Reaction Energetics — Internal Energy and Enthalpy

Theoretical Background

  • First Law of Thermodynamics:
    \(\Delta U = q + w\)

  • Enthalpy definition:
    \(H = U + pV \;\Rightarrow\; \Delta H = \Delta U + \Delta(pV)\)

  • For reactions at constant $(T,p)$:
    \(\Delta H = q_p\)

  • Connection between $\Delta H$ and $\Delta U$ (ideal gases):
    \(\Delta H = \Delta U + (\Delta n_{\text{gas}})RT \qquad \Rightarrow \qquad \Delta U = \Delta H - (\Delta n_{\text{gas}})RT\)

Validity note (ideal-gas assumption).
The identities above use $pV = nRT$; the correction $(\Delta n_{\text{gas}})RT$ applies when the gas phase behaves ideally. For real gases, use an appropriate equation of state (the simple $RT\,\Delta n$ term no longer holds exactly).

Sign convention used here.
$w>0$ = work done on the system (compression); $w<0$ = work done by the system (expansion). The First Law is written as $\Delta U = q + w$.

where $\Delta n_{\text{gas}}$ is the change in the number of moles of gas.

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:

\[\Delta H^{\circ} = -566.0\,\text{kJ mol}^{-1}\]

Tasks:

  1. Calculate the standard internal energy change $\Delta U^{\circ}$.
  2. 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: \(\Delta n_{\text{gas}} = n_{\text{prod}} - n_{\text{reag}} = 2 - 3 = -1\)

Step 2. Relation between $\Delta H$ and $\Delta U$
For ideal gases: \(\Delta U = \Delta H - (\Delta n_{\text{gas}})RT\)

Step 3. Insert data
With $T = 298\,\text{K}$ and $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$: \((\Delta n_{\text{gas}})RT = (-1)(8.314)(298) = -2.48 \times 10^{3}\,\text{J} = -2.48\,\text{kJ}\)

Step 4. Final value
\(\Delta U^{\circ} = (-566.0)\,\text{kJ mol}^{-1} - (-2.48)\,\text{kJ mol}^{-1} = -563.5\,\text{kJ mol}^{-1}.\)

Answer:
\(\Delta U^{\circ} = -563.5\,\text{kJ mol}^{-1}\)

Notes

  • The difference between $\Delta U$ and $\Delta H$ is small here because only 1 mol of gas disappears.
  • If $\Delta n_{\text{gas}} = 0$, then $\Delta H = \Delta U$.
  • The correction $(\Delta n_{\text{gas}})RT$ grows with $T$ and with large changes in gas moles.
  • Pedagogical point: $\Delta H$ is often tabulated (easy to measure at constant $p$), while $\Delta U$ is the central state function in the First Law.
  • Real-gas reminder: for non‑ideal gases the simple $RT\,\Delta n$ correction is only approximate; use the appropriate equation of state if accuracy is required.