Hess's Law: MDCAT Chemistry notes

Statement

Hess's law of constant heat summation: if a chemical change takes place by several different routes, the overall enthalpy change is the same, whichever route is followed, provided the initial and final states are the same.

$$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \cdots$$

Hess's law follows from the first law of thermodynamics (conservation of energy), because enthalpy is a state function.

Why it is useful

Hess's law lets us find $\Delta H$ indirectly for reactions that cannot be measured directly, for example:

• formation of CO (burning carbon always gives some $\mathrm{CO_2}$ as well)
• conversion of graphite into diamond
• lattice energy of ionic crystals (the Born-Haber cycle is an application of Hess's law)
• enthalpies of formation of many organic compounds from combustion data

Rules for combining equations

• Reverse an equation: change the sign of $\Delta H$.
• Multiply an equation by a number: multiply $\Delta H$ by the same number.
• Add equations: add their $\Delta H$ values. Cancel species that appear on both sides.
• Physical states matter: $\mathrm{I_2(s)}$ and $\mathrm{I_2(g)}$ have different enthalpies.

Worked example 1: enthalpy of formation of CO

Given:

• $\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$, $\Delta H = -393.5\ \mathrm{kJ\,mol^{-1}}$
• $\mathrm{CO(g) + \tfrac12 O_2(g) \rightarrow CO_2(g)}$, $\Delta H = -283.0\ \mathrm{kJ\,mol^{-1}}$

Reverse the second and add to the first:

$$\mathrm{C(s) + \tfrac12 O_2(g) \rightarrow CO(g)}\qquad \Delta H = -393.5 + 283.0 = -110.5\ \mathrm{kJ\,mol^{-1}}$$

Turned around: the heat of combustion of CO equals $\Delta H_f(\mathrm{CO_2}) - \Delta H_f(\mathrm{CO})$.

Worked example 2: graphite to diamond

Combustion enthalpies: graphite $-393.5$, diamond $-395.4\ \mathrm{kJ\,mol^{-1}}$. For $\mathrm{C(graphite) \rightarrow C(diamond)}$: $\Delta H = -393.5 - (-395.4) = +1.9\ \mathrm{kJ\,mol^{-1}}$. The conversion is slightly endothermic, so graphite is the more stable form.

Worked example 3: sublimation of iodine

Sublimation, $\mathrm{I_2(s) \rightarrow I_2(g)}$, is endothermic; the standard value is about $+62\ \mathrm{kJ\,mol^{-1}}$. So a reaction using $\mathrm{I_2(g)}$ always releases about 62 kJ more (or absorbs 62 kJ less) than the same reaction using $\mathrm{I_2(s)}$. When combining such equations, subtract carefully and check that the sign comes out positive for sublimation.

Key formulas

• $\Delta H_{\text{overall}} = \sum \Delta H_{\text{steps}}$
• $\Delta H_{\text{reaction}} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
• $\Delta H_{\text{reaction}} = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$

Common MDCAT traps

• The Born-Haber cycle applies Hess's law, not Le Chatelier's principle or Henry's law.
• Hess's law is a consequence of the first law of thermodynamics.
• Reversing an equation flips the sign of $\Delta H$; forgetting this is the commonest error.
• Graphite to diamond is $+1.9$ kJ/mol, not negative.
• Heat of combustion of CO is about $-283$ kJ/mol (or $-67.6$ kcal/mol), not the formation value of CO.

Quick revision

• Overall $\Delta H$ does not depend on the route.
• Hess's law is used to find $\Delta H$ indirectly.
• $\Delta H_f$ of CO is about $-110.5$ kJ/mol.
• Sublimation of iodine is endothermic, about +62 kJ/mol.