An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V3) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
卷一 - 選擇題
Answer all 40 questions by choosing the single best option (A, B, C, or D) on the separate answer sheet.
40 題目 · 40 分
題目 1 · 選擇題
1 分
The Born-Haber cycle for magnesium fluoride, \(\text{MgF}_2(s)\), can be constructed using the experimental data below:
- Enthalpy of formation of \(\text{MgF}_2(s) = -1124\text{ kJ mol}^{-1}\) - Atomisation enthalpy of \(\text{Mg}(s) = +148\text{ kJ mol}^{-1}\) - First ionisation energy of \(\text{Mg}(g) = +738\text{ kJ mol}^{-1}\) - Second ionisation energy of \(\text{Mg}(g) = +1451\text{ kJ mol}^{-1}\) - Atomisation enthalpy of fluorine, \(\frac{1}{2}\text{F}_2(g) \rightarrow \text{F}(g) = +79\text{ kJ mol}^{-1}\) - First electron affinity of fluorine, \(\text{F}(g) \rightarrow \text{F}^-(g) = -328\text{ kJ mol}^{-1}\)
What is the lattice energy, \(\Delta H_{latt}^\ominus\), of \(\text{MgF}_2(s)\)?
Using the Nernst equation, \(E = E^\ominus + \frac{0.059}{z}\log_{10}[\text{ion}]\), what is the non-standard cell potential, \(E_{cell}\), of this cell?
A.\(0.46\text{ V}\)
B.\(0.37\text{ V}\)
C.\(0.40\text{ V}\)
D.\(0.55\text{ V}\)
查看答案詳解收起答案詳解
解題
1. Calculate the non-standard electrode potential for the copper half-cell: \(E_{\text{Cu}} = E^\ominus_{\text{Cu}} + \frac{0.059}{2}\log_{10}[\text{Cu}^{2+}]\) \(E_{\text{Cu}} = +0.34 + 0.0295\log_{10}(0.10) = +0.34 - 0.0295 = +0.3105\text{ V}\)
2. Calculate the non-standard electrode potential for the silver half-cell: \(E_{\text{Ag}} = E^\ominus_{\text{Ag}} + \frac{0.059}{1}\log_{10}[\text{Ag}^+]\) \(E_{\text{Ag}} = +0.80 + 0.059\log_{10}(0.010) = +0.80 - 0.118 = +0.682\text{ V}\)
1 mark for correctly applying the Nernst equation to both half-cells with correct z-values, and calculating the final cell potential as cathode potential minus anode potential.
題目 3 · 選擇題
1 分
Initial rates of reaction data were collected for the reaction:
What is the initial rate of reaction when \([\text{A}] = 0.30\text{ mol dm}^{-3}\), \([\text{B}] = 0.30\text{ mol dm}^{-3}\), and \([\text{C}] = 0.15\text{ mol dm}^{-3}\) at the same temperature?
1. Determine the order with respect to each reactant: - Comparing Exp 1 and Exp 2: \([\text{A}]\) doubles, rate quadruples (\(2.0 \times 10^{-4}\) to \(8.0 \times 10^{-4}\)). Order with respect to A = 2. - Comparing Exp 1 and Exp 3: \([\text{B}]\) doubles, rate remains constant. Order with respect to B = 0. - Comparing Exp 2 and Exp 4: \([\text{C}]\) doubles, rate doubles (\(8.0 \times 10^{-4}\) to \(16.0 \times 10^{-4}\)). Order with respect to C = 1.
2. Establish the rate equation: \(\text{rate} = k[\text{A}]^2[\text{C}]\)
1 mark for the correct stoichiometric calculation of remaining acid and conjugate base moles, and the correct application of the Henderson-Hasselbalch equation.
題目 5 · 選擇題
1 分
Which statement best explains why aqueous copper(II) ions, \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), are blue?
A.d-orbitals split into two energy levels; d-electrons absorb blue light as they are promoted to a higher energy level.
B.d-orbitals split into two energy levels; d-electrons absorb light in the orange-red region as they are promoted to a higher energy level, and the complementary blue light is transmitted.
C.d-orbitals split into two energy levels; d-electrons emit blue light as they transition from a higher to a lower energy level.
D.Water molecules coordinate to the \(\text{Cu}^{2+}\) ion and undergo excitation, releasing blue light as they return to the ground state.
查看答案詳解收起答案詳解
解題
In a transition metal complex, coordinate bonding by the ligands splits the 3d orbitals into two different energy levels. When visible light shines on the complex, an electron from the lower energy level is promoted to the higher level (d-d transition) by absorbing energy corresponding to a specific wavelength. For \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), light in the orange-red region is absorbed. The transmitted light consists of the remaining unabsorbed wavelengths, which represent the complementary colour, blue.
評分準則
1 mark for identifying that colour is due to d-d orbital splitting, absorption of orange-red light, and transmission of the complementary blue light.
題目 6 · 選擇題
1 分
Methylbenzene is subjected to two different reaction conditions:
- **Condition 1**: Boiled with chlorine gas in the presence of anhydrous iron(III) chloride catalyst, in the dark. - **Condition 2**: Boiled with chlorine gas under ultraviolet (UV) light.
Which row correctly describes the type of mechanism occurring and the major monosubstituted organic product formed for each condition?
- Under **Condition 1** (FeCl3 catalyst, dark), electrophilic substitution occurs on the aromatic ring. Because the methyl group is an electron-donating group, it is 2,4-directing. The major monosubstituted products are 2-chlorotoluene and 4-chlorotoluene. - Under **Condition 2** (UV light), free-radical substitution occurs on the aliphatic side-chain (methyl group), yielding (chloromethyl)benzene.
評分準則
1 mark for the row containing electrophilic substitution with ring chlorination for Condition 1 and free-radical substitution with side-chain chlorination for Condition 2.
題目 7 · 選擇題
1 分
What is the major ionic species of the amino acid lysine, \(\text{H}_2\text{N-(CH}_2)_4\text{-CH(NH}_2)\text{-COOH}\), present in an aqueous solution of pH 1?
Lysine contains two basic amine groups (one \(\alpha\)-amino group and one side-chain amine group) and one carboxylic acid group.
- At pH 1 (highly acidic), both amine groups (\(-\text{NH}_2\)) are protonated to form positive ammonium ions (\(-\text{NH}_3^+\)). - The carboxylic acid group (\(-\text{COOH}\)) remains fully protonated as a neutral group.
Therefore, the major ionic species is \(\text{H}_3\text{N}^+\text{-(CH}_2)_4\text{-CH(NH}_3^+\text{-)-COOH}\).
評分準則
1 mark for identifying the correct structure with both amine groups protonated and the carboxylic acid group intact.
題目 8 · 選擇題
1 分
An organic compound, Z, contains carbon, hydrogen, and one halogen atom.
In the mass spectrum of Z: - The ratio of the abundance of the molecular ion peak \([\text{M}]^+\) to the \([\text{M}+1]^+\) peak is \(100 : 6.6\). - The ratio of the abundance of the \([\text{M}]^+\) peak to the \([\text{M}+2]^+\) peak is approximately \(3 : 1\).
What is the molecular formula of Z?
A.\(\text{C}_6\text{H}_{13}\text{Cl}\)
B.\(\text{C}_6\text{H}_{13}\text{Br}\)
C.\(\text{C}_5\text{H}_{11}\text{Cl}\)
D.\(\text{C}_5\text{H}_{11}\text{Br}\)
查看答案詳解收起答案詳解
解題
1. Determine the number of carbon atoms, \(n\): \(n = \frac{100}{1.1} \times \frac{\text{abundance of } [M+1]^+}{\text{abundance of } [M]^+}\) \(n = \frac{100}{1.1} \times \frac{6.6}{100} = 6\)
2. Determine the halogen: An \([M]^+ : [M+2]^+\) abundance ratio of \(3 : 1\) is characteristic of a compound containing a single chlorine atom, due to the natural abundance ratio of \({}^{35}\text{Cl}\) and \({}^{37}\text{Cl}\).
Therefore, Z must contain 6 carbon atoms and 1 chlorine atom, matching \(\text{C}_6\text{H}_{13}\text{Cl}\).
評分準則
1 mark for calculating the correct number of carbon atoms and identifying the halogen from isotopic ratios to determine the molecular formula.
題目 9 · 選擇題
1 分
The table shows kinetic data for a reaction between reactants X and Y at constant temperature. Experiment 1: [X] = 0.10 mol dm\(^{-3}\), [Y] = 0.10 mol dm\(^{-3}\), Initial rate = 2.0 \(\times\) 10\(^{-4}\) mol dm\(^{-3}\) s\(^{-1}\). Experiment 2: [X] = 0.20 mol dm\(^{-3}\), [Y] = 0.10 mol dm\(^{-3}\), Initial rate = 4.0 \(\times\) 10\(^{-4}\) mol dm\(^{-3}\) s\(^{-1}\). Experiment 3: [X] = 0.10 mol dm\(^{-3}\), [Y] = 0.30 mol dm\(^{-3}\), Initial rate = 1.8 \(\times\) 10\(^{-3}\) mol dm\(^{-3}\) s\(^{-1}\). What is the value and units of the rate constant, k, for this reaction?
A.0.20 mol\(^{-2}\) dm\(^6\) s\(^{-1}\
B.0.20 mol\(^{-1}\) dm\(^3\) s\(^{-1}\
C.2.0 mol\(^{-2}\) dm\(^6\) s\(^{-1}\
D.2.0 mol\(^{-1}\) dm\(^3\) s\(^{-1}\
查看答案詳解收起答案詳解
解題
Step 1: Determine the order with respect to X. Comparing Exp 1 and Exp 2, [Y] is constant while [X] is doubled. The rate doubles, so the reaction is first-order with respect to X. Step 2: Determine the order with respect to Y. Comparing Exp 1 and Exp 3, [X] is constant while [Y] is tripled. The rate increases by a factor of 9, so the reaction is second-order with respect to Y. Step 3: Write the rate equation: Rate = k[X][Y]\(^2\). Step 4: Calculate k. Using Exp 1: 2.0 \(\times\) 10\(^{-4}\) = k(0.10)(0.10)\(^2\), which gives k = 0.20. Step 5: Units of k = (mol dm\(^{-3}\) s\(^{-1}\)) / (mol\(^3\) dm\(^{-9}\)) = mol\(^{-2}\) dm\(^6\) s\(^{-1}\).
評分準則
1 mark for the correct calculation of k as 0.20 and determination of the units as mol\(^{-2}\) dm\(^6\) s\(^{-1}\).
題目 10 · 選擇題
1 分
How many stereoisomers (including cis-trans and optical isomers) exist for the octahedral complex ion [Co(en)\(_2\)Cl\(_2\)]\(^+\) (where 'en' represents the bidentate ligand ethane-1,2-diamine)?
A.2
B.3
C.4
D.5
查看答案詳解收起答案詳解
解題
The complex [Co(en)\(_2\)Cl\(_2\)]\(^+\) exhibits both cis-trans and optical isomerism. The trans-isomer has a plane of symmetry, making it achiral (1 isomer). The cis-isomer lacks a plane of symmetry and is chiral, existing as a pair of non-superimposable mirror images or enantiomers (2 isomers). Total number of stereoisomers = 1 (trans) + 2 (cis-enantiomers) = 3.
評分準則
1 mark for identifying that there is 1 trans-isomer and a pair of cis-enantiomers, making 3 stereoisomers in total.
題目 11 · 選擇題
1 分
A standard silver electrode has a potential of E\(^\ominus\) = +0.80 V for the half-reaction Ag\(^+\)(aq) + e\(^-\) \(\rightleftharpoons\) Ag(s). What is the electrode potential of this half-cell at 298 K when the concentration of Ag\(^+\)(aq) is decreased to 0.010 mol dm\(^{-3}\)? Use the Nernst equation: E = E\(^\ominus\) + (0.059/z) log[M\(^{z+}\)]
A.+0.68 V
B.+0.74 V
C.+0.86 V
D.+0.92 V
查看答案詳解收起答案詳解
解題
Using the Nernst equation: E = +0.80 + (0.059/1) * log(0.010). Since log(0.010) = -2, E = 0.80 + 0.059 * (-2) = 0.80 - 0.118 = +0.682 V, which rounds to +0.68 V.
評分準則
1 mark for the correct application of the Nernst equation with z=1 and log(0.010) = -2 to get +0.68 V.
題目 12 · 選擇題
1 分
A buffer solution contains [HA] = 0.100 mol dm\(^{-3}\) of a weak acid and [A\(^-\)] = 0.200 mol dm\(^{-3}\) of its conjugate base. The acid dissociation constant, K\(_a\), of HA is 2.00 \(\times\) 10\(^{-5}\) mol dm\(^{-3}\) at 298 K. What is the pH of this buffer solution? (Assume log\(_{10}\) 2.00 = 0.301)
1 mark for calculating pK\(_a\) correctly and using the Henderson-Hasselbalch equation to find a pH of 5.00.
題目 13 · 選擇題
1 分
Which statement correctly describes and explains the trend in the lattice energy values of the Group 2 oxides, MgO, CaO, and SrO?
A.Lattice energy becomes more negative from MgO to SrO because the ionic radius of the metal ion increases.
B.Lattice energy becomes less negative from MgO to SrO because the ionic radius of the metal ion increases.
C.Lattice energy becomes more negative from MgO to SrO because the charge density of the metal ion increases.
D.Lattice energy becomes less negative from MgO to SrO because the first ionisation energy of the metal increases.
查看答案詳解收起答案詳解
解題
Lattice energy depends on ionic charge and ionic radii. Going down Group 2 from Mg\(^{2+}\) to Sr\(^{2+}\), the cationic charge remains 2+ but the ionic radius increases. Therefore, the distance between the centers of the cation and anion increases, reducing the strength of the electrostatic attraction. This causes the lattice energy to become less negative (less exothermic) down the group.
評分準則
1 mark for identifying that lattice energy becomes less negative down the group because the ionic radius of the metal ion increases.
題目 14 · 選擇題
1 分
Methylbenzene (toluene) is reacted with chlorine under two different sets of conditions. Reaction 1: chlorine gas in the presence of ultraviolet (UV) light. Reaction 2: chlorine gas in the dark, with an anhydrous AlCl\(_3\) catalyst. What are the major organic products for Reaction 1 and Reaction 2?
Reaction 1 involves free-radical substitution of the methyl side chain under UV light, producing (chloromethyl)benzene. Reaction 2 involves electrophilic aromatic substitution on the benzene ring catalyzed by AlCl\(_3\). Because the methyl group is 2,4-directing, the major products are 2-chlorotoluene and 4-chlorotoluene. This makes option A the correct answer.
評分準則
1 mark for identifying (chloromethyl)benzene as the side-chain product and 2-chlorotoluene as the ring-substituted product.
題目 15 · 選擇題
1 分
Which list shows ammonia, ethylamine, and phenylamine in order of increasing basic strength in aqueous solution (weakest base first)?
A.phenylamine < ammonia < ethylamine
B.ethylamine < ammonia < phenylamine
C.ammonia < phenylamine < ethylamine
D.phenylamine < ethylamine < ammonia
查看答案詳解收起答案詳解
解題
Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom is delocalised into the benzene ring, making it less available to accept a proton. Ethylamine is the strongest base because the ethyl group is electron-donating due to the positive inductive effect, increasing availability of the lone pair. Ammonia is intermediate. Thus, the order of increasing basic strength is phenylamine < ammonia < ethylamine.
評分準則
1 mark for the correct explanation of delocalisation and inductive effects resulting in the order phenylamine < ammonia < ethylamine.
題目 16 · 選擇題
1 分
Which statement correctly describes the reaction that occurs when phenol is treated with aqueous bromine (bromine water) at room temperature?
A.Phenol reacts rapidly without a catalyst to form a white precipitate of 2,4,6-tribromophenol.
B.Phenol reacts only in the presence of an AlBr\(_3\) catalyst to form 2-bromophenol.
C.Phenol reacts with bromine water to form a colorless solution of bromobenzene.
D.Phenol does not react with bromine water unless heated under reflux.
查看答案詳解收起答案詳解
解題
The -OH group in phenol strongly activates the benzene ring by delocalising an oxygen lone pair into the pi-system, making it highly nucleophilic. It reacts rapidly with bromine water at room temperature without any catalyst, forming a white precipitate of 2,4,6-tribromophenol and decolourising the bromine.
評分準則
1 mark for identifying that phenol reacts rapidly without a catalyst to form a white precipitate of 2,4,6-tribromophenol.
題目 17 · 選擇題
1 分
The rate equation for a reaction between \(P\) and \(Q\) is shown below.
$$\text{Rate} = k[P][Q]^2$$
In a series of experiments, the initial concentration of \(P\) is doubled and the initial concentration of \(Q\) is halved.
What is the ratio of the new initial rate of reaction to the original initial rate of reaction?
A.0.25
B.0.50
C.1.00
D.2.00
查看答案詳解收起答案詳解
解題
The rate equation is given by \(\text{Rate} = k[P][Q]^2\). Let the initial rate be \(R_1 = k[P]_0 [Q]_0^2\). If \([P]\) is doubled to \(2[P]_0\) and \([Q]\) is halved to \(0.5[Q]_0\), the new rate \(R_2\) is:
Here, \(E^\ominus = +0.34\text{ V}\), \(z = 2\) (since two electrons are transferred), and \([\text{Cu}^{2+}] = 0.010\text{ mol dm}^{-3}\).
\[E = +0.34 + \frac{0.059}{2}\log(0.010)\]
\[E = +0.34 + 0.0295(-2)\]
\[E = +0.34 - 0.059 = +0.281\text{ V}\]
Hence, the correct option is B.
評分準則
1 mark for correctly substituting \(z = 2\), \([\text{Cu}^{2+}] = 0.010\), and \(E^\ominus = +0.34\text{ V}\) into the Nernst equation, and calculating the final potential of \(+0.281\text{ V}\).
題目 20 · 選擇題
1 分
Equal amounts of four different Period 3 chlorides are added to separate, equal volumes of water at \(298\text{ K}\).
The chlorides used are: 1. \(\text{NaCl}\) 2. \(\text{MgCl}_2\) 3. \(\text{AlCl}_3\) 4. \(\text{SiCl}_4\)
Which sequence shows these chlorides in order of the decreasing pH (from highest pH to lowest pH) of their resulting mixtures?
\(\text{NaCl}\) is an ionic chloride that simply dissolves in water to form a neutral solution with pH \(\approx 7\).
\(\text{MgCl}_2\) undergoes slight hydrolysis due to the relatively high charge density of the \(\text{Mg}^{2+}\) ion, giving a weakly acidic solution (pH \(\approx 6.5\)).
\(\text{AlCl}_3\) contains the small, highly charged \(\text{Al}^{3+}\) ion which causes significant hydrolysis, giving a strongly acidic solution (pH \(\approx 3\)).
\(\text{SiCl}_4\) reacts violently and completely with water to produce hydrogen chloride gas, which dissolves to form a highly acidic solution (pH \(\approx 1\text{ to } 2\)).
Therefore, the order of decreasing pH (highest pH to lowest pH) is \(\text{NaCl} \rightarrow \text{MgCl}_2 \rightarrow \text{AlCl}_3 \rightarrow \text{SiCl}_4\).
評分準則
1 mark for correctly ordering the chlorides based on the extent of hydrolysis and the pH of their resulting solutions.
題目 21 · 選擇題
1 分
Some enthalpy data for calcium and oxygen are given below:
* Enthalpy change of atomisation of \(\text{Ca}(s) = +178\text{ kJ mol}^{-1}\) * First ionisation energy of \(\text{Ca}(g) = +590\text{ kJ mol}^{-1}\) * Second ionisation energy of \(\text{Ca}(g) = +1145\text{ kJ mol}^{-1}\) * Enthalpy change of atomisation of oxygen, \(\Delta H_{\text{at}}^\ominus [\frac{1}{2}\text{O}_2(g)] = +249\text{ kJ mol}^{-1}\) * First electron affinity of \(\text{O}(g) = -141\text{ kJ mol}^{-1}\) * Second electron affinity of \(\text{O}(g) = +798\text{ kJ mol}^{-1}\) * Standard enthalpy change of formation of \(\text{CaO}(s) = -635\text{ kJ mol}^{-1}\)
What is the lattice energy of calcium oxide, \(\text{CaO}(s)\)?
A.\(-1858\text{ kJ mol}^{-1}\)
B.\(-3205\text{ kJ mol}^{-1}\)
C.\(-3454\text{ kJ mol}^{-1}\)
D.\(-3703\text{ kJ mol}^{-1}\)
查看答案詳解收起答案詳解
解題
By applying Hess's Law to the Born-Haber cycle of \(\text{CaO}(s)\):
1 mark for correctly applying Hess's Law to the Born-Haber cycle and calculating the lattice energy as \(-3454\text{ kJ mol}^{-1}\).
題目 22 · 選擇題
1 分
Methylbenzene is treated with bromine under two different reaction conditions.
* **Reaction 1:** Methylbenzene is reacted with bromine at room temperature in the dark in the presence of an anhydrous \(\text{AlBr}_3\) catalyst. * **Reaction 2:** Methylbenzene is heated to boiling with bromine in the presence of ultraviolet (UV) light.
Which row correctly identifies the type of reaction and the major organic product formed in each case?
In **Reaction 1**, bromine reacts with methylbenzene in the presence of an \(\text{AlBr}_3\) catalyst in the dark via **electrophilic substitution**. The methyl group is an electron-donating group and is 2,4-directing, meaning the bromine electrophile (\(\text{Br}^+\)) substitutes onto the benzene ring to yield 1-bromo-4-methylbenzene (and 1-bromo-2-methylbenzene) as the major products.
In **Reaction 2**, reaction with bromine under boiling/UV conditions occurs via a **free-radical substitution** on the aliphatic methyl side-chain rather than the aromatic ring, yielding (bromomethyl)benzene.
Therefore, **Row A** is correct.
評分準則
1 mark for correctly identifying electrophilic substitution on the ring for Reaction 1 yielding 1-bromo-4-methylbenzene, and free-radical substitution on the side chain for Reaction 2 yielding (bromomethyl)benzene.
題目 23 · 選擇題
1 分
Three chlorine-containing compounds are added separately to equal volumes of aqueous silver nitrate at \(50^\circ\text{C}\).
What is the correct order of the ease of hydrolysis of these compounds, from easiest (fastest precipitate formation) to hardest (slowest or no precipitate formation)?
1. **Benzoyl chloride** is an acyl chloride. The carbonyl carbon is highly electron-deficient (due to electronegative oxygen and chlorine atoms), making it extremely susceptible to nucleophilic attack. It hydrolyses rapidly even with cold water, forming a precipitate of \(\text{AgCl}\) almost instantly.
2. **(Chloromethyl)benzene** is an alkyl chloride (specifically a benzylic halide). The carbon-chlorine bond is polar and can undergo nucleophilic substitution (hydrolysis) upon warming at \(50^\circ\text{C}\), forming a precipitate of \(\text{AgCl}\) within a few minutes.
3. **Chlorobenzene** is an aryl chloride. The lone pair of electrons on the chlorine atom overlaps with the \(\pi\)-system of the benzene ring, giving the \(\text{C}-\text{Cl}\) bond partial double-bond character. Thus, the \(\text{C}-\text{Cl}\) bond is very strong and resistant to hydrolysis, producing no precipitate at \(50^\circ\text{C}\).
Therefore, the order of ease of hydrolysis is benzoyl chloride > (chloromethyl)benzene > chlorobenzene.
評分準則
1 mark for correctly comparing the reactivity of acyl chlorides, alkyl chlorides, and aryl chlorides towards nucleophilic substitution (hydrolysis).
題目 24 · 選擇題
1 分
Phenylamine is reacted with a mixture of sodium nitrite and dilute hydrochloric acid at \(5^\circ\text{C}\) to form compound \(X\).
An alkaline solution of phenol is then added to compound \(X\) at \(5^\circ\text{C}\) to form an azo dye, compound \(Y\).
What are the formulas of compound \(X\) and compound \(Y\)?
When phenylamine reacts with nitrous acid (generated in situ from \(\text{NaNO}_2\) and \(\text{HCl}\)) at a temperature below \(10^\circ\text{C}\) (typically \(5^\circ\text{C}\)), it undergoes diazotisation to form the benzenediazonium salt, \(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-\). This is compound \(X\).
When compound \(X\) is coupled with alkaline phenol, electrophilic substitution occurs at the para-position (4-position) of the phenoxide ring relative to the \(-\text{OH}\) group, forming an orange azo dye, 4-(phenylazo)phenol, which has the formula \(\text{C}_6\text{H}_5\text{N}=\text{NC}_6\text{H}_4\text{OH}\). This is compound \(Y\).
Therefore, Row A is correct.
評分準則
1 mark for identifying the correct formulas for both the diazonium salt (Compound X) and the coupled azo dye (Compound Y).
題目 25 · 選擇題
1 分
The Born-Haber cycle for lithium oxide, \(\text{Li}_2\text{O(s)}\), includes the following steps:
- Standard enthalpy change of formation of \(\text{Li}_2\text{O(s)} = -598\text{ kJ mol}^{-1}\) - Standard enthalpy change of atomisation of lithium, \(\text{Li(s)} \rightarrow \text{Li(g)} = +159\text{ kJ mol}^{-1}\) - First ionisation energy of lithium = \(+520\text{ kJ mol}^{-1}\) - Standard enthalpy change of atomisation of oxygen, \(\frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{O(g)} = +249\text{ kJ mol}^{-1}\) - First electron affinity of oxygen = \(-141\text{ kJ mol}^{-1}\) - Second electron affinity of oxygen = \(+798\text{ kJ mol}^{-1}\)
What is the lattice energy, \(\Delta H_{\text{latt}}^\ominus\), of \(\text{Li}_2\text{O(s)}\)?
A.\(-2862\text{ kJ mol}^{-1}\)
B.\(-2183\text{ kJ mol}^{-1}\)
C.\(-1544\text{ kJ mol}^{-1}\)
D.\(-3518\text{ kJ mol}^{-1}\)
查看答案詳解收起答案詳解
解題
The Born-Haber cycle equation for \(\text{Li}_2\text{O(s)}\) is:
- Option A is correct. - Option B is incorrect because it fails to double the atomisation and ionisation energies of lithium. - Option C is incorrect due to sign errors in applying the electron affinities. - Option D represents an arithmetic error.
評分準則
1 mark: Correctly setting up the Born-Haber cycle equation taking into account the stoichiometry of lithium (factor of 2) and oxygen (no doubling needed for atomisation energy as the equation given is already per mole of O atoms), and calculating the correct value of -2862 kJ mol^-1.
題目 26 · 選擇題
1 分
The reaction between three reactants, A, B, and C, was investigated at a constant temperature. The initial rates of reaction were measured for various starting concentrations:
1. Determine the order with respect to each reactant: - Comparing Experiment 1 and 2: \([\text{B}]\) and \([\text{C}]\) are constant, \([\text{A}]\) doubles. The rate quadruples (\(1.2 \times 10^{-3}\) to \(4.8 \times 10^{-3}\)). Thus, the reaction is second order with respect to A: order = 2. - Comparing Experiment 3 and 4: \([\text{A}]\) and \([\text{B}]\) are constant, \([\text{C}]\) halves. The rate remains unchanged (\(2.4 \times 10^{-3}\)). Thus, the reaction is zero order with respect to C: order = 0. - Comparing Experiment 1 and 4: \([\text{A}]\) and \([\text{C}]\) are constant, \([\text{B}]\) doubles. The rate doubles (\(1.2 \times 10^{-3}\) to \(2.4 \times 10^{-3}\)). Thus, the reaction is first order with respect to B: order = 1.
2. Write the rate equation: \(\text{Rate} = k[\text{A}]^2[\text{B}]\)
3. Calculate the rate constant, \(k\), using the data from Experiment 1: \(k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]} = \frac{1.2 \times 10^{-3}}{(0.10)^2 \times (0.10)} = 1.2\)
1 mark: Correctly deducing the orders of reaction for A, B, and C, calculating the rate constant value of 1.2, and determining the units as dm^6 mol^-2 s^-1.
題目 27 · 選擇題
1 分
The following standard electrode potentials are given:
Which reaction will occur spontaneously under standard conditions?
A.Iron(III) ions oxidise chloride ions to chlorine gas.
B.Chlorine gas oxidises iodine to iodate(V) ions in acidic solution.
C.Iron(II) ions reduce sulfate(VI) ions to peroxodisulfate ions (\(\text{S}_2\text{O}_8^{2-}\)).
D.Iodine reduces iron(III) ions to iron(II) ions in acidic solution.
查看答案詳解收起答案詳解
解題
For a reaction to be spontaneous under standard conditions, the cell potential \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}\) must be positive (\(E^\ominus_{\text{cell}} > 0\)).
Let's evaluate each option: - Option A: \(\text{Fe}^{3+}\) oxidising \(\text{Cl}^-\). Here, \(\text{Fe}^{3+}\) is reduced (\(E^\ominus = +0.77\text{ V}\)) and \(\text{Cl}^-\) is oxidised (\(E^\ominus = +1.36\text{ V}\)). \(E^\ominus_{\text{cell}} = 0.77 - 1.36 = -0.59\text{ V}\) (non-spontaneous). - Option B: \(\text{Cl}_2\) oxidising \(\text{I}_2\) to \(\text{IO}_3^-\). Here, \(\text{Cl}_2\) is reduced (\(E^\ominus = +1.36\text{ V}\)) and \(\text{I}_2\) is oxidised to \(\text{IO}_3^-\) (\(E^\ominus = +1.19\text{ V}\)). \(E^\ominus_{\text{cell}} = 1.36 - 1.19 = +0.17\text{ V}\) (spontaneous). - Option C: \(\text{Fe}^{2+}\) reducing \(\text{SO}_4^{2-}\) to \(\text{S}_2\text{O}_8^{2-}\). Here, \(\text{SO}_4^{2-}\) is reduced to \(\text{S}_2\text{O}_8^{2-}\) (\(E^\ominus = +2.01\text{ V}\)) and \(\text{Fe}^{2+}\) is oxidised (\(E^\ominus = +0.77\text{ V}\)). Since the standard reduction potential of \(\text{S}_2\text{O}_8^{2-}/\text{SO}_4^{2-}\) is \(+2.01\text{ V}\), \(\text{S}_2\text{O}_8^{2-}\) is a very powerful oxidising agent and will spontaneously oxidise \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\). Thus, the reverse reaction has \(E^\ominus_{\text{cell}} = 0.77 - 2.01 = -1.24\text{ V}\) (non-spontaneous). - Option D: \(\text{I}_2\) reducing \(\text{Fe}^{3+}\) to \(\text{Fe}^{2+}\). Here, \(\text{Fe}^{3+}\) is reduced (\(E^\ominus = +0.77\text{ V}\)) and \(\text{I}_2\) is oxidised to \(\text{IO}_3^-\) (\(E^\ominus = +1.19\text{ V}\)). \(E^\ominus_{\text{cell}} = 0.77 - 1.19 = -0.42\text{ V}\) (non-spontaneous).
評分準則
1 mark: Correctly identifying that Option B represents a positive standard cell potential (+0.17 V) and is therefore spontaneous, while the other processes have negative cell potentials.
題目 28 · 選擇題
1 分
The stability constants, \(K_{\text{stab}}\), for two complex ions of copper(II) are given:
The target reaction is: \([\text{CuCl}_4]^{2-} + 4\text{NH}_3 + 2\text{H}_2\text{O} \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{Cl}^-\)
We can obtain this reaction by subtracting Reaction 1 from Reaction 2:
Target Reaction = Reaction 2 - Reaction 1
Therefore, the equilibrium constant \(K_c\) for this reaction is given by dividing the stability constant of the product complex (Reaction 2) by the stability constant of the starting complex (Reaction 1):
Since both stability constants have the same units (\(\text{dm}^{12}\text{ mol}^{-4}\)), the units cancel out, leaving \(K_c\) as a dimensionless value of \(2.5 \times 10^7\).
評分準則
1 mark: Identifying that the target reaction is equivalent to subtracting reaction 1 from reaction 2, and calculating the equilibrium constant Kc = K2 / K1 = 2.5 x 10^7.
What is the correct order of increasing acid strength (weakest acid first)?
A.III < I < II < IV
B.I < III < II < IV
C.IV < II < I < III
D.III < I < IV < II
查看答案詳解收起答案詳解
解題
The acidity of phenols is determined by the stability of the phenoxide ion formed upon deprotonation. Electron-withdrawing groups stabilise the negative charge on the phenoxide oxygen, increasing acidity, while electron-donating groups destabilise the charge, decreasing acidity.
- **4-Methylphenol (III)**: The methyl group is electron-donating by positive inductive effect, which destabilises the phenoxide ion relative to phenol. Thus, it is the weakest acid. - **Phenol (I)**: Has no substituent groups on the ring. - **4-Nitrophenol (II)**: The nitro group is strongly electron-withdrawing by induction and resonance, which stabilises the phenoxide ion and makes it significantly more acidic than phenol. - **2,4,6-Trichlorophenol (IV)**: The three highly electronegative chlorine atoms exert a combined powerful electron-withdrawing inductive effect, stabilising the phenoxide ion even more than a single nitro group does. Thus, IV is the strongest acid.
Therefore, the correct order is III < I < II < IV.
評分準則
1 mark: Correctly ordering the compounds by analyzing the electron-donating/withdrawing effects of the substituents on the stability of the phenoxide ion, yielding the order III < I < II < IV.
題目 30 · 選擇題
1 分
Ethanol is oxidised to ethanoic acid by acidified potassium dichromate(VI). The unbalanced ionic equation for the reaction is shown below:
3. **Equalise the number of electrons (LCM of 4 and 6 is 12):** Multiply the oxidation half-equation by 3: \(3\text{CH}_3\text{CH}_2\text{OH} + 3\text{H}_2\text{O} \rightarrow 3\text{CH}_3\text{COOH} + 12\text{H}^+ + 12\text{e}^-\)
Multiply the reduction half-equation by 2: \(2\text{Cr}_2\text{O}_7^{2-} + 28\text{H}^+ + 12\text{e}^- \rightarrow 4\text{Cr}^{3+} + 14\text{H}_2\text{O}\)
5. **Simplify by cancelling common terms:** Subtract \(12\text{H}^+\) and \(3\text{H}_2\text{O}\) from both sides: \(2\text{Cr}_2\text{O}_7^{2-} + 3\text{CH}_3\text{CH}_2\text{OH} + 16\text{H}^+ \rightarrow 4\text{Cr}^{3+} + 3\text{CH}_3\text{COOH} + 11\text{H}_2\text{O}\)
This gives the lowest whole-number coefficients. The stoichiometric coefficient for \(\text{H}^+\), \(r\), is 16.
評分準則
1 mark: Correctly balancing the full redox equation using half-equations to find the lowest whole-number coefficient for H^+ as 16.
題目 31 · 選擇題
1 分
Paracetamol (4-acetamidophenol) has the structure shown below:
A sample of paracetamol is heated under reflux with an excess of aqueous sodium hydroxide. Which pair of organic ions is present in the resulting reaction mixture?
A.\(\text{CH}_3\text{COO}^-\) and \(\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{O}^-\)
B.\(\text{CH}_3\text{COOH}\) and \(\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{OH}\)
C.\(\text{CH}_3\text{COO}^-\) and \(\text{H}_3\text{N}^+-\text{C}_6\text{H}_4-\text{OH}\)
D.\(\text{CH}_3\text{CONH}_2\) and \(\text{O}^--\text{C}_6\text{H}_4-\text{O}^-\)
查看答案詳解收起答案詳解
解題
When paracetamol is heated under reflux with aqueous sodium hydroxide, the amide linkage is cleaved by alkaline hydrolysis:
Since the reaction mixture contains an excess of the strong base \(\text{NaOH}\): 1. Ethanoic acid exists as the ethanoate anion, \(\text{CH}_3\text{COO}^-\). 2. The phenol group (\(-\text{OH}\)) on the 4-aminophenol molecule is weakly acidic and undergoes deprotonation in the presence of excess hydroxide ions to form the phenoxide ion, \(\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{O}^-\). 3. The basic amine group (\(-\text{NH}_2\)) remains unprotonated in alkaline conditions.
Therefore, the organic ions present in the mixture are \(\text{CH}_3\text{COO}^-\) and \(\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{O}^-\).
評分準則
1 mark: Correctly predicting the products of alkaline amide hydrolysis and recognizing that both the carboxylic acid and phenol functional groups will be deprotonated under strongly alkaline conditions.
題目 32 · 選擇題
1 分
An organic compound \(Y\) with molecular formula \(\text{C}_5\text{H}_{10}\text{O}\) is analysed by mass spectrometry. The mass spectrum shows a prominent peak at \(m/z = 57\) but no significant peak at \(m/z = 43\).
What is the identity of compound \(Y\)?
A.pentan-2-one
B.pentan-3-one
C.3-methylbutan-2-one
D.pentanal
查看答案詳解收起答案詳解
解題
Let's analyse the fragmentation of each possible isomer:
- **Pentan-3-one (\(\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3\))**: Fragmentation occurs primarily adjacent to the carbonyl group (\(\alpha\)-cleavage). Cleaving either \(\text{C}-\text{C}\) bond next to the \(\text{C}=\text{O}\) group results in the loss of an ethyl group (\(M - 29\)) to yield the propionyl cation, \([\text{CH}_3\text{CH}_2\text{CO}]^+\), with \(m/z = 57\). Due to the symmetry of the molecule, this is the main cleavage pathway. It is impossible to form a fragment of \(m/z = 43\) (such as \([\text{CH}_3\text{CO}]^+\)) by a single simple bond cleavage.
- **Pentan-2-one (\(\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3\))**: \(\alpha\)-cleavage on the methyl side yields the highly stable acetyl cation, \([\text{CH}_3\text{CO}]^+\) (\(m/z = 43\)), which is typically the most prominent peak in its spectrum.
- **3-Methylbutan-2-one (\(\text{(CH}_3)_2\text{CHCOCH}_3\))**: Like pentan-2-one, this contains a methyl group directly bonded to the carbonyl, yielding a major peak at \(m/z = 43\) due to the \([\text{CH}_3\text{CO}]^+\) fragment.
- **Pentanal (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}\))**: Commonly yields a strong fragment peak at \(m/z = 43\) due to the propyl cation, \([\text{C}_3\text{H}_7]^+\), formed by fragmentation of the alkyl chain.
Therefore, only pentan-3-one matches the description.
評分準則
1 mark: Explaining the fragmentation patterns of pentan-3-one to show why it produces a prominent peak at m/z = 57 and no significant peak at m/z = 43, unlike the other isomers.
題目 33 · multiple_choice
1 分
For the reaction \(2\text{X} + \text{Y} \rightarrow \text{Z}\), the rate equation is \(\text{rate} = k[\text{X}]^2[\text{Y}]\). When the concentration of \(\text{X}\) is increased by a factor of 3 and the concentration of \(\text{Y}\) is halved, by what factor does the initial rate of reaction change?
A.It is multiplied by 1.5
B.It is multiplied by 4.5
C.It is multiplied by 6.0
D.It is multiplied by 9.0
查看答案詳解收起答案詳解
解題
Using the rate equation: \(\text{rate}_1 = k[\text{X}]^2[\text{Y}]\). If \([\text{X}]\) becomes \(3[\text{X}]\) and \([\text{Y}]\) becomes \(0.5[\text{Y}]\), the new rate is \(\text{rate}_2 = k(3[\text{X}])^2(0.5[\text{Y}]) = 9 \times 0.5 \times k[\text{X}]^2[\text{Y}] = 4.5 \times \text{rate}_1\). Therefore, the initial rate of reaction is multiplied by 4.5.
評分準則
1 mark: Correctly calculates the rate change factor using the order of reaction with respect to each reactant.
題目 34 · multiple_choice
1 分
Phenylamine is reacted with nitrous acid at \(5\ ^\circ\text{C}\) to form benzenediazonium chloride. This intermediate is then reacted with alkaline phenol to produce an azo dye. Which row correctly identifies the formula of the azo dye and the type of mechanism occurring in the coupling step?
Diazotisation of phenylamine produces the benzenediazonium ion, \(\text{C}_6\text{H}_5\text{N}_2^+\). Coupling this electrophile with phenol (specifically the phenoxide ion in alkaline conditions) yields the azo dye 4-(phenyldiazenyl)phenol, which has the formula \(\text{C}_6\text{H}_5\text{N}=\text{NC}_6\text{H}_4\text{OH}\). The reaction mechanism is electrophilic substitution on the benzene ring of the phenol molecule.
評分準則
1 mark: Identifies both the correct formula of the azo dye and the electrophilic substitution mechanism.
題目 35 · multiple_choice
1 分
The standard electrode potential for the \(\text{Cu}^{2+}/\text{Cu}\) half-cell is \(E^\ominus = +0.34\text{ V}\). What is the electrode potential, \(E\), of this half-cell at \(298\text{ K}\) when the concentration of \(\text{Cu}^{2+}\text{(aq)}\) ions is \(0.010\text{ mol dm}^{-3}\)?
A.\(+0.222\text{ V}\)
B.\(+0.281\text{ V}\)
C.\(+0.399\text{ V}\)
D.\(+0.458\text{ V}\)
查看答案詳解收起答案詳解
解題
Using the Nernst equation at \(298\text{ K}\): \(E = E^\ominus + \frac{0.059}{z}\log[\text{Cu}^{2+}]\). For the reduction \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\), the number of transferred electrons \(z = 2\). Substituting the values: \(E = +0.34 + \frac{0.059}{2}\log(0.010) = +0.34 + 0.0295 \times (-2) = +0.34 - 0.059 = +0.281\text{ V}\).
評分準則
1 mark: Correct application of Nernst equation with \(z = 2\) and the given concentrations to find the electrode potential.
題目 36 · multiple_choice
1 分
Which statement correctly explains why the transition metal complex \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) is light blue whereas \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) is a very deep blue-violet?
A.The ammonia ligands cause a smaller d-orbital splitting than water ligands, meaning light of a longer wavelength is absorbed.
B.The ammonia ligands cause a larger d-orbital splitting than water ligands, meaning light of a shorter wavelength (higher frequency) is absorbed.
C.Ammonia acts as a reducing agent, converting \(\text{Cu}^{2+}\) to \(\text{Cu}^+\) which shifts the absorption spectrum.
D.The addition of ammonia increases the coordination number of the copper center, causing more d-electrons to be promoted.
查看答案詳解收起答案詳解
解題
Ammonia is a stronger-field ligand than water in the spectrochemical series. Consequently, when water ligands are replaced by ammonia ligands, the d-orbital splitting energy (\(\Delta E\)) increases. A larger \(\Delta E\) value corresponds to the absorption of shorter-wavelength, higher-energy light (yellow/orange instead of red), resulting in a complementary color of deeper blue-violet.
評分準則
1 mark: Explains the color difference based on stronger-field ligand action causing larger d-orbital splitting and absorption of higher frequency/energy light.
題目 37 · multiple_choice
1 分
Using the following thermodynamic data, what is the lattice energy, \(\Delta H_{\text{latt}}\), of lithium fluoride, \(\text{LiF(s)}\)?
- Enthalpy change of formation of \(\text{LiF(s)}\) = \(-617\text{ kJ mol}^{-1}\) - Enthalpy change of atomisation of \(\text{Li(s)}\) = \(+159\text{ kJ mol}^{-1}\) - First ionisation energy of \(\text{Li(g)}\) = \(+520\text{ kJ mol}^{-1}\) - Enthalpy change of atomisation of fluorine, \(\frac{1}{2}\text{F}_2\text{(g)} \rightarrow \text{F(g)}\) = \(+79\text{ kJ mol}^{-1}\) - First electron affinity of \(\text{F(g)}\) = \(-328\text{ kJ mol}^{-1}\)
A.\(-1047\text{ kJ mol}^{-1}\)
B.\(-889\text{ kJ mol}^{-1}\)
C.\(-447\text{ kJ mol}^{-1}\)
D.\(+1047\text{ kJ mol}^{-1}\)
查看答案詳解收起答案詳解
解題
According to the Born-Haber cycle and Hess's Law: \(\Delta H_f = \Delta H_{\text{at}}(\text{Li}) + \text{IE}_1(\text{Li}) + \Delta H_{\text{at}}(\text{F}) + \text{EA}_1(\text{F}) + \Delta H_{\text{latt}}(\text{LiF})\). Substituting the given values: \(-617 = 159 + 520 + 79 + (-328) + \Delta H_{\text{latt}}\) \(-617 = 430 + \Delta H_{\text{latt}}\) \(\Delta H_{\text{latt}} = -617 - 430 = -1047\text{ kJ mol}^{-1}\).
評分準則
1 mark: Correctly constructs the Born-Haber cycle equation and calculates the value of the lattice energy.
題目 38 · multiple_choice
1 分
How many stereoisomers exist for the compound 4-chlorohex-2-ene?
A.2
B.3
C.4
D.8
查看答案詳解收起答案詳解
解題
4-chlorohex-2-ene is \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH(Cl)}-\text{CH}_2-\text{CH}_3\). It contains: 1. A carbon-carbon double bond between C2 and C3 that can exhibit geometrical (cis-trans or E/Z) isomerism because each carbon of the double bond is bonded to two different groups (C2 has \(-\text{H}\) and \(-\text{CH}_3\); C3 has \(-\text{H}\) and \(-\text{CH(Cl)CH}_2\text{CH}_3\)). 2. One chiral carbon atom at C4, which is bonded to four different groups (\(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{CH}=\text{CHCH}_3\)). Thus, there are \(2^1 = 2\) geometrical isomers, each of which has \(2^1 = 2\) optical isomers. The total number of stereoisomers is \(2 \times 2 = 4\).
評分準則
1 mark: Correctly identifies the number of chiral centers and double bonds displaying stereoisomerism to calculate the total number of stereoisomers as 4.
題目 39 · multiple_choice
1 分
A buffer solution is prepared by mixing \(50.0\text{ cm}^3\) of \(0.100\text{ mol dm}^{-3}\) propanoic acid (\(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)) with \(50.0\text{ cm}^3\) of \(0.050\text{ mol dm}^{-3}\) sodium propanoate. What is the pH of this buffer solution?
A.4.57
B.4.87
C.5.17
D.9.43
查看答案詳解收起答案詳解
解題
The total volume of the mixture is \(100.0\text{ cm}^3\). Concentration of propanoic acid, \([\text{HA}] = 0.100 \times \frac{50.0}{100.0} = 0.050\text{ mol dm}^{-3}\). Concentration of propanoate ions, \([\text{A}^-] = 0.050 \times \frac{50.0}{100.0} = 0.025\text{ mol dm}^{-3}\). Using the Henderson-Hasselbalch equation: \(\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}\) \(\text{p}K_a = -\log(1.35 \times 10^{-5}) = 4.87\) \(\text{pH} = 4.87 + \log\left(\frac{0.025}{0.050}\right) = 4.87 + \log(0.5) = 4.87 - 0.30 = 4.57\).
評分準則
1 mark: Accurately calculates buffer concentrations and applies the Henderson-Hasselbalch equation to find pH.
題目 40 · multiple_choice
1 分
An organic compound \(\text{Y}\) contains carbon, hydrogen, and a single chlorine atom. In the mass spectrum of \(\text{Y}\), the molecular ion peak \([\text{M}]^+\) is at \(m/z = 78\) with a relative abundance of 51.0%. What is the predicted relative abundance of the \([\text{M}+2]^+\) peak, and what is the molecular formula of \(\text{Y}\)?
A.\([\text{M}+2]^+\) abundance = 17.0%; molecular formula = \(\text{C}_3\text{H}_7\text{Cl}\)
B.\([\text{M}+2]^+\) abundance = 17.0%; molecular formula = \(\text{C}_4\text{H}_9\text{Cl}\)
C.\([\text{M}+2]^+\) abundance = 51.0%; molecular formula = \(\text{C}_3\text{H}_7\text{Cl}\)
D.\([\text{M}+2]^+\) abundance = 51.0%; molecular formula = \(\text{C}_4\text{H}_9\text{Cl}\)
查看答案詳解收起答案詳解
解題
Chlorine has two main stable isotopes, \(^{35}\text{Cl}\) and \(^{37}\text{Cl}\), in an approximate ratio of 3:1. Therefore, for a compound containing a single chlorine atom, the ratio of the \([\text{M}]^+\) peak (containing \(^{35}\text{Cl}\)) to the \([\text{M}+2]^+\) peak (containing \(^{37}\text{Cl}\)) is 3:1. Given that the abundance of the \([\text{M}]^+\) peak is 51.0%, the predicted abundance of the \([\text{M}+2]^+\) peak is \(51.0\% / 3 = 17.0\%\). To find the molecular formula: the molecular mass of the \(^{35}\text{Cl}\)-containing isotopologue is 78. Subtracting the mass of one chlorine atom (35) leaves a mass of 43 for the remaining alkyl group (\(\text{C}_n\text{H}_{2n+1}\)). Since \(12n + (2n+1) = 43 \Rightarrow 14n = 42 \Rightarrow n=3\), the alkyl group is \(\text{C}_3\text{H}_7\). Thus, the molecular formula is \(\text{C}_3\text{H}_7\text{Cl}\).
評分準則
1 mark: Correctly identifies the 3:1 ratio of M to M+2 peaks for mono-chlorinated compounds and calculates the correct molecular formula based on isotopic mass subtraction.
卷二 - AS Structured
Answer all structured questions in the spaces provided. Show all calculations and diagrams.
6 題目 · 60 分
題目 1 · Short Answer & 結構題
10 分
This question is about atomic structure and ionisation energies.
(a) Define the term first ionisation energy of an element. [3]
(b) Explain why the first ionisation energy of sulfur is lower than that of phosphorus, even though sulfur has a higher nuclear charge. [3]
(c) An element, X, is in Period 3 of the Periodic Table. The successive ionisation energies (in \(\text{kJ mol}^{-1}\)) for X are given below: - \(I_1 = 1012\) - \(I_2 = 1903\) - \(I_3 = 2912\) - \(I_4 = 4957\) - \(I_5 = 6274\) - \(I_6 = 21269\) - \(I_7 = 25430\)
(i) Identify element X and justify your choice using the data. [3] (ii) Write the equation, including state symbols, for the process corresponding to the fourth ionisation energy of X. [1]
查看答案詳解收起答案詳解
解題
(a) The first ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous 1+ ions under standard conditions.
(b) Phosphorus has the outer electronic configuration \(3s^2 3p_x^1 3p_y^1 3p_z^1\) (half-filled subshell), whereas sulfur has \(3s^2 3p_x^2 3p_y^1 3p_z^1\). In sulfur, one of the \(3p\) orbitals contains a paired set of electrons. The mutual repulsion between these two paired electrons in the same orbital makes it easier to remove one of them from sulfur, despite its higher nuclear charge.
(c)(i) Element X is Phosphorus (P). This is indicated by the extremely large jump in energy between the fifth ionisation energy (\(I_5 = 6274\text{ kJ mol}^{-1}\)) and the sixth ionisation energy (\(I_6 = 21269\text{ kJ mol}^{-1}\)). This large increase shows that the sixth electron is removed from an inner shell closer to the nucleus, which experiences significantly less shielding. Thus, X has five valence electrons and belongs to Group 15. (ii) The fourth ionisation energy corresponds to removing an electron from a 3+ ion to form a 4+ ion: \(\text{X}^{3+}(\text{g}) \rightarrow \text{X}^{4+}(\text{g}) + \text{e}^-\). Since X is P, the equation is: \(\text{P}^{3+}(\text{g}) \rightarrow \text{P}^{4+}(\text{g}) + \text{e}^-\).
評分準則
(a) - Energy required to remove one mole of electrons from one mole of gaseous atoms [1] - To form one mole of gaseous 1+ ions [1] - Under standard conditions [1]
(b) - P has a half-filled p-subshell / \(3s^2 3p_x^1 3p_y^1 3p_z^1\) AND S has one paired set of electrons in a p-orbital / \(3s^2 3p_x^2 3p_y^1 3p_z^1\) [1] - In sulfur, the paired electrons in the same orbital repel each other [1] - This orbital spin-pair repulsion makes the outer electron easier to remove in S than in P [1]
(c)(i) - Identity: Phosphorus / P [1] - Jump between 5th and 6th ionisation energy [1] - Indicates 5 electrons in the outer shell (Group 15) [1]
(c)(ii) - \(\text{P}^{3+}(\text{g}) \rightarrow \text{P}^{4+}(\text{g}) + \text{e}^-\) [1] for correct species and state symbols.
題目 2 · Short Answer & 結構題
10 分
This question is about chemical energetics and enthalpy determinations.
(a) Define the standard enthalpy change of combustion, \(\Delta H_c^\ominus\). [2]
(b) A student carried out a laboratory experiment to determine the enthalpy change of combustion of propan-1-ol (\(M_r = 60.1\)). A spirit burner containing propan-1-ol was weighed and used to heat \(150.0\text{ g}\) of water in a copper calorimeter.
The following measurements were recorded: - Initial mass of burner + propan-1-ol = \(120.45\text{ g}\) - Final mass of burner + propan-1-ol = \(119.85\text{ g}\) - Initial temperature of water = \(19.5\text{ }^\circ\text{C}\) - Final temperature of water = \(48.1\text{ }^\circ\text{C}\) - Specific heat capacity of water, \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\)
(i) Calculate the heat energy released, \(q\), in kJ. [2] (ii) Calculate the number of moles of propan-1-ol burned. [1] (iii) Calculate the experimental standard enthalpy change of combustion of propan-1-ol, in \(\text{kJ mol}^{-1}\). Give your answer to 3 significant figures and include a sign. [2]
(c) Suggest two reasons, other than student misreadings, why this experimental value is significantly less exothermic than the data book value of \(-2021\text{ kJ mol}^{-1}\). Explain how these reasons lead to a smaller temperature rise. [3]
查看答案詳解收起答案詳解
解題
(a) Standard enthalpy change of combustion is the enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions (298 K, 1 atm).
(b)(i) \(\Delta T = 48.1 - 19.5 = 28.6\text{ }^\circ\text{C} = 28.6\text{ K}\) \(q = m c \Delta T = 150.0 \times 4.18 \times 28.6 = 17932.2\text{ J} = 17.93\text{ kJ}\) (or \(17.9\text{ kJ}\))
(b)(ii) Mass of propan-1-ol burned = \(120.45 - 119.85 = 0.60\text{ g}\) \(n = \frac{0.60}{60.1} = 0.009983\text{ mol}\)
(b)(iii) \(\Delta H = -\frac{q}{n} = -\frac{17.9322\text{ kJ}}{0.009983\text{ mol}} = -1796.2\text{ kJ mol}^{-1}\) Rounding to 3 significant figures gives \(-1800\text{ kJ mol}^{-1}\). (If using \(q = 17.9\text{ kJ}\), \(\Delta H = -1793 \approx -1790\text{ kJ mol}^{-1}\). Both are acceptable).
(c) Possible reasons include: 1. Heat loss to the surrounding air, copper calorimeter, or spirit burner. 2. Incomplete combustion of propan-1-ol, which produces less heat per mole than complete combustion. 3. Evaporation of the alcohol from the wick after weighing but before lighting, which overestimates the mass burned. These factors mean that less of the chemical energy released is transferred as heat to the water, resulting in a lower temperature rise than expected.
評分準則
(a) - Enthalpy change when 1 mole of substance [1] - Is burned completely in excess oxygen under standard conditions [1]
(b)(i) - \(\Delta T = 28.6\text{ K}\) used in formula [1] - \(q = 17.9\text{ kJ}\) (or \(17.93\text{ kJ}\)) [1]
(b)(iii) - Use of \(\Delta H = -\frac{q}{n}\) [1] - Value in range \(-1790\) to \(-1800\text{ kJ mol}^{-1}\) with negative sign and to 3 significant figures [1]
(c) - Any two reasons: heat loss to surroundings / incomplete combustion / evaporation of fuel [2] - Explanation: results in less heat energy being transferred to the water, leading to a smaller recorded temperature rise and a less exothermic value [1]
題目 3 · Short Answer & 結構題
10 分
This question is about reversible reactions and dynamic chemical equilibria.
(a) State two characteristics of a system in dynamic equilibrium. [2]
(b) A mixture of \(2.00\text{ mol}\) of \(\text{H}_2(\text{g})\) and \(1.50\text{ mol}\) of \(\text{I}_2(\text{g})\) is placed in a closed vessel of volume \(V\text{ dm}^3\) at a constant temperature of \(700\text{ K}\). The mixture is allowed to reach equilibrium according to the equation:
At equilibrium, \(2.40\text{ mol}\) of \(\text{HI}(\text{g})\) is found to be present.
(i) Construct an expression for the equilibrium constant, \(K_c\), for this reaction. [1] (ii) Calculate the equilibrium amounts, in moles, of \(\text{H}_2(\text{g})\) and \(\text{I}_2(\text{g})\). [2] (iii) Calculate the value of \(K_c\) at \(700\text{ K}\). Show your working. [2] (iv) Deduce the units of \(K_c\) for this reaction. [1]
(c) Predict, with a reason, the effect of increasing the pressure on the position of equilibrium at constant temperature. [2]
查看答案詳解收起答案詳解
解題
(a) Two characteristics of dynamic equilibrium are: 1. The rate of the forward reaction is equal to the rate of the backward reaction. 2. The concentrations of all reactants and products remain constant over time in a closed system.
(b)(ii) Since \(2.40\text{ mol}\) of \(\text{HI}\) is formed, \(1.20\text{ mol}\) of \(\text{H}_2\) and \(1.20\text{ mol}\) of \(\text{I}_2\) must have reacted. - Equilibrium amount of \(\text{H}_2 = 2.00 - 1.20 = 0.80\text{ mol}\) - Equilibrium amount of \(\text{I}_2 = 1.50 - 1.20 = 0.30\text{ mol}\)
Substituting these into the \(K_c\) expression: \(K_c = \frac{(2.40/V)^2}{(0.80/V)(0.30/V)} = \frac{2.40^2}{0.80 \times 0.30} = \frac{5.76}{0.24} = 24.0\)
(b)(iv) Since the total number of moles of gaseous reactants is equal to the total number of moles of gaseous products (2 on both sides), the units cancel out, so \(K_c\) has no units.
(c) Increasing the pressure will have no effect on the position of the equilibrium. This is because there is an equal number of moles of gas (2 moles) on both sides of the equation, so a change in pressure does not favour either the forward or reverse direction.
評分準則
(a) - Rate of forward reaction = rate of backward reaction [1] - Concentrations of reactants and products remain constant (in a closed system) [1]
(c) - Position of equilibrium does not change / shift [1] - There are equal numbers of moles of gas on both sides of the equation [1]
題目 4 · Short Answer & 結構題
10 分
This question is about Period 3 elements and their compounds.
(a) Sodium reacts vigorously with water. - State two observations during this reaction. [1] - Write a balanced chemical equation for this reaction, including state symbols. [2]
(b) Phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), is a molecular covalent oxide. (i) Write a balanced chemical equation for the reaction of \(\text{P}_4\text{O}_{10}\) with water. [1] (ii) State the pH of the resulting solution and classify the acid-base nature of \(\text{P}_4\text{O}_{10}\). [2]
(c) Silicon(IV) chloride, \(\text{SiCl}_4\), reacts rapidly with water. (i) Describe what is observed during the reaction of \(\text{SiCl}_4\) with water. [1] (ii) Write a balanced chemical equation for the reaction of \(\text{SiCl}_4\) with water. [1] (iii) Explain why silicon(IV) chloride hydrolyses in water whereas tetrachloromethane, \(\text{CCl}_4\), does not. [2]
查看答案詳解收起答案詳解
解題
(a) When sodium reacts with water, the following are observed: the sodium melts into a shiny ball, floats on the surface, moves rapidly, and effervescence (bubbles of hydrogen gas) is produced. Eventually, the sodium disappears. The chemical equation with state symbols is: \(2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)}\)
(b)(i) \(\text{P}_4\text{O}_{10}\text{(s)} + 6\text{H}_2\text{O(l)} \rightarrow 4\text{H}_3\text{PO}_4\text{(aq)}\) (ii) The resulting phosphoric(V) acid is a strong acid, so the pH of the solution is approximately 1 to 2. Hence, \(\text{P}_4\text{O}_{10}\) is classified as an acidic oxide.
(c)(i) Vigorous reaction with the evolution of dense white fumes (of \(\text{HCl}\)) and the formation of a white precipitate/solid (silicon dioxide). (ii) \(\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(g or aq)}\) (iii) Silicon has empty, low-lying \(3d\) orbitals in its valence shell. These vacant orbitals can accept a lone pair of electrons from water molecules to initiate hydrolysis. Carbon (in \(\text{CCl}_4\)) does not have vacant \(d\)-orbitals in its outer shell (valence shell \(n = 2\)), and is also sterically shielded by the chlorine atoms, so it cannot undergo hydrolysis.
評分準則
(a) - Any two observations: sodium melts into a ball, floats, moves around, effervescence/fizzing, gets smaller/disappears [1] - Balanced chemical equation: \(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\) [1] - State symbols: \(\text{(s)}\), \(\text{(l)}\), \(\text{(aq)}\), \(\text{(g)}\) [1]
(c)(iii) - Silicon has vacant \(3d\) orbitals [1] - Carbon does not have available \(d\)-orbitals [1]
題目 5 · Short Answer & 結構題
10 分
This question is about alkenes and addition reactions.
(a) Alkenes react with hydrogen halides via electrophilic addition. (i) Define the term electrophile. [1] (ii) Propene reacts with hydrogen bromide, \(\text{HBr}\), to yield a major and a minor product. Identify the major organic product and draw its structural formula. [1] (iii) Outline the mechanism for the reaction of propene with \(\text{HBr}\) to form the major product. Include curly arrows to show the movement of electron pairs, and indicate all relevant dipoles and charges. [4]
(b) Explain, in terms of the stability of the carbocation intermediates, why the major product is formed in preference to the minor product. [2]
(c) Propene can also undergo polymerisation. (i) State the type of polymerisation reaction that propene undergoes. [1] (ii) Draw the repeat unit of the polymer formed. [1]
查看答案詳解收起答案詳解
解題
(a)(i) An electrophile is an electron-pair acceptor (which attacks areas of high electron density). (ii) The major product is 2-bromopropane. Structural formula: \(\text{CH}_3\text{CH(Br)CH}_3\). (iii) Mechanism: - Step 1: The double bond of propene attacks the partially positive hydrogen atom of \(\text{H}^{\delta+}-\text{Br}^{\delta-}\). Draw a curly arrow from the double bond to the H of H-Br. Show the dipole on H-Br and draw a curly arrow from the H-Br bond to the Br. - Step 2: This forms a secondary carbocation intermediate, \(\text{CH}_3\text{C}^+\text{HCH}_3\), and a bromide ion, \(\text{Br}^-\). - Step 3: Draw a curly arrow from a lone pair on \(\text{Br}^-\) to the carbon atom with the positive charge in the carbocation to form the product, 2-bromopropane.
(b) The major product is formed via the secondary carbocation intermediate (\(\text{CH}_3\text{C}^+\text{HCH}_3\)), while the minor product is formed via a primary carbocation (\(\text{CH}_3\text{CH}_2\text{C}^+\text{H}_2\)). The secondary carbocation has two electron-donating alkyl groups (methyl groups) attached to the positively charged carbon. These groups push electron density toward the carbon atom (positive inductive effect), which disperses and stabilises the positive charge more effectively than the single alkyl group in the primary carbocation.
(c)(i) Addition polymerisation. (ii) The repeat unit of poly(propene) is \(\text{-[CH(CH}_3\text{)-CH}_2\text{]-}\) with open-ended bonds through brackets.
評分準則
(a)(i) - Electron-pair acceptor [1]
(a)(ii) - 2-bromopropane / correct structural/displayed formula of 2-bromopropane [1]
(a)(iii) - Curly arrow from the \(\text{C}=\text{C}\) double bond to the H of H-Br [1] - Correct dipole on \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) and curly arrow from the H-Br bond to the Br atom [1] - Correct structure of the secondary carbocation \(\text{CH}_3\text{C}^+\text{HCH}_3\) and \(\text{Br}^-\)[1] - Curly arrow from a lone pair on the bromide ion to the positive carbon atom of the carbocation [1]
(b) - Secondary carbocation is more stable than primary carbocation [1] - Due to the electron-donating inductive effect of two alkyl (methyl) groups dispersing the positive charge (compared to only one in the primary) [1]
(c)(i) - Addition (polymerisation) [1]
(c)(ii) - Correct structure of repeat unit showing continuation bonds / brackets: \(\text{-[CH(CH}_3\text{)-CH}_2\text{]-}\) [1]
題目 6 · Short Answer & 結構題
10 分
This question is about the carbonyl compounds propanal and propanone.
(a) Propanal and propanone are structural isomers. (i) State the specific type of structural isomerism shown by propanal and propanone. [1] (ii) Give a chemical test, including reagent and observations, that can be used to distinguish between propanal and propanone. [3]
(b) Propanal reacts with hydrogen cyanide, \(\text{HCN}\), in the presence of a sodium cyanide, \(\text{NaCN}\), catalyst. (i) Identify the mechanism of this reaction. [1] (ii) Draw the structural formula of the organic product formed. [1] (iii) The product molecule exhibits optical isomerism. Explain why the product exhibits optical isomerism, and why the reaction mixture produced is optically inactive. [2]
(c) Propanone can be reduced to an alcohol. - State the necessary reagent and conditions for this reduction. [1] - State the IUPAC name of the organic product formed. [1]
查看答案詳解收起答案詳解
解題
(a)(i) Functional group isomerism. (ii) Reagent: Tollens' reagent (ammoniacal silver nitrate solution). - Observation with propanal: Silver mirror formed on the walls of the tube / grey precipitate. - Observation with propanone: No change / solution remains colourless. (Alternatively, Fehling's solution: propanal turns from blue solution to red precipitate; propanone remains blue. Or acidified potassium dichromate(VI): propanal turns from orange to green; propanone remains orange).
(b)(i) Nucleophilic addition. (ii) The addition of HCN to propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) yields 2-hydroxybutanenitrile. Structural formula: \(\text{CH}_3\text{CH}_2\text{CH(OH)CN}\). (iii) The product molecule contains a chiral centre (C2 carbon atom is bonded to four different groups: \(-\text{H}\), \(-\text{OH}\), \(-\text{CN}\), and \(-\text{CH}_2\text{CH}_3\)), allowing for optical isomerism. The reaction mixture is optically inactive because the reactant's carbonyl group is planar. The nucleophile (\(\text{CN}^-\)) can attack the planar carbon atom from above or below with equal probability. This yields a racemic mixture (equal/equimolar amounts of both enantiomers), and the optical rotations of the enantiomers cancel each other out.
(c) Reagent: \(\text{NaBH}_4\) in aqueous/alcoholic solution (or \(\text{LiAlH}_4\) in dry ether, or \(\text{H}_2\) with nickel catalyst). Name of product: propan-2-ol.
評分準則
(a)(i) - Functional group isomerism [1]
(a)(ii) - Tollens' reagent (or Fehling's solution / acidified potassium dichromate(VI)) [1] - Positive result with propanal: silver mirror / red precipitate / green solution [1] - Negative result with propanone: remains colourless / remains blue / remains orange [1]
(b)(iii) - Carbon has 4 different groups attached / is chiral [1] - Planar carbonyl group leads to equal probability of attack from above or below, forming a racemic mixture [1]
(c) - Reagent and conditions: \(\text{NaBH}_4\) in aqueous / alcohol solution (or \(\text{LiAlH}_4\) in dry ether) [1] - Product name: propan-2-ol [1]
Paper 3 - Practical
Complete all practical tasks, recording quantitative results with appropriate precision. Answer qualitative tests using the analytical tables provided.
3 題目 · 40 分
題目 1 · practical
12 分
In this experiment, you will determine the number of water molecules of crystallisation, \(x\), in a sample of hydrated iron(II) sulfate, \(\text{FeSO}_4 \cdot x\text{H}_2\text{O}\), by redox titration with potassium manganate(VII).
FA 1 is a solution prepared by dissolving \(13.90\text{ g}\) of the hydrated iron(II) sulfate in dilute sulfuric acid and making up to \(1.00\text{ dm}^3\) with distilled water. FA 2 is \(0.0100\text{ mol dm}^{-3}\) potassium manganate(VII), \(\text{KMnO}_4\).
Method 1. Pipette \(25.0\text{ cm}^3\) of FA 1 into a clean conical flask. 2. Add about \(10\text{ cm}^3\) of \(1.0\text{ mol dm}^{-3}\) sulfuric acid using a measuring cylinder. 3. Titrate this solution with FA 2 until a permanent pale pink colour is obtained. 4. Record your results in a suitable table below and calculate the mean titre.
Results Assume that your average concordant titre of FA 2 is \(25.00\text{ cm}^3\).
Calculations Show all your working. State your answers to 3 significant figures where appropriate.
(a) Present your titration results in a table, showing initial and final burette readings and volume of FA 2 added for rough and accurate runs, with appropriate headings and units. Calculate your mean titre. (b) Calculate the number of moles of \(\text{MnO}_4^-\right.\) ions in your mean titre. (c) Use the ionic equation below to determine the number of moles of \(\text{Fe}^{2+}\) ions present in \(25.0\text{ cm}^3\) of FA 1: \(5\text{Fe}^{2+}(\text{aq}) + \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\) (d) Calculate the concentration of \(\text{Fe}^{2+}\) ions in \(\text{mol dm}^{-3}\) in FA 1. (e) Calculate the formula mass (\(M_r\)) of the hydrated iron(II) sulfate, assuming the sample was pure, and use this to determine the integer value of \(x\). (f) Calculate the percentage error in measuring the volume of FA 1 using the \(25.0\text{ cm}^3\) pipette (absolute error of pipette = \(\pm 0.06\text{ cm}^3\)).
查看答案詳解收起答案詳解
解題
(a) A suitable titration table is drawn with appropriate column headings (Initial burette reading / \(\text{cm}^3\), Final burette reading / \(\text{cm}^3\), Titre / \(\text{cm}^3\)). All readings are recorded to \(0.05\text{ cm}^3\). Two concordant titres within \(0.10\text{ cm}^3\) are used to calculate the mean titre of \(25.00\text{ cm}^3\). (b) Moles of \(\text{MnO}_4^-\right.\) = \(0.0100\text{ mol dm}^{-3} \times (25.00 / 1000)\text{ dm}^3 = 2.50 \times 10^{-4}\text{ mol}\). (c) From the equation, \(1\text{ mol}\) of \(\text{MnO}_4^-\right.\) reacts with \(5\text{ mol}\) of \(\text{Fe}^{2+}\). Moles of \(\text{Fe}^{2+}\) = \(5 \times 2.50 \times 10^{-4}\text{ mol} = 1.25 \times 10^{-3}\text{ mol}\). (d) Concentration of \(\text{Fe}^{2+}\) in FA 1 = \(1.25 \times 10^{-3}\text{ mol} / 0.0250\text{ dm}^3 = 0.0500\text{ mol dm}^{-3}\). (e) Concentration of hydrated salt = \(13.90\text{ g dm}^{-3}\). \(M_r = 13.90\text{ g dm}^{-3} / 0.0500\text{ mol dm}^{-3} = 278\text{ g mol}^{-1}\). \(M_r(\text{FeSO}_4) = 55.8 + 32.1 + (4 \times 16.0) = 151.9\text{ g mol}^{-1}\). Mass of water in 1 mole of salt = \(278 - 151.9 = 126.1\text{ g}\). \(x = 126.1 / 18.0 = 7.01 \approx 7\). (f) Percentage error = \((0.06 / 25.0) \times 100\% = 0.240\%\).
評分準則
M1: Presentation of titration results (Table format with correct headings and units, e.g., / \(\text{cm}^3\)) [1] M2: Burette readings recorded to \(0.05\text{ cm}^3\) precision for all accurate runs [1] M3: Concordancy of titres (two titres within \(0.10\text{ cm}^3\)) [1] M4: Correct calculation of mean titre showing selected values [1] M5: Correct calculation of moles of \(\text{MnO}_4^-\right.\): \(2.50 \times 10^{-4}\text{ mol}\) [1] M6: Moles of \(\text{Fe}^{2+}\) correctly calculated by multiplying M5 by 5: \(1.25 \times 10^{-3}\text{ mol}\) [1] M7: Concentration of \(\text{Fe}^{2+}\) correctly calculated: \(0.0500\text{ mol dm}^{-3}\) [1] M8: Molar mass of \(\text{FeSO}_4 \cdot x\text{H}_2\text{O}\) calculated as \(278\text{ g mol}^{-1}\) (allow ECF) [1] M9: Accurate calculation of \(126.1\text{ g}\) water per mole of salt [1] M10: Correct integer value \(x = 7\) obtained with showing appropriate working [1] M11: Percentage error calculated correctly to 2 or 3 sig figs: \(0.24\%\) or \(0.240\%\) [1] M12: All calculations (b to e) shown to 3 significant figures [1]
題目 2 · practical
12 分
In this experiment, you will determine the enthalpy change of hydration of anhydrous copper(II) sulfate, \(\text{CuSO}_4\), to form hydrated copper(II) sulfate, \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\).
FA 1 is anhydrous copper(II) sulfate powder, \(\text{CuSO}_4\). FA 2 is hydrated copper(II) sulfate crystals, \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\).
Method Experiment 1: Enthalpy change of solution of anhydrous copper(II) sulfate 1. Measure \(50.0\text{ cm}^3\) of distilled water into a polystyrene cup. 2. Measure and record the initial temperature of the water to the nearest \(0.1\text{ }^\circ\text{C}\). 3. Weigh a sample bottle containing \(4.79\text{ g}\) of FA 1. 4. Add FA 1 to the water in the cup, stir continuously, and record the maximum temperature reached. 5. Weigh the empty bottle.
Experiment 2: Enthalpy change of solution of hydrated copper(II) sulfate 1. Measure \(50.0\text{ cm}^3\) of distilled water into a polystyrene cup. 2. Measure and record the initial temperature of the water to the nearest \(0.1\text{ }^\circ\text{C}\). 3. Weigh a sample bottle containing \(7.49\text{ g}\) of FA 2. 4. Add FA 2 to the water, stir, and record the minimum temperature reached. 5. Weigh the empty bottle.
Results Assume the following experimental results are obtained: - Experiment 1: Mass of FA 1 used = \(4.79\text{ g}\); initial temperature = \(21.5\text{ }^\circ\text{C}\); maximum temperature = \(30.8\text{ }^\circ\text{C}\). - Experiment 2: Mass of FA 2 used = \(7.49\text{ g}\); initial temperature = \(22.0\text{ }^\circ\text{C}\); minimum temperature = \(20.2\text{ }^\circ\text{C}\).
Calculations Assume the specific heat capacity of the solution is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\) and its density is \(1.00\text{ g cm}^{-3}\).
(a) Present the results of both experiments clearly in a single table, showing the masses used, temperature changes, and appropriate units. (b) Calculate the heat energy change, \(q\), in joules for both Experiment 1 and Experiment 2. (c) Calculate the number of moles of copper(II) sulfate dissolved in both experiments. (d) Calculate the enthalpy change of solution for anhydrous \(\text{CuSO}_4\) (\(\Delta H_1\)) and for hydrated \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) (\(\Delta H_2\)) in \(\text{kJ mol}^{-1}\). Include the correct signs. (e) Use Hess's Law and your values from (d) to calculate the enthalpy change of hydration, \(\Delta H_{\text{hydration}}\), for the reaction: \(\text{CuSO}_4(\text{s}) + 5\text{H}_2\text{O}(\text{l}) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(\text{s})\).
查看答案詳解收起答案詳解
解題
(a) Result Table: Experiment 1: Mass of FA 1 = \(4.79\text{ g}\), \(\Delta T_1 = +9.3\text{ }^\circ\text{C}\). Experiment 2: Mass of FA 2 = \(7.49\text{ g}\), \(\Delta T_2 = -1.8\text{ }^\circ\text{C}\).
(b) Heat energy calculations: \(q = m \cdot c \cdot \Delta T\) For Experiment 1: \(q_1 = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 9.3\text{ K} = 1943.7\text{ J}\) For Experiment 2: \(q_2 = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 1.8\text{ K} = 376.2\text{ J}\)
(c) Moles calculated: For FA 1 (\(\text{CuSO}_4\)): \(M_r = 159.6\text{ g mol}^{-1}\) Moles = \(4.79\text{ g} / 159.6\text{ g mol}^{-1} = 0.0300\text{ mol}\) For FA 2 (\(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)): \(M_r = 249.6\text{ g mol}^{-1}\) Moles = \(7.49\text{ g} / 249.6\text{ g mol}^{-1} = 0.0300\text{ mol}\)
M1: Results for both experiments presented in a single, clear table with headings and units (e.g., mass / g, temperature / °C) [1] M2: All temperature readings recorded to \(0.1\text{ }^\circ\text{C}\) and masses to \(0.01\text{ g}\) [1] M3: Correct calculation of temperature changes (\(\Delta T_1 = +9.3\text{ }^\circ\text{C}\) and \(\Delta T_2 = -1.8\text{ }^\circ\text{C}\)) [1] M4: Heat energy for Exp 1 calculated correctly: \(1943.7\text{ J}\) (or \(1.94\text{ kJ}\)) [1] M5: Heat energy for Exp 2 calculated correctly: \(376.2\text{ J}\) (or \(0.376\text{ kJ}\)) [1] M6: Calculation of molar mass of \(\text{CuSO}_4\) (\(159.6\)) and \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\) (\(249.6\)) [1] M7: Correct calculation of moles for both samples (both equal \(0.0300\text{ mol}\)) [1] M8: Correct calculation of \(\Delta H_1\) with negative sign (\(-64.8\text{ kJ mol}^{-1}\)) [1] M9: Correct calculation of \(\Delta H_2\) with positive sign (\(+12.5\text{ kJ mol}^{-1}\)) [1] M10: Construction or description of a correct energy cycle relating the three enthalpy terms [1] M11: Enthalpy of hydration calculated correctly with negative sign: \(-77.3\text{ kJ mol}^{-1}\) (allow ECF) [1] M12: Final answers for enthalpy changes reported to 3 significant figures [1]
題目 3 · practical
16 分
FA 6 is an aqueous solution containing two cations and two anions, all of which are listed in the standard Qualitative Analysis Notes. Carry out the following tests and record your observations. (a) Carry out these four tests and record your detailed observations: (i) Test 1: To a 1 cm depth of FA 6 in a test-tube, add aqueous sodium hydroxide dropwise until in excess. Then, warm the mixture gently and test any gas evolved using damp red litmus paper. (ii) Test 2: To a 1 cm depth of FA 6 in a test-tube, add aqueous ammonia dropwise until in excess. (iii) Test 3: To a 1 cm depth of FA 6 in a test-tube, add a 1 cm depth of aqueous barium nitrate, followed by dilute nitric acid. (iv) Test 4: To a 1 cm depth of FA 6 in a test-tube, add a 1 cm depth of aqueous silver nitrate, followed by aqueous ammonia. (b) Identify the two cations and two anions present in FA 6. For each ion, state its formula and provide the specific evidence from your observations in (a) that supports your identification.
查看答案詳解收起答案詳解
解題
Observations: Test 1: Blue precipitate is formed, which is insoluble in excess sodium hydroxide. On warming, a gas is evolved that turns damp red litmus paper blue. Test 2: Blue precipitate is formed, which dissolves in excess ammonia to give a deep yellow-brown/brown solution. Test 3: White precipitate is formed, which remains insoluble on addition of dilute nitric acid. Test 4: White precipitate is formed, which dissolves completely upon the addition of aqueous ammonia. Identification and Evidence: - Cation 1 is cobalt(II), \(\mathrm{Co}^{2+}\): Supported by the formation of a blue precipitate with both sodium hydroxide and ammonia, which is insoluble in excess sodium hydroxide but dissolves in excess ammonia to form a brown solution. - Cation 2 is ammonium, \(\mathrm{NH}_4^+\): Supported by the production of an alkaline gas (ammonia) that turns damp red litmus paper blue when warmed with sodium hydroxide. - Anion 1 is sulfate, \(\mathrm{SO}_4^{2-}\): Supported by the formation of a white precipitate with aqueous barium nitrate that does not dissolve in dilute nitric acid. - Anion 2 is chloride, \(\mathrm{Cl}^-\): Supported by the formation of a white precipitate with aqueous silver nitrate that dissolves on adding aqueous ammonia.
評分準則
Total Marks: 16. Observations (8 marks): - M1 (Test 1): Blue precipitate with aqueous sodium hydroxide [1] - M2 (Test 1): Blue precipitate is insoluble in excess AND gas evolved on warming turns damp red litmus paper blue [1] - M3 (Test 2): Blue precipitate with aqueous ammonia [1] - M4 (Test 2): Blue precipitate dissolves in excess aqueous ammonia to form a brown / yellow-brown / dark solution [1] - M5 (Test 3): White precipitate with aqueous barium nitrate [1] - M6 (Test 3): White precipitate is insoluble in dilute nitric acid [1] - M7 (Test 4): White precipitate with aqueous silver nitrate [1] - M8 (Test 4): White precipitate dissolves in aqueous ammonia [1] Ion Identification and Evidence (8 marks): - M9: Identifies \(\mathrm{Co}^{2+}\) [1] - M10: Evidence for \(\mathrm{Co}^{2+}\): Blue precipitate with NaOH/ammonia, dissolving in excess ammonia to a brown solution [1] - M11: Identifies \(\mathrm{NH}_4^+\) [1] - M12: Evidence for \(\mathrm{NH}_4^+\): Alkaline gas/ammonia evolved on warming with NaOH [1] - M13: Identifies \(\mathrm{SO}_4^{2-}\) [1] - M14: Evidence for \(\mathrm{SO}_4^{2-}\): White precipitate with Ba2+(aq) insoluble in dilute HNO3 [1] - M15: Identifies \(\mathrm{Cl}^-\) [1] - M16: Evidence for \(\mathrm{Cl}^-\): White precipitate with Ag+(aq) soluble in ammonia [1]
Paper 4 - A Level Structured
Answer all structured A Level chemistry questions. Meticulous working and logical mechanism steps are expected.
10 題目 · 100 分
題目 1 · structured
10 分
The reaction between peroxodisulfate(VI) ions, \(\text{S}_2\text{O}_8^{2-}\), and iodide ions, \(\text{I}^-\), is represented by the following equation:
Explain how this mechanism is consistent with your rate equation. [2]
(e) State how the rate constant, \(k\), changes, if at all, when the temperature is increased. [1]
查看答案詳解收起答案詳解
解題
(a) The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the experimental rate equation.
(b) Comparing Run 1 and Run 2: \([\text{S}_2\text{O}_8^{2-}]\) is kept constant. \([\text{I}^-]\) is doubled (from \(0.0200\) to \(0.0400 \text{ mol dm}^{-3}\)), and the initial rate doubles (from \(1.50 \times 10^{-5}\) to \(3.00 \times 10^{-5} \text{ mol dm}^{-3}\text{ s}^{-1}\)). Thus, the order of reaction with respect to \(\text{I}^-\) is 1. Comparing Run 2 and Run 3: \([\text{I}^-]\) is kept constant. \([\text{S}_2\text{O}_8^{2-}]\) is doubled (from \(0.0100\) to \(0.0200 \text{ mol dm}^{-3}\)), and the initial rate doubles (from \(3.00 \times 10^{-5}\) to \(6.00 \times 10^{-5} \text{ mol dm}^{-3}\text{ s}^{-1}\)). Thus, the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) is 1.
(c) The rate equation is: \[ \text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-] \] Using data from Run 1 to calculate \(k\): \[ 1.50 \times 10^{-5} = k (0.0100)(0.0200) \] \[ k = \frac{1.50 \times 10^{-5}}{2.00 \times 10^{-4}} = 0.0750 \text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1} \] Units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)
(d) The rate of the reaction is determined by the slowest step (the rate-determining step), which is Step 1. The rate-determining step involves one \(\text{S}_2\text{O}_8^{2-}\) ion and one \(\text{I}^-\) ion. This matches the experimentally determined rate equation where both reactants have an order of 1.
(e) The rate constant, \(k\), increases when the temperature is increased.
評分準則
(a) [1] For stating that it is the power to which the concentration of a reactant is raised in the rate equation.
(b) [1] For deduction of 1st order with respect to \(\text{I}^-\), citing comparison between Run 1 and 2. [1] For deduction of 1st order with respect to \(\text{S}_2\text{O}_8^{2-}\), citing comparison between Run 2 and 3. [1] For clear mathematical showing of the proportional relationships (e.g. \(2^x = 2\)).
(d) [1] For stating that Step 1 is the slowest step / rate-determining step. [1] For linking the reactants in Step 1 (one of each species) to the orders in the rate equation (first order for both).
(e) [1] For stating that \(k\) increases.
題目 2 · structured
10 分
Cobalt forms several different complex ions.
(a) Define the terms: (i) *Ligand* [1] (ii) *Coordination number* [1]
(b) An aqueous solution of cobalt(II) chloride is pink due to the presence of octahedral \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) ions. When excess concentrated hydrochloric acid is added, the solution turns a deep blue color as the tetrahedral complex ion \([\text{CoCl}_4]^{2-}\) is formed. (i) Write a balanced equation for this ligand substitution reaction. [1] (ii) Explain, in terms of ligand size, why there is a change in the coordination number of cobalt in this reaction. [1] (iii) Describe how the split d-orbitals differ in energy level arrangement between an octahedral complex like \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) and a tetrahedral complex like \([\text{CoCl}_4]^{2-}\). [2]
(c) Cobalt(III) forms an octahedral complex with the bidentate ligand ethane-1,2-diamine, \(\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2\) (abbreviated as 'en'). (i) Complete the formula of the complex ion containing cobalt(III) and three 'en' ligands, including its charge. [1] (ii) State the type of stereoisomerism exhibited by this complex ion, and draw 3D representations of both isomers to show their relationship. [3]
查看答案詳解收起答案詳解
解題
(a) (i) A ligand is a species containing a lone pair of electrons that can form a dative/coordinate bond to a central metal atom/ion. (ii) The coordination number is the total number of coordinate bonds formed between the ligands and the central transition metal ion.
(b) (i) \([\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CoCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\) (ii) Chloride ligands, \(\text{Cl}^-\), are larger than water ligands, \(\text{H}_2\text{O}\). Therefore, fewer chloride ligands can pack around the central cobalt ion due to steric hindrance and repulsion. (iii) In an octahedral complex, the d-orbitals split into a lower energy level containing three d-orbitals (\(\text{d}_{xy}\), \(\text{d}_{yz}\), \(\text{d}_{xz}\)) and a higher energy level containing two d-orbitals (\(\text{d}_{x^2-y^2}\), \(\text{d}_{z^2}\)). In a tetrahedral complex, this splitting pattern is reversed, with two orbitals at the lower energy level and three orbitals at the higher energy level.
(c) (i) \([\text{Co}(\text{en})_3]^{3+}\) (ii) This complex exhibits optical isomerism (enantiomerism). The isomers are non-superimposable mirror images. [Drawings should show a central Co ion, with three bidentate loops forming octahedral coordination, reflecting across a vertical mirror plane.]
評分準則
(a) (i) [1] Species with a lone pair forming a dative coordinate bond to a metal ion. (ii) [1] The total number of coordinate bonds to the central metal ion.
(b) (i) [1] Balanced equation: \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\). (ii) [1] Chloride ions are larger / have more repulsion, so fewer fit around cobalt. (iii) [1] Octahedral: split into 3 lower and 2 higher d-orbitals. [1] Tetrahedral: split into 2 lower and 3 higher d-orbitals (reversed pattern).
(c) (i) [1] \([\text{Co}(\text{en})_3]^{3+}\). (ii) [1] Identifies optical isomerism. [2] Draws two 3D structures that are clearly non-superimposable mirror images of each other, displaying wedge and dashed bonds correctly for octahedral geometry.
題目 3 · structured
10 分
Standard electrode potentials can be used to predict the feasibility of redox reactions.
(a) Describe the three essential features of a Standard Hydrogen Electrode (SHE). [3]
(b) A student constructs an electrochemical cell under standard conditions. One half-cell consists of a platinum electrode in a mixture of \(1.0 \text{ mol dm}^{-3} \text{ Fe}^{3+}(\text{aq})\) and \(1.0 \text{ mol dm}^{-3} \text{ Fe}^{2+}(\text{aq})\). The other half-cell consists of a silver electrode in \(1.0 \text{ mol dm}^{-3} \text{ Ag}^+(\text{aq})\).
(i) Calculate the standard cell potential, \(E^\ominus_{\text{cell}}\). [1] (ii) Write the overall ionic equation for the spontaneous cell reaction. [1] (iii) State the direction of electron flow in the external circuit. [1]
(c) The concentration of \(\text{Fe}^{2+}(\text{aq})\) in the iron half-cell is decreased to \(0.010 \text{ mol dm}^{-3}\) while all other concentrations remain at \(1.0 \text{ mol dm}^{-3}\) at \(298 \text{ K}\). (i) State and explain the effect of this change on the electrode potential of the iron half-cell. [2] (ii) Calculate the new non-standard electrode potential, \(E\), of the iron half-cell using the Nernst equation:
\[ E = E^\ominus + \frac{0.059}{z} \log\left(\frac{[\text{oxidised}]}{[\text{reduced}]}\right) \] [2]
查看答案詳解收起答案詳解
解題
(a) The three essential features of a Standard Hydrogen Electrode are: 1. Hydrogen gas (\(\text{H}_2\)) at a pressure of \(1.0 \text{ bar}\) (or \(100 \text{ kPa}\) / \(1.0 \text{ atm}\)). 2. An aqueous solution containing hydrogen ions (\(\text{H}^+\)) at a concentration of \(1.0 \text{ mol dm}^{-3}\). 3. A platinum electrode covered in platinum black at a temperature of \(298 \text{ K}\) (\(25^\circ\text{C}\)).
(b) (i) \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.80 - (+0.77) = +0.03 \text{ V}\). (ii) \(\text{Fe}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag}(\text{s})\). (iii) Electrons flow from the more negative electrode (the iron half-cell, anode) to the more positive electrode (the silver half-cell, cathode).
(c) (i) If \([\text{Fe}^{2+}]\) is decreased, the equilibrium \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})\) shifts to the right to oppose the change. This increases the electrode potential (it becomes more positive). (ii) Using the Nernst equation: \[ E = +0.77 + \frac{0.059}{1} \log\left(\frac{1.0}{0.010}\right) \] \[ E = +0.77 + 0.059 \log(100) \] \[ E = +0.77 + 0.059(2) = +0.77 + 0.118 = +0.888 \text{ V} \approx +0.89 \text{ V} \]
評分準則
(a) [1] For hydrogen gas at \(1.0 \text{ bar}\). [1] For \(1.0 \text{ mol dm}^{-3} \text{ H}^+(\text{aq})\). [1] For Platinum electrode and temperature of \(298 \text{ K}\).
(b) (i) [1] Correct \(E^\ominus_{\text{cell}} = +0.03 \text{ V}\) (must include sign). (ii) [1] Correct ionic equation: \(\text{Fe}^{2+} + \text{Ag}^+ \rightarrow \text{Fe}^{3+} + \text{Ag}\). (iii) [1] From iron (or \(\text{Fe}^{3+}/\text{Fe}^{2+}\)) half-cell to silver half-cell.
(c) (i) [1] States that the potential becomes more positive / increases. [1] Explains by stating the equilibrium shifts to the right as \([\text{Fe}^{2+}]\) is reduced. (ii) [1] Correct substitution into the equation: \(E = 0.77 + 0.059 \log(100)\). [1] Correct final answer: \(+0.89 \text{ V}\) (or \(+0.888 \text{ V}\)).
題目 4 · structured
10 分
Propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\), is a weak monoprotic acid with \(K_{\text{a}} = 1.35 \times 10^{-5} \text{ mol dm}^{-3}\) at \(298 \text{ K}\).
(a) Write the expression for the acid dissociation constant, \(K_{\text{a}}\), of propanoic acid. [1]
(b) Calculate the \(\text{pH}\) of a \(0.150 \text{ mol dm}^{-3}\) solution of propanoic acid at \(298 \text{ K}\). State two assumptions you made in this calculation. [4]
(c) A buffer solution is prepared by mixing \(60.0 \text{ cm}^3\) of \(0.150 \text{ mol dm}^{-3}\) propanoic acid with \(40.0 \text{ cm}^3\) of \(0.100 \text{ mol dm}^{-3}\) sodium hydroxide, \(\text{NaOH}\). (i) Calculate the amount, in moles, of propanoic acid and sodium propanoate present in the resulting buffer solution. [2] (ii) Calculate the \(\text{pH}\) of this buffer solution. [2] (iii) Explain how this buffer solution is able to resist changes in pH when a small amount of dilute acid is added. [1]
(iii) When a small amount of dilute acid is added, the added \(\text{H}^+\) ions react with the conjugate base (propanoate ions) to form undissociated propanoic acid: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\). This keeps the concentration of \(\text{H}^+\) (and therefore the pH) relatively constant.
評分準則
(a) [1] Correct expression for \(K_{\text{a}}\).
(b) [1] For both correct assumptions. [1] For calculating \([\text{H}^+] = 1.423 \times 10^{-3} \text{ mol dm}^{-3}\). [1] For \(\text{pH} = 2.85\) (2 decimal places required).
(c) (i) [1] Calculating initial moles of acid (0.00900) and NaOH (0.00400). [1] Determining correct final moles in the buffer: propanoic acid = 0.00500 mol; sodium propanoate = 0.00400 mol. (ii) [1] Correct calculation of \(\text{p}K_{\text{a}} = 4.87\). [1] Correct calculation of \(\text{pH} = 4.77\). (iii) [1] Explains with a chemical equation or description that propanoate ions react with added \(\text{H}^+\).
題目 5 · structured
10 分
Benzene and methylbenzene undergo electrophilic substitution reactions.
(a) Benzene can be nitrated using a mixture of concentrated nitric acid and concentrated sulfuric acid at \(50^\circ\text{C}\). (i) Write a chemical equation for the generation of the active electrophile, \(\text{NO}_2^+\), from the acid mixture. [1] (ii) Draw the complete mechanism for the nitration of benzene by this electrophile. Include all necessary curly arrows, the positive charge on the intermediate, and the loss of the proton. [3] (iii) Write an equation to show how the sulfuric acid catalyst is regenerated. [1]
(b) Methylbenzene reacts significantly faster than benzene when nitrated. (i) Explain why the methyl group increases the rate of electrophilic substitution. [2] (ii) Draw the structural formulas of the two major organic isomers formed when methylbenzene is mononitrated. [2]
(c) State the reagent and conditions required to oxidize the methyl group of methylbenzene to form benzoic acid. [1]
查看答案詳解收起答案詳解
解題
(a) (i) \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\) (ii) Mechanism steps: 1. A curly arrow starts from the delocalised \(\pi\) ring system of benzene and points to the nitrogen atom of the \(\text{NO}_2^+\) electrophile. 2. The intermediate is a cyclohexadienyl cation with a positive charge inside a broken ring (open towards the \(\text{sp}^3\) carbon). The \(\text{sp}^3\) carbon is bonded to both a hydrogen atom and a nitro group. 3. A curly arrow starts from the \(\text{C-H}\) bond of the \(\text{sp}^3\) carbon and points back inside the broken ring to regenerate the aromatic system. (iii) \(\text{H}^+ + \text{HSO}_4^- \rightarrow \text{H}_2\text{SO}_4\)
(b) (i) The methyl group is electron-donating due to a positive inductive effect (+I). This increases the electron density of the benzene ring, making it more attractive to electrophiles. (ii) The major isomers are 2-nitromethylbenzene (2-nitrotoluene) and 4-nitromethylbenzene (4-nitrotoluene).
(c) Acidified potassium manganate(VII), \(\text{KMnO}_4\), heated under reflux.
評分準則
(a) (i) [1] Balanced equation for electrophile generation. (ii) [1] Correct curly arrow from benzene ring to \(\text{NO}_2^+\). [1] Correct structure of intermediate (horseshoe positive charge open to the \(\text{sp}^3\) carbon). [1] Correct curly arrow from \(\text{C-H}\) bond back to the ring system. (iii) [1] Equation for catalyst regeneration: \(\text{H}^+ + \text{HSO}_4^- \rightarrow \text{H}_2\text{SO}_4\).
(b) (i) [1] States that the methyl group is electron-donating / positive inductive effect. [1] Explains that this increases the electron density of the ring (making it more susceptible to electrophilic attack). (ii) [1] Correctly draws the structure of 2-nitromethylbenzene. [1] Correctly draws the structure of 4-nitromethylbenzene.
(c) [1] Potassium manganate(VII) / \(\text{KMnO}_4\) and heat / reflux.
題目 6 · structured
10 分
This question concerns nitrogen-containing organic compounds.
(a) Phenylamine, ammonia, and ethylamine are all weak bases. (i) Arrange these three bases in order of increasing basic strength (weakest base first). [1] (ii) Explain this order of basic strength in terms of their electronic structures. [3]
(b) Phenylamine can be converted into a benzenediazonium salt at low temperatures using nitrous acid, \(\text{HNO}_2\). (i) State the reagents and conditions needed to generate nitrous acid *in situ* and carry out this reaction. [2] (ii) Explain why the temperature of this reaction must be kept between \(0^\circ\text{C}\) and \(10^\circ\text{C}\). [1]
(c) The benzenediazonium salt can be reacted with an alkaline solution of phenol to form an azo dye. (i) State the type of reaction that occurs. [1] (ii) Draw the structure of the azo dye formed. [2]
查看答案詳解收起答案詳解
解題
(a) (i) Phenylamine < ammonia < ethylamine (ii) Explanation: - In phenylamine, the lone pair of electrons on the nitrogen atom is delocalised into the aromatic \(\pi\)-system. This makes the lone pair much less available to accept a proton. - In ethylamine, the ethyl group is electron-donating via a positive inductive effect. This increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton than in ammonia. - Ammonia has no electron-donating or electron-withdrawing groups attached.
(b) (i) Reagents: Sodium nitrite (\(\text{NaNO}_2\)) and dilute hydrochloric acid (\(\text{HCl}\)). Conditions: Temperature between \(0^\circ\text{C}\) and \(10^\circ\text{C}\). (ii) If the temperature is below \(0^\circ\text{C}\), the reaction is too slow. If the temperature is above \(10^\circ\text{C}\), the benzenediazonium salt is unstable and decomposes to form phenol and nitrogen gas.
(c) (i) Coupling reaction (or electrophilic substitution). (ii) The structure is 4-hydroxyphenylazobenzene: \(\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{OH}\) (with the azo group para to the OH group of the phenol ring).
評分準則
(a) (i) [1] Phenylamine < ammonia < ethylamine (correct order). (ii) [1] Phenylamine: Lone pair on N delocalised into ring, less available. [1] Ethylamine: Alkyl group is electron-donating (+I), making N lone pair more available. [1] Ammonia is intermediate because it has no reinforcing or withdrawing groups.
(b) (i) [1] Sodium nitrite / \(\text{NaNO}_2\) and dilute hydrochloric acid / \(\text{HCl}\). [1] Cool / ice bath / temperature of \(0-10^\circ\text{C}\). (ii) [1] Above \(10^\circ\text{C}\) the diazonium salt decomposes / is unstable.
(c) (i) [1] Coupling (reaction) / electrophilic substitution. (ii) [2] Correctly drawn structure of the azo dye (1 mark for azo group \(-\text{N}=\text{N}-\), 1 mark for correct links to both aromatic rings, with \(-\text{OH}\) para to the azo group).
題目 7 · structured
10 分
Acyl chlorides are versatile intermediates in organic synthesis.
(a) Explain why acyl chlorides are much more reactive towards nucleophilic substitution than alkyl chlorides. [2]
(b) Write balanced equations for the reaction of propanoyl chloride, \(\text{CH}_3\text{CH}_2\text{COCl}\), with: (i) Ethanol, and name the organic product. [2] (ii) Excess concentrated ammonia, and name the organic product. [2]
(c) The relative ease of hydrolysis of halogen-containing compounds varies dramatically. (i) Arrange the following compounds in order of their ease of hydrolysis (most rapid first):
(ii) Explain the resistance of chlorobenzene, \(\text{C}_6\text{H}_5\text{Cl}\), to hydrolysis. [2]
(d) State the reagent needed to prepare propanoyl chloride from propanoic acid. [1]
查看答案詳解收起答案詳解
解題
(a) In acyl chlorides, the carbon atom of the carbonyl group is bonded to two highly electronegative atoms (oxygen and chlorine). This draws electron density away from the carbon, making it highly electron-deficient (\(\delta+\)) and very susceptible to nucleophilic attack. Additionally, the chloride ion is a very stable leaving group.
(c) (i) \(\text{CH}_3\text{COCl} > \text{CH}_3\text{CH}_2\text{Cl} > \text{C}_6\text{H}_5\text{Cl}\) (ii) In chlorobenzene, a lone pair of electrons on the chlorine atom is delocalised into the \(\pi\)-system of the benzene ring. This gives the \(\text{C-Cl}\) bond partial double-bond character, making it much stronger and harder to break. Additionally, the high electron density of the ring repels incoming nucleophiles.
(a) [1] The carbonyl carbon is bonded to two electronegative atoms (O and Cl), making it highly electron-deficient. [1] Accept: ease of nucleophilic attack due to planar carbonyl group.
(b) (i) [1] Balanced equation producing ethyl propanoate and \(\text{HCl}\). [1] Correct name: ethyl propanoate. (ii) [1] Balanced equation producing propanamide and \(\text{NH}_4\text{Cl}\). [1] Correct name: propanamide.
(c) (i) [1] Correct order: \(\text{CH}_3\text{COCl} > \text{CH}_3\text{CH}_2\text{Cl} > \text{C}_6\text{H}_5\text{Cl}\). (ii) [1] Delocalisation of chlorine lone pair into the \(\pi\) ring system. [1] Results in partial double-bond character / stronger \(\text{C-Cl}\) bond.
(d) [1] Correct reagent: \(\text{PCl}_5\), \(\text{PCl}_3\), or \(\text{SOCl}_2\).
題目 8 · structured
10 分
Lattice energy values provide a measure of ionic bond strength.
(a) (i) Define the term *lattice energy*. [2] (ii) Explain why lattice energy cannot be measured directly. [1]
(b) Use the following thermodynamic data to construct a Born-Haber cycle and calculate the standard lattice energy, \(\Delta H^\ominus_{\text{latt}}\), of calcium chloride, \(\text{CaCl}_2(\text{s})\).
- Enthalpy change of formation of \(\text{CaCl}_2(\text{s}) = -796 \text{ kJ mol}^{-1}\) - Enthalpy change of atomisation of \(\text{Ca}(\text{s}) = +178 \text{ kJ mol}^{-1}\) - First ionisation energy of \(\text{Ca}(\text{g}) = +590 \text{ kJ mol}^{-1}\) - Second ionisation energy of \(\text{Ca}(\text{g}) = +1145 \text{ kJ mol}^{-1}\) - Bond enthalpy of \(\text{Cl-Cl}(\text{g}) = +242 \text{ kJ mol}^{-1}\) - First electron affinity of \(\text{Cl}(\text{g}) = -349 \text{ kJ mol}^{-1}\)
Show your working clearly. [4]
(c) Standard lattice energies of sodium chloride, \(\text{NaCl}\), and magnesium chloride, \(\text{MgCl}_2\), are \(-787 \text{ kJ mol}^{-1}\) and \(-2526 \text{ kJ mol}^{-1}\) respectively. Explain why the lattice energy of magnesium chloride is much more exothermic than that of sodium chloride. [3]
查看答案詳解收起答案詳解
解題
(a) (i) Lattice energy is the enthalpy change when one mole of an ionic compound is formed from its constituent gaseous ions under standard conditions. (ii) It is impossible to form a solid lattice directly from isolated gaseous ions in a single, measurable step.
Note: The standard enthalpy change of atomisation of chlorine is half the bond enthalpy of \(\text{Cl-Cl}\), which is \(+121 \text{ kJ mol}^{-1}\). The process of producing \(2\text{Cl}(\text{g})\) from \(\text{Cl}_2(\text{g})\) requires exactly the bond enthalpy of \(\text{Cl-Cl}\), which is \(+242 \text{ kJ mol}^{-1}\).
(c) The lattice energy of \(\text{MgCl}_2\) is much more exothermic because: 1. The magnesium ion, \(\text{Mg}^{2+}\), has a higher charge (\(2+\)) than the sodium ion, \(\text{Na}^+\) (\(1+\)). 2. The ionic radius of \(\text{Mg}^{2+}\) is smaller than that of \(\text{Na}^+\). 3. This results in a much higher charge density for \(\text{Mg}^{2+}\), leading to much stronger electrostatic attractions between \(\text{Mg}^{2+}\) and the chloride ions, releasing more energy when the lattice is formed.
評分準則
(a) (i) [1] Enthalpy change when one mole of an ionic compound is formed. [1] From its constituent gaseous ions. (ii) [1] It is impossible to measure the direct transition from gas-phase ions to a solid.
(b) [1] For using correct stoichiometry for chlorine atomisation (\(2 \times 121\) or simply \(242\)) and electron affinity (\(2 \times -349 = -698\)). [1] Correct algebraic setup of Hess's Law equation. [1] Correct arithmetic calculation. [1] Correct value: \(-2253 \text{ kJ mol}^{-1}\) (unit required).
(c) [1] \(\text{Mg}^{2+}\) has a higher ionic charge than \(\text{Na}^+\). [1] \(\text{Mg}^{2+}\) has a smaller ionic radius than \(\text{Na}^+\). [1] Thus, \(\text{Mg}^{2+}\) has higher charge density, leading to stronger electrostatic forces of attraction in the lattice.
題目 9 · A2 Physical, Inorganic & Organic 結構題
10 分
An electrochemical cell is set up to study the effect of concentration on electrode potential.
(a) Define the term standard electrode potential, \(E^\theta\). [2]
(b) The cell consists of a standard \(\text{Ag}^+/\text{Ag}\) electrode (\(E^\theta = +0.80\text{ V}\)) connected via a salt bridge to a non-standard \(\text{Cu}^{2+}/\text{Cu}\) electrode (\(E^\theta = +0.34\text{ V}\)). The concentration of \(\text{Cu}^{2+}(aq)\) in this half-cell is \(0.0050\text{ mol dm}^{-3}\) at \(298\text{ K}\).
(i) State the Nernst equation for the \(\text{Cu}^{2+}/\text{Cu}\) half-cell. [1]
(ii) Calculate the electrode potential, \(E\), of this non-standard \(\text{Cu}^{2+}/\text{Cu}\) electrode at \(298\text{ K}\). [2]
(iii) Calculate the cell potential, \(E_{\text{cell}}\), for this combined cell under these conditions and state which electrode acts as the anode. [2]
(c) Explain, with reference to the position of equilibrium, why the electrode potential of the copper electrode changes when the concentration of \(\text{Cu}^{2+}(aq)\) is decreased. [3]
查看答案詳解收起答案詳解
解題
(a) Standard electrode potential, \(E^\theta\), is the potential difference (voltage) of a cell consisting of a half-cell connected to a standard hydrogen electrode (SHE) under standard conditions of \(298\text{ K}\), \(1\text{ atm}\) pressure, and all solution concentrations at \(1.0\text{ mol dm}^{-3}\).
(b)(i) The Nernst equation for the copper electrode is: \(E = E^\theta + \frac{0.059}{z}\log[\text{Cu}^{2+}]\) or \(E = E^\theta + \frac{RT}{zF}\ln[\text{Cu}^{2+}]\) where \(z = 2\).
(b)(iii) The cell potential is: \(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}\) Since \(E_{\text{Ag}^+/\text{Ag}} = +0.80\text{ V}\) is more positive than \(E_{\text{Cu}^{2+}/\text{Cu}} = +0.27\text{ V}\), silver is the cathode and copper is the anode. \(E_{\text{cell}} = +0.80 - (+0.272) = +0.528\text{ V}\) (or \(+0.53\text{ V}\)). The anode is the copper electrode.
(c) The equilibrium of the half-reaction is: \(\text{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \text{Cu}(s)\) When the concentration of \(\text{Cu}^{2+}(aq)\) is decreased, according to Le Chatelier's principle, the position of equilibrium shifts to the left to oppose the change. This shift means the system is less likely to undergo reduction (or more likely to release electrons), which causes the electrode potential to become less positive (decreases).
評分準則
(a) - M1: Potential difference measured between a half-cell and a standard hydrogen electrode [1] - M2: Under standard conditions of 298 K, 1 atm, and 1.0 mol dm-3 solutions [1]
(b)(iii) - M1: Correct calculation of cell potential: \(E_{\text{cell}} = +0.53\text{ V}\) (allow \(+0.528\text{ V}\) or ecf from b(ii)) [1] - M2: Correctly identifies the copper electrode as the anode [1]
(c) - M1: Decreasing \([\text{Cu}^{2+}]\) shifts the position of equilibrium \(\text{Cu}^{2+} + 2e^- \rightleftharpoons \text{Cu}\) to the left [1] - M2: This makes the electrode more willing to release electrons / less willing to undergo reduction [1] - M3: Thus, the electrode potential decreases (becomes less positive) [1]
題目 10 · A2 Physical, Inorganic & Organic 結構題
10 分
Azo dyes are manufactured via a two-stage process. Compound X, 4-methylaniline, is converted into an intense orange azo dye, Compound Y, using phenol as a coupling agent.
(a) Stage 1 involves the conversion of 4-methylaniline into a diazonium salt.
(i) State the reagents and temperature condition required for Stage 1. [2]
(ii) Draw the structural formula of the diazonium cation intermediate formed in Stage 1. [1]
(iii) Explain why the temperature in Stage 1 must be kept below \(10\text{ }^\circ\text{C}\). [1]
(b) In Stage 2, the diazonium salt is reacted with an alkaline solution of phenol to form the azo dye, Compound Y.
(i) Draw the skeletal structure of Compound Y and state the name of the reaction type. [3]
(ii) Outline the mechanism for this coupling reaction. In your answer, show the curly arrows representing the initial attack on the phenoxide ion, the structure of the intermediate formed, and the subsequent loss of a proton to restore aromaticity. [3]
查看答案詳解收起答案詳解
解題
(a)(i) Reagents: Sodium nitrite (\(\text{NaNO}_2\)) and dilute hydrochloric acid (\(\text{HCl}\)) (or nitrous acid, \(\text{HNO}_2\)). Condition: Temperature between \(0\text{ }^\circ\text{C}\) and \(10\text{ }^\circ\text{C}\) (or ice bath).
(a)(ii) The diazonium cation is \(4\text{-CH}_3\text{-C}_6\text{H}_4\text{-N}_2^+\). Structure: A benzene ring with a methyl (\(-\text{CH}_3\)) group at position 1 and a diazonium (\(-\text{N}^+\equiv\text{N}\)) group at position 4.
(a)(iii) Above \(10\text{ }^\circ\text{C}\), the diazonium salt is unstable and decomposes to form nitrogen gas (\(\text{N}_2\)) and 4-methylphenol (or reacts with water).
(b)(i) Compound Y is formed by coupling the diazonium cation with the phenoxide ion at the para position (opposite the \(-\text{OH}\) group). Structure: \(4\text{-CH}_3\text{-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-4-OH}\). Reaction type: Electrophilic substitution (or electrophilic aromatic substitution / coupling reaction).
(b)(ii) Mechanism steps: 1. The nucleophilic phenoxide ion (specifically the carbon para to the oxygen atom) attacks the terminal nitrogen atom of the diazonium cation. A curly arrow is drawn from the C=C double bond (or the ring delocalised system) to the terminal nitrogen, and another arrow from the \(\text{N}\equiv\text{N}\) triple bond onto the adjacent nitrogen atom. 2. This forms a resonance-stabilised intermediate with a positive charge in the ring, where the hydrogen atom and the azo group are both attached to the same carbon (sp3 hybridised). 3. A curly arrow is drawn from the C-H bond back into the ring system to restore aromaticity, releasing \(\text{H}^+\).
(a)(ii) - M1: Correct structural formula showing \(4\text{-CH}_3\text{-C}_6\text{H}_4\text{-N}_2^+\) or equivalent display (positive charge must be located on the nitrogen) [1]
(a)(iii) - M1: High temperature causes the diazonium salt to decompose, releasing \(\text{N}_2\) gas [1]
(b)(i) - M1: Correct skeletal structure showing the methyl group para to the azo group, and the OH group para to the azo group on the second ring [1] - M2: Correct azo linkage (\(-\text{N=N}-\)) [1] - M3: Reaction type: Electrophilic substitution / electrophilic aromatic substitution / coupling [1]
(b)(ii) - M1: Curly arrow from the ring system of the phenoxide/phenol (para carbon) to the terminal nitrogen of the diazonium ion, and curly arrow from the triple bond to the inner nitrogen [1] - M2: Correct structure of the cationic intermediate with a positive charge in the ring and sp3 carbon showing both H and the azo group [1] - M3: Curly arrow from the C-H bond back into the ring to restore the delocalised system [1]
Paper 5 - Planning, Analysis & Evaluation
Answer all analysis and planning tasks, focusing on experimental setup, data handling, and line-of-best-fit evaluation.
2 題目 · 30 分
題目 1 · Experimental Planning & Hazard Evaluation
16 分
An elegant clock reaction can be used to study the kinetics of the reaction between peroxodisulfate ions, \(\text{S}_2\text{O}_8^{2-}\), and iodide ions, \(\text{I}^-\):
In the presence of a small, fixed amount of sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3\), and starch indicator, the iodine produced is immediately reduced back to iodide:
Once all the thiosulfate ions are consumed, the excess iodine reacts with starch, turning the solution blue-black. The time, \(t\), for this color change to occur is measured.
You are to plan an experiment to determine the activation energy, \(E_a\), of this reaction by varying the temperature, \(T\).
**(a) Hazard and Safety** (i) Potassium peroxodisulfate, \(\text{K}_2\text{S}_2\text{O}_8\), is used to prepare the reactant solution. State one hazard associated with \(\text{K}_2\text{S}_2\text{O}_8\) and a specific safety precaution (other than wearing eye protection or a laboratory coat) that must be taken. [1] (ii) Suggest why heating the reaction mixture directly using a Bunsen burner flame above \(60\ ^\circ\text{C}\) should be avoided, and suggest a safer heating method to maintain temperature control. [2]
**(b) Experimental Planning** You are required to prepare \(250\text{ cm}^3\) of a \(0.150\text{ mol dm}^{-3}\) standard solution of potassium peroxodisulfate, \(\text{K}_2\text{S}_2\text{O}_8\), from solid \(\text{K}_2\text{S}_2\text{O}_8\). (i) Calculate the mass of solid potassium peroxodisulfate, \(\text{K}_2\text{S}_2\text{O}_8\), required to prepare this solution. Show your working. [1] (ii) Describe the precise laboratory procedure you would use to prepare this \(250\text{ cm}^3\) standard solution using your calculated mass of solid. [4]
**(c) Data Analysis and Processing** The rate of the reaction is assumed to be proportional to \(1/t\). The Arrhenius equation can be expressed as:
\[\ln(1/t) = -\frac{E_a}{RT} + \text{constant}\]
(i) In one of the trials, the reaction was carried out at \(45.0\ ^\circ\text{C}\) (\(318.1\text{ K}\)) and the blue-black color appeared at \(28.0\text{ s}\). Calculate the values of \(\frac{1}{T}\) (with its appropriate unit) and \(\ln(1/t)\). State both values to 3 significant figures. [2] (ii) State the quantities that should be plotted on the x-axis and y-axis of a linear graph to determine the activation energy, \(E_a\). [1] (iii) Explain how the activation energy, \(E_a\), is calculated from the gradient of this graph. State the value and unit of the gas constant, \(R\), that should be used. [2]
**(d) Evaluation and Improvement** At higher temperatures (e.g. above \(55\ ^\circ\text{C}\)), the blue-black color appears very rapidly (\(t < 5\text{ s}\)), leading to a high percentage uncertainty in the time measured using a manual stopwatch. (i) Suggest a modification to the apparatus to obtain a more accurate start-to-finish time measurement. [1] (ii) Suggest a modification to the reaction mixture that would increase the time taken, \(t\), for the blue-black color to appear without changing the total volume of the reaction mixture. Explain your answer. [2]
查看答案詳解收起答案詳解
解題
**Part (a) Hazard and Safety** (i) Potassium peroxodisulfate, \(\text{K}_2\text{S}_2\text{O}_8\), is a strong oxidiser and skin/respiratory sensitiser. *Precaution*: Wear protective gloves (nitrile gloves) to prevent direct skin contact, or keep far away from flammable solvents and organic substances. (ii) Directly heating peroxodisulfate solution with a Bunsen flame can lead to thermal decomposition of the peroxodisulfate ions, altering reactant concentration, or lead to unsafe local boiling. A thermostatically controlled water bath (or hot water bath) provides even, controlled heating to maintain a stable, uniform temperature.
**Part (b) Experimental Planning** (i) Molar mass of \(\text{K}_2\text{S}_2\text{O}_8 = 2(39.1) + 2(32.1) + 8(16.0) = 270.4\text{ g mol}^{-1}\) \(\text{Moles required} = 0.150\text{ mol dm}^{-3} \times 0.250\text{ dm}^3 = 0.0375\text{ mol}\) \(\text{Mass required} = 0.0375\text{ mol} \times 270.4\text{ g mol}^{-1} = 10.14\text{ g}\) (allow \(10.1\text{ g}\) if using \(M_r = 270\)) (ii) Procedure: 1. Weigh \(10.14\text{ g}\) of potassium peroxodisulfate solid on a balance using a weighing bottle/boat. 2. Transfer the solid to a clean beaker and dissolve it completely in approximately \(100 - 150\text{ cm}^3\) of distilled water, stirring with a clean glass rod. 3. Pour the dissolved solution through a funnel into a clean \(250\text{ cm}^3\) volumetric flask. Rinse the beaker, glass rod, and funnel several times with small volumes of distilled water, transferring all washings quantitatively into the flask. 4. Add distilled water carefully until the level is just below the graduation mark. Use a teat pipette to add distilled water dropwise until the bottom of the meniscus is exactly aligned with the graduation mark on the neck of the flask. Stopper the flask and invert it at least 10 times to ensure thorough mixing.
**Part (d) Evaluation and Improvement** (i) Connect a colorimeter to a data logger to continuously monitor the absorbance of the solution. The data logger can capture the precise millisecond of the absorbance spike corresponding to the blue-black starch-iodine complex. (ii) Increase the volume of sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3\) solution added (and correspondingly decrease the volume of distilled water to keep the total reaction volume constant). Because more thiosulfate is present, it will take longer to be completely oxidized by the iodine produced, extending the time \(t\) before the blue-black starch-iodine color appears.
評分準則
**(a) Hazard and Safety (3 marks)** * **Mark 1**: Correct hazard-precaution pair for potassium peroxodisulfate. Acceptable hazards: oxidizer, skin/eye irritant, sensitizer. Precaution: wear protective gloves (if irritant), handle away from organic/flammable materials (if oxidizer), or avoid inhaling dust (if sensitizer). Reject general terms like "wear safety goggles/lab coat". * **Mark 2**: States that peroxodisulfate decomposes thermally at high temperatures / direct heating prevents precise temperature control. * **Mark 3**: Suggests a thermostatically controlled water bath (or hot water bath).
**(b) Experimental Planning (5 marks)** * **Mark 1**: Calculation: \(10.14\text{ g}\) (or \(10.1\text{ g}\)) with working shown (moles \(= 0.0375\), \(M_r = 270.4\)). * **Mark 2**: Dissolving step: Dissolves the weighed solid completely in distilled water in a beaker using a volume significantly less than \(250\text{ cm}^3\) (e.g. \(100 - 150\text{ cm}^3\)), stirring with a glass rod. * **Mark 3**: Quantitative transfer: Transfers solution into a \(250\text{ cm}^3\) volumetric flask and rinses the beaker, rod, and funnel with distilled water, adding all washings into the flask. * **Mark 4**: Making up to volume: Fills the volumetric flask to the graduation mark (using a pipette for the final drops) so that the bottom of the meniscus aligns with the mark. * **Mark 5**: Homogenisation: Stoppers the flask and inverts multiple times to ensure uniform concentration.
**(c) Data Analysis and Processing (5 marks)** * **Mark 1**: Calculate \(\frac{1}{T} = 3.14 \times 10^{-3}\) (or \(0.00314\)) and state unit as \(\text{K}^{-1}\). * **Mark 2**: Calculate \(\ln(1/t) = -3.33\) (must be negative, stated to 3 significant figures). * **Mark 3**: Identifies the axes correctly: \(x\)-axis is \(\frac{1}{T}\) and \(y\)-axis is \(\ln(1/t)\). * **Mark 4**: Explains that \(E_a = -\text{gradient} \times R\). * **Mark 5**: States correct constant \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\) (units required).
**(d) Evaluation and Improvement (3 marks)** * **Mark 1**: Suggests using a colorimeter (connected to a data logger/light sensor) to monitor the color change automatically. * **Mark 2**: Suggests increasing the volume/concentration of sodium thiosulfate used (or decreasing the concentration of reactants) while maintaining the same total volume. * **Mark 3**: Explains that more thiosulfate will take longer to react completely with the produced iodine, thus increasing the clock time, \(t\), before the color change occurs.
題目 2 · Data Analysis & Graphical Interpretation
14 分
An investigation is carried out into the rate of decomposition of aqueous benzenediazonium chloride, \(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-(aq) \rightarrow \text{C}_6\text{H}_5\text{OH}(aq) + \text{N}_2(g) + \text{HCl}(aq)\), at different temperatures.
The rate constant, \(k\), for this first-order reaction was determined at several temperatures, \(T\). The results of the experiment are shown in the table below.
**(a)** Calculate the missing values for **(a)** and **(b)** in the table. Ensure your answers are written to an appropriate number of significant figures and decimal places. [2]
**(b)** To determine the activation energy, \(E_a\), a graph of \(\ln(k / \text{s}^{-1})\) against \(1/T\) is plotted. (i) Identify which variable should be plotted on the x-axis and which on the y-axis. [1] (ii) Identify the anomalous data point from the table. State whether the rate constant at this temperature is faster or slower than expected. Explain your reasoning. [2]
**(c)** A line of best fit is drawn through the non-anomalous points. Two points on this line of best fit are selected: - Point 1: \((2.95 \times 10^{-3} \text{ K}^{-1}, -4.30)\) - Point 2: \((3.40 \times 10^{-3} \text{ K}^{-1}, -10.30)\)
Calculate the gradient of this line. Show your working and state the units of the gradient. [3]
**(d)** Use your gradient from **(c)** to calculate the activation energy, \(E_a\), of the reaction in \(\text{kJ mol}^{-1}\). (The gas constant, \(R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}\)). [3]
**(e)** (i) Explain why a thermostatically controlled water bath is used to heat the reaction mixtures rather than direct heating with a Bunsen burner. [1] (ii) The reaction progress can be monitored by measuring the volume of \(\text{N}_2(g)\) evolved over time. State how the total, maximum volume of gas collected at the completion of the reaction would change, if at all, if the experiment was repeated at 313 K instead of 293 K at the same pressure using the same starting amount of reactant. Explain your answer. [2]
查看答案詳解收起答案詳解
解題
**(a)** - For **(a)**: \(1/T = 1 / 303 = 3.30 \times 10^{-3} \text{ K}^{-1}\). (Must be written to 3 significant figures to match other entries in that column). - For **(b)**: \(\ln(k) = \ln(5.3 \times 10^{-4}) = -7.54\). (Must be written to 2 decimal places to match other entries in that column).
**(b)** (i) x-axis: \(1/T\) (reciprocal temperature in \(\text{K}^{-1}\)); y-axis: \(\ln(k / \text{s}^{-1})\). (ii) The anomalous point is at \(T = 323 \text{ K}\) (or \(1/T = 3.10 \times 10^{-3} \text{ K}^{-1}\), \(\ln k = -6.57\)). The rate is *slower* than expected. Reasoning: At \(1/T = 3.10 \times 10^{-3} \text{ K}^{-1}\), the line of best fit gives an expected \(\ln k\) value of approximately \(-6.30\). Since the experimental \(\ln k = -6.57\) is more negative than \(-6.30\), \(k\) is smaller than expected, meaning the reaction rate was slower.
**(d)** According to the Arrhenius equation: \(\text{gradient} = -\frac{E_a}{R}\) \(E_a = -\text{gradient} \times R = -(-13333) \times 8.31 = 110797 \text{ J mol}^{-1}\) In \(\text{kJ mol}^{-1}\): \(E_a = 111 \text{ kJ mol}^{-1}\) (accept \(110\) to \(112\) depending on rounding of gradient).
**(e)** (i) A thermostatically controlled water bath ensures a constant and uniform temperature throughout the reaction flask during the entire course of the run, whereas direct heating with a Bunsen burner causes localized hot spots and temperature fluctuations. (ii) The final volume of gas collected would *increase*. According to Charles's Law (or the ideal gas equation \(pV = nRT\)), gas volume is directly proportional to temperature in Kelvin at constant pressure. At 313 K, the same number of moles of \(\text{N}_2\) gas will occupy a larger volume than at 293 K.
評分準則
**(a)** [2 marks] - Award 1 mark for calculating **(a)** = \(3.30 \times 10^{-3}\) (must be precisely 3 sig figs, i.e., 3.30, not 0.0033 or 3.3). - Award 1 mark for calculating **(b)** = \(-7.54\) (must be 2 decimal places, i.e., -7.54, not -7.5 or -7.542).
**(b)** [3 marks] - (i) Award 1 mark for stating x-axis is \(1/T\) and y-axis is \(\ln(k)\). - (ii) Award 1 mark for identifying the anomalous point at \(T = 323 \text{ K}\) or coordinates \((3.10 \times 10^{-3}, -6.57)\). - Award 1 mark for explaining that the rate is slower than expected because the experimental \(\ln k\) value (\(-6.57\)) is more negative than the expected line-of-best-fit value (approx \(-6.30\)).
**(c)** [3 marks] - Award 1 mark for showing correct substitution into the gradient formula: \((-10.30 - (-4.30)) / ((3.40 - 2.95) \times 10^{-3})\). - Award 1 mark for the calculated value: \(-13333\) (or \(-13300\) or \(-1.33 \times 10^4\)). - Award 1 mark for correct unit: \(\text{K}\).
**(d)** [3 marks] - Award 1 mark for equating gradient to \(-E_a / R\). - Award 1 mark for calculating value in J/mol: \(110797 \text{ J mol}^{-1}\) (or value consistent with candidate's gradient). - Award 1 mark for converting to \(\text{kJ mol}^{-1}\) to 3 sig figs (e.g., \(111 \text{ kJ mol}^{-1}\) or \(110 \text{ kJ mol}^{-1}\)) and ensuring the sign is positive.
**(e)** [3 marks] - (i) Award 1 mark for stating that a water bath keeps temperature uniform / constant / prevents temperature fluctuations (do not accept safety/flammability as the primary kinetic reason, though allow as a secondary benefit if linked to temperature control). - (ii) Award 1 mark for stating the final volume of gas *increases*. - Award 1 mark for explanation: temperature is higher (313 K vs 293 K) so gas expands / volume is proportional to temperature (\(V \propto T\) or \(pV=nRT\)).
想知道自己有幾分把握?
Thinka 是 DSE 學生用的 AI 練習應用程式,有無限量練習題、即時自動批改和詳細解題步驟。逾 100,000 名學生用它確認自己真的識,而不只是「以為識」。