Hess's Law: Calculating Enthalpy Changes Indirectly

What Is Hess's Law?

Hess's Law states that the total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same. This is a direct consequence of enthalpy being a state function.

In practice, Hess's Law lets us calculate the enthalpy change ( $\Delta H$ ) for a reaction that is difficult or impossible to measure directly, by constructing an energy cycle through intermediate steps whose enthalpy changes are known.

Learning Goals: By the end of this guide, you should be able to:

State Hess's Law and explain why it works.

Construct energy cycles using standard enthalpies of formation or combustion.

Calculate unknown $\Delta H$ values from tabulated data.

Apply Hess's Law to multi-step industrial and biological processes.

The Two Key Formulas

Using Standard Enthalpies of Formation ( $\Delta_f H°$ )

\Delta_r H° = \sum \Delta_f H°(\text{products}) - \sum \Delta_f H°(\text{reactants})

The standard enthalpy of formation of any element in its standard state is zero by definition.

Using Standard Enthalpies of Combustion ( $\Delta_c H°$ )

\Delta_r H° = \sum \Delta_c H°(\text{reactants}) - \sum \Delta_c H°(\text{products})

Note the subtraction is reversed compared to the formation formula — reactants minus products.

Method	Data used	Subtraction order
Formation route	$\Delta_f H°$	Products − Reactants
Combustion route	$\Delta_c H°$	Reactants − Products

Energy Cycle Construction

Step 1: Write the target reaction

C(s) + 2H_2(g) \rightarrow CH_4(g) \quad \Delta_f H° = ?

Step 2: Identify alternate routes via combustion

All reactants and products can be burned to form $CO_2$ and $H_2O$ :

$C(s) + O_2(g) \rightarrow CO_2(g)$ — $\Delta_c H° = -393\ kJ/mol$
$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$ — $\Delta_c H° = -286\ kJ/mol$
$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ — $\Delta_c H° = -890\ kJ/mol$

Step 3: Apply Hess's Law

\Delta_f H°(CH_4) = \Delta_c H°(C) + 2 \times \Delta_c H°(H_2) - \Delta_c H°(CH_4)

= (-393) + 2(-286) - (-890) = -393 - 572 + 890 = -75\ kJ/mol

Worked Examples

Example 1: Enthalpy of Formation of Ethanol

Calculate $\Delta_f H°$ for $C_2H_5OH(l)$ given:

Substance	$\Delta_c H°$ (kJ/mol)
$C(s)$	$-393$
$H_2(g)$	$-286$
$C_2H_5OH(l)$	$-1367$

Target: $2C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)$

\Delta_f H° = 2(-393) + 3(-286) - (-1367)

= -786 - 858 + 1367 = -277\ kJ/mol

Example 2: Using Formation Data

Calculate $\Delta_r H°$ for the thermite reaction:

2Al(s) + Fe_2O_3(s) \rightarrow Al_2O_3(s) + 2Fe(s)

Given: $\Delta_f H°(Fe_2O_3) = -824\ kJ/mol$ , $\Delta_f H°(Al_2O_3) = -1676\ kJ/mol$

\Delta_r H° = [-1676 + 0] - [-824 + 0] = -1676 + 824 = -852\ kJ/mol

Elements in their standard states ( $Al$ , $Fe$ ) have $\Delta_f H° = 0$ .

Example 3: Multi-step Neutralisation

Calculate $\Delta H$ for dissolving $NaOH(s)$ in water, given:

$NaOH(s) \rightarrow Na^+(aq) + OH^-(aq)$ — $\Delta H = ?$
$NaOH(s) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$ — $\Delta H_1 = -76\ kJ/mol$
$NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$ — $\Delta H_2 = -57\ kJ/mol$

By Hess's Law: $\Delta H_1 = \Delta H_{dissolution} + \Delta H_2$

\Delta H_{dissolution} = -76 - (-57) = -19\ kJ/mol

Common Mistakes

Reversing the subtraction order — For formation data it's products − reactants; for combustion data it's reactants − products. Mixing these up inverts the sign.
Forgetting stoichiometric coefficients — If the balanced equation has 2 mol of $H_2$ , you must multiply $\Delta_c H°(H_2)$ by 2.
Including elements in formation calculations — The standard enthalpy of formation of any element in its standard state is zero. Don't look up a value for $O_2(g)$ or $C(\text{graphite})$ .
Not checking the state symbols — $H_2O(l)$ and $H_2O(g)$ have different enthalpies. Ensure your data matches the states in the reaction.

Exam Tips (A-Level / AP / IB)

Always draw the energy cycle before calculating. It prevents sign errors and impresses examiners.
State Hess's Law in words: "The total enthalpy change is independent of the route taken, as enthalpy is a state function."
For IB: both the formation and combustion route formulas are in the data booklet — learn which to apply and when.
For AP: Hess's Law problems often give you 2-3 equations to combine algebraically. Reverse equations (flip signs) and multiply (scale enthalpies) as needed.

Frequently Asked Questions

Why does Hess's Law work?

Because enthalpy is a state function — it depends only on the initial and final states, not on the path. This is a consequence of the first law of thermodynamics (conservation of energy).

When should I use the combustion route vs the formation route?

Use whichever data you are given. If the question gives $\Delta_c H°$ values, use the combustion route. If it gives $\Delta_f H°$ values, use the formation route. Both give the same answer.

Can Hess's Law be applied to Gibbs free energy and entropy?

Yes! Since $G$ and $S$ are also state functions, the same principle applies: $\Delta_r G° = \sum \Delta_f G°(\text{products}) - \sum \Delta_f G°(\text{reactants})$ .

Born-Haber Cycles — Apply Hess's Law to calculate lattice energies of ionic compounds.
Gibbs Free Energy — Extend thermodynamic analysis to include entropy and spontaneity.
Le Chatelier's Principle — How temperature changes (related to $\Delta H$ ) shift equilibrium position.