What is Standard reference state, standard enthalpy of formation? — Chemistry

Definition 1.7 University Chemistry — Year 2 · Chapter 1 — Enthalpies of Reaction

The standard reference state of an element at temperature $T$ is its most stable form at $T$ under $p^\circ$ : $\ce{H2}$ , $\ce{O2}$ , $\ce{N2}$ and $\ce{Cl2}$ as perfect gases, carbon as graphite, sodium as the metal, bromine as the liquid (at $298\,\mathrm{K}$ ). The standard enthalpy of formation $\Delta_f H^\circ$ of a compound is the standard reaction enthalpy of its formation reaction, the one that makes one mole of the compound from its elements in their reference states. By construction, $\Delta_f H^\circ = 0$ for an element in its reference state.

Enthalpy levels for the combustion of methane at 298.15\, K. The elements in their reference states are the zero. Methane lies 74.9\, kJ/ mol below them, the products 965.2\, kJ/ mol below: the combustion enthalpy is the difference, -890.3\, kJ/ mol, whatever the path actually followed by the flame. — Enthalpy levels for the combustion of methane at $298.15\,\mathrm{K}$ . The elements in their reference states are the zero. Methane lies $74.9\,\mathrm{kJ}/\mathrm{mol}$ below them, the products $965.2\,\mathrm{kJ}/\mathrm{mol}$ below: the combustion enthalpy is the difference, $-890.3\,\mathrm{kJ}/\mathrm{mol}$ , whatever the path actually followed by the flame.

Examples

Example 1.8 (Formation reactions)

The formation reaction of water is $\ce{H2(g) + 1/2O2(g) -> H2O(l)}$ , with $\Delta_f H^\circ = -285.83\,\mathrm{kJ}/\mathrm{mol}$ at $298.15\,\mathrm{K}$ ; that of methane is $\ce{C(s) + 2H2(g) -> CH4(g)}$ , $\Delta_f H^\circ = -74.9\,\mathrm{kJ}/\mathrm{mol}$ ; that of nitrogen monoxide, $\ce{1/2N2(g) + 1/2O2(g) -> NO(g)}$ , $\Delta_f H^\circ = +90.3\,\mathrm{kJ}/\mathrm{mol}$ : an endothermic compound, which the hot gases of an engine nevertheless make. The half coefficients are not a problem: the formation reaction is a bookkeeping device, not a mechanism.

Example 1.11 (Combustion of methane)

For $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$ , $\Delta_r H^\circ = -393.51 + 2(-285.83) - (-74.87) - 0 = -890.3\,\mathrm{kJ}/\mathrm{mol}$ at $298.15\,\mathrm{K}$ . The value measured in a calorimeter is $-890.7\,\mathrm{kJ}/\mathrm{mol}$ : the table and the flame agree within the uncertainty of the data. If the water leaves as vapour ( $\Delta_f H^\circ = -241.83\,\mathrm{kJ}/\mathrm{mol}$ ), $\Delta_r H^\circ = -802.3\,\mathrm{kJ}/\mathrm{mol}$ : the difference, $88.0\,\mathrm{kJ}$ , is the enthalpy of vaporisation of two moles of water, which a gas boiler recovers by condensing its fumes.

Example 1.13 (Carbon to carbon monoxide)

Burning carbon always makes some carbon dioxide, so the enthalpy of $\ce{C(s) + 1/2O2(g) -> CO(g)}$ is not measured directly. The two combustions $\ce{C(s) + O2(g) -> CO2(g)}$ ( $-393.51\,\mathrm{kJ}/\mathrm{mol}$ ) and $\ce{CO(g) + 1/2O2(g) -> CO2(g)}$ ( $-283.0\,\mathrm{kJ}/\mathrm{mol}$ , measured) are; the target is the first minus the second: $\Delta_r H^\circ = -393.51 + 283.0 = -110.5\,\mathrm{kJ}/\mathrm{mol}$ , the formation enthalpy of carbon monoxide.

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