4. Thermochemistry
Thermochemistry tracks heat flow. The system is what you study; the surroundings are everything else. Enthalpy change, $\Delta H$, tells you which way the heat went.
Enthalpy and Sign Conventions
$\Delta H < 0$: exothermic — heat leaves the system (combustion). $\Delta H > 0$: endothermic — heat enters the system (melting ice). Heat absorbed or released by a substance follows $$q = mc\Delta T$$
Example: Heat 250 g of water from 22.0 °C to 98.0 °C ($c = 4.184\ \text{J/g·°C}$):
$$q = (250\ \text{g})(4.184\ \text{J/g·°C})(76.0\ \text{°C}) = 7.95 \times 10^4\ \text{J} \approx 79.5\ \text{kJ}$$
Hess's Law
Enthalpy is a state function — the total $\Delta H$ depends only on start and end points, so reaction steps can be added like algebra.
Example: Find $\Delta H$ for $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$ from
(1) $\text{C} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO}$, $\Delta H = -110.5$ kJ/mol
(2) $\text{CO} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO}_2$, $\Delta H = -283.0$ kJ/mol
Adding (1) + (2) gives the target reaction, so $\Delta H = -110.5 + (-283.0) = -393.5$ kJ/mol.
(1) $\text{C} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO}$, $\Delta H = -110.5$ kJ/mol
(2) $\text{CO} + \tfrac{1}{2}\text{O}_2 \rightarrow \text{CO}_2$, $\Delta H = -283.0$ kJ/mol
Adding (1) + (2) gives the target reaction, so $\Delta H = -110.5 + (-283.0) = -393.5$ kJ/mol.
Standard Enthalpies of Formation
$$\Delta H^\circ_{rxn} = \sum n\,\Delta H^\circ_f(\text{products}) - \sum n\,\Delta H^\circ_f(\text{reactants})$$
Example: For $\text{CH}_4 + 2\,\text{O}_2 \rightarrow \text{CO}_2 + 2\,\text{H}_2\text{O}(l)$, with $\Delta H^\circ_f$ values −74.8 (CH$_4$), 0 (O$_2$), −393.5 (CO$_2$), −285.8 (H$_2$O, kJ/mol):
$$\Delta H^\circ = [-393.5 + 2(-285.8)] - [-74.8] = -890.3\ \text{kJ/mol}$$
Practice Problems
- How much heat is released when 100 g of water cools from 90.0 °C to 25.0 °C?
- Given N2 + O2 → 2 NO, ΔH = +180 kJ, and 2 NO + O2 → 2 NO2, ΔH = −114 kJ, find ΔH for N2 + 2 O2 → 2 NO2.
- Use ΔH°f (−285.8 for H2O(l)) to find ΔH° for 2 H2 + O2 → 2 H2O(l).
- Is the combustion of methane endothermic or exothermic? How do you know?
Answers: 1. 27.2 kJ released (q = −27.2 kJ) 2. +66 kJ 3. −571.6 kJ 4. Exothermic — its ΔH° is −890.3 kJ/mol (negative).