科目：化学 Chemistry (9701)
年份：2024 冬季
试卷：Paper 43
总页数：19
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, 29 Organic synthesis, 30 Analytical chemistry
1. (a) Disodium phosphate, $(Na^{+})_{2}(HPO_{4}^{2-})$, reacts with an acid to form monosodium phosphate, $Na^{+}(H_{2}PO_{4}^{-})$.
(i) Identify the ions that are a conjugate acid–base pair in this reaction, using the formulae of the species involved.
conjugate acid conjugate base
[1]
(ii) Define buffer solution.
[2]
(iii) Write two equations to show how a mixture of $(Na^{+})_{2}(H PO_{4}^{2-})$ and $Na^{+}(H_{2} PO_{4}^{-})$ can act as a buffer solution.
equation 1 ....
equation 2 ... [2]
(iv) Identify one inorganic ion that acts as a buffer in blood.
[1]
(b) 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.
(i) Show that the concentration of (c) Compound E is much more soluble than magnesium hydroxide.
As saturated solution of magnesium hydroxide in water has a concentration of $1.40 \times 10^{-4} \text{ mol dm}^{-3}$ at 298 K.
Calculate the solubility product, $K_{\text{sp}}$, of magnesium hydroxide. Include units.
$$K_{\text{sp}} = \dots\dots\dots\dots\quad\text{units}\quad\dots\dots\dots...$$
(d) Explain why compound E is much more soluble than magnesium Hydroxide.
[Total: 16]2 (a) Predict and explain the variation in enthalpy change of hydration for the ions F⁻, Cl⁻, Br⁻ and I⁻. [2] (b) Fig. 2.1 shows an incomplete energy cycle involving calcium fluoride, CaF₂. Fig. 2.1 (i) Complete line D. Include state symbols. [1] (ii) 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 Fig. 2.1. process 1: Ca(s) + F₂(g) → Ca²⁺(g) + 2F⁻(g) Identify these five other enthalpy changes, using either names or symbols. [2]
(iii) Define lattice energy, $\Delta H_{\text{latt}}$.
[2]
(iv) Complete the expression to give the mathematical relationship between $\Delta H_{\text{latt}}$, of calcium fluoride and the enthalpy changes for processes 1 and 3.
$$\Delta H_{\text{latt}} =$$
[1]
(c) Use data from Table 2.1 to calculate a value for the hydration energy, $\Delta H_{\text{hyd}}$, of fluoride ions, $F^{-}(g)$.
Table 2.1
| | value/kJ mol$^{-1}$ |
| :--- | :--- |
| enthalpy change of solution of calcium fluoride, CaF$_2$(s) | +13 |
| overall enthalpy change of process 1 in Fig. 2.1 | +1395 |
| enthalpy change of formation of calcium fluoride | -1214 |
| enthalpy change of hydration of Ca$^{2+}$(g) | -1650 |
$\Delta H_{\text{hyd}} F^{-}(g) = \dots\dots\dots\dots\kappa J \text{mol}^{-1}$ [2]
(d) Define entr(e) At 298 K, the Gibbs free energy change, $\Delta G$, for the solution of compound $T$ is +6.00 kJ mol$^{-1}$.
The enthalpy change of solution, $\Delta H_{\text{sol}}$, of compound $T$ is +30.0 kJ mol$^{-1}$ at 298 K.
Calculate the value of the entropy change, $\Delta S$, for the solution of compound $T$ at 298 K.
$$\Delta S = \dots\dots\dots\dots\quad JK^{-1}mol^{-1}$$
(f) Predict whether compound $T$ becomes more or less soluble as the water is heated from 298 K to 360 K. Explain your answer.
[Total: 14]3 (a) A and B react together to give product AB.
$$A + B \rightarrow AB$$
When the concentrations of A and B are both 0.0100 mol dm$^{-3}$, the rate of formation of AB is $7.62 \times 10^{-4} \text{mol dm}^{-3}\text{s}^{-1}$. When the concentrations of A and B are both 1.0200 mol dm$^{-3}$,the rate of formation of AB is $3.05 \times 10^{-3} \text{mol dm}^{-2}\text{s}^{-1}$.
(i) Complete the three possible rate equations that are consistent with these data.
rate = ...
rate = ...
rate = ...
[2]
(ii) Choose one of the rate equations you have written in (i), and calculate the value of the rate constant, $k$. Include the units of $k$.
$$k = \text{units}$$
[2]
(iii) Explain why it is not possible to calculate a value for the half-life, $\frac{t_1}{2}$, of this reaction using the value(b) Catalysts may be homogeneous or heterogeneous.
(i) Identify two metals that act as heterogeneous catalysts in the removal of NO$_2$ from the exhaust gases of car engines.
... and ... [1]
(ii) Iron acts as a heterogeneous catalyst in the Haber process.
Describe the mode of action of this iron catalyst.
... ... ... ... ... ... ... ... ... ... ... ..... [2]
(iii) Fe$^{2+}$ ions act as a homogeneous catalyst in the reaction between I$^{-}$ (aq) and S$_2$O$_8$$^{2-}$ (aq).
Write equations for the two reactions that occur when Fe$^{2+}$ (aq) is added to a mixture of I$^{-}$ (aq) and S $_2$O$_8$ $^{2-}$ (aq).
equation 1: S$_2$O$_8$ $^{2-$} + ...
equation 2 ... [2]
(iv) Explain the difference between a homogeneous catalyst and a heterogeneous catalyst.(c) Fe$^{2+}$ ions can be oxidised to Fe$^{3+}$ ions under alkaline conditions by suitable oxidising agents.
(i) Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states.
(ii) The half-equation for the reduction of Fe$^{3+}$ under alkaline conditions, and its $E^\ominus$ value, are shown.
$$\text{Fe(OH)}_3 + \text{e}^- \rightleftharpoons \text{Fe(OH)}_2 + \text{OH}^- \quad E^\ominus = -0.56\text{ V}$$
Four more half-equations for reactions under alkaline conditions, and their $E^\ominus$ values, are shown.
$$\text{Al(OH)}_4^- + 3\text{e}^- \rightleftharrow \text{Al} + 4\text{OH}^- \quad E^\ominuss = -2.35\text{ V}$$
$$\text{ClO}^- + \text{H}_2\text{O} + 2\text{e}^- \rightlefthrow \text{Cl}^- + 2\text{OH}^- \quad E^\Omega = +0.894 Transition metal atoms and transition metal ions form complexes by combining with ligands.
(a) Explain why transition elements form complex ions.
[1]
(b) Co$^{2+}$ ions form complex ion G.
Each G ion contains two Co$^{2+}$ ions, both of which are octahedrally coordinated.
Each G ion contains one O$_2$ molecule, which donates one pair of electrons to each Co$^{2+}$ ion, and one NH$_2^-$ ion, which donates one pair of electrons到each Co$^{2+}$ ion.
The remaining ligands are NH$_3$ molecules.
(i) Deduce the formula of complex ion G. Include its overall charge.
formula of G ... [2]
(ii) The d-orbitals of the Co$^{2+}$ 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 areat a lower energy level i(c) Cadmium forms complex ion X, $[Cd(NH_3)_4]^{2+}$.
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 NH$_3$ ligands and no H$_2$O 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.
(i) Write an ionic equation for this reaction, using the formulae of the complex ions.
(ii) 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(aq) 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, $K_{\text{5 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 (a) Label one bond angle on each of complexes P, Q, R and S, and identify the size of the angle in degrees. [2] (b) Identify the shapes of complexes P, Q, R andS. P ..... [2] Q ..... [2] R ..... [2] S ..... [2] (c) Two L ligands are exchanged with two different monodentate ligands X and Y in each of complexes P, Q, R和S. Identify all the complexes which form new complexes that show geometrical isomerism. [1] (d) 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新 complexes
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6 Benzene, $C_6H_6$, reacts with chloroethane, $C_2H_5Cl$, in the presence of a suitable catalyst to form ethylbenzene, $C_6H_5C_2H_5$. In the presence of the catalyst, the ion $C_2H_5^+$ is formed. This ion reacts with benzene. (a) Complete the equation for the reaction of $C_2H_5Cl$ with this catalyst to form $C_2H_5^+$$ as one product. $$C_2H_5Cl + \rightarrow C_2H_5^+ +$$ [1] (b) Ethylbenzene reacts with more $C_2H_5Cl$, forming a mixture containing 1,2-diethylbenzene and 1,4-diethylbenzene. (i) Draw the structures of 1,2-diethylbenzone and 1,4-diethylbenzeine. 1,2-diethylbenzene 1,4-diethylbenzene [1] (ii) Explain why there is very little 1,3-diethylbenzene in the product mixture. ... [1]
(c) 1,2-diethylbenzene can be oxidised to benzene-1,2-dioic acid, $C_6H_4(COOH)_2$.
benzene-1,2-dioic酸
(i) State the reagent and conditions used for this reaction.
[1]
(ii) Complete the overall equation for this reaction.
An atom of oxygen from the oxidising agent is represented as [O].
All of the atoms in the two ethyl groups are fully oxidised in this reaction.
$$\text{+ } [\text{O}] \rightarrow C_6H_4(COOH)__2 + \text{+ } (1,2\text{-diethylbenzene})$$
[2]
(iii) Predict the number of peaks in the carbon-13 NMR spectrum of benzene-1,2-dioIC acid.(d) The proton (1H) NMR spectra of ethylbenzene, C$_6$H$_5$C$_2$H$_5$, in CDCl$_3$ and of benzene-1,2-dioic acid, C$_6$H$_4$(COOH)$_2$, in CDCl$_3$ are shown. They have not been identified. Fig. 6.1 Fig. 6.2 (i) Explain the use of CDCl$_3$, instead of CHCl$_3$, as the solvent when obtaining these spectra. [1]
(ii) Identify the substance shown by the spectrum in Fig. 6.1, and complete Table 6.1. substance ..... [3] Table 6.1 | | peak at $\delta = 1.2$ | peak at $\delta = 2.6$ | | :--- | :---: | :---: | | name of splitting pattern | | | | group responsible for peak | | | | explanation of splitting pattern | | | (iii) Identify the substance shown by the spectrum在Fig. 6.2, and complete Table 6.2. substance ..... [1] Table 6.2 | | peak at $\delta = 7.8$ | peak at $\delta = 13.1$ | | :--- | :---: | | group responsible for peak | | | (iv) When $D_2O$ is used as a solvent, the spectrum obtained is different from the spectrum in Fig. 6.2. Describe this difference and explain your answer. [1] (e) Benzene-1,2-dioic acid can be used to produce K. Suggest the name of this type of reaction. [1]
7 A reaction scheme is shown in Fig. 7.1. The reagents needed for reaction 2 and reaction 3 are stated. Reaction 5 takes place when $C_2H_5NH_2$ is mixed with compound V. No special conditions are required. Fig. 7.1 (a) Identify compound U which contains only three elements. [1] (b) Describe the reagents and conditions for reaction 1. [1] (c) Identify compound V. [1] (d) Complete the equation for reaction 3. $$CH_3COOH + SOCl_2 \rightarrow $$ [1] (e) Identify compound W. [1] (f) Describe the conditions for reaction 4. [1]
(g) Suggest the reagent needed for reaction 6. (h) Complete Table 7.1 by adding the reaction numbers, 1, 2, 3, 4, 5 and 6, to the right-hand column. Use the reaction numbers given in Fig. 7.1. Each of the numbers 1, 2, 3,
8 (a) An aqueous solution of phenol, C$_6$H$_5$OH, is acidic at 298 K.
Explain why phenol is more acidic than water.
[2]
(b) (i) Name the two products formed when phenol reacts with an excess of Br$_2$(aq).
and [1]
(ii) Draw the structures of the two isomeric organic products, with $M_r = 139$, that are formed when phenol reacts with HNO$_3$(aq) at room temperature.
[1]
(iii) Write the equation for the reaction between phenol, C$_6$H$_五$OH, and sodium metal.
[1]
(c) Phenol can be produced from phenylamine in a two-step synthesis.
$$\text{phenylamine} \xrightarrow{\text{step one}} \text{intermediate compound} \xrightarrow{\text{step two}} \text{phenol}$$
Describe the reagents and conditions needed in each step.
step one:
reagents ...
conditions ...
step two:
reagents ...
cond