Reaction Energetics — Internal Energy and Enthalpy
Theoretical Background
-
First Law of Thermodynamics:
\(\Delta U = q + w\) -
Enthalpy definition:
\(H = U + pV \;\Rightarrow\; \Delta H = \Delta U + \Delta(pV)\) -
At constant pressure, when the only mechanical work is pressure–volume work: $ \Delta H=q_p. $
-
For an ideal-gas reaction at a fixed temperature: $ \Delta_r H^\circ =\Delta_r U^\circ+\Delta\nu_{\text{gas}}RT, \qquad \Delta_r U^\circ =\Delta_r H^\circ-\Delta\nu_{\text{gas}}RT, $ where $\Delta\nu_{\text{gas}}$ is the sum of the gaseous stoichiometric coefficients of the products minus that of the reactants.
Validity note.
The relation containing $\Delta\nu_{\text{gas}}RT$ follows from the ideal-gas equation for the gaseous species. It is exact for the all-gas ideal reaction considered below. Real gases require an appropriate equation of state.
Sign convention used here.
$w>0$ denotes work done on the system, while $w<0$ denotes work done by the system. With this convention, the First Law is $\Delta U=q+w$.
Exercise
Consider the combustion of carbon monoxide:
\[2\,\mathrm{CO}(g) + \mathrm{O}_2(g) \;\longrightarrow\; 2\,\mathrm{CO}_2(g)\]At $T = 298\,\text{K}$ and $p = 1\,\text{bar}$, the standard enthalpy of reaction is:
\[\Delta_r H^{\circ} = -566.0\,\text{kJ mol}^{-1}\]Tasks:
- Calculate the standard molar internal-energy change $\Delta_r U^{\circ}$.
- Explain the relation between $\Delta U$ and $\Delta H$ for reactions involving gases.
Step-by-Step Solution
Step 1. Count moles of gas
- Reactants: $n_{\text{gas}} = 2 + 1 = 3$
- Products: $n_{\text{gas}} = 2$
So: \(\Delta\nu_{\text{gas}} = \sum\nu_{\text{products}}-\sum\nu_{\text{reactants}} =2-(2+1)=-1\)
Step 2. Relation between $\Delta H$ and $\Delta U$
For ideal gases:
\(\Delta_r U^\circ
=\Delta_r H^\circ-\Delta\nu_{\text{gas}}RT\)
Step 3. Insert data
With $T = 298\,\text{K}$ and $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$:
\(\Delta\nu_{\text{gas}}RT
=(-1)(8.314)(298)
=-2.48 \times 10^{3}\,\text{J mol}^{-1}
=-2.48\,\text{kJ mol}^{-1}\)
Step 4. Final value
\(\Delta_r U^{\circ}
=(-566.0)\,\text{kJ mol}^{-1}
-(-2.48)\,\text{kJ mol}^{-1}
=-563.5\,\text{kJ mol}^{-1}.\)
Answer:
\(\Delta_r U^{\circ} = -563.5\,\text{kJ mol}^{-1}\)
Notes
- The difference between $\Delta_r U^\circ$ and $\Delta_r H^\circ$ is small here because $\Delta\nu_{\text{gas}}=-1$.
- If $\Delta\nu_{\text{gas}}=0$ for an ideal-gas reaction, then $\Delta_r H^\circ=\Delta_r U^\circ$.
- The magnitude of the correction $\Delta\nu_{\text{gas}}RT$ grows with temperature and with the change in gaseous stoichiometric coefficients.
- Standard reaction enthalpies are commonly tabulated, while the First Law is written directly in terms of internal energy.
- For non-ideal gases, the simple $RT\Delta\nu_{\text{gas}}$ correction is only approximate.