6.9 Hess's Law

Hess's Law

Step through intermediate reactions on a multi-level energy diagram. Cancelled species strike through and ΔH values accumulate to match the direct path.

Energy LevelsAnimated StepsPath Independence5 Examples
Topic 6.9

Hess’s Law

Represent a chemical or physical process as a sequence of steps; explain the relationship between the enthalpy of a process and the sum of the enthalpies of the individual steps.

Many processes can be broken down into a series of steps, each with its own energy change.

EK 6.9.B.1 gives the justification, and it is worth reading closely because it is a first-law argument: because total energy is conserved, and each individual reaction in a sequence transfers thermal energy to or from the surroundings, the net thermal energy transferred in the sequence equals the sum of the transfers in each step. Those transfers result from potential-energy changes among the species, so at constant pressure the enthalpy change of the overall process equals the sum of the enthalpy changes of the individual steps.

EK 6.9.B.2 lists the two manipulation rules:

  • Reversing a reaction keeps the magnitude of ΔH but reverses its sign.
  • Multiplying a reaction by a factor c multiplies ΔH by the same factor c.

The strategy for solving a Hess's law problem:

  1. Write the target equation and identify where each species must end up.
  2. Find the given equation containing each species. Reverse it if the species is on the wrong side.
  3. Scale it so the coefficient matches the target.
  4. Apply the same operations to each ΔH.
  5. Add all the manipulated equations; everything not in the target must cancel.
  6. Sum the manipulated ΔH values.

Practical anchor: start with a species that appears in only one of the given equations. That fixes the required operation on that equation with no ambiguity, and the rest follows.

The same logic reappears for equilibrium constants (7.6) — except that combining equilibria multiplies K rather than adding it.

Key points

  • Reverse a reaction → flip the sign of ΔH. Multiply a reaction → multiply ΔH.
  • Everything not in the target equation must cancel when the steps are added.
  • Enthalpies of formation (6.8) are just a pre-tabulated application of Hess’s law.
  • Start with a species that appears in only one given equation.

Equations

  • not on the sheet

Common mistakes

  • Forgetting to flip the sign when a reaction is reversed.
  • Scaling the equation but not ΔH (or the reverse).
  • Leaving an uncancelled species. If something extra survives, an operation was wrong.
  • Adding K values for combined equilibria. Enthalpies add; equilibrium constants multiply (7.6).

Worked example

Calculate ΔH for 2 C(graphite) + H₂(g) → C₂H₂(g) given: (1) C₂H₂(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(l) ΔH = −1299.6 kJ (2) C(graphite) + O₂(g) → CO₂(g) ΔH = −393.5 kJ (3) H₂(g) + ½ O₂(g) → H₂O(l) ΔH = −285.8 kJ

Target: 2 C(graphite) + H₂(g) → C₂H₂(g)

Place C₂H₂: it must be a product, but in equation (1) it is a reactant. Reverse (1):
2 CO₂(g) + H₂O(l) → C₂H₂(g) + 5/2 O₂(g) ΔH = +1299.6 kJ

Place C(graphite): needed as a reactant with coefficient 2. Equation (2) has it as a reactant with coefficient 1. Multiply (2) by 2:
2 C(graphite) + 2 O₂(g) → 2 CO₂(g) ΔH = 2(−393.5) = −787.0 kJ

Place H₂: needed as a reactant with coefficient 1. Equation (3) already has it that way. Use as is:
H₂(g) + ½ O₂(g) → H₂O(l) ΔH = −285.8 kJ

Add and cancel: 2 CO₂ cancels; H₂O(l) cancels; oxygen: 2 + ½ = 5/2 on the left cancels 5/2 on the right ✓
Result: 2 C(graphite) + H₂(g) → C₂H₂(g) ✓

ΔH = 1299.6 − 787.0 − 285.8 = +226.8 kJ

Positive, meaning acetylene is less stable than its constituent elements — consistent with its high reactivity.

Full notes for topic 6.9 →