. conjugate acid 1 conjugate base 1 conjugate acid 2 conjugate base 2 conjugate acid 3 conjugate base 3 The partition coefficient, Kpc, of a substance, Q, between hexane and water is 7.84 at 298 K. Q is more soluble in hexane than it is in water. Define partition coefficient, Kpc. 5.00 g of Q is shaken with a mixture of 100.0 cm3 of water and 100.0 cm3 of hexane at 298 K and left until there is no further change in concentrations. Calculate the mass of Q dissolved in the water. mass of Q = g A sample of Q is shaken with a different mixture of water and hexane and left until there is no further change in concentrations. It is found that the mass of Q dissolved in each solvent is the same. Use the Kpc value to suggest possible values for the volume of water used and the volume of hexane used. volume of water = cm3 volume of hexane = cm3 Q is more soluble in hexane than it is in water. It is suggested that Q is one of KCl, CH3(CH2)4OH or HCOOH. Identify Q. Explain your answer.
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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Hydrogen peroxide is a liquid at 298 K. It is moderately stable under room conditions but will decompose quickly if a catalyst is added. reaction 1 2H2O22H2O + O2 Define entropy. Predict the sign of the standard entropy change of reaction 1. Explain your answer. sign explanation Some bond energy data are shown in Table 3.1. Table 3.1 type of bond bond energy / kJ mol–1 O–O O–H O=O Use the data in Table 3.1 to show that the enthalpy change of the following reaction is –196 kJ mol–1. 2H2O22H2O+ O2 Some standard entropies, S o , are shown in Table 3.2. Table 3.2 substance S o / J K–1 mol–1 H2O2+102 H2O+70 The enthalpy change and Gibbs free energy change for the following reaction are shown. 2H2O22H2O+ O2ΔH o = –196 kJ mol–1 ΔG o = –238 kJ mol–1 Use the data given to calculate the standard entropy of oxygen, S o , O2. S o , O2= J K–1 mol–1 The decomposition of H2O2is catalysed by aqueous iron(chloride and by silver metal. Identify which of these two catalysts is acting as a homogeneous catalyst. Explain your answer. homogeneous catalyst explanation The E o values for two electrode reactions are given. H2O2 + 2H+ + 2e− 2H2O E o = +1.77 V Cr3+ + e− Cr2+ E o = −0.41 V An electrochemical cell is constructed with the following half-cells : • an acidified solution of H2O2, a platinum wire • Cr2+ mixed with Cr3+, a platinum wire. Identify the positive half-cell and calculate the standard cell potential, E o cell. positive half-cell E o cell = V Calculate the value of ΔG o for the cell reaction that occurs, per mole of H2O2. ΔG o = kJ mol–1 The E o values for two electrode reactions are given. H2O2 + 2H+ + 2e– 2H2O E o = +1.77 V Co3+ + e– Co2+ E o = +1.82 V An electrochemical cell is constructed with the following half-cells. half-cell 1 an acidified solution of H2O2 under standard conditions, a platinum wire half-cell 2 a solution containing 0.020 mol dm–3 Co3+ and 2.0 mol dm–3 Co2+, a platinum wire Use the Nernst equation to calculate the value of E, the electrode potential of half-cell 2 under these conditions. E = V Write an equation for the cell reaction that occurs in this cell under these conditions. Define enthalpy change of hydration, ΔHhyd. Aluminium fluoride, Al F3, is an ionic solid. Complete and label the energy cycle to show the relationship between: ● the enthalpy change of solution of Al F3, ΔH o sol ● the lattice energy of Al F3, ΔH o latt ● the enthalpy changes of hydration of Al 3+ and F−, ΔH o hyd. Include state symbols for all substances and ions. …………………………… …………………………… …………………………… …………… ΔH o …………… ΔH o …………… ΔH o Relevant data for this question are given. ΔH o sol Al F3 = –209 kJ mol−1 ΔH o hyd Al 3+ = – 4690 kJ mol−1 ΔH o hyd F– = –506 kJ mol−1 Use these data and your energy cycle in to calculate the ΔH o latt of Al F3. ΔH o latt of Al F3 = kJ mol−1
5. Chemical energetics  →  5.1. Enthalpy change, \(\Delta H\)
14. Hydrocarbons  →  14.2. Alkenes
11. Group 17  →  11.4. The reactions of chlorine
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Cobalt(nitrate, Co(NO3)2, is a reddish-brown crystalline solid. It dissolves in water to form a solution containing [Co(H2O)6]2+ complex ions. Complete Table 4.1 giving the formula of the cobalt-containing species that is formed in each of the three reactions described. Table 4.1 reaction reagent added to [Co(H2O)6]2+cobalt-containing species formed NaOH2 an excess of NH33 an excess of conc. HCl Describe the colour change seen in reaction 3. original colour of [Co(H2O)6]2+final colour after addition of an excess of conc. HCl Calcium nitrate, Ca(NO3)2, is a white crystalline solid. When heated, it starts to decompose at approximately 500°C. Write an equation for the decomposition of Ca(NO3)2. Suggest temperatures at which Mg(NO3)2 and Ba(NO3)2 start to decompose. Explain your answer. temperature at which Mg(NO3)2 starts to decompose °C temperature at which Ba(NO3)2 starts to decompose °C explanation
11. Group 17  →  11.4. The reactions of chlorine
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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Transition elements behave as catalysts and can form complex ions. Explain why transition elements behave as catalysts. Silver forms the linear complex ion [Ag(CN)2]–. Copper forms the tetrahedral complex ion [Cu(CN)4]3−. Titanium forms the complex [TiCl42], where diars is a neutral bidentate ligand. State the oxidation state and the coordination number of titanium in [TiCl42]. oxidation state coordination number Draw three-dimensional diagrams to show the shapes of [Ag(CN)2]– and [Cu(CN)4]3−, in the boxes. Label one bond angle on each diagram. Ag Cu The numerical value of the stability constant, Kstab, of the copper(complex [Cu(CN)4]3– is 2.0 × 1027. Write an expression for the Kstab of [Cu(CN)4]3−. Kstab = In a solution the concentrations of CN− and [Cu(CN)4]3− are both 0.0010 mol dm−3. Use your expression from and the value of Kstab to calculate the concentration of Cu+in this solution. concentration of Cu+= mol dm–3 A piece of a copper-containing alloy has a mass of 0.567 g. It is dissolved in an acid giving 100.0 cm3 of a blue solution in which all the copper is present as Cu2+ ions. An excess of KIis added to a 25.0 cm3 sample of this solution. All of the copper is precipitated as white CuI. Cu2+ ions are the only component in the solution that react with KI. This is reaction 1. reaction 1 2Cu2+ + 4I− 2CuI + I2 The liberated I2 is then titrated with 0.0200 mol dm–3 S2O3 2−. This is reaction 2. reaction 2 I2 + 2S2O3 2– 2I− + S4O6 2– The titration requires 20.10 cm3 of 0.0200 mol dm–3 S2O3 2– to reach the end-point. Calculate the number of moles of I2 that are reduced in this titration. number of moles of I2 = mol Calculate the number of moles of copper in the original piece of alloy. number of moles of copper = mol Calculate the percentage of copper in the alloy. percentage of copper = % Suggest why a solution of Cu2+ is coloured but solid CuI is white.
11. Group 17  →  11.3. Some reactions of the halide ions
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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Five ligands are listed in Table 6.1. Table 6.1 ligand type of ligand NH3 EDTA4− CN− H2NCH2CH2NHCH2CH2NH2 tridentate C2O4 2− Complete Table 6.1 using the words monodentate, bidentate and polydentate only. Each of these three words may be used once, more than once, or not at all. The molecule H2NCH2CH2NHCH2CH2NH2 is a tridentate ligand. Suggest the meaning of tridentate ligand. Suggest how H2NCH2CH2NHCH2CH2NH2 acts as a tridentate ligand. Nickel forms the octahedral complex [Ni2(H2O)2]2+. This complex can exist in three isomeric forms, listed in Table 6.2. One of these forms is a trans isomer, the other forms are two different cis isomers. Table 6.2 isomer polarity trans isomer cis isomer 1 cis isomer 2 Complete Table 6.2 using the terms polar or non-polar. Each term may be used once, more than once, or not at all. Describe the difference between cis isomer 1 and cis isomer 2.
11. Group 17  →  11.3. Some reactions of the halide ions
13. An introduction to AS Level organic chemistry  →  13.1. Formulas, functional groups and the naming of organic compounds
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
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Sunset Yellow is an additive used for colouring foods. A synthetic route for making Sunset Yellow is shown. Molecules E and G each contain one –SO – 3 group. These groups are unchanged in the formation of Sunset Yellow. HOC10H6SO3 – – O3SC6H4NH2 E F G step 1 step 2 HO –O3S SO3 – Sunset Yellow anion N N State the molecular formula of the Sunset Yellow anion. Deduce the structures of E, F and G and draw them in the boxes in . Suggest suitable reagents and conditions for step 1 and 2. step 1 step 2 Predict the number of peaks in the carbon-13 NMR spectrum of the Sunset Yellow anion.
21. Organic synthesis  →  21.1. Organic synthesis
13. An introduction to AS Level organic chemistry  →  13.1. Formulas, functional groups and the naming of organic compounds
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Capsaicin is found in chilli peppers. CH3O HO O N H capsaicin You should assume the CH3O group is unreactive in the reactions involved in this question. Name all the functional groups in capsaicin in addition to the CH3O group. Complete the equation for the reaction of capsaicin with an excess of Br2in the dark. Draw the structure of the organic product in the labelled box. CH3O HO O capsaicin organic product + ………………… + ………Br2 N H Capsaicin is heated with an excess of hydrogen gas in the presence of platinum metal. The six-membered ring reacts in the same way as benzene under these conditions. Draw the structure of the organic product formed. When capsaicin is treated with reagent J under suitable conditions one of the products is methylpropanoic acid, CH3CH(CH3)COOH. Identify reagent J and any necessary conditions. There are three different peaks in the proton (1H) NMR spectrum of CH3CH(CH3)COOH in CDCl 3. 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 Table 8.1 to complete Table 8.2 and state: ● the typical proton (1H) chemical shift values (δ) for the protons ● the splitting pattern (singlet, doublet, triplet, quartet or multiplet) shown by each peak ● the explanation for the splitting patterns of the CH3 protons and the CH proton. Table 8.2 environment δ / ppm splitting pattern explanation for splitting pattern CH3 CH COOH Capsaicin is heated with an excess of hot aqueous NaOH. CH3O HO O N H capsaicin Draw the structures of the two organic products H and K. H C8H10NO2Na K C10H17O2Na Name the two types of reaction occurring in . Draw the structure of the organic product L formed when capsaicin is treated with LiAl H4 in dry ether. L
14. Hydrocarbons  →  14.1. Alkanes
14. Hydrocarbons  →  14.2. Alkenes
17. Carbonyl compounds  →  17.1. Aldehydes and ketones
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Benzoyl chloride, C6H5COCl, can be made from ethyl benzene in a two-step process. A reaction scheme is shown. ethyl benzene benzoyl chloride Cl step 1 step 2 compound M O CH2CH3 Draw the intermediate organic compound M in the box. Suggest suitable reagents and conditions for step 1 and step 2. step 1 step 2 Identify the type of reaction in step 1 and step 2. step 1 step 2 C6H5COCl reacts with phenol, C6H5OH, to give the ester phenyl benzoate, C6H5COOC6H5. An incomplete description of the mechanism of this reaction is shown in . box one reactants Cl C O O H intermediate box three products and O C O box two Cl C O O H Complete the mechanism in and 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. Benzoyl chloride, chlorobenzene and chloroethane differ in their rates of hydrolysis when each compound is added separately to water at 25 °C. Suggest the relative ease of hydrolysis of these three compounds. Explain your answer. hardest to hydrolyse easiest to hydrolyse explanation
16. Hydroxy compounds  →  16.1. Alcohols
21. Organic synthesis  →  21.1. Organic synthesis
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