Equilibrium Extent Computation of the Decomposition of Pure Nitrogen Tetroxide
Author's Data: Housam BINOUS
Department of Chemical Engineering
National Institute of Applied Sciences and Technology
Tunis, TUNISIA
Email: binoushousam@yahoo.com

Reference: See Appendix II and IV for constant pressure heat capacities and enthalpies and free energies of formation in book by S. I. Sandler, Chemical and Engineering Thermodynamics, 3rd Edition, Wiley 1999.

Off[General::"spell1"]

Standard state Gibbs free energy of reaction and standard state heat of reaction at 298.15K and 1 bar. Obtained from enthalpies and free energies of formation.

Reaction is decomposition of pure nitrogen tetroxide into nitrogen dioxide:        N\(TraditionalForm\`\_2\)O\(TraditionalForm\`\_4\)(g)  = 2 NO\(TraditionalForm\`\_2\) (g)

Δ(Subscript[H, 298]) = 2 33.2 -  9.2

Δ(Subscript[G, 298]) = 2 51.3 - 97.9

Equilibrium constant at 298.15K.

(Subscript[K, a, 298]) = 
      Exp[-4700/(R 298.15)] /. R ->  8.314

Standard state heat of reaction at T K and 1 bar. Obtained from expressions of constant pressure heat capacities of N\(TraditionalForm\`\_2\)O\(TraditionalForm\`\_4\) and NO\(TraditionalForm\`\_2\).

ΔH[T_] := 
        56189 + 12.80 T - 3.62 10^-2 T^2 + 1.434 10^-5 T^3 + 
          3.933 10^-9 T^4;

ΔH[298.15]

Mole fractions versus extent of reaction X

(Subscript[y, NO2]) = 2 X/(1 + X)

(Subscript[y, N2O4]) = (1 - X)/(1 + X)

X (the extent of reaction) versus the equilibrium constant (K\(TraditionalForm\`\_a\)) and the total pressure (P).

Eq1 = (Subscript[K1, a]) == (Subscript[y, NO2])^2 P/(Subscript[y, N2O4])

sol = Solve[Eq1, X]

X = sol[[2, 1, 2]]

Plotting the standard heat of reaction as a function of temperature

Plot[ΔH[T]/1000, {T, 200, 400}, 
      PlotRange -> {50, 60}, 
      PlotStyle -> {Thickness[0.01], RGBColor[1, 0, 0]}, 
      Frame -> True]

Plotting the equilibrium constant versus temperature

(Subscript[K1, a]) = 
      (Subscript[K, a, 298]) Exp[-6758.4 (1/T - 1/298.15) + 
            1.54 Log[T/298.15] - 0.435 10^-2 (T - 298.15) + 
            0.862 10^-6 (T^2 - 298.15^2) + 
            0.1577 10^-9 (T^3 - 298.15^3)]

f[T_] := ΔH[T]/(R T^2) /. R ->  8.314 // 
        Expand

Plot[{Log[10, (Subscript[K, a, 298]) Exp[NIntegrate[f[x], {x, 298.15, T}]]], 
        Log[10, (Subscript[K1, a])]}, {T, 200, 400}, PlotRange -> {-5, 2}, 
      PlotStyle -> {{Thickness[0.01], 
            RGBColor[0, 1, 0]}, {Thickness[0.01], RGBColor[1, 0, 0]}}, 
      Frame -> True]

Plotting extend of reaction as a function of temperature and pressure (P varies from 0.1 bars to 10 bars). We find that since the reaction is endothermic it is favored by high temperatures. In addition, since the number of mole increases, the extend of reaction is higher at lower pressures in agreement with LeChatelier's rule.

Plot[{X /. P ->  0.1, X /. P ->  1, 
        X /. P ->  10}, {T, 200, 400}, 
      PlotStyle -> {{Thickness[0.01], 
            RGBColor[1, 0, 0]}, {Thickness[0.01], 
            RGBColor[0, 1, 0]}, {Thickness[0.01], RGBColor[0, 0, 1]}}, 
      Frame -> True]