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 4 of 5: Repeat at 348 K and solve for

\Delta H^{\circ}

\Delta S^{\circ}

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x_{348} = \dfrac{1.886}{1.184} - 1 = 0.593 \Rightarrow K_{348} = 4.09,\; \Delta G^{\circ}_{348} = -4.08\ \text{kJ mol}^{-1} \Rightarrow \Delta S^{\circ} = 176\ \text{J K}^{-1}\text{mol}^{-1},\; \Delta H^{\circ} = 57.2\ \text{kJ mol}^{-1}

Analysis

At 348 K, $p_{ini} = RT/V = 1.184$ bar so $x = 0.593$ and $K_{348} = 4.09$ , giving $\Delta G^{\circ}_{348} = -4.08$ kJ mol $^{-1}$ . Subtracting $\Delta G^{\circ}_{348} = \Delta H - 348\Delta S$ from $\Delta G^{\circ}_{298} = \Delta H - 298\Delta S$ yields $\Delta S^{\circ} = 0.176$ kJ K $^{-1}$ mol $^{-1}$ and $\Delta H^{\circ} = 4.72 + 298\times0.176 = 57.2$ kJ mol $^{-1}$ .