Dinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide: N2O4(g) ⇌ 2 NO2(g) 1.00 mol — Physical Chemistry — Thermodynamics Chemistry Question
Dissociating gas cycle
Dinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:
N2O4(g) ⇌ 2 NO2(g)
1.00 mol of N2O4 was put into an empty vessel with a fixed volume of 24.44 dm3. The equilibrium gas pressure at 298 K was found to be 1.190 bar. When heated to 348 K, the gas pressure increased to its equilibrium value of 1.886 bar.
The tendency of N2O4 to dissociate reversibly into NO2 enables its potential use in advanced power generation systems. A simplified scheme for one such system is shown below in Figure (a).
Initially, "cool" N2O4 is compressed (1→2) in a compressor (X), and heated (2→3). Some N2O4 dissociates into NO2. The hot mixture is expanded (3→4) through a turbine (Y), resulting in a decrease in both temperature and pressure. The mixture is then cooled further (4→1) in a heat sink (Z), to promote the reformation of N2O4. This recombination reduces the pressure, thus facilitates the compression of N2O4 to start a new cycle. All these processes are assumed to take place reversibly.
Calculate ∆G0 of the reaction at 298K, assuming the gases are ideal.
Model Answer
1.593 = 1 + x
x = 0.593 mol
At equilibrium:
∆G0 298 = 4.72 kJ = ∆H – 298 ∆S
Calculate ∆H0 and ∆S0 of the reaction, assuming that they do not change significantly with temperature.
Note: If you cannot calculate ∆H0, use ∆H0 = 30.0 kJ mol–1 for further calculations.
Model Answer
At 348 K,
∆Go = – RT lnK348 = – 8.3145 × 348 × ln 4.0897 = – 4075 J mol-1 = – 4.08 kJ mol-1
For ∆S0:
∆G0 348 = – 4.08 kJ = ∆H – 348 ∆S (1)
∆G0 298 = 4.72 kJ = ∆H – 298 ∆S (2)
(2) – (1) → ∆S = 0.176 kJ mol–1 K–1
For ∆H0:
∆H0 = 4.720 + 298 × 0.176 = 57.2 (kJ mol–1)
Give the equation to calculate the work done by the system w(air) during the reversible adiabatic expansion for 1 mol of air during stage 3 → 4. Assume that Cv,m(air) (the isochoric molar heat capacity of air) is constant, and the temperature changes from T3 to T4.
Model Answer
∆U = q + w; work done by turbine wair = – w
q = 0, thus wair = ∆U = Cv,m (air) [T3 – T4]
Estimate the ratio w(N2O4)/w(air), in which w(N2O4) is the work done by the gas during the reversible adiabatic expansion process 3 → 4 with the cycle working with 1 mol of N2O4, T3 and T4 are the same as in part 2. Take the conditions at stage 3 to be T3 = 440 K and p3 = 12.156 bar and assume that: (i) the gas is at its equilibrium composition at stage 3; (ii) Cv,m for the gas is the same as for air; (iii) the adiabatic expansion in the turbine takes place in a way that the composition of the gas mixture (N2O4 + NO2) is unchanged until the expansion is completed.
Model Answer
→ K440 = 255.2
N2O4 ⇌ 2 NO2 (1)
Initial molar number: 1 0
At equilibrium 1 – x 2x
ntotal = 1 – x + 2 x = (1 + x) mol; ptotal = 12.156 bar
At equilibrium:
(p0 = 1 bar) →
4x2 = 20.99 – 20.99 x2 → 24.99 x2 = 20.99 → x = 0.92;
ntotal = 1 + x = 1.92