Nitrogen oxides play a critical role in atmospheric chemistry. They are produced in internal combust — Physical Chemistry — Thermodynamics Chemistry Question
Nitrogen Oxides: Chemistry, Reaction Equilibrium and Thermodynamics
Nitrogen oxides play a critical role in atmospheric chemistry. They are produced in internal combustion engines from the high-temperature combination of O2 and N2 in air, and contribute to photochemical smog in many large cities. In the stratosphere, nitrogen oxides contribute to the photochemical degradation of ozone that maintains a steady state of this ultraviolet-absorbing gas. Some of the chemistry of nitrogen oxides is described below.
A. Interconversion of Nitrogen Oxides
A colorless, gaseous, paramagnetic nitrogen oxide A is allowed to react with excess O2, and the mixture passed through a trap at –120 ºC, in which condenses a colorless solid B. A sample of B (2.00 g) is introduced into a 1.00 dm3 evacuated container and its red-brown vapor equilibrated at various temperatures, giving rise to the pressures recorded below.
T, ºC p, atm
25.0 0.653
50.0 0.838
8.1 Identify compounds A and B.
Model Answer
8.1 A = NO, B = N2O4
8.2 What chemical reaction takes place when B is introduced into the evacuated container? Give ∆Hº and ∆Sº values for this reaction.
Model Answer
8.2 N2O4 ⇌ 2 NO2
2.00 g N2O4 = 0.0217 mol N2O4 which would exert a pressure of 0.532 atm at 298 K in a volume of 1.00 dm3. Each atm of N2O4 that dissociates increases the total pressure by 1 atm, so if p = 0.653 atm, then (0.653 – 0.532) = 0.121 atm is the partial pressure of N2O4 that has dissociated, giving p(N2O4) = 0.532 – 0.121 = 0.411 atm, p(NO2) = 0.242 atm, and Kp(298 K) = 0.142. The analogous calculation at 323 K gives Kp(323 K) = 0.859.
From the van't Hoff equation, ln(K) = –(∆Hº/RT) + (∆Sº/R). Substituting the values of K at temperatures of 298 and 323 K give the two simultaneous equations in ∆Hº and ∆Sº:
–1.952 = –∆H°(4.034 · 10–4 mol J–1) + ∆S°(0.1203 mol K J–1)
–0.152 = –∆H°(3.722 · 10–4 mol J–1) + ∆S°(0.1203 mol K J–1)
Solving gives:
∆H° = +57.7 kJ mol–1
∆S° = +177 J mol–1 K–1
B. Reactivity of Nitrogen Oxides
Compound B (from Part A above) reacts with F2 to form a colorless gas C.
Compound C reacts with gaseous boron trifluoride to form a colorless solid D. A 1.000 g sample of compound D is dissolved in water and titrated with a NaOH solution (0.5000 mol dm–3) to a phenolphthalein endpoint, which requires 30.12 cm3 of the titrant.
8.3 Give structural formulas for compounds C and D, and explain the results of the titration of D.
Model Answer
8.3
C = FNO2,
D = [NO2][BF4],
When nitronium tetrafluoroborate dissolves in water, it reacts to make two moles of H3O+ per mole of salt:
[NO2][BF4] + 3 H2O → 2 H3O+ + NO3– + BF4–
1.000 g [NO2][BF4] = 7.530 · 10–3 mol is equivalent to 0.01506 mol H3O+ which is equivalent to 30.12 cm3 of NaOH solution (0.5000 mol dm-3)
8.4 Compound D reacts with excess nitrobenzene to give a major organic product E. Give the structural formula of E.
Model Answer
8.4 Nitronium tetrafluoroborate (compound D) is a very powerful nitrating agent, capable of nitrating even deactivated aromatics such as nitrobenzene (see, e.g., Olah, G. A.; Lin, H. C. J. Am. Chem. Soc. 1974, 96, 549–553). Since –NO2 is a meta-directing substituent, the major (> 85%) product E is m-dinitrobenzene.
E =