A-LevelChemistryReaction kinetics9701/42 · Oct/Nov 2018 · Q814 Marks

(a) Chloramine, NH2Cl, can be used in the treatment of drinking water to kill bacteria. Excess chloramine in water is destroyed using UV light. The mechanism for this involves free radicals. The initiation step in this process is shown. NH2Cl UV •NH2 + •Cl

(a)(i)What is meant by the term free radical?[1]
(a)(ii)The equation for a possible propagation step in the process is shown. NH2Cl + •Cl → •NHCl + HCl Suggest an equation for a possible termination step in this process.[1]
(b)(i)Draw the 'dot-and-cross' diagram of NH2Cl. Show outer electrons only.[1]
(b)(ii)State the hybridisation of the nitrogen atom and suggest the H-N-Cl bond angle in the NH2Cl molecule. hybridisation of N H-N-Cl bond angle[1]
(c)Some values for standard enthalpy changes of formation, ΔH°f, and standard entropies, S°, are given in the table. ΔH°f / kJ mol⁻¹ | S° / JK⁻¹ mol⁻¹ --- | --- NH2Cl(g) | +80.1 | +241 NH3(g) | -45.9 | +198 N2H4(g) | +95.4 | +237 HCl(g) | -92.3 | +187
(c)(i)Define the meaning of the term entropy.[1]
(c)(ii)Hydrazine, N2H4, can be produced from chloramine and ammonia as shown. NH2Cl(g) + NH3(g) → N2H4(g) + HCl(g) Calculate the standard entropy change, ΔS°, for this reaction. ΔS° =[1]
(c)(iii)Calculate the standard enthalpy change, ΔH°, for this reaction. ΔH° =[1]
(c)(iv)Calculate the standard Gibbs free energy change, ΔG°, for this reaction at 298 K. ΔG° =[2]
(c)(v)Explain, with reference to ΔG°, why this reaction becomes less feasible at higher temperatures.[1]
(d)Compare and explain the basicities of ammonia, ethylamine and phenylamine.[4]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[1] species with an unpaired electron
[1] NH2 + Cl → NH2Cl
[1] (b)(i) dot-and-cross diagram of NH2Cl
[1] sp³ AND 100-107°
[1] (entropy) is a measure of the disorder/randomness of a system
[1] △S° = 237 + 187 – (241 + 198) = −15.0 (J K⁻¹ mol⁻¹)
[1] ΔH° = 95.4–92.3 – (80.1–45.9) = −31.1 (kJ mol⁻¹)
[1] ∆G° = ∆H° – T∆S°
[1] ∆G° = -31.1 – (298 × -0.015) = −26.6 (kJ mol⁻¹)
[1] (at higher temperatures) T∆S° becomes more negative so ∆G° becomes more positive OR (at higher temperatures) –T∆S is becomes more positive so ∆G becomes more positive
[1] ethylamine > ammonia > phenylamine
[1] ethyl group is electron donating group
[1] p-orbital from N in phenylamine overlaps with π-ring system OR lone pair on N is delocalised into benzene ring
[1] basicity linked to ability of N to accept a proton

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