Bubbling gaseous NH3 through a solution of SCl2 generates a red explosive solid, S4N4. Its structure — Physical Chemistry — Kinetics Chemistry Question
Explosive S4N4
Bubbling gaseous NH3 through a solution of SCl2 generates a red explosive solid, S4N4. Its structure can be represented in a number of ways; one way is as shown below.
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Additional data:
E(S–S) = 226 kJ mol–1 E(N≡N) = 946 kJ mol–1
E(S–N) = 273 kJ mol–1 E(S=N) = 328 kJ mol–1
∆Hvap(S8) = 77 kJ mol–1 ∆Hvap(S4N4) = 88 kJ mol–1
∆fH(NH3) = – 45.9 kJ mol–1 ∆fH(SCl2) = – 50.0 kJ mol–1
∆fH(HCl) = – 92.3 kJ mol–1
Write a balanced equation for the formation of S4N4 from NH3 and SCl2
Model Answer
4 NH3 + 6 SCl2 → S4N4 + 12 HCl + 1/4 S8
could also be written with the extra ammonia molecules needed to react with the product HCl. i.e.:
12 NH3 + 6 SCl2 → S4N4 + 12 NH4Cl + 1/4 S8
Construct a Born-Haber cycle for the formation of S4N4 and use the data below to detine the enthalpy of formation of S4N4
Model Answer
To form one mole of S4N4 from the elements requires breaking four S–S bonds, two N≡N bonds and forming four S=N bonds and four S–N bonds:
∆fH° = (4 × 226) + (2 × 946) – (4 × 328) – (4 × 273) = 392 kJ mol–1
(This value is somewhat out due to the imprecise nature of the bond strengths.)
Use the additional data and your answer to part 19.1 to determine the enthalpy change for the reaction of NH3 with SCl2
Model Answer
For the reaction as first written in 19.1:
∆rH° = ∆fH°(S4N4) + 12 ∆fH°(HCl) + 1/4 ∆fH°(S8) – 4 ∆fH°(NH3) – 6 ∆fH°(SCl2)
392 + 12×(–92.3) + 0 – 4×(– 45.9) – 6×(– 50.0) = –232 kJ mol–1
The S4N4 molecule has a rich reaction chemistry including both oxidation and reduction reactions. Treatment of S4N4 with an excess of AsF5 in sulfur dioxide generates the salt [S4N4][AsF6]2 whereas treatment with excess SnCl2×2 H2O in methanol yields S4N4H4
Write balanced equations for these two reactions
Model Answer
i) S4N4 + 3 AsF5 → (S4N4)2+ 2 AsF6– + AsF3
Further complexation occurs with the AsF3 with AsF5 and AsF6– so the reaction may also be written:
S4N4 + 4 AsF5 → (S4N4)2+ AsF6– + [As3F14]–
ii) During this reaction, the Sn(II) becomes oxidised to Sn(IV):
S4N4 + 2 SnCl2 + 4 MeOH → S4N4H4 + 2 SnCl2(MeO)2