9701_m24_qp_42
A paper of Chemistry, 9701
Questions:
6
Year:
2024
Paper:
4
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2
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1
Potassium iodide, KI, is used as a reagent in both inorganic and organic chemistry. KI forms an ionic lattice that is soluble in water. Define enthalpy change of solution, ΔHsol. KIhas a high solubility in water although its enthalpy change of solution is endothermic. Explain how this high solubility is possible. Table 1.1 gives some data about the halide ions, Cl –, Br – and I–, and their potassium salts. Table 1.1 halide ion enthalpy change of hydration, ΔHhyd / kJ mol–1 lattice energy of potassium halide, ΔHlatt / kJ mol–1 Cl – –364 –701 Br – –335 –670 I– –293 –629 Explain the trend in the enthalpy change of hydration of the halide ions. The ΔHsol values of these potassium halides are almost constant. Use the ΔHhyd and ΔHlatt data in Table 1.1 to suggest why. The enthalpy change of solution of KIis +21.0 kJ mol–1. Use this information and the data in Table 1.1 to calculate the enthalpy change of hydration of the potassium ion, K+. ΔHhyd of K+= kJ mol–1 Solid PbI2 forms when KIis mixed with Pb2+ions. The solubility product, Ksp, of PbI2 is 7.1 × 10–9 mol3 dm–9 at 25 °C. Calculate the solubility, in mol dm–3, of PbI2. solubility of PbI2= mol dm–3 The ionic radius of Pb2+ is 0.120 nm compared to 0.133 nm for K+. Suggest how the ΔH o latt of PbI2differs from ΔH o latt of KI. Explain your answer. KI slowly oxidises in air, forming I2. reaction 1 4KI+ 2CO2+ O22K2CO3+ 2I2ΔH o = –203.4 kJ mol–1 Table 1.2 shows some data relevant to this question. Table 1.2 substance standard entropy, S o / J K–1 mol–1 CO2213.6 I2116.1 K2CO3155.5 KI106.3 O2205.2 Calculate the standard entropy change, ΔS o , of reaction 1. ΔS o = J K–1 mol–1 Use your answer to to show that reaction 1 is spontaneous at 298 K. The Group 1 carbonates are much more thermally stable than the Group 2 carbonates. State and explain the trend in the thermal stability of the Group 2 carbonates. A student electrolyses a solution of KIfor 8 minutes using a direct current. The half-equation for the reaction that occurs at the anode is given. 2I–I2+ 2e– Write a half-equation for the reaction that occurs at the cathode. Include state symbols. After the electrolysis, the I2produced requires 21.35 cm3 of 0.100 mol dm–3 Na2S2O3to react completely. I2+ 2Na2S2O32NaI+ Na2S4O6Calculate the average current used in 8 minutes during the electrolysis. current = A KI is used as a source of I– ions in organic synthesis. One example of this is shown in the synthetic route in . NO2 NH2 N2 + N2 I I step 1 A B C D E step 2 step 3 step 4 NaNO2 and HCl Identify the reagents required for steps 1 and 2. step 1 step 2 Step 3 occurs in two stages. stage I NaNO2 and HCl undergo an acid–base reaction to produce HNO2. stage II HNO2 reacts with C, C6H5NH2, to produce D, C6H5N2 +. Complete the equations for stage I and for stage II. stage I NaNO2 + HCl stage II The I– from KI reacts with D in step 4. The mechanism is shown in . Suggest the name for this mechanism.
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  →  11.2. The chemical properties of the halogen elements and the hydrogen halides
11. Group 17  →  11.3. Some reactions of the halide ions
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Water is an amphoteric compound that also acts as a good solvent of polar and ionic compounds. Equation 1 shows water acting as a Brønsted–Lowry acid. equation 1 H2O + NO2 – HNO2 + OH– Identify the two conjugate acid–base pairs in equation 1. acid I H2O conjugate base of acid I acid II conjugate base of acid II Water also behaves as a Brønsted–Lowry acid when it dissolves CH3NH2. Explain the ability of CH3NH2 to act as a base. Write an equation to show water acting as a base with CH3COOH. The ionic product of water, Kw, measures the extent to which water dissociates. H2OH++ OH–shows how Kw varies with temperature. 0.00 1.00 2.00 3.00 4.00 5.00 6.00 temperature / °C Kw / 10–14 mol2 dm–6 Write an expression for Kw. Use information from to deduce whether the dissociation of water is an exothermic or an endothermic process. Explain your answer. An aqueous solution has pH = 7.00 at 30 °C. Use information from to explain why this solution can be considered to be alkaline at 30 °C. The three physical states of H2O have different standard entropies, S o , associated with them. Table 2.1 shows these S o values. Table 2.1 state of H2O standard entropy, S o / J K–1 mol–1 solid +48.0 liquid +70.1 gas +188.7 Explain the difference in the S o values of H2Oand H2O. Explain why the increase in S o is much greater when H2O boils than when it melts. The energy changes for H2O→ H2Oare shown. ΔG = 0.00 kJ mol–1 ΔH = +6.03 kJ mol–1 Use these data to show that the melting point of H2Ois 0 °C. Metal–air batteries are electrochemical cells that generate electrical energy from the reaction of metal anodes with air. The standard electrode potentials for the zinc–air battery are shown. [Zn(OH)4]2– + 2e– Zn + 4OH– E o = –1.22 V 2 O2 + H2O + 2e– 2OH– E o = +0.40 V Calculate the standard cell potential, E o cell, of the zinc–air battery. E o cell = V The zinc–air battery usually operates at pH 11 and 298 K. The overall cell potential is dependent on [OH–]. The Nernst equation shows how the electrode potential at the cathode changes with [OH–]. E = 0.40 – (0.059 z ) log([OH–]2) Calculate the electrode potential, E, at pH 11. E = V
7. Equilibria  →  7.2. Brønsted–Lowry theory of acids and bases
12. Nitrogen and sulfur  →  12.1. Nitrogen and sulfur
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Iron is a transition metal in Group 8 of the Periodic Table. Explain why iron has variable oxidation states. Complete the shorthand electronic configurations of Fe and Fe3+. Fe [Ar] Fe3+ [Ar] An aqueous solution of Fe(NO3)3 contains the complex [Fe(H2O)6]3+. When solutions of KSCNand [Fe(H2O)6]3+are mixed, a colour change is observed. The red complex [Fe(H2O)5SCN]2+ forms. Define complex. State the coordination number of Fe in [Fe(H2O)6]3+. The H—O—H bond angle in water is 104.5°. Suggest the H—O—H bond angle in [Fe(H2O)6]3+. Explain your answer. Explain why iron complexes are coloured. Aqueous solutions of complexes [Fe(H2O)6]3+ and [Fe(H2O)5SCN]2+ are different colours. Explain why these complexes are different colours. Table 3.1 gives values for the stability constants, Kstab, of different complexes of iron. Table 3.1 complex stability constant, Kstab [Fe(H2O)5(H2PO4)]2+ 5.90 × 101 [Fe(H2O)5SCN]2+ 1.30 × 102 [Fe(H2O)5(H2PO4)]2+ can form when H3PO4 reacts with [Fe(H2O)6]3+. Write an equation for this reaction. Write an expression for Kstab of [Fe(H2O)5SCN]2+ and give its units. Kstab = units = Use the stability constant data in Table 3.1 to calculate the value of the equilibrium constant, Kc, for the following equilibrium. [Fe(H2O)5(H2PO4)]2+ + SCN– [Fe(H2O)5SCN]2+ + H2PO4 – value of Kc =
1. Atomic structure  →  1.3. Electrons, energy levels and atomic orbitals
9. The Periodic Table: chemical periodicity  →  9.2. Periodicity of chemical properties of the elements in Period 3
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Ruthenium and osmium are transition metals below iron in Group 8 of the Periodic Table. Two different complex ions, X and Y, can form when anhydrous RuCl 3 reacts with water under certain conditions. X and Y have octahedral geometry. Aqueous samples of X and Y react separately with an excess of AgNO3. Different amounts of AgCl are precipitated: • 1 mole of complex ion X produces 2 moles of AgCl • 1 mole of complex ion Y produces 1 mole of AgCl. Complete Table 4.1 to suggest formulae for X and Y. Table 4.1 X Y formula of complex Both complexes react with an excess of bipyridine, bipy, to form a mixture of two stereoisomers of [Ru3]3+. N N bipy Bipyridine is a bidentate ligand. Draw three-dimensional diagrams of the two stereoisomers of [Ru3]3+. Use N N to represent the bipy ligand in your structures. Ru Ru shows another ruthenium complex. N Ru(NH3)5 5+ (H3N)5Ru N This complex contains the neutral ligand pyrazine. pyrazine N N Suggest how pyrazine is able to bond to two separate ruthenium ions. Pyrazine is an aromatic compound. The bonding and structure of pyrazine is similar to that of benzene. Describe and explain the shape of pyrazine. In your answer, include: • the hybridisation of the nitrogen and carbon atoms • how orbital overlap forms π bonds between the atoms in the ring. Predict the number of peaks seen in the carbon–13 NMR spectrum of pyrazine. Explain your answer. The overall charge of the ruthenium complex in is 5+. Deduce the possible oxidation states of the two ruthenium ions in the complex. Osmium tetroxide, OsO4, reacts with alkenes in a similar manner to cold dilute acidified MnO4 –. shows a proposed synthesis of a condensation polymer G. COOH step 1 HOOC COCl G OsO4 C6H12O2 ClOC step 2 step 3 Suggest a reagent for step 1. Draw the structure of exactly one repeat unit of the condensation polymer G. The ester linkage should be shown fully displayed.
3. Chemical bonding  →  3.4. Covalent bonding and coordinate (dative covalent) bonding
11. Group 17  →  11.3. Some reactions of the halide ions
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Compound Q can be synthesised from chlorobenzene in seven steps, using the route shown in . step 1 step 2 step 3 [O] step 4 step 5 acidified CH3OH step 6 step 7 SOCl2 Cl Cl AlCl3 and CH3Cl CH3 Cl CHO Cl CN chlorobenzene J K L M N OH Cl COOH OH Cl COOCH3 OH P Cl COOCH3 Cl Q Cl COOCH3 N S HN S and K2CO3Write an equation for the formation of the electrophile for step 1. Complete the mechanism in for step 1, the alkylation of chlorobenzene. Include all relevant curly arrows and charges. Draw the structure of the intermediate. Cl Cl intermediate CH3 + Step 2 is an oxidation reaction. Construct an equation for the reaction in step 2. Use [O] to represent an atom of oxygen from an oxidising agent. Suggest reagents for the conversion of K to M in steps 3 and 4. step 3 step 4 Identify the type of reaction that occurs in step 5. Step 7 takes place when P is heated with a weak base such as K2CO3. P Cl COOCH3 Cl Q Cl COOCH3 N S step 7 HN S and K2CO3Suggest why a strong base such as NaOHis not used for this reaction. Q is optically active. Explain the meaning of optically active. Give two reasons why it might be desirable to synthesise a single optical isomer of Q for use as a drug. Q is commonly used in conjunction with aspirin. COOH aspirin O O Aspirin is a weak Brønsted–Lowry acid. The pKa of aspirin is 3.49. 75 mg of aspirin dissolves in water to form 100 cm3 of an aqueous solution. Calculate the pH of this solution. [Mr: aspirin, 180.0] pH = Aspirin undergoes acid hydrolysis in the stomach. Give the structures of the organic products of this acid hydrolysis.
15. Halogen compounds  →  15.1. Halogenoalkanes
21. Organic synthesis  →  21.1. Organic synthesis
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Amino acids are molecules that contain —NH2 and —COOH functional groups. Glycine, H2NCH2COOH, is the simplest stable amino acid. The isoelectric point of glycine is 6.2. Define isoelectric point. Draw the structure of glycine at pH 4. shows two syntheses starting with glycine. hippuric acid O COOH N H N C2H5Br glycine U reaction 1 H2N COOH COOH reaction 2 reaction 3 an excess of LiAl H4 State the essential conditions for reaction 1. Identify the reagent used in reaction 2. Draw the structure of the organic product U that forms when hippuric acid reacts with an excess of LiAl H4 in reaction 3. A molecule of phenylalanine, R, can react with a molecule of glycine to form two dipeptides, S and T. S and T are structural isomers. glycine S and T R H2N + COOH COOH H2N Draw the structures of these dipeptides. The peptide bond formed should be shown fully displayed. S T A student proposes a synthesis of hippuric acid by the reaction of benzamide, C6H5CONH2, and chloroethanoic acid, Cl CH2COOH. The reaction does not work well because benzamide is a very weak base. Explain why amides are weaker bases than amines. The pKa of chloroethanoic acid is 2.86 whereas the pKa of ethanoic acid is 4.76. Explain the difference between these two pKa values. Compound V is another amino acid. The proton (1H) NMR spectrum of V shows hydrogen atoms in five different environments, a, b, c, d and e, as shown in . H H H H HOOC CH2CH2NH2 b V a c d e Table 6.1 environment of proton example chemical shift range, d / 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, –CH2–N 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 Complete Table 6.2 for the proton (1H) NMR spectrum of V taken in CDCl 3. Table 6.1 gives some relevant data. Table 6.2 proton a b c d e chemical shift range, d / ppm name of splitting pattern multiplet Complete Table 6.3 by placing a tick (✓) to indicate any protons whose peaks are still present in the proton (1H) NMR spectrum of V taken in D2O. Table 6.3 proton a b c d e present in D2O
14. Hydrocarbons  →  14.2. Alkenes
21. Organic synthesis  →  21.1. Organic synthesis
17. Carbonyl compounds  →  17.1. Aldehydes and ketones
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