Questions 1
(a) Topic -8.1 Rate of reaction
(b)Topic – 8.1 Rate of reaction
(c)Topic -8.1 Rate of reaction
(d)Topic -8.1 Rate of reaction
(e)Topic -8.1 Rate of reaction
Propanone, CH₃COCH₃, reacts with iodine, I₂, in the presence of an acid catalyst.
CH₃COCH₃ + I₂ \(\to\) CH₃COCH₂I + H⁺ + I⁻
The rate equation for this reaction is shown.
\(rate = k[CH_3COCH_3][H^+]\)
(a) Complete Table 1.1 to describe the order of the reaction.
(b) An experiment is performed using a large excess of CH₃COCH₃ and a large excess of H⁺(aq). The initial concentration of I₂ is 1.00 × 10⁻⁵ moldm⁻³. The initial rate of decrease in the \(I_2\) concentration is 2.27 × 10⁻⁷ moldm⁻³ s⁻¹.
(i) Use the axes to draw a graph of [\(I_2\)] against time for the first 10 seconds of the reaction.
(ii) State whether it is possible to calculate the numerical value of the rate constant, k, for this reaction from your graph. Explain your answer.
(c) The experiment is repeated at a different temperature. The initial concentrations of \(H^+\) ions, I₂, and CH₃COCH₃ are all 0.200 mol/dm³. The value of k at this temperature is 2.31 × 10⁻⁵ mol⁻¹ dm³ s⁻¹. Calculate the initial rate of this reaction.
(d) The experiment is repeated using an excess of H+(aq). The new rate equation is shown.
rate = k₁[CH₃COCH₃]
(i) The value of k₁ is 1.1 × 10⁻³ s⁻¹. Calculate the value of the half-life, \(t_\frac{1}{2}\).
(ii) Use your answer to (i) to draw a graph of [CH₃COCH₃] against time for this reaction. The initial value of [CH₃COCH₃] on your graph should be 0.200 mol/dm³. The final value of
[CH₃COCH₃] on your graph should be 0.0250 moldm⁻³.
(e) A four-step mechanism is suggested for the overall reaction.
CH₃COCH₃ + I₂ \(\to\) CH₃COCH₂I + H⁺ + I⁻ rate = k [CH₃COCH₃][H⁺]
Part of this mechanism is shown.
(i) Write an equation for step 3.
(ii) Suggest the slowest step of the mechanism. Explain your answer.
(iii) Identify one conjugate acid-conjugate base pair in the mechanism.
conjugate acid …………………………………… conjugate base ………………………………….
▶️Answer/Explanation
Ans:
(a) 1
0 [1]
1
2 [1]
(b)(i) The line should start at 1 × 10⁻⁵ and decrease to 0.773 × 10⁻⁵/ 7.73 × 10⁻⁶ and be straight for 10 s [1]
(ii) its not possible AND the concentrations in the rate law are not known [1]
(c) 9.24 × 10⁻⁷ [1]
(d)(i) 630 [1]
(ii) Three halvings taking equal times, these times to agree with their answer to di.[1]
(e)(i) CH₃COH=CH₂ + I₂ → CH₃C⁺(OH)CH₂I + I⁻ [1]
(ii) step 1 AND has (both) substances in rate law.[1]
(iii) CH₃C+OHCH₃ and CH₃COCH₃ OR
CH₃C+(OH)CH₃ and CH₃C(OH)=CH₂ OR
CH₃C+(OH)CH₂I and CH₃COCH₂I [1]
Questions 2
(a) Topic -25.1 Acids and bases
(b)Topic – 25.1 Acids and bases
(c)Topic -25.1 Acids and bases
(d)Topic -25.1 Acids and bases
(e)Topic -25.1 Acids and bases
Benzoic acid, C₆H₅COOH, is a weak acid. The Ka of benzoic acid is 6.31 × 10⁻⁵ moldm⁻³ at 298K. A 1.00 dm³ buffer solution is made at 298 K containing 1.00 g of \(C_6H_5COOH\) and a slightly greater mass of sodium benzoate, C₆H₅COO⁻Na⁺. This buffer solution has a pH of 4.15.
(a) Define buffer solution.(b) Write equations to show how this solution acts as a buffer solution when the named substances are added to it:
(i) dilute aqueous sodium hydroxide
(ii) dilute aqueous nitric acid.
(c) Calculate the H⁺ concentration and the C₆H₅COOH concentration in the buffer solution described. Use the expression for the \(K_a\) of C₆H₅COOH to calculate the concentration of
C₆H₅COO⁻Na⁺ in the buffer solution. Show your working and give each answer to a minimum of three significant figures.
(d) A 10.0 cm³ sample of the buffer solution is mixed with 10.0 cm³ of \(1.00moldm^{–3}\) KOH. Both solutions are at 298K. A reaction is allowed to occur without stirring. Two observations are recorded:
● the temperature, after the reaction is complete, is fractionally above 298K
● the pH, after the reaction, is greater than 13.
Explain these two observations.
(e) Magnesium benzoate, Mg(C₆H₅COO)₂, has a solubility in water of less than 1.00 g/dm³ at 298K.
\(K_{sp} = [Mg^{2+}][C_6H_5COO^–]^2 = 1.76 × 10^{–7}\) at 298K
(i) Calculate the solubility of Mg(C₆H₅COO)₂ in water at 298K. Give your answer in \(gdm^{–3}\). Show your working.
\([M_r: Mg(C_6H_5COO)_2, 266.3]\)
(ii) An excess of Mg(C₆H₅COO)₂ is added to a sample of 0.50 mol/dm³ MgSO₄ at 298 K. State whether the equilibrium concentration of Mg(C₆H₅COO)₂ is higher than, the same as, or lower than your answer to (i). Explain your answer. The concentration is …………………………………………………………… the concentration in (i).
explanation …………………………………………………………………………………………………………
▶️Answer/Explanation
Ans:
(a) resists pH change when small amount of acid or alkali is added [1]
(b)(i) C₆H₅COOH + NaOH → C₆H₅COO⁻Na⁺ + H₂O [1]
(ii) C₆H₅COO⁻Na⁺ + HNO₃ → C₆H₅COOH + NaNO₃ [1]
(c) [H⁺] = 7.08 × 10⁻⁵ [1]
[C₆H₅COOH] = 8.20 × 10⁻³ [1]
[C₆H₅COO⁻Na⁺] = 7.31 × 10⁻³ [1]
(d) neutralisation is exothermic [1]
All of the C₆H₅COOH has reacted AND excess KOH 1]
(i) If [X] = [\(Mg^{2+}\)] = [\(Mg(C_6H_5COO^–)_2\)]
\(4X^3\) = 1.76 × 10⁻⁷, so X = 3.53 × 10⁻³ [1]
0.940 g dm⁻³ [1]
(ii) lower than AND common ion effect [1]
Questions 3
(a) Topic -24.1 Electrolysis
(b)Topic – 24.1 Electrolysis
(c)Topic -24.2 Standard electrode potentials, standard cell potentials and the Nernst equation
(d)Topic -24.2 Standard electrode potentials, standard cell potentials and the Nernst equation
(e)Topic -23.4 Gibbs free energy change, ΔG
(f) Topic- 23.4 Gibbs free energy change, ΔG
Some electrode potentials are shown in Table 3.1.
(a) (i) Complete the diagram to show a standard hydrogen electrode. Label your diagram. Identify all substances. You do not need to state standard conditions.
(ii) An electrochemical cell is set up using an Fe³⁺/Fe²⁺ electrode and a standard hydrogen electrode. Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit.
positive electrode ………………………………………………………………………………………………..
Electrons flow from the ………………………… electrode to the ………………………… electrode
(b) The vanadium-containing species in the electrode reactions given in Table 3.1 are \(V, V^{2+}, V^{3+}, VO^{2+}\) and \(VO_2^{+}\).
(i) Identify one vanadium-containing species that does not react with Fe²⁺ ions under standard conditions. Use data from Table 3.1 to explain your answer.
(ii) Identify all the vanadium-containing species that will react with Fe²⁺ ions under standard conditions.
(iii) Write an equation for one of the possible reactions identified in (ii)
(c) Another electrochemical cell is set up using an Fe³⁺/Fe²⁺ electrode and an alkaline ClO⁻/Cl⁻ electrode. The concentration of Fe³⁺ is 1000 times greater than the concentration of Fe²⁺+ in the Fe³⁺/Fe²⁺ electrode. All other conditions are standard.
(i) Use the Nernst equation to calculate the E value of the Fe³⁺/Fe²⁺ electrode. Show your working.
(ii) Write an equation for the reaction that occurs in the cell, under these conditions.
(d) Another electrochemical cell is set up using an Fe²⁺/Fe electrode and an alkaline ClO⁻/Cl⁻ electrode under standard conditions. Calculate the value of \(\Delta G^0\) for the cell.
(e) A solution of iron(II) sulfate, FeSO₄(aq), is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode. A current of 0.640A is passed for 17.0 minutes. The mass of the cathode increases by 0.185 g. Use these results to calculate an experimental value for the Avogadro constant, L. Show your working.
(f) Iron(II) chloride, FeCl₂, is oxidised by chlorine to form iron(III) chloride, FeCl₃, under standard conditions.
\(2FeCl_2(s) + Cl_2(g) \to 2FeCl_3(s)\) \(\Delta H^0 = –128kJmol^{–1}\)
(i) Use Table 3.2 and other data to calculate the Gibbs free energy change, \(\Delta G^0\), for this reaction. Show your working.
(ii) Predict whether this reaction becomes more or less feasible at a higher temperature. Explain your answer.
The reaction becomes ………………………… feasible.
explanation …………………………………………………………..
▶️Answer/Explanation
Ans:
(a)(i) hydrogen, delivery system, \(H^+\), platinum, [1]
(ii) iron . . . . . hydrogen . . . . . iron [1]
(b)(i) (for specified \(V^{2+}, V^{3+} or VO^{2+}\)) \(E^o\) is more positive than / above –0.44 AND more negative than / below 0.77 V [1]
(ii) V and VO₂⁺ [1]
(iii) V + Fe²⁺ → V²⁺ + Fe OR
VO₂⁺ + 2H⁺ + Fe²⁺ → VO²⁺ + H₂O + Fe³⁺ [1]
(c)(i) Nernst: E = 0.77 + (0.059 / z)log[ox] / [red] [1]
0.947 [1]
(ii) 2Fe³⁺ + Cl⁻ + 2OH⁻ → 2Fe²⁺ + ClO⁻ + H₂O [1]
(d) \(E^o_{cell}\) = 1.33 V [1]
\(\Delta G^o = –nE^o_{cell} F\) [1]
–257 [1]
(e) 0.64 × 17 × 60 = 653 / 652.8 Coulombs [1]
652.8 ÷ 1.6×10⁻¹⁹ = 4.08 × 10²¹ (number of electrons)
4.08 × 10²¹ ÷ 2 = 2.04 × 10²¹ (number of atoms Fe) [1]
0.185 ÷ 55.8 = 3.31 × 10⁻³
(number of moles Fe atoms
2.04 × 10²¹ ÷ 3.31 × 10⁻³ = L = 6.153 × 10²³ [1]
(f)(i) ΔS = –179 [1]
ΔG = ΔH – TΔS [1]
–74.7 [1]
(ii) less AND ΔG becomes more positive [1]
Questions 4
(a) Topic -7.1 Chemical equilibria: reversible reactions, dynamic equilibrium
(b)Topic – 7.1 Chemical equilibria: reversible reactions, dynamic equilibrium
(c)Topic -7.1 Chemical equilibria: reversible reactions, dynamic equilibrium
(d)Topic -7.1 Chemical equilibria: reversible reactions, dynamic equilibrium
(e)Topic -7.1 Chemical equilibria: reversible reactions, dynamic equilibrium
(f) Topic-7.1 Chemical equilibria: reversible reactions, dynamic equilibrium
The structure of the polydentate ligand, \(EDTA^{4–}\), is shown in Fig. 4.1.
The stability constants, at 298K, of five octahedral complexes are given in Table 4.1.
(a) Define stability constant.
(b) Calculate the oxidation states of Cu in \([Cu(EDTA)]^{2–}\) and Cr in \([Cr(EDTA)]^–\).
Cu …………………………
Cr …………………………
(c) Deduce the number of lone pairs donated by each \(EDTA^{4–}\) ligand in a single \([Fe(EDTA)]^{2–}\) complex ion.
(d) Identify the most stable complex in Table 4.1. Explain your choice.
(e) In a solution at equilibrium at 298K, [[Cu(H₂O)₆]²⁺] = 3.00 × 10⁻¹⁰ moldm⁻³ and [EDTA⁴⁻] = 5.00 × 10⁻¹² moldm⁻³. Use the expression for \(K_{stab}\) to calculate the concentration of \([Cu(EDTA)]^{2–}\) in this solution. Show your working.
(f) A solution of \([Cu(EDTA)]^{2–}\) ions is pale blue while a solution of \([Cu(NH_3)_4(H_2O)_2]^{2+}\) ions is deep blue. Explain this difference in colour
▶️Answer/Explanation
Ans:
(a) equilibrium constant for the formation of a complex in a solvent /from its constituents [1]
(b) +2 AND + 3 [1]
(c) Six [1]
(d) [Fe(EDTA)]– AND largest \(K_{stab}\) [1]
(e) \([[CuEDTA]^{2–}] ÷ ([[Cu(H_2O)_6]^{2+}] \times [EDTA^{2–}]) = 6.31\times 10^{19} [1]\)
0.095 [1]
(f) different ΔE OR different energy gap between d-orbitals [1]
absorption of different wavelength OR absorption of different frequency [1]
Questions 5
(a) Topic – 10.1 Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
(b) Topic – 10.1 Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
(c) Topic – 10.1 Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
Some of the ionic compounds of Group 2 elements undergo thermal decomposition. Thermal decomposition of solid anhydrous magnesium ethanedioate, MgC₂O₄, occurs above 650°C. The products are magnesium oxide and a mixture of two different gases, one of which gives a white precipitate with saturated calcium hydroxide solution.
(a) Complete the equation for the thermal decomposition of MgC₂O₄.
(b) Suggest which of MgC₂O₄ or CaC₂O₄ undergoes thermal decomposition at a lower temperature. Explain your answer.
(c) The ethanedioate ion is oxidised by acidified KMnO₄.
\(5C_2O_4^{2–} + 2MnO4^– + 16H^+ \to 10CO_2 + 2Mn^{2+} + 8H_2O\)
An experiment is performed to find the solubility of MgC₂O₄ in water. A 40.0 cm³ sample of saturated aqueous MgC₂O₄ requires 27.05 cm³ of 0.00200 mol/dm³ acidified KMnO₄ to oxidise all the C₂O₄²⁻ ions. Calculate the solubility, in mol/dm³, of MgC₂O₄ in water. Show your working.
▶️Answer/Explanation
Ans:
(a) MgC₂O₄ → MgO + CO₂ + CO [1]
(b) magnesium ethanedioate decomposes at lower T because Mg²⁺ has smaller radius than Ca²⁺. [1]
so anion is more polarised by Mg²⁺. [1]
(c) (27.05 ÷ 1000) × 0.02 = 5.41 × 10⁻⁵ moles MnO₄⁻ [1]
5.41 × 10⁻⁵ × 5/2 = 1.3525 × 10⁻⁴ moles C₂O₄²⁻ [1]
0.00338 mol dm⁻³ [1]
Questions 6
(a) Topic -28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(b)Topic – 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(a) Phosphine, :PH₃, and carbon monoxide, :CO, are monodentate ligands found in some transition element complexes.
(i) Define monodentate ligand.
(ii) Define transition element complex.
(iii) Explain why transition elements form complexes.
(b) The formulae of six complexes are given in Table 6.1. The abbreviation en is used for 1,2-diaminoethane. The abbreviation dien is used for the tridentate ligand H₂NCH₂CH₂NHCH₂CH₂NH₂. The dien ligand forms three bonds to the gold ion in [Au(dien)(H₂O)₂Cl]²⁺ and Au(dien)Cl₃. These three bonds all lie in the same plane. The CO ligand coordinates through the carbon atom in [Rh(CO)₂Cl₂]⁺.
(i) Complete Table 6.1 to state the geometry of the first three complexes. Each complex is either square planar, tetrahedral or octahedral. [1]
(ii) Use complexes [Au(dien)(H₂O)₂Cl]₂⁺ and Au(dien)Cl₃ to write an equation showing ligand exchange.
(iii) Draw the three-dimensional structure of Au(dien)Cl₃ in the box. The dien ligand can be drawn as
(iv) Draw the three-dimensional structure of Ni(PH₃)₂Cl₂ in the box.
(v) One of the complexes, [Rh(en)₂Cl₂]⁺ or [Rh(CO)₂Cl₂]⁺, can exist in three isomeric forms. Identify this complex and the types of isomerism shown.
(vi) Draw the three-dimensional structures of the two isomers of [Ni(H₂O)₂(NH₃)₄]²⁺ in the boxes and identify the type of isomerism shown.
▶️Answer/Explanation
Ans:
(a)(i) donates one lp to metal atom or ion [1]
(ii) metal atom or ion bonded to one or more ligands [1]
(iii) has vacant d-orbitals which are energetically accessible [1]
(b)(i) octahedral
square planar
octahedral [1]
(ii) \([Au(dien)(H_2O)_2Cl]^{2+}\) + \(2Cl^–\) → [Au(dien)Cl₃] + 2H₂O [1]
OR [Au(dien)Cl₃] + 2H₂O → \([Au(dien)(H_2O)_2Cl]^{2+}\)+ \(2Cl^–\)
(v) \([Rh(en)_2Cl_2]^+\) AND
optical AND cis / trans OR geometric [1]
Questions 7
(a) Topic -21.1 Organic synthesis
(b)Topic – 21.1 Organic synthesis
(c)Topic -21.1 Organic synthesis
Benzene can be used to make benzoic acid in the two-step process shown in Fig. 7.1.
(a) Give the reagents and conditions for step 1 and step 2.
(b) Methylbenzene and benzoic acid each have five different peaks in the carbon \((^{13}C)\) NMR spectrum.
Use Table 7.1 to complete the two sentences to suggest descriptions of these two spectra.
The carbon \((^{13}C)\) NMR spectrum of methylbenzene:
● has …………….. peak(s) in the chemical shift range of ………………… and
● has …………….. peak(s) in the chemical shift range of ………………… .
The carbon \((^{13}C)\) NMR spectrum of benzoic acid:
● has …………….. peak(s) in the chemical shift range of ………………… and
● has …………….. peak(s) in the chemical shift range of ………………… .
(c) (i) When treated with Cl₂ under suitable conditions, methylbenzene forms compound J. When treated with Cl₂ under different conditions with different reagents, methylbenzene
forms compound K. Suggest and draw structures of compounds J and K in the boxes. The molecular formulam of each compound is given.
State the reagents and conditions required to form each product.
to form compound J …………………………………………………………………………………………….
to form compound K …………………………………………………………………………………………….
(ii) When treated with a chlorine-containing reagent under suitable conditions, benzoic acid forms compound L. When treated with a different chlorine-containing reagent under different conditions, benzoic acid forms compound M. Suggest and draw structures of compounds L and M in the boxes. The molecular formula of each product is given.
State the reagents and conditions to form compound M from benzoic acid.
▶️Answer/Explanation
Ans:
(a) CH₃Cl + AlCl₃ [1]
hot alkaline) \(MnO_4^–\) [1]
(b) 4, 110-160,
1, 25-50 either order [1]
4, 110-160,
1, 160-185 either order [1]
Questions 8
(a) Topic -21.1 Organic synthesis
(b)Topic – 21.1 Organic synthesis
(c)Topic -21.1 Organic synthesis
Lactic acid, CH₃CH(OH)COOH, is the only monomer needed to form the polymer polylactic acid, PLA.
(a) (i) Draw a short length of the PLA polymer chain, including a minimum of two monomer residues. The methyl groups may be written as –CH₃, but all other bonds should be shown fully displayed. Label one repeat unit of polylactic acid on your diagram.
(ii) Give the name of the type of polymerisation involved in the formation of PLA and the name of the functional group that forms between the monomers.
type of polymerisation ………………………………………………………………………………………….
functional group …………………………………………………………………………………………………..
(iii) Predict whether PLA is readily biodegradable. Explain your answer.
(b) The proton (1H) NMR spectrum of CH₃CH(OH)COOH in CDCl₃ is shown in Fig. 8.1. The proton NMR chemical shift ranges are shown in Table 8.1.
(i) Use Fig. 8.1 and Table 8.1 to complete Table 8.2.
(ii) Name the substance responsible for the peak at δ = 0.0.
(iii) Explain why CDCl₃ is a better solvent than CHCl₃ for use in proton NMR.
(c) An impure sample of CH₃CH(OH)COOH contains pentan-3-one as the only contaminant. The mixture is analysed using gas/liquid chromatography. The pentan-3-one is found to have a longer retention time than the lactic acid.
(i) Explain what is meant by retention time.
(ii) Suggest suitable substances, or types of substances, that could be used as the mobile and stationary phases.
mobile phase ………………………………………………………………………………………………………
stationary phase ………………………………………………………………………………………………….
(iii) Describe how the percentage composition of the mixture can be determined from the gas /liquid chromatogram.
▶️Answer/Explanation
Ans:
(ii) condensation AND ester [1]
(iii) it is biodegradable because it can hydrolyse [1]
(b)(i) d values:
12, 4.4 [1]
5.0, 1.6 [1]
splitting patterns:
singlet, quartet / quadruplet, singlet, doublet [1]
(ii) TMS / tetramethylsilane [1] 1
(iii) CDCl₃ will not give an absorption / peak OR
CHCl₃ will give an absorption / peak [1]
(c)(i) time between injection and detection [1]
(ii) an unreactive gas AND a non-polar liquid [1]
(iii) area of peak divided by total area of all peaks × 100% [1]
Questions 9
(a) Topic -21.1 Organic synthesis
(b)Topic – 21.1 Organic synthesis
(c)Topic -21.1 Organic synthesis
(d)Topic -21.1 Organic synthesis
(a) State the reactants and conditions for two different types of reactions that both produce diethylamine, CH₃CH₂NHCH₂CH₃.
(b) Describe the relative basicities of diethylamine, phenylamine and ammonia in aqueous solution. Explain your answer in terms of structure
(c) Phenylamine reacts with HNO₂(aq) at 4°C to form compound P. Compound P reacts with phenol under alkaline conditions at 4°C. The product of this reaction is acidified, forming azo compound Q. Draw the structure of compound Q. Circle the azo group on your structure. State one use of an azo compound such as Q.
compound Q:
An azo compound can be used …………………………………………………………………………………. .
(d) CH₃CH₂NHCH₂CH₃ reacts with ethanoyl chloride, CH₃COCl, to give the amide N,N-diethylethanamide, CH₃CON(C₂H₅)₂. An incomplete description of the mechanism of this reaction is shown in Fig. 9.1.
(i) Complete the mechanism in Fig. 9.1. You should include:
● all relevant dipoles (δ+ and δ–) and full electric charges (+ and –) on the species in box one and in box two
● all relevant lone pairs on the species in box one and in box two
● all relevant curly arrows to show the movement of electron pairs in box one and in box two
● the formula of the second product in box three.
(ii) Name this mechanism.
▶️Answer/Explanation
Ans:
(a) chloroethane and ethylamine [1]
heat, pressure, ethanol [1]
N-ethyl ethanamide CH₃CONHC₂H₅ [1]
LiAlH₄ [1]
Other syntheses that work also got credit /
(b) phenylamine, ammonia, diethylamine [1]
• Availability of lone pair on N to receive \(H^+\)
• phenylamine – LP, delocalised / overlaps into ring [1]
• diethylamine – ethyl are electron donating [1]
Two bullet points for one mark, three bullet points for two marks.
box one:
1 dipole on C=O
2 LP on N of amine [1]
3 CA from N of amine to C of acyl chloride
4 CA from = to O of acyl chloride [1]
box two:
5 + on what was N of amine and nowhere else
6 – on what was O of acyl chloride and nowhere else
7 LP on what was O of acyl chloride [1]
8 CA from N-H bond to what was N of amine
9 CA from C-Cl bond to what was Cl of acyl chloride
10 CA from (LP on) O to reform C=O [1]
box three:
11 HCl
(ii) Addition-elimination