Problem 2

Dissociating gas cycle. Dinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:

\ce{N2O4(g) <=> 2 NO2(g)}

. 1.00 mol of

\ce{N2O4}

was placed in an empty vessel of fixed volume 24.44 dm

^3

. The equilibrium gas pressure at 298 K was 1.190 bar; when heated to 348 K the equilibrium pressure rose to 1.886 bar. (a) Calculate

\Delta G^{\circ}

of the reaction at 298 K, assuming ideal gases. (b) Calculate

\Delta H^{\circ}

and

\Delta S^{\circ}

, assuming they do not vary significantly with temperature. The reversible dissociation of

\ce{N2O4}

can be exploited in power cycles: in step

3 \to 4

the hot gas mixture expands reversibly and adiabatically through a turbine. (c) Give the equation for the work done by 1 mol of an inert gas (air) during the reversible adiabatic expansion

3 \to 4

, assuming constant

C_{V,m}

and a temperature drop from

T_3

T_4

Step 2 of 5: Equilibrium constant at 298 K

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K_p = \dfrac{p^2(\ce{NO2})}{p(\ce{N2O4})\,p^{\circ}} = \dfrac{4x^2}{1-x^2}\cdot\dfrac{p_{eq}}{p^{\circ}} = \dfrac{4(0.174)^2}{1-(0.174)^2}\times1.190 = 0.149

Analysis

Partial pressures: $p(\ce{N2O4}) = p_{eq}(1-x)/(1+x)$ and $p(\ce{NO2}) = p_{eq}\,2x/(1+x)$ , which combine to $K_p = (4x^2/(1-x^2))(p_{eq}/p^{\circ}) = 0.149$ at 298 K.