科目：化学 Chemistry (9701)
年份：2023 冬季
试卷：Paper 42
总页数：23
章节：19 Lattice energy, 20 Electrochemistry, 21 Further aspects of equilibria, 22 Reaction kinetics, 23 Entropy and Gibbs free energy, 24 Transition elements, 25 Benzene and its compounds, 26 Carboxylic acids and their derivatives, 27 Organic nitrogen compounds, 28 Polymerisation, 30 Analytical chemistry
1. Propanone, CH$_3$COCH$_3$, reacts with iodine, I$_2$, in the presence of an acid catalyst.
$$\text{CH}_3\text{COCH}_3 + \text{I}_2 \rightarrow \text{CH}_3\text{COCH}_{2}\text{I} + \text{H}^+ + \text{I}^{-}$$
The rate equation for this reaction is shown.
$$\text{rate} = k[\text{CH}_3\text{COCH_3}][\text{H}^+]$$
(a) Complete Table 1.1 to describe the order of the reaction.
| order of the reaction with respect to [CH$_3$COCH$_3]$ | |
| :--- | :--- |
| order of the reaction with respect to [I$_2$] | |
| order of the reaction with respect to[H$^+$] | |
| overall order of the reaction | |
(b) An experiment is performed using a large excess of CH$_3$COCH$_3$ and a large excess of H$^{+}$(aq). The initial concentration of I$_2$ is $1.00 \times 10^{-5}$ mol dm$^{-3}$. The initial rate of d(ii) State whether it is possible to calculate the numerical value of the rate constant, $k$, for this reaction from your graph. Explain your answer.
(c) The experiment is repeated at a different temperature. The initial concentrations of H$^+$ ions, I$_2$ and CH$_3$COCH$_3$ are all 0.200 mol dm$^{-3}$.
The value of $k$ at this temperature is $2.31 \times 10^{-5} \text{mol}^{-1}\text{dm}^3\text{s}^{-1}$.
Calculate the initial rate of this reaction.
$$\text{rate} = 2.31 \times 10^{-5} \text{mol}^{-1]\text{dm}^3\text{s}^{{-1}}$$
(d) The experiment is repeated using an excess of H$^+$(aq). The new rate equation is shown.
$$\text{rate} = k_1[\text{CH}_3\text{COCH}_3]$$
(i) The value of $k_1$ is $1.1 \times 10^{-3} \text{s}^{{-1}}$. Calculate the value of the half-life, $t_{\frac{1}{2}}$(e) A four-step mechanism is suggested for the overall reaction.
$$\text{CH}_3\text{COCH}_3 + \text{I}_2 \rightarrow \text{CH}_3\text{COCH}_{2}\text{I} + \text{H}^+ + \text{I}^{-}$$
rate = $k[\text{CH}_3\text{COCH_3}][\text{H}^+]$
Part of this mechanism is shown.
step 1: $\text{CH}_3\text{COCH}_{\text{3}} + \text{H}^+ \rightarrow \text{CH}_3\mathrm{C}^{+}(\text{OH})\text{CH}_3$
step 2: $\text{CH}_3\mathrm{C}^{\dagger}(\text{OH})\text{CH}_{\text{3}} \rightarrow \text{CH}_3 \mathrm{C}(\text{OH})=\text{CH}_2 + \text{H}^+$
step 3: $\rightarrow$
step 4: $\text{CH}_3\mathrm C}^{\dagger}(\text{OH})\mathrm{CH}_2\mathrm{I} \rightarrow \text{CH}_3{\text{COCH}}_{2}\mathrm{I} + \text{H}^+\)
(i) Write an equation for step 3.
[1]
(ii) Suggest the slowest step of the mechanism. Explain your ansBLANK PAGE
2 Benzoic acid, $C_6H_5COOH$, is a weak acid. The $K_a$ of benzoic acid is $6.31 \times 10^{-5} \text{ mol dm}^{-3}$ at 298 K.
A 1.00 $\text{dm}^3$ buffer solution is made at 298 K containing 1.00 g of $C_6H_5COOH$ and a slightly greater mass of sodium benzoate, $C_6H_5COO^-$Na$^{+}$.
This buffer solution has a pH of 4.15.
(a) Define buffer solution.
[1]
(b) Write equations to show how this solution acts as a buffer solution when the named substances are added to it:
(i) dilute aqueous sodium hydroxide
[1]
(ii) dilute aqueous nitric acid.
[1]
(c) Calculate the $H^+$ concentration and the $C_6H_5COOH$-concentration in the buffer solution described. Use the expression for the $K_a$ of $C_6H_5CO OH$ to calculate the concentration of $C_6H_5COO^{-}\text{Na}^+$ in the buffer solution.(d) A 10.0 cm$^3$ sample of the buffer solution is mixed with 10.0 cm$^3$/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.
[2]
(e) Magnesium benzoate, Mg(C$_6$H$_5$COO)$_2$, has a solubility in water of less than 1.00 g dm$^{-3}$ at 298 K.
$$K_{\text{sp}} = [\text{Mg}^{2+}][\text{C}_6\text{H}_5\text{COO}^-]^2 = 1.76 \times 10^{-7}$$ at $298\text{K}$$
(i) Calculate the solubility of Mg(C$_6$H$_5$ COO)$_2$ in water at 298 K. Give your answer in g dm$^{-3}$.
Show your working.
$$M_r: \text{Mg(C}_6\text{H}_5\text{ COO)}_2, 266.3$$
solubilit3 Some electrode potentials are shown in Table 3.1.
Table 3.1
| electrode reaction | $E^\circ$/V |
| :--- | :--- |
| $V^{2+} + 2e^- \rightleftharpoons V$ | -1.20 |
| $V^{3+} + e^- \rightleftharpoons W^{2+}$ | -0.26 |
| $VO^{2+} + 2H^+ + e^- \rightleftharrow V^{3+} + H_2O$ | +0.34 |
| $VO_{2}^+ + 2H^+ + e^-\rightleftharpoons VO^{2+} + H_2O$ |+1.00 |
| $Fe^{2+} + 2e^-\rightleftharpoons Fe$ | -0.44 |
| $Fe^{3+} + 3e^-\rightleftharpoons F_e$ | -0.04 |
| $Fe^{3+}+e^-\rightleftharpoons FE^{2+}$ | +0.77 |
| $2H^+ + 2e^-\rightlefsharpoons H_2$ | 0.00 |
| $ClO^- + H_2O + 2e^- \rightleftarrow Cl^- + 2OH^-$ | +0.89 |
(a) (i) Complete the diagram to show a standard hydrogen electrode.
Label your diagram. Identify all substances. You do not need to state standard conditions.
(ii) An electrochemical(b) The vanadium-containing species in the electrode reactions given in Table 3.1 are V, $V^{2+}$, $V^{3+}$, $VO^{2+}$ and $VO_{2}^{+}$.
(i) Identify one vanadium-containing species that does not react with $Fe^{2+}$ ions under standard conditions.
Use data from Table 3.1 to explain your answer.
[1]
(ii) Identify all the vanadium-containing species that will react with $Fe^{2+}$ iones under standard conditions.
[1]
(iii) Write an equation for one of the possible reactions identified in (ii).
[1]
(c) Another electrochemical cell is set up using an $Fe^{3+}/Fe^{2+}$ electrode and an alkaline $ClO^{-}/Cl^{-}$ electrode.
The concentration of $Fe^{3+}$ is 1000 times greater than the concentration of $Fe^{2+}$ in the $Fe^{3+}/Fe^{2}+$ electrode. All other conditions are standard.
(i) U(e) A solution of iron(II) sulfate, FeSO$_4$(aq) is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode.
A current of 0.640A is passed for 17.0 minutes. The mass of the cathode increases by 0.185 g.
Use these results to calculate an experimental value for the Avogadro constant, $L$.
Show your working.
$$L = \dots\dots\dots\dots\dot{m}ol^{-1}$$
(f) Iron(II) chloride, FeCl$_2$, is oxidised by chlorine to form iron(III) chloride, FeCl$_3$, under standard conditions.
$$2\text{FeCl}_2(\text{s}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{FeCl}_3(\text{s})$$
$$\Delta H^\circ = -128\text{kJ mol}^{-1}$$
Table 3.2
| species | $S^\circ$/JK$^{-1}$mol$^{-1}$ |
| :--- | :--- |
| $\text{Cl}_2(\text{g })$ | 223 |
| $\text{FeCl}_2(\text{s })$ | 120 |
| $\(ii) Predict whether this reaction becomes more or less feasible at a higher temperature. Explain your answer. The reaction becomes ... feasible. explanation ... [1] [Total: 18]
4 The structure of the polydentate ligand, EDTA$^{4-}$, is shown in Fig. 4.1.
Fig. 4.1
The stability constants, at 298 K, of five octahedral complexes are given in Table 4.1.
Table 4.1
| complex | $K_{\text{stab}}$ |
| :--- | :---: |
| $\text{[Cu(EDTA)]}^{2-}$ | $6.31 \times 10^{19}$ |
| $\text{[Cr(EDTA)]}^{2-}$ | 1.00 $\times 10^{13}$ |
| $\text{[Cr(EdTA)]}^-$ | 1.00 $\times 10^{24}$ |
| $\text{[Fe(EDTA)]}^{2-}$ | $$2.00 \times 10^{14}$$ |
| $$\text{[Fe(EDTA)]}^-$ | $$1.26 \times 10^{25}$$ |
(a) Define stability constant.
(b) Calculate the oxidation states of Cu in [Cu(EDTA)]$^{2-}$ and Cr in [Cr(EDTA)]$^{-}$.
Cu ......
Cr ......
(c) Deduce the number of lone pairs donated by each EDTA$^{4-}$ ligand in a single [Fe(EDTA)]$^{2-}$ complex ion.
(d) Identify the most stable complex(e) In a solution at equilibrium at 298 K, $[\text{Cu(H}_2\text{O})_6]^{2+} = 3.00 \times 10^{-10}\text{mol dm}^{-3}$ and $[\text{EDTA}^4-] = 5.00 \times 10^{-{12}}\text{mol dm}^{-3}$.
Use the expression for $K_{\text{stab}}$ to calculate the concentration of $[\text{Cu(EDTA)}]^{2-}$ in this solution.
Show your working.
$$[\text{Cu(EDTA)}]^2- = \dots\dots\dots\dots\mathrm{mol dm}^{-3}$$
(f) A solution of $[\text{Cu(EDTA)]}^{2-}$ ions is pale blue while a solution of $[\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ ions is deep blue.
Explain this difference in colour.
[Total: 8]5 Some of the ionic compounds of Group 2 elements undergo thermal decomposition.
Thermal decomposition of solid anhydrous magnesium ethanedioate, MgC$_2$O$_4$, occurs above 650°C. The products are magnesium oxide and a mixture of two different gases, one of which gives a white precipitate with saturated calcium hydroxide solution.
(a) Complete the equation for the thermal decomposition of MgC$_2$O$_4$.
$$\text{MgC}_2\text{O}_4 \rightarrow [1]$$
(b) Suggest which of MgC$_2$O$_4$, or CaC$_2$O$_4$, undergoes thermal decomposition at a lower temperature. Explain your answer.
[2]
(c) The ethanedioate ion is oxidised by acidified KMnO$_4$.
$$5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 10\text{CO}_2 + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$$
An experiment is per6 (a) Phosphine, :PH$_3$, and carbon monoxide, :CO, are monodentate ligands found in some transition element complexes.
(i) Define monodentate ligand.
[1]
(ii) Define transition element complex.
[1]
(iii) Explain why transition elements form complexes.
[1]
(b) The formulae of six complexes are given in Table 6.1.
The abbreviation en is used for 1,2-diaminoethane.
The abbreviation dien is used for the tridentate ligand H$_2$NCH$_2$CH$_2$NHCH$_2$CH$_2$NH$_2$.
The dien ligand forms three bonds to the gold ion in [Au(dien)(H$_2$O)$_2$Cl]$^{2+}$ and Au(dien)Cl$_3$.
These three bonds all lie in the same plane.
The CO ligand coordinates through the carbon atom in [Rh(CO)$_2$Cl$_2$$^+$].
Table 6.1
| formula | isomerism shown | geometry |
| :--- | :---: | :---: |
| $[\text{Rh}(en)_2\text(iii) Draw the three-dimensional structure of Au(dien)Cl$_3$ in the box. The dien ligand can be drawn as
$$\text{N} \xrightarrow{\cdot} \text{N}$$
[1]
(iv) Draw the three-dimensional structure of Ni(PH$_3$)$_2$Cl$_2$ in the box.
$$\text{Ni}$$
[1]
(v) One of the complexes, [Rh(en)$_2$Cl$_2$]$^+$ or [Rh(CO)$_2$Cl$_2$]$^{+}$, can exist in three isomeric forms. Identify this complex and the types of isomerism shown.
... [1](vi) Draw the three-dimensional structures of the two isomers of $[\text{Ni}(\text{H}_2\text{O})_2(\text{NH}_3)_4]^{2+}$ in the boxes and identify the type of isomerism shown.
type of isomerism shown ..... [2]
[Total: 10]7 Benzene can be used to make benzoic acid in the two-step process shown in Fig. 7.1.
(a) Give the reagents and conditions for step 1 and step 2.
step 1 ...
step 2 ... [2]
(b) Methylbenzene and benzoic acid each have five different peaks in the carbon (^{13}C) NMR spectrum.
Table 7.1
| hybridisation of the carbon atom | environment of carbon atom | example | chemical shift range /ppm |
| :--- | :--- | :--- | :--- |
| sp³ | alkyl | CH₃-, -CH₂-, -CH<, >C< | 0–50 |
| sp³ | next to alkene/arene | -C-C=C, -C-Ar | 25–50 |
| sp³ |next to carbonyl/carboxyl | C-COR, C-O₂R | 30–65 |
| sp³ | next to halogen | C-X | 30–60 |
| sp³ | next to oxygen | C-O | 50–70 |
| sp² | alkene or arene | >C=C<, c-c-c | 110–160 |
| sp² | carboxyl | R-COOH, R-COOR | 160–185 |
| sp² | carbonyl | RThe carbon (13C) NMR spectrum of benzoic acid: • has ... peak(s) in the chemical shift range of ... and • has ... peak(s) in the chemicalShift range of ... (c) (i) 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. State the reagents and conditions required to form each product. to form compound J to form compound K (ii) 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 com
8 Lactic acid, CH$_3$CH(OH)COOH, is the only monomer needed to form the polymer polylactic acid, PLA.
(a) (i) Draw a short length of the PLA polymer chain, including a minimum of two monomer residues. The methyl groups may be written as $-CH_3$ but all other bonds should be shown fully displayed.
Label one repeat unit of polylactic acid on your diagram.
[2]
(ii) 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 ..... [1]
functional group ..... [1]
(iii) Predict whether PLA is readily biodegradable. Explain your answer.
[1]
(b) The proton ($^1\text{H}$) NMR spectrum of CH$_3$CH(OH)COOH in CDC$l_3$ is shown in Fig. 8.1. The proton NMR chemical shift ranges are showTable 8.1 | environment of proton | example | chemical shift range $\delta$/ppm | | :--- | :--- | :--- | | alkane | $-CH_3$, $-CH_2-$, $>CH-$ | 0.9–1.7 | | alkyl next to C=O | $CH_3$-C=O, $-CH_2$-C=O, $>CH-C=O$ | 2.2–3.0 | | alkyl next to aromatic ring | $CH_3$-Ar, $-CH_2$-Ar, $>CH-Ar$ | 2.3–3.0 | | alkyl next到 electronegative atom | $CH_3-O$, $-CH_2-O$, $-CH_2-Cl$ | 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 | (i) Use Fig. 8.1 and Table 8.1 to complete Table 8.2. Table 8.2 | proton environment | chemical shift ($\delta$) | name of splitting pattern | | :--- | :--- | :-- | | $-COOH$ | | | | $>>CH$ | | | | $-OH
(c) An impure sample of CH$_3$CH(OH)COOH contains pentan-3-one as the only contaminant. The mixture is analysed using gas/liquid chromatography. The pentan-3-one is found to have a longer retention time than the lactic acid. (i) Explain what is meant by retention time. [1] (ii) Suggest suitable substances, or types of substances, that could be used as the mobile and stationary phases. mobile phase stationary phase [1] (iii) Describe how the percentage composition of the mixture can be determined from the gas/liquid chromatogram. [1] [Total: 12]
9. (a) State the reactants and conditions for two different types of reactions that both produce diethylamine, CH$_3$CH$_2$NHCH$_2$CH$_3$. reaction one ..... [4] reaction two ..... [4] (b) Describe the relative basicities of diethylamine, phenylamine and ammonia in aqueous solution. Explain your answer in terms of structure. least basic most basic explanation ..... [3] (c) Phenylamine reacts with HNO$_2$(aq) at 4°C to form compound P. Compound P reacts with phenol under alkaline conditions at 4°C. The product of this reaction is acidified, forming azo compound Q. Draw the structure of compound Q. Circle the azo group on your structure. State one use of an azo compound such as Q. compound Q: An azo compound can be used ..... [2]
(d) CH$_3$CH$_2$NHCH$_2$CH$_3$ reacts with ethanoyl chloride, CH$_3$COCl, to give the amide N,N-diethylethanamide, CH$_3$CON(C$_2$H$_5$)$_2$. An incomplete description of the mechanism of this reaction is shown in Fig. 9.1. Fig. 9.1 (i) Complete the mechanism in Fig. 9.1. You should include: - all relevant dipoles ($\delta$+ and $\delta$-) 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 boxtwo - 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. [4] (ii) Name this mechanism. ... [Total: 14]