How Thermochemical Equations Scale and Reverse ΔH

ΔH belongs to the reaction as written

A thermochemical equation combines a balanced chemical equation with an enthalpy change. The coefficients define the amount of reaction to which that $\Delta H$ applies. Because enthalpy is an extensive property, doubling every coefficient doubles $\Delta H$; halving the equation halves $\Delta H$.

For example,

$$\mathrm{2H_2(g)+O_2(g)\rightarrow2H_2O(l)} \qquad \Delta H=-571.6\,\mathrm{kJ}.$$

This statement means that forming $2\,\mathrm{mol\ H_2O(l)}$ from the stated reactants releases $571.6\,\mathrm{kJ}$ under the specified conditions.

Scale coefficients and enthalpy together

For one mole of liquid water, divide the entire equation and $\Delta H$ by 2:

$$\mathrm{H_2(g)+\tfrac12O_2(g)\rightarrow H_2O(l)} \qquad \Delta H=-285.8\,\mathrm{kJ}.$$

The value may also be written as $-285.8\,\mathrm{kJ\,mol^{-1}\ H_2O}$ when the per-mole basis is stated explicitly. A bare $\Delta H$ for the balanced equation and a molar enthalpy are related, but they are not interchangeable without naming the stoichiometric basis.

Reverse the reaction, reverse the sign

Reversing the equation reverses the direction of energy transfer:

$$\mathrm{2H_2O(l)\rightarrow2H_2(g)+O_2(g)} \qquad \Delta H=+571.6\,\mathrm{kJ}.$$

The magnitude stays the same because the initial and final states are exchanged; the sign changes because an exothermic forward reaction becomes endothermic in reverse.

Preserve states and conditions

Physical states are part of a thermochemical equation. Forming $\mathrm{H_2O(g)}$ does not have the same $\Delta H$ as forming $\mathrm{H_2O(l)}$, because condensation itself has an enthalpy change. Temperature, pressure, and phase should remain attached to a reported value.

Before using $q/n$, identify what $n$ counts—moles of reaction, reactant consumed, or product formed—and verify the limiting reactant when amounts of multiple reactants are supplied.