AI SearchTopicalsYearlysNotesDashboardPricing
9701/42

Chemistry 9701/42 — May/June 2020

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

8
questions
100
marks
120
minutes

Topics Equilibria · Transition Elements · Chemical Energetics · Electrochemistry · Hydrocarbons · Carboxylic Acids and Derivatives · +6 more

Q1Transition ElementsChemical EnergeticsFree sample

EDTA4−^{4-}, is a polydentate ligand.

(a)
(i)

Explain what is meant by the term polydentate ligand.

2M
(ii)

When a solution containing EDTA4−^{4-} is added to a solution containing [Cd(H2_2O)6_6]2+^{2+} a new complex is formed, [CdEDTA]2−^{2-}.

equilibrium 1[Cd(H2O)6]2++EDTA4−⇌[CdEDTA]2−+6H2O\text{equilibrium 1} \quad [\text{Cd}(\text{H}_2\text{O})_6]^{2+} + \text{EDTA}^{4-} \rightleftharpoons [\text{CdEDTA}]^{2-} + 6\text{H}_2\text{O}

Circle, on the structure of EDTA4−^{4-}, the six atoms that form bonds with the metal ion.

1M
(iii)

Write an expression for the stability constant, Kstab1K_{\text{stab1}}, for equilibrium 1, and state its units.

Kstab1=K_{\text{stab1}} =

units=\text{units} =
2M
(b)

Cadmium ions form complexes with methylamine, CH3_3NH2_2, and with 1,2-diaminoethane, H2_2NCH2_2CH2_2NH2_2, as shown in equilibriums 2 and 3. 1,2-diaminoethane is shown as en.

equilibrium 2[Cd(H2O)6]2++4CH3NH2⇌[Cd(CH3NH2)4(H2O)2]2++4H2OKstab2=3.60×106equilibrium 3[Cd(H2O)6]2++2en⇌[Cd(en)2(H2O)2]2++4H2OKstab3=4.20×1010\begin{aligned} &\text{equilibrium 2} && [\text{Cd}(\text{H}_2\text{O})_6]^{2+} + 4\text{CH}_3\text{NH}_2 \rightleftharpoons [\text{Cd}(\text{CH}_3\text{NH}_2)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O} && K_{\text{stab2}} = 3.60 \times 10^6 \\ &\text{equilibrium 3} && [\text{Cd}(\text{H}_2\text{O})_6]^{2+} + 2\text{en} \rightleftharpoons [\text{Cd}(\text{en})_2(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O} && K_{\text{stab3}} = 4.20 \times 10^{10} \end{aligned}

An equilibrium is set up between these two complexes as shown in equilibrium 4.

equilibrium 4[Cd(CH3NH2)4(H2O)2]2++2en⇌[Cd(en)2(H2O)2]2++4CH3NH2ΔH⊖=+0.840 kJ mol−1ΔS⊖=+80.9 J K−1 mol−1\begin{aligned} \text{equilibrium 4} \quad [\text{Cd}(\text{CH}_3\text{NH}_2)_4(\text{H}_2\text{O})_2]^{2+} + 2\text{en} \rightleftharpoons [\text{Cd}(\text{en})_2(\text{H}_2\text{O})_2]^{2+} + 4\text{CH}_3\text{NH}_2 \quad &\Delta H^\ominus = +0.840 \text{ kJ mol}^{-1} \\ &\Delta S^\ominus = +80.9 \text{ J K}^{-1} \text{ mol}^{-1} \end{aligned}
(i)

Keq4K_{\text{eq4}} is the equilibrium constant for equilibrium 4.

Write an expression for Keq4K_{\text{eq4}} in terms of Kstab2K_{\text{stab2}} and Kstab3K_{\text{stab3}}.

Keq4=K_{\text{eq4}} =

1M
(ii)

Calculate the value of the standard Gibbs free energy change, ΔG⊖\Delta G^\ominus, for equilibrium 4 at 298 K.

ΔG⊖=.............................. kJ mol−1\Delta G^\ominus = \text{.............................. kJ mol}^{-1}
2M
(iii)

State how the value of ΔG⊖\Delta G^\ominus changes as the temperature increases. Explain your answer.

1M

The rest of this paper

7 more questions
  • Q2Group 2 · Equilibria · Electrochemistry · Reaction Kinetics19M
  • Q3Transition Elements · Equilibria9M
  • Q4Hydrocarbons · Carboxylic Acids and Derivatives · Hydroxy Compounds · Equilibria · Nitrogen Compounds19M
  • Q5Equilibria · Hydrocarbons · Carboxylic Acids and Derivatives · Analytical Techniques15M
  • Q6Polymerisation · Nitrogen Compounds · Analytical Techniques12M
  • Q7Chemical Energetics10M
  • Q8Electrochemistry7M
Loading the full paper…