Reaction Enthalpy (Hess's Law)

Add up the standard enthalpies of formation of the products, add up those of the reactants and take the difference: ΔH°$R = \Sigma{}\Delta{}H$°f(products) − ΣΔH°f(reactants). Each value is multiplied by its stoichiometric coefficient; elements in their standard state count as zero. Example methane combustion CH₄ + 2 O₂ → CO₂ + 2 H₂O(l): products $-393.5 + 2\cdot{}(-285.8) = -\frac{965.1 kJ}{mol}$, reactants $-74.9 + 0 = -\frac{74.9 kJ}{mol}$, so ΔH°$R = -965.1 - (-74.9) = -\frac{890.2 kJ}{mol}$. The negative sign indicates an exothermic reaction. The tabulated values refer to 298 K and 1 bar; mind the stated physical state of the substances.

The total heat exchanged in a reaction depends only on the initial and final state, not on the path in between. Whether carbon burns directly to CO₂ or first to CO and then on to CO₂: the sum of the reaction enthalpies is the same in both cases, namely $-\frac{393.5 kJ}{mol}$. The reason is that enthalpy is a state function, comparable to altitude in mountaineering: the height difference between valley and summit is independent of the chosen route. Practically this allows unmeasurable reactions to be assembled from measurable ones. The formation of CO from the elements can hardly be measured cleanly, but from two combustion enthalpies it can be calculated exactly: $-393.5 - (-283.0) = -\frac{110.5 kJ}{mol}$.

Three patterns dominate. First, the order: it is products minus reactants; calculating the other way round turns every exothermic reaction into an endothermic one. Second, reversing partial equations: if a reaction is used backwards, the sign of its enthalpy flips; when combining given equations this is the most frequent error. Third, subtracting negative values: −965.1 − (−74.9) is −890.2, not −1040; the double negative becomes an addition. Add to this the classic of wrongly assigning an enthalpy of formation to elements such as O₂; by definition they have ΔH°$f = 0$. An order-of-magnitude and sign check at the end (combustions are exothermic!) catches most of these errors.

Because changing the physical state itself costs or releases energy. Liquid water has ΔH°$f = -\frac{285.8 kJ}{mol}$, water vapour only $-\frac{241.8 kJ}{mol}$; the difference of $\frac{44.0 kJ}{mol}$ is exactly the molar enthalpy of vaporization. If you calculate a combustion with H₂O(g) instead of H₂O(l), the result is therefore 44 kJ less exothermic per mole of water. Precisely this difference lies behind the terms gross calorific value (water liquid, heat of condensation used) and net calorific value (water gaseous). In tasks you must therefore consistently use the tabulated values for the stated state and take the state symbols (s), (l), (g) in the equation seriously.

The reaction enthalpy ΔH is the energy balance between reactants and products: it states how much heat is released or absorbed overall and thus determines the thermodynamics. The activation energy E_A, by contrast, is the energy barrier on the way to the transition state: it determines how fast the reaction proceeds, i.e. the kinetics. The two are independent: the oxyhydrogen reaction is strongly exothermic at $\Delta{}H = -\frac{286 kJ}{mol}$, yet does not proceed noticeably at room temperature because of the high barrier, until a spark or catalyst starts it. A catalyst lowers only E_A, never ΔH. In the energy diagram ΔH is the height difference of the plateaus, E_A the hill in between.