Energy, heat and work
Thermochemistry tracks energy changes in chemical and physical change. The system is the part we study; the surroundings are everything else. Heat q is energy transferred because of a temperature difference, and work w is energy transferred by a force moving something. The first law of thermodynamics says energy is conserved: the change in internal energy is ΔU = q + w, where q and w are counted positive when energy flows into the system. A system that absorbs 300 J of heat and does 120 J of work on the surroundings has w = −120 J, so ΔU = +180 J.
Internal energy, enthalpy and temperature are state functions: they depend only on the current state, not the path. Heat and work are not.
Enthalpy
At constant pressure, the heat exchanged equals the enthalpy change ΔH. A reaction with ΔH < 0 is exothermic (the surroundings warm up); ΔH > 0 is endothermic (the surroundings cool). ΔH is proportional to the amount reacting, and reversing a reaction flips its sign.
Calorimetry
Heat absorbed or released by a substance with no phase change is q = m c ΔT, where c is the specific heat (water is 4.184 J/(g·°C)). In a coffee-cup calorimeter, heat lost by the reaction is gained by the solution: q_reaction = −q_solution.
Hess's law and formation data
Because enthalpy is a state function, you can add reactions and add their ΔH values. The standard enthalpy of formation ΔHf° is the enthalpy change to form one mole of a compound from its elements in their standard states; for an element in its standard state it is zero. Then ΔH_reaction = Σ n ΔHf°(products) − Σ n ΔHf°(reactants).
Worked example
You mix 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH in a calorimeter. The temperature rises by 6.7 °C. Assume the 100.0 g of solution has c = 4.184 J/(g·°C) and the calorimeter absorbs nothing.
- Heat absorbed by the solution: q = (100.0)(4.184)(6.7) = 2803 J = 2.80 kJ.
- The reaction released this heat, so q_reaction = −2.80 kJ.
- Moles of water formed equal moles of the limiting reactant: 0.0500 L × 1.00 mol/L = 0.0500 mol.
- ΔH per mole = −2.80 kJ / 0.0500 mol = −56 kJ/mol (two sig figs from ΔT).
The negative sign matches what we saw: the solution warmed, so the reaction was exothermic. When asked for an answer per mole, always divide by moles, not by grams, and decide the sign by asking whether the surroundings gained energy.