Chemistry Labs

Problem 2

Dissociating gas cycle. Dinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide: NX2OX4(g)⇌2 NOX2(g)\ce{N2O4(g) <=> 2 NO2(g)}. 1.00 mol of NX2OX4\ce{N2O4} was placed in an empty vessel of fixed volume 24.44 dm3^3. 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 ΔG∘\Delta G^{\circ} of the reaction at 298 K, assuming ideal gases. (b) Calculate ΔH∘\Delta H^{\circ} and ΔS∘\Delta S^{\circ}, assuming they do not vary significantly with temperature. The reversible dissociation of NX2OX4\ce{N2O4} can be exploited in power cycles: in step 3→43 \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→43 \to 4, assuming constant CV,mC_{V,m} and a temperature drop from T3T_3 to T4T_4.
Step 3 of 5: Gibbs energy at 298 K
ΔG298∘=−RTln⁡K298=−8.3145×298×ln⁡0.149=+4.72 kJ mol−1\Delta G^{\circ}_{298} = -RT\ln K_{298} = -8.3145 \times 298 \times \ln 0.149 = +4.72\ \text{kJ mol}^{-1}
Analysis

Direct application of ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT\ln K gives +4.72+4.72 kJ mol−1^{-1}; the positive sign reflects that NX2OX4\ce{N2O4} is only partly dissociated under these conditions.