Hess's Law: Finding ΔH by Adding Reactions

Why a state function makes Hess's law work

Enthalpy is a state function: its value depends only on the current state of a system, not on how the system got there. The change ΔH between a starting state and an ending state is therefore the same for every route between them. Elevation works the same way. Whether you climb a hill by the steep trail or the long switchbacks, your change in height is identical.

Hess's law follows directly. If a reaction can be written as the sum of two or more steps, its ΔH is the sum of the steps' ΔH values. The steps do not have to happen in reality; they only need to be valid chemical equations whose enthalpies are known. That freedom is what makes the law useful for reactions you cannot run cleanly in a calorimeter.

Three rules for manipulating thermochemical equations

Each rule follows from the fact that ΔH is an extensive state function. Reversing a process returns the system to its starting state, so the enthalpy change must be equal and opposite. Doubling the amount of reaction doubles the energy involved.

Before combining anything, write the target equation at the top of your work. Then find, in the given equations, each substance that appears only once. Its position and coefficient tell you whether that equation must be reversed and by what factor it must be multiplied.

• Reverse an equation and you change the sign of its ΔH.
• Multiply an equation by a factor and you multiply its ΔH by the same factor. Fractions such as ½ are allowed.
• Add equations and you add their ΔH values. Species that appear on both sides of the sum cancel, but only if their physical states match.

Worked example: reversing and adding two equations

Find ΔH° for converting graphite to diamond, C(graphite) → C(diamond). The direct conversion needs extreme pressure and cannot be run in a calorimeter, but both forms of carbon burn. Use (1) C(graphite) + O₂(g) → CO₂(g), ΔH° = −393.51 kJ, and (2) C(diamond) + O₂(g) → CO₂(g), ΔH° = −395.40 kJ. The first value is from OpenStax Chemistry 2e Appendix G; the second was calculated from the same table.

Graphite is a reactant in (1) and in the target, so keep (1). Diamond is a reactant in (2) but a product in the target, so reverse (2): CO₂(g) → C(diamond) + O₂(g), ΔH° = +395.40 kJ. Adding the two, O₂ and CO₂ appear once on each side and cancel, leaving exactly the target. ΔH° = −393.51 kJ + 395.40 kJ = +1.89 kJ.

Interpret the answer. The positive sign means diamond lies slightly higher in enthalpy than graphite, consistent with graphite, not diamond, being the reference form of carbon. The small size of the number is a reminder that subtracting two large, nearly equal values magnifies their uncertainty.

Worked example: three equations and a cross-check

Find ΔH° for the hydrogenation C₂H₄(g) + H₂(g) → C₂H₆(g) from combustion data. Use (1) C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l), ΔH° = −1411.1 kJ; (2) H₂(g) + ½ O₂(g) → H₂O(l), ΔH° = −285.83 kJ; and (3) C₂H₆(g) + 7/2 O₂(g) → 2 CO₂(g) + 3 H₂O(l), ΔH° = −1560.5 kJ. These combustion values were calculated from Appendix G formation data so that you can reproduce them.

C₂H₄ and H₂ are reactants in both the target and equations (1) and (2), so keep those as written. C₂H₆ is a product in the target but a reactant in (3), so reverse (3) and change its sign to +1560.5 kJ. Now tally: oxygen is 3 + ½ = 7/2 O₂ on the left and 7/2 O₂ on the right; carbon dioxide is 2 on each side; water is 2 + 1 = 3 on the right and 3 on the left. Everything cancels except the target. ΔH° = −1411.1 − 285.83 + 1560.5 = −136.4 kJ, reported to one decimal place because the least precise values have one.

Cross-check with formation enthalpies: ΔH°f of C₂H₆(g) is −84.0 kJ/mol and of C₂H₄(g) is +52.4 kJ/mol, and H₂(g) is zero, so ΔH° = −84.0 − 52.4 = −136.4 kJ. Two different routes give the same answer, which is exactly what a state function guarantees.

Standard enthalpies of formation and reference states

The standard enthalpy of formation, ΔH°f, is the enthalpy change when one mole of a compound forms from its elements, each in its reference form, under standard conditions. The reference form is the most stable form with rare exceptions: phosphorus uses white P₄(s), which Appendix G lists at zero. IUPAC sets standard pressure at 1 bar; many tables still use 1 atm, and the difference is negligible for most enthalpy work. Tabulated values are usually given at 298.15 K.

A formation equation always makes exactly one mole of product, so fractional coefficients are normal: ½ N₂(g) + 3/2 H₂(g) → NH₃(g), ΔH°f = −45.9 kJ/mol. For an element already in its reference form, such as O₂(g), H₂(g), graphite, or liquid bromine, ΔH°f is zero by definition. Other forms of an element are not zero: Appendix G lists diamond at +1.89 kJ/mol relative to graphite.

Calculating ΔH°rxn from formation data

Any reaction can be imagined as two steps: break the reactants down into their elements, then build the products from those elements. The first step is the reverse of the reactants' formation reactions and the second is the products' formation reactions. That gives the working formula ΔH°rxn = Σ n ΔH°f(products) − Σ m ΔH°f(reactants), where n and m are the coefficients in the balanced equation.

Worked example. Find ΔH° for 2 N₂H₄(l) + N₂O₄(g) → 3 N₂(g) + 4 H₂O(g), a hydrazine reaction of the kind used in some rocket engines. From Appendix G: N₂H₄(l) +50.63 kJ/mol, N₂O₄(g) +11.1 kJ/mol, H₂O(g) −241.82 kJ/mol, and N₂(g) 0. Products: 3(0) + 4(−241.82) = −967.28 kJ. Reactants: 2(50.63) + 11.1 = 112.36 kJ. ΔH° = −967.28 − 112.36 = −1079.6 kJ, reported to one decimal place because 11.1 has one. That is −539.8 kJ per mole of hydrazine.

The state of water matters. If the equation produced liquid water, using −285.83 kJ/mol, the answer would be −1255.7 kJ. The difference is four times the enthalpy released when a mole of water vapor condenses.

Errors that break a Hess's law calculation

Most errors in these problems are bookkeeping errors, and a written tally of every species catches nearly all of them. When the cancellation does not reproduce the target exactly, an equation has been reversed or scaled incorrectly.

A second check is to estimate the sign. Combustions and the formation of most stable compounds are exothermic, so a positive answer for burning a fuel signals a reversed step.

• Reversing an equation without changing the sign of its ΔH.
• Multiplying an equation by 2 or ½ but leaving its ΔH unchanged.
• Subtracting in the wrong order: the formula is products minus reactants.
• Forgetting the coefficients when summing formation enthalpies.
• Using ΔH°f for the wrong state, such as H₂O(g) when the equation has H₂O(l).
• Giving an element in its reference form a nonzero ΔH°f, or giving a compound zero.