Thermochemistry and Energetics
Sign conventions and calorimetry
ΔH is negative for exothermic reactions (energy released, products lower in enthalpy) and positive for endothermic ones. Calorimetry uses q = mcΔT, where m is the mass of the solution, c is its specific heat capacity (4.18 J g-1 K-1 for water) and ΔT is the temperature change. In an insulated cup, q(reaction) = -q(solution). A bomb calorimeter measures ΔH at constant volume.
Hess's law and formation enthalpies
Enthalpy is a state function, so the route does not matter: ΔH(total) = sum of the steps, and reversing a step changes the sign. Standard enthalpy of formation, ΔHf°, is the enthalpy change when one mole of a compound forms from its elements in their standard states; ΔHf° of an element in its standard state is zero. Then ΔHrxn° = ΣΔHf°(products) - ΣΔHf°(reactants).
Bond energies give another route: ΔH = Σ(bonds broken) - Σ(bonds formed).
Entropy and spontaneity
A reaction is spontaneous when ΔG = ΔH - TΔS is negative. Melting and dissolution of most solids have positive ΔS; forming gas from liquid or solid strongly increases entropy.
Worked example
The combustion of methane, CH4 + 2O2 -> CO2 + 2H2O(l), has ΔH = -890.3 kJ/mol. Given ΔHf°(CO2) = -393.5 and ΔHf°(H2O,l) = -285.8 kJ/mol, find ΔHf°(CH4).
-890.3 = [-393.5 + 2(-285.8)] - [ΔHf°(CH4) + 0] -890.3 = -965.1 - ΔHf°(CH4), so ΔHf°(CH4) = -74.8 kJ/mol.
Traps
Forgetting to multiply ΔHf° by the stoichiometric coefficient. Assigning a nonzero formation enthalpy to an element in its standard state. Reporting q(reaction) as positive in an exothermic combustion. Forgetting that bond breaking costs energy and bond forming releases it.