Reaction Energetics — Internal Energy and Enthalpy

By Prof. Marco Ruzzi.

Theoretical background

For ideal gases:

\[H=U+pV=U+nRT.\]

At fixed temperature, the change in gaseous amount gives:

\[dH=dU+d(nRT)=dU+RT\,dn.\] \[\Delta H=\Delta U+RT\Delta n.\]

For molar reaction quantities, use Δν_gas, the gaseous stoichiometric coefficients of products minus those of reactants:

\[\Delta_rU^\circ=\Delta_rH^\circ-RT\Delta\nu_{\mathrm{gas}}.\]

Exercise 2 — Oxidation of nitric oxide

Consider the reaction under standard conditions at T=298 K:

\[2\mathrm{NO}(g)+\mathrm{O}_2(g)\longrightarrow2\mathrm{NO}_2(g).\]

The standard formation enthalpies are +90.2 kJ mol⁻¹ for NO(g) and +33.2 kJ mol⁻¹ for NO₂(g). Calculate the standard molar internal-energy change of the reaction.

1. Change in gaseous amount

\[\Delta\nu_{\mathrm{gas}}=2-(2+1)=-1.\]

Thus:

\[\Delta_rU^\circ=\Delta_rH^\circ+RT.\]

2. Reaction enthalpy

Use formation enthalpies, taking products minus reactants. Oxygen in its standard elemental reference state has zero standard formation enthalpy by convention:

\[\begin{aligned} \Delta_rH^\circ={}&2\Delta_fH^\circ(\mathrm{NO}_2)\\ &-2\Delta_fH^\circ(\mathrm{NO})\\ &-\Delta_fH^\circ(\mathrm{O}_2). \end{aligned}\] \[\Delta_rH^\circ=2(33.2)-2(90.2)-0.\] \[\Delta_rH^\circ=-114\ \mathrm{kJ\,mol^{-1}}.\]

3. Internal-energy change

With R=8.314 J mol⁻¹ K⁻¹:

\[\Delta_rU^\circ=-114\times10^3+(8.314)(298)\ \mathrm{J\,mol^{-1}}.\] \[\Delta_rU^\circ=-111522.428\ \mathrm{J\,mol^{-1}}.\] \[\Delta_rU^\circ\simeq-111.5\ \mathrm{kJ\,mol^{-1}}.\]

The enthalpy change must be distinguished from the internal-energy change.