Energetics and Thermochemistry

Exothermic: $\Delta H < 0$ (energy released, temperature rises) Endothermic: $\Delta H > 0$ (energy absorbed, temperature falls)

Standard Enthalpy Changes

• $\Delta H_c^\ominus$: combustion of 1 mol
• $\Delta H_f^\ominus$: formation of 1 mol from elements
• $\Delta H_{neut}^\ominus$: formation of 1 mol H₂O from acid + base
• $\Delta H_{at}^\ominus$: formation of 1 mol gaseous atoms

$q = mc\Delta T$

$\Delta H = -\frac{q}{n}$

Sources of error: heat loss, incomplete combustion, assumptions about specific heat capacity.

$\Delta H$ is independent of the route taken.

Using formation enthalpies: $\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$

Using combustion enthalpies: $\Delta H_r = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$

$\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$

Exothermic: more energy released making bonds than used breaking them Endothermic: more energy used breaking bonds

Bond enthalpies are averages → approximate values.

Apply Hess's law to ionic compound formation:

$\Delta H_f = \Delta H_{at}(\text{metal}) + \Delta H_{at}(\text{non-metal}) + IE + EA + \Delta H_{LE}$

Lattice enthalpy depends on:

• Ion charge (higher charge → more exothermic)
• Ion size (smaller ions → more exothermic)

Entropy ($S$): measure of disorder. $S_{gas} \gg S_{liquid} > S_{solid}$

$\Delta S = \sum S(\text{products}) - \sum S(\text{reactants})$

Gibbs free energy: $\Delta G = \Delta H - T\Delta S$

• $\Delta G < 0$: spontaneous/feasible
• $\Delta G = 0$: equilibrium
• $\Delta G > 0$: non-spontaneous

Calculate $\Delta H$ for $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$

Bonds broken: 4(C-H) + 2(O=O) = 4(414) + 2(498) = 2652 kJ Bonds made: 2(C=O) + 4(O-H) = 2(804) + 4(463) = 3460 kJ

$\Delta H = 2652 - 3460 = -808$ kJ/mol

• $\Delta H < 0$ exothermic; $\Delta H > 0$ endothermic
• Hess's law: $\Delta H$ is route-independent
• Bond enthalpies: $\Delta H = \text{broken} - \text{made}$
• HL: Born-Haber cycles for lattice enthalpy; $\Delta G = \Delta H - T\Delta S$