Standard Enthalpy of Reaction from Enthalpies of Formation
The standard enthalpy of reaction, \( \Delta_{r}H^{\circ} \), is the heat change when a
reaction written with a specific balanced equation proceeds from reactants in their
standard states to products in their standard states (1 bar, usually \(25^\circ\text{C}\)).
Using Hess’s law and tabulated standard enthalpies of formation
\( \Delta_{f}H^{\circ} \) (kJ·mol\(^{-1}\)), we can compute \( \Delta_{r}H^{\circ} \)
without performing calorimetry.
The key relation (use coefficients \(n\), not Greek \(\nu\))
\[
\boxed{\;
\Delta_{r}H^{\circ}
\;=\;
\sum_{\text{products}} n\,\Delta_{f}H^{\circ}
\;-\;
\sum_{\text{reactants}} n\,\Delta_{f}H^{\circ}
\;}
\]
- \(n\) = the stoichiometric coefficient from the balanced equation (pure numbers).
- \(\Delta_{f}H^{\circ}\) is per mole of species as written (kJ·mol\(^{-1}\)).
- By convention, \(\Delta_{f}H^{\circ}\) of any element in its standard state
is zero (e.g., \( \mathrm{O_{2}(g)}\), \( \mathrm{H_{2}(g)}\), C(graphite)).
How this tool works
- Enter an unbalanced reaction. The tool balances it and lists each species with its coefficient \(n\).
- Enter \( \Delta_{f}H^{\circ} \) values for each species (kJ·mol\(^{-1}\)). Phases matter—use the tabulated value for the phase in your equation (e.g., \( \mathrm{H_{2}O(l)} \) vs \( \mathrm{H_{2}O(g)} \)).
- Choose the task:
- Calculate \( \Delta_{r}H^{\circ}\) from the table; or
- Solve for one unknown \( \Delta_{f}H^{\circ}\) when \( \Delta_{r}H^{\circ}\) and the other formation enthalpies are given.
- Press Calculate. The steps show the symbolic sum, substitution, and the numerical result with your chosen significant figures.
Worked patterns
1) Find \( \Delta_{r}H^{\circ} \) (combustion-like example)
Balanced equation:
\[
\mathrm{C_{2}H_{6}(g)} + \tfrac{7}{2}\,\mathrm{O_{2}(g)}
\longrightarrow 2\,\mathrm{CO_{2}(g)} + 3\,\mathrm{H_{2}O(l)}
\]
Apply the formula (remember \( \Delta_{f}H^{\circ}[\mathrm{O_{2}(g)}]=0\)):
\[
\Delta_{r}H^{\circ}
= \big(2\,\Delta_{f}H^{\circ}[\mathrm{CO_{2}(g)}] + 3\,\Delta_{f}H^{\circ}[\mathrm{H_{2}O(l)}]\big)
- \big(1\,\Delta_{f}H^{\circ}[\mathrm{C_{2}H_{6}(g)}] + \tfrac{7}{2}\cdot 0\big).
\]
2) Solve for an unknown \( \Delta_{f}H^{\circ} \) (decomposition example)
Balanced equation:
\[
2\,\mathrm{NaHCO_{3}(s)} \longrightarrow \mathrm{Na_{2}CO_{3}(s)} + \mathrm{H_{2}O(l)} + \mathrm{CO_{2}(g)}
\]
Rearrange to isolate the unknown (if, say, \( \Delta_{f}H^{\circ}[\mathrm{NaHCO_{3}(s)}] \) is unknown and \( \Delta_{r}H^{\circ} \) known):
\[
\Delta_{r}H^{\circ}
= \big(\Delta_{f}H^{\circ}[\mathrm{Na_{2}CO_{3}(s)}] + \Delta_{f}H^{\circ}[\mathrm{H_{2}O(l)}] + \Delta_{f}H^{\circ}[\mathrm{CO_{2}(g)}]\big)
- 2\,\Delta_{f}H^{\circ}[\mathrm{NaHCO_{3}(s)}]
\]
\[
\Rightarrow\quad
\Delta_{f}H^{\circ}[\mathrm{NaHCO_{3}(s)}]
= \frac{ \big(\Delta_{f}H^{\circ}[\mathrm{Na_{2}CO_{3}(s)}] + \Delta_{f}H^{\circ}[\mathrm{H_{2}O(l)}] + \Delta_{f}H^{\circ}[\mathrm{CO_{2}(g)}]\big)
- \Delta_{r}H^{\circ} }{2}.
\]
Tips & common pitfalls
- Balance first. Coefficients \(n\) must match the final balanced equation shown by the tool.
- Zero for elements. Enter 0 for elements in their standard states
(e.g., \( \mathrm{O_{2}(g)}\), \( \mathrm{N_{2}(g)}\), \( \mathrm{H_{2}(g)}\), C(graphite), \( \mathrm{Na(s)} \)).
- Correct phase. Use \( \mathrm{H_{2}O(l)} \) for combustion tables unless a gas-phase value is specifically requested.
- Signs. Negative \( \Delta \) means heat released (exothermic), positive means absorbed (endothermic).
- Units. Input and output are in kJ (and kJ·mol\(^{-1}\) for formation values). The result for \( \Delta_{r}H^{\circ} \) is per the balanced equation as written.
- Significant figures. The tool formats the final value using your selected sig figs.
Why Hess’s law works here
Enthalpy is a state function. Writing each compound as formed from elements,
adding product formations and subtracting reactant formations yields the same net path
as the reaction itself; therefore the enthalpy change is the same—precisely the
difference of the weighted sums above.
