Heat of Reaction

Enthalpy is an extensive quantity: run a reaction twice and you get twice the heat. That is the entire justification for $q = n\Delta H$, and it is why tabulating a single per-mole figure is enough to describe a reaction at any scale from a test tube to a boiler. The sign convention runs from the system's point of view — ΔH negative for an exothermic reaction, because the system's enthalpy falls as heat leaves it. A negative $q$ is not an error message; it is the answer telling you the heat came out.

Something concrete. Heating 200 L of water from 10 °C to 60 °C takes $200 \times 4.186 \times 50 = 41\,860$ kJ. Methane burns at ΔH = −890.3 kJ/mol, so the amount required is $n = 41\,860/890.3 = 47.0$ mol — about 754 g of methane, or 1.05 m³ at STP, before any consideration of how much of that heat actually reaches the water. Chain those three pages together and you have most of a combustion calculation.

The relation's real power is that enthalpy is a state function, which is Hess's law: the heat of a reaction depends only on where it starts and ends, not on the route. So you can add reactions like algebra, and you can build any ΔH you need from tabulated standard enthalpies of formation — products minus reactants — without ever running the reaction. One caution on what is being measured: ΔH is the heat at constant pressure, which is what an open vessel or a flowing burner delivers. A bomb calorimeter holds volume constant and measures ΔU instead, and the two differ by the work done pushing the atmosphere aside, $\Delta H = \Delta U + \Delta n_{gas}RT$.

The question this page most needs you to ask is "per mole of what?" ΔH belongs to the balanced equation as written, not to any one substance in it. $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ has ΔH = −571.6 kJ, but $\text{H}_2 + \tfrac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}$ has ΔH = −285.8 kJ. Same chemistry, same physical world, different bookkeeping — and a factor of two waiting for anyone who reads a table without reading the equation above it. Write the equation first, decide which species your $n$ counts, and make the two agree.

Then the trap that costs more marks than any other in stoichiometry. The limiting reagent is found by moles, never by mass. Given 100 g of hydrogen and 100 g of oxygen, the masses are equal and the amounts are not remotely: 49.6 mol of H₂ against 3.13 mol of O₂, and the reaction needs two hydrogens per oxygen. Oxygen limits by a factor of eight, and 100 g of hydrogen looks generous only because hydrogen is light. Convert everything to moles, divide each by its coefficient in the balanced equation, and the smallest quotient is the limiter. The $n$ that goes into $q = n\Delta H$ is then the extent of reaction that limiter permits — not the amount of whatever you happened to weigh out.

One last note for anyone comparing fuel figures. Combustion enthalpies come in two flavours depending on whether the product water is counted as liquid or as vapour, differing by 44 kJ per mole of water. That is the higher-heating-value and lower-heating-value distinction, and for methane it is about a 10% gap. Two sources can disagree by that much while both being correct.