Consider the dimerization of nitrogen dioxide: 2 NO2(g) ⇌ N2O4(g) Thermodynamic data are given below — Thermodynamics Chemistry Question
Problem Context
Consider the dimerization of nitrogen dioxide:
2 NO2(g) ⇌ N2O4(g)
Thermodynamic data are given below for the gaseous species at 298 K, except for the heats of vaporization, which are for the liquids:
Calculate ∆Hºrxn, ∆Sºrxn, and ∆Gºrxn for the dimerization of NO2(g) at 298 K.
Model Answer
∆Hº = 11.1 kJ mol-1 – 2(33.2 kJ mol-1) = –55.3 kJ mol-1
∆Sº = 304.4 J mol-1 K-1 – 2(240.1 J mol-1 K-1) = –175.8 J mol-1 K-1
∆Gº = ∆Hº – T∆Sº = –55.3 kJ mol-1 –(298 K)(–0.1758 kJ mol-1 K-1) = –2.9 kJ mol-1
Calculate Keq for the dimerization of NO2(g) at 298 K.
Model Answer
∆Gº = –RTln(Keq)
ln(Keq) = –(–2900 J mol-1)/(8.314 J mol-1 K-1)(298 K) = 1.17
Keq = 3.2
Would Keq for the dimerization at 308 K be greater than, less than, or equal to Keq at 298 K? Justify your answer based on the above data.
Model Answer
The reaction is exothermic, so Keq will decrease as the temperature is increased.
Would ∆Hºrxn at 308 K be greater than, less than, or equal to ∆Hºrxn at 298 K? Justify your answer based on the above data.
Model Answer
From the thermochemical cycle:
∆Hºrxn (308 K) = ∆Hºrxn (298 K) + 10 K(Cp[N2O4] – 2Cp[NO2])
Since Cp[N2O4] – 2Cp[NO2] = 4.8 J mol-1 K-1 is positive, then ∆Hºrxn (308 K) will be greater (more positive) than ∆Hºrxn (298 K), although the difference is rather small.
Would ∆Hºrxn in the liquid phase be greater than, less than, or equal to ∆Hºrxn in the gas phase? Justify your answer based on the above data.
Model Answer
An analogous thermochemical cycle . . .
. . . establishes that ∆Hºrxn (l) = ∆Hºrxn (g) + 2∆Hvap(NO2) – ∆Hvap(N2O4).
Since 2∆Hvap(NO2) – ∆Hvap(N2O4) = 17.3 kJ mol-1 is positive, then ∆Hºrxn (l) will be greater (more positive) than ∆Hºrxn (g).