Hess’s Law Calculations

Hess’s law states that the total enthalpy change of a reaction is the same whether it takes place in one step or in several steps, so the enthalpy changes of known reactions can be added to find the enthalpy of an unknown one. In brief: the enthalpy of formation of CO, which is hard to measure directly, follows from C + O2 → CO2 (−393.5 kJ) and CO + ½O2 → CO2 (−283.0 kJ): ΔH = −393.5 − (−283.0) = −110.5 kJ.

Many reactions cannot be carried out cleanly in a calorimeter. They may be too slow, give mixtures of products or require extreme conditions. Hess’s law gets around this by using enthalpy as a state function. This article explains the law, gives the rules for manipulating equations, and works through four problems in increasing difficulty, including one with an enthalpy cycle.

What does Hess’s law say?

Enthalpy is a state function: its change depends only on the initial and final states and not on the path. If a reaction can be written as the sum of several steps, then:

ΔHtotal = ΔH1 + ΔH2 + ΔH3 + ...

An analogy is the height difference between the bottom and top of a mountain, which is independent of the path chosen. The law is a consequence of the first law of thermodynamics, and it is valid for all state functions.

What are the rules for combining equations?

Manipulating thermochemical equations
Operation on the equation Effect on ΔH Example
Reverse the equationChange the signCO2 → C + O2 has ΔH = +393.5 kJ
Multiply by a factor nMultiply ΔH by n2C + 2O2 → 2CO2 has ΔH = −787.0 kJ
Add two equationsAdd the ΔH valuesSee the worked examples
Cancel species on both sidesNo changeAn intermediate disappears

How do you solve a Hess’s law problem?

  1. Write the target equation.
  2. Look at each species in the target and find the given equation that contains it; reverse or multiply that equation so that the species is on the correct side and in the correct amount.
  3. Add the adjusted equations and cancel the species that appear on both sides.
  4. Check that the sum is the target equation.
  5. Add the adjusted enthalpies.

Example 1: carbon monoxide

Find ΔH for C(s) + ½O2(g) → CO(g), given:

CO is a product in the target, so reverse (b): CO2 → CO + ½O2, ΔH = +283.0 kJ. Adding (a) and the reversed (b), the CO2 cancels and the net O2 is ½: C + ½O2 → CO. ΔH = −393.5 + 283.0 = −110.5 kJ.

Enthalpy level diagram for carbon, carbon monoxide and carbon dioxide C + O₂ CO + ½O₂ CO₂ ΔH = −110.5 kJ −283.0 kJ −393.5 kJ Enthalpy, H
Enthalpy diagram for Example 1 (levels to scale). Burning carbon straight to CO2 releases 393.5 kJ. The two-step route first forms CO (−110.5 kJ) and then burns it (−283.0 kJ). Both routes end at the same level, and −110.5 + (−283.0) = −393.5 kJ, which is Hess’s law.

Example 2: ethyne

Find ΔH for 2C(s) + H2(g) → C2H2(g), given:

  1. C2H2 must be a product, so reverse (a): 2CO2 + H2O → C2H2 + 5/2 O2, ΔH = +1299.6 kJ.
  2. Two moles of carbon are needed, so multiply (b) by 2: ΔH = −787.0 kJ.
  3. Use (c) as written: ΔH = −285.8 kJ.
  4. Adding the three equations cancels CO2, H2O and O2.
  5. ΔH = +1299.6 − 787.0 − 285.8 = +226.8 kJ.

The positive value shows that ethyne is energy-rich and unstable relative to its elements, which explains why it burns at such a high temperature.

−a +1299.6 2b −787.0 c −285.8 ΔH +226.8 kJ

Example 3: nitrogen monoxide

Find ΔH for N2(g) + O2(g) → 2NO(g), given:

Use (a) as written and reverse (b): 2NO2 → 2NO + O2, ΔH = +114.2 kJ. The sum is N2 + 2O2 + 2NO2 → 2NO2 + 2NO + O2, which simplifies to N2 + O2 → 2NO. ΔH = 66.4 + 114.2 = +180.6 kJ. This reaction takes place in the cylinders of engines and in lightning, which is the source of nitrogen oxides in the air.

a +66.4 −b +114.2 ΔH +180.6 kJ

Example 4: graphite and diamond

Find ΔH for C(graphite) → C(diamond), given that the enthalpies of combustion are −393.5 kJ/mol for graphite and −395.4 kJ/mol for diamond.

  1. Graphite + O2 → CO2, ΔH = −393.5 kJ.
  2. CO2 → diamond + O2 (reversed), ΔH = +395.4 kJ.
  3. ΔH = −393.5 + 395.4 = +1.9 kJ/mol.

Graphite is slightly more stable than diamond at normal pressure, and the small difference, +1.9 kJ/mol, could not be measured directly because the conversion takes place only at high pressure. This does not mean that diamonds will turn into graphite: the transformation is extremely slow at room temperature.

Who was Hess?

Germain Henri Hess (1802–1850) was a Swiss-born Russian chemist and physician who taught in Saint Petersburg. In 1840 he reported that the heat released in a chemical process is the same whether the process takes place in one stage or in several stages. He arrived at this empirically, through measurements of the heats of neutralization of sulfuric acid with different amounts of water and ammonia, a few years before the first law of thermodynamics was established. His work is regarded as the foundation of thermochemistry.

Which mistakes are common?