How to Measure the Heat of a Reaction That Never Happens Cleanly

Some reactions can be run in an insulated container and their temperature change measured directly. Others — like burning carbon to carbon monoxide only — never happen cleanly, because some carbon dioxide always forms too.

For those, chemists combine reactions that can be measured, using the fact that enthalpy is a state function.

This part covers calorimetry, standard enthalpies of formation, Hess's law, and the main kinds of enthalpy change. Take $R=8.314\times 10^{-3}$ kJ/mol K.

**A bomb calorimeter works at constant volume, so the heat it measures gives $\Delta U$; a constant-pressure calorimeter gives $\Delta H$ directly; in both, the heat released by the reaction equals the heat gained by the calorimeter, $q=C\Delta T$.

Worked example 1 — bomb calorimeter.** $0.30$ g of graphite ($0.025$ mol) burns in excess oxygen. The calorimeter's heat capacity is $10.0$ kJ/K and its temperature rises by $0.98$ K.

$$q=10.0\times 0.98=9.8\text{ kJ absorbed by the calorimeter}$$

$$\Delta U=-\frac{9.8}{0.025}=-392\text{ kJ/mol}$$

For C(s) + O${}_2$(g) $\to$ CO${}_2$(g), $\Delta n_g=0$, so $\Delta H=\Delta U=-392$ kJ/mol.

Worked example 2 — constant pressure. $50$ mL of $1.0$ M HCl is mixed with $50$ mL of $1.0$ M NaOH in an insulated cup, and the temperature rises by $6.8$ K. Taking $100$ g of solution with specific heat $4.18$ J/g K:

$$q=100\times 4.18\times 6.8\approx 2842\text{ J}\Delta H=-\frac{2.842}{0.050}\approx -56.8\text{ kJ/mol}$$

An everyday example. An insulated flask with a thermometer makes a simple constant-pressure calorimeter for reactions in solution.

The substance. The sign is negative for the reaction because the heat that warms the calorimeter came out of the reacting system.

**The standard enthalpy of formation, ${\Delta }_f{H}^{\circ }$, is the enthalpy change when one mole of a compound forms from its elements in their standard states at $1$ bar; a reaction's standard enthalpy is ${\Delta }_r{H}^{\circ }=\sum {\Delta }_f{H}^{\circ }(\text{products})-\sum {\Delta }_f{H}^{\circ }(\text{reactants})$.**

Elements in their standard states — O${}_2$(g), H${}_2$(g), C(graphite) — have ${\Delta }_f{H}^{\circ }=0$.

Worked example 1 — burning methane. CH${}_4$(g) + 2O${}_2$(g) $\to$ CO${}_2$(g) + 2H${}_2$O(l), with ${\Delta }_f{H}^{\circ }$ values of $-74.8$ (CH${}_4$), $-393.5$ (CO${}_2$) and $-285.8$ (H${}_2$O) kJ/mol:

$${\Delta }_r{H}^{\circ }=[-393.5+2(-285.8)]-[-74.8+0]=-965.1+74.8=-890.3\text{ kJ/mol}$$

Worked example 2 — decomposing limestone. CaCO${}_3$(s) $\to$ CaO(s) + CO${}_2$(g), with ${\Delta }_f{H}^{\circ }$ values of $-1206.9$, $-635.1$ and $-393.5$ kJ/mol:

$${\Delta }_r{H}^{\circ }=(-635.1-393.5)-(-1206.9)=+178.3\text{ kJ/mol}$$

The reaction is endothermic, which is why limestone must be strongly heated.

An everyday example. CNG buses and LPG stoves run on reactions like methane's combustion, releasing about $890$ kJ for every mole burned.

The substance. **Multiply each ${\Delta }_f{H}^{\circ }$ by its coefficient** in the balanced equation before adding.

Hess's law of constant heat summation says the total enthalpy change of a reaction is the same whether it happens in one step or several, so known reactions can be added, reversed or scaled to give an unknown one.

Rules: reversing a reaction changes the sign of $\Delta H$; multiplying a reaction by a number multiplies $\Delta H$ by that number.

Worked example 1 — carbon monoxide. Find $\Delta H$ for C(graphite) + $\frac{1}{2}$O${}_2$(g) $\to$ CO(g), given:

- C(graphite) + O${}_2$(g) $\to$ CO${}_2$(g), $\Delta H=-393.5$ kJ/mol
- CO(g) + $\frac{1}{2}$O${}_2$(g) $\to$ CO${}_2$(g), $\Delta H=-283.0$ kJ/mol

Keep the first and reverse the second:

$$\Delta H=-393.5-(-283.0)=-110.5\text{ kJ/mol}$$

Worked example 2 — formation of methane from combustion data. With ${\Delta }_{c}H$ of $-393.5$ (C), $-285.8$ (H${}_2$) and $-890.3$ (CH${}_4$) kJ/mol, for C + 2H${}_2$ $\to$ CH${}_4$:

$${\Delta }_{f}H=-393.5+2(-285.8)-(-890.3)=-74.8\text{ kJ/mol}$$

An everyday example. Reaching a hill village by the direct road or by a longer bus route gives the same rise in height — just as every route between reactants and products gives the same $\Delta H$.

The substance. Hess's law works because enthalpy is a state function, depending only on start and end states.

Each is the enthalpy change for one particular process per mole — burning completely, splitting into gaseous atoms, breaking one kind of bond, separating an ionic solid into gaseous ions, dissolving, diluting, or changing phase — and a Born-Haber cycle uses Hess's law to find lattice enthalpy from measurable steps.

- Combustion: complete burning in oxygen, always negative — methane $-890.3$ kJ/mol
- Atomisation: H${}_2$(g) $\to$ 2H(g), $+435.0$ kJ/mol
- Bond enthalpy: CH${}_4$ needs about $1.66\times 10^3$ kJ/mol to break into atoms, so the average C–H bond is about $416$ kJ/mol
- Phase transition: melting ice, ${\Delta }_{fus}H=+6.01$ kJ/mol
- Dilution: the heat change as more solvent is added to a solution

Worked example 1 — Born-Haber cycle for NaCl (kJ/mol): sublimation of Na $+108.4$, ionisation of Na $+496$, half the Cl–Cl bond $+121$, electron gain by Cl $-348.6$, and ${\Delta }_{f}H$ of NaCl $-411.2$.

$${\Delta }_{lattice}H=108.4+496+121-348.6-(-411.2)=+788\text{ kJ/mol}$$

Worked example 2 — solution. ${\Delta }_{sol}H={\Delta }_{lattice}H+{\Delta }_{hyd}H$. For NaCl, $788+(-784)=+4$ kJ/mol, so salt dissolves with a very slight cooling.

An everyday example. Instant cold packs used for sports injuries contain a salt whose dissolving absorbs heat, so the pack feels cold.

The substance. Whether dissolving is exothermic or endothermic depends on which is larger — the lattice enthalpy or the hydration enthalpy.