IChO 2014 · Theory problem 2

Dissociating gas cycle

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Dinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:

Figure 1

1.00 mol of \(\ce{N2O4}\) was put into an empty vessel with a fixed volume of 24.44 dm\(^{3}\). 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.

2.1 Calculate \(\Delta G^0\) of the reaction at 298K, assuming the gases are ideal.

2.2 Calculate \(\Delta H^0\) and \(\Delta S^0\) of the reaction, assuming that they do not change significantly with temperature.

Note: If you cannot calculate \(\Delta H^0\), use \(\Delta H^0 = 30.0\ \mathrm{kJ\,mol^{-1}}\) for further calculations.

The tendency of \(\ce{N2O4}\) to dissociate reversibly into \(\ce{NO2}\) enables its potential use in advanced power generation systems. A simplified scheme for one such system is shown below in Figure (a). Initially, "cool" \(\ce{N2O4}\) is compressed (\(1\to2\)) in a compressor (X), and heated (\(2\to3\)). Some \(\ce{N2O4}\) dissociates into \(\ce{NO2}\). The hot mixture is expanded (\(3\to4\)) through a turbine (Y), resulting in a decrease in both temperature and pressure. The mixture is then cooled further (\(4\to1\)) in a heat sink (Z), to promote the reformation of \(\ce{N2O4}\). This recombination reduces the pressure, thus facilitates the compression of \(\ce{N2O4}\) to start a new cycle. All these processes are assumed to take place reversibly.

Figure 2

To understand the benefits of using reversible dissociating gases such as \(\ce{N2O4}\), we will focus on step \(3\to4\) and consider an ideal gas turbine working with 1 mole of air (which we assume to be an inert, non-dissociating gas). During the reversible adiabatic expansion in the turbine, no heat is exchanged.

2.3 Give the equation to calculate the work done by the system \(w(\mathrm{air})\) during the reversible adiabatic expansion for 1 mol of air during stage \(3\to4\). Assume that \(C_{v,m}(\mathrm{air})\) (the isochoric molar heat capacity of air) is constant, and the temperature changes from \(T_3\) to \(T_4\).

2.4 Estimate the ratio \(w_{\ce{N2O4})}/w_{(\mathrm{air})}\), in which \(w_{(\ce{N2O4})}\) is the work done by the gas during the reversible adiabatic expansion process \(3\to4\) with the cycle working with 1 mol of \(\ce{N2O4}\), \(T_3\) and \(T_4\) are the same as in part 2. Take the conditions at stage 3 to be \(T_3 = 440\ \mathrm{K}\) and \(p_3 = 12.156\ \mathrm{bar}\) and assume that:

(i) the gas is at its equilibrium composition at stage 3;

(ii) \(C_{v,m}\) 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 (\(\ce{N2O4 + NO2}\)) is unchanged until the expansion is completed.

Official English text of the problem.

Answers

  • 2.1: \(\boxed{\Delta G^\circ_{298}=+4.73\ \mathrm{kJ\,mol^{-1}}}\).
  • 2.2: \(\boxed{\Delta H^\circ=+57.2\ \mathrm{kJ\,mol^{-1}}}\), \(\boxed{\Delta S^\circ=+176\ \mathrm{J\,mol^{-1}\,K^{-1}}}\).
  • 2.3: For the closed-system expansion-work interpretation, \[ \boxed{w(\mathrm{air})=(1.00\ \mathrm{mol})\,C_{v,m}(\mathrm{air})(T_3-T_4)}, \] positive for work done by the gas.
  • 2.4: \(\boxed{w_{(\ce{N2O4})}/w_{(\mathrm{air})}\approx1.92}\).

Solution

2.1 — Standard Gibbs energy of dissociation

The balanced reaction is

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

The species are dinitrogen tetroxide, O=[N+]([O-])[N+](=O)[O-], and nitrogen dioxide, O=[N+][O-]. Nitrogen dioxide is an odd-electron radical; its SMILES represents one Lewis form and does not depict resonance delocalization.

Use

\[ R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}} =0.08314\ \mathrm{bar\,dm^3\,mol^{-1}\,K^{-1}}, \qquad p^\circ=1\ \mathrm{bar}. \]

Let \(\alpha\) be the fraction of the original \(n_0=1.00\ \mathrm{mol}\) of \(\ce{N2O4}\) that dissociates. At equilibrium,

\[ n(\ce{N2O4})=n_0(1-\alpha),\qquad n(\ce{NO2})=2n_0\alpha, \]

so

\[ n_{\mathrm{tot}}=n_0(1+\alpha). \]

The ideal-gas equation therefore gives

\[ \alpha=\frac{pV}{n_0RT}-1. \]

At 298 K,

\[ \alpha_{298} = \frac{(1.190\ \mathrm{bar})(24.44\ \mathrm{dm^3})} {(1.00\ \mathrm{mol})(0.08314\ \mathrm{bar\,dm^3\,mol^{-1}\,K^{-1}}) (298\ \mathrm{K})}-1 =0.173875. \]

The partial pressures are

\[ p(\ce{NO2})=\frac{2\alpha}{1+\alpha}p =0.352527\ \mathrm{bar}, \]

\[ p(\ce{N2O4})=\frac{1-\alpha}{1+\alpha}p =0.837473\ \mathrm{bar}. \]

The dimensionless equilibrium constant is

\[ K_p =\frac{[p(\ce{NO2})/p^\circ]^2}{p(\ce{N2O4})/p^\circ} =\frac{4\alpha^2}{1-\alpha^2}\frac{p}{p^\circ}. \]

Thus,

\[ K_p(298)=\frac{(0.352527)^2}{0.837473}=0.148393. \]

Consequently,

\[ \Delta G^\circ_{298} =-RT\ln K_p(298) =-(8.314)(298)\ln(0.148393)\ \mathrm{J\,mol^{-1}}, \]

\[ \boxed{\Delta G^\circ_{298}=+4.73\ \mathrm{kJ\,mol^{-1}}}. \]

2.2 — Standard enthalpy and entropy of dissociation

At 348 K, the same calculation gives

\[ \alpha_{348} =\frac{(1.886)(24.44)}{(1.00)(0.08314)(348)}-1 =0.593139, \]

and hence

\[ K_p(348) =\frac{4(0.593139)^2}{1-(0.593139)^2} \frac{1.886\ \mathrm{bar}}{1\ \mathrm{bar}} =4.09463. \]

For temperature-independent \(\Delta H^\circ\), the integrated van ’t Hoff equation is

\[ \ln\frac{K_p(T_2)}{K_p(T_1)} = \frac{\Delta H^\circ}{R} \left(\frac1{T_1}-\frac1{T_2}\right). \]

Therefore,

\[ \Delta H^\circ = \frac{(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}) \ln(4.09463/0.148393)} {\dfrac1{298\ \mathrm{K}}-\dfrac1{348\ \mathrm{K}}} =57207.8\ \mathrm{J\,mol^{-1}}, \]

\[ \boxed{\Delta H^\circ=+57.2\ \mathrm{kJ\,mol^{-1}}}. \]

Using \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\),

\[ \Delta S^\circ =\frac{\Delta H^\circ-\Delta G^\circ_{298}}{298\ \mathrm{K}} =\frac{(57207.8-4726.93)\ \mathrm{J\,mol^{-1}}}{298\ \mathrm{K}}, \]

\[ \boxed{\Delta S^\circ=+176\ \mathrm{J\,mol^{-1}\,K^{-1}}}. \]

Dissociation is endothermic and increases entropy, consistent with converting one gas molecule into two.

2.3 — Adiabatic expansion work

Take \(w>0\) for work done by the gas. For a closed system, the first law is

\[ \Delta U=q-w. \]

Since the expansion is adiabatic, \(q=0\), so \(w=-\Delta U\). For a fixed amount of ideal gas with constant molar heat capacity,

\[ \Delta U=nC_{v,m}(T_4-T_3). \]

For \(n=1.00\ \mathrm{mol}\) of air,

\[ \boxed{ w(\mathrm{air}) =(1.00\ \mathrm{mol})\,C_{v,m}(\mathrm{air})(T_3-T_4) }. \]

The temperature decreases during expansion, so this work is positive.

2.4 — Work ratio for the dissociating gas

First calculate the equilibrium constant at stage 3:

\[ \ln\frac{K_p(440)}{K_p(298)} = \frac{\Delta H^\circ}{R} \left(\frac1{298\ \mathrm{K}}-\frac1{440\ \mathrm{K}}\right). \]

Using the unrounded results above,

\[ K_p(440) = 0.148393\exp\!\left[ \frac{57207.8}{8.314} \left(\frac1{298}-\frac1{440}\right) \right] =255.690. \]

Let \(\alpha_3\) be the dissociation fraction at \(p_3=12.156\ \mathrm{bar}\). From the equilibrium expression derived in 2.1,

\[ K_p(440)=\frac{4\alpha_3^2}{1-\alpha_3^2}\frac{p_3}{p^\circ}. \]

Solving,

\[ \alpha_3 = \sqrt{\frac{K_p(440)} {K_p(440)+4p_3/p^\circ}} = \sqrt{\frac{255.690}{255.690+4(12.156)}} =0.916634. \]

Thus, starting with 1.00 mol of \(\ce{N2O4}\), the stage-3 mixture contains

\[ n(\ce{N2O4})=0.083366\ \mathrm{mol}, \qquad n(\ce{NO2})=1.833268\ \mathrm{mol}, \]

and

\[ n_3=(1.00\ \mathrm{mol})(1+\alpha_3) =1.916634\ \mathrm{mol}. \]

The composition is unchanged during expansion, so there is no reaction contribution to the change in internal energy during \(3\to4\). With the same molar heat capacity and temperature change as for air,

\[ w_{(\ce{N2O4})}=n_3C_{v,m}(\mathrm{air})(T_3-T_4). \]

Therefore,

\[ \frac{w_{(\ce{N2O4})}}{w_{(\mathrm{air})}} = \frac{n_3}{1.00\ \mathrm{mol}} =1+\alpha_3, \]

\[ \boxed{\frac{w_{(\ce{N2O4})}}{w_{(\mathrm{air})}} =1.916634\approx1.92}. \]

The increased work comes from the larger number of gas molecules produced by dissociation.

The solution was checked against the official solution.