1.8.5: Enthalpies- Solutions- Equilibrium and Frozen Partial Molar Enthalpies
- Page ID
- 374779
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A given system at fixed \(\mathrm{T}\) and \(\mathrm{p}\) is at thermodynamic equilibrium. The enthalpy of the system is perturbed by adding \(\delta \mathrm{n}_{j}\) moles of chemical substance \(j\). We imagine two possible limiting changes to the system. In one limit the enthalpy of the system changes to a neighbouring state where the extent of chemical reaction remains constant; i.e. at fixed \(\xi\). In another limit the enthalpy of the system changes to a neighbouring state where the affinity for spontaneous change \(\mathrm{A}\) remains constant. The two differential changes in enthalpy are related.
\[\left(\frac{\partial H}{\partial n_{j}}\right)_{A}=\left(\frac{\partial H}{\partial n_{j}}\right)_{\xi}-\left(\frac{\partial \mathrm{A}}{\partial n_{j}}\right)_{\xi} \,\left(\frac{\partial \xi}{\partial A}\right)_{n_{j}} \,\left(\frac{\partial H}{\partial \xi}\right)_{n_{j}} \nonumber \]
We identify the state being perturbed as the equilibrium state where \(\mathrm{A} = 0\) and the composition-organisation is represented by \(\xi^{\mathrm{eq}\). We identify two quantities describing the impact of adding \(\delta \mathrm{n}_{j}\) moles of chemical substance \(j\).
Equilibrium partial molar enthalpy,
\[\mathrm{H}_{\mathrm{j}}(\mathrm{A}=0)=\left(\frac{\partial \mathrm{H}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{A}=0} \nonumber \]
Frozen partial molar enthalpy,
\[\mathrm{H}_{\mathrm{j}}\left(\xi^{\mathrm{eq}}\right)=\left(\frac{\partial \mathrm{H}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \xi^{\mathrm{eq}}} \nonumber \]
Because the triple product term on the r.h.s. of equation (a) is not zero at equilibrium (i.e. at \(\mathrm{A} = \text { zero}\) and \(\xi = \xi^{\mathrm{eq}}\)), then \(\mathrm{H}_{\mathrm{j}}(\mathrm{A}=0)\) is not equal to. By convention, the term ‘ partial molar enthalpy is taken to mean \(\mathrm{H}_{\mathrm{j}}(\mathrm{A}=0)\).


