Hess’s Law

Enthalpies can be measured experimentally using calorimetry. Sometimes, though, it is difficult or impossible to measure the enthalpy of a reaction directly. Hess’s law provides a solution: the total enthalpy change of a reaction is independent of the sequence of steps taken. This is a consequence of enthalpy being a state function.

A photograph of Germain Henri Hess

Germain Henri Hess: A Swiss-Russian chemist (1802-1850) who formulated Hess’s law, an early principle of thermochemistry. (Source: Wikipedia)

Why Hess’s Law Works

Hess’s law relies on enthalpy being a state function, meaning the value of ΔH depends only on the initial and final states of a system, not on the pathway taken between them.

Think of altitude when hiking a mountain. Whether you take a direct steep path or a winding gradual trail, the total change in elevation from base to summit is the same. Similarly, whether a reaction proceeds directly or through multiple intermediate steps, the total enthalpy change must be identical.

Transformation Rules for Enthalpy

Because enthalpy is a state function, chemical equations can be manipulated algebraically. Three rules govern how ΔH changes when you transform a reaction:

  1. Reversing a reaction changes the sign of ΔH

    If A → B has ΔH = +100 kJ, then B → A has ΔH = −100 kJ

  2. Scaling a reaction (multiplying by a coefficient) scales ΔH by the same factor

    If A → B has ΔH = +100 kJ, then 2 A → 2 B has ΔH = +200 kJ

  3. Adding reactions adds their ΔH values

    The ΔH for the overall reaction equals the sum of ΔH values for the individual steps

These rules allow us to calculate the enthalpy change for any reaction by combining known reactions, regardless of whether the direct path actually occurs in nature.

Practice

What is the enthalpy of reaction (in kJ mol−1) if the following reaction was reversed and multiplied by 2.5?

\[\mathrm{NH_4NO_3(s)} \longrightarrow \mathrm{N_2O(g)} + 2~\mathrm{H_2O(g)} \qquad \Delta_{\mathrm{r}}H^{\circ} = 205~\mathrm{kJ~mol^{-1}}\]

Solution

The transformed reaction is:

\[\mathrm{2.5~N_2O(g) + 5~H_2O(g)} \longrightarrow \mathrm{2.5~NH_4NO_3(s)}\]

Applying the transformation rules:

  • Reversing changes the sign: +205 → −205 kJ mol−1
  • Multiplying by 2.5 scales ΔH by 2.5

\[ \begin{align*} \Delta_{\mathrm{r}} H^{\circ} &= 2.5 \times -20\bar{5}~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -51\bar{2}.5~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -512~\mathrm{kJ~mol^{-1}} \end{align*} \]

The final answer uses banker’s rounding (512.5 rounds to 512, the nearest even number).

Applying Hess’s Law

Take, for example, the formation of carbon dioxide under standard conditions and 298.15 K.

\[ \begin{alignat*}{3} && \mathrm{C(s, graphite)} + \mathrm{O_2(g)} &&~\longrightarrow \mathrm{CO_2(g)} &\qquad& \Delta_{\mathrm{f}}H^{\circ} &&= -393.5~\mathrm{kJ~mol^{-1}} \end{alignat*} \]

We could examine this process as a two-step process where in Step 1, CO is formed from its constituent elemental forms and, in Step 2, CO2 is produced from the CO(g) intermediate and oxygen gas. Each individual step has its own heat of formation. Combining these steps gives the heat of formation of CO2(g).

\[ \begin{array}{lcrllll} \text{Step 1:} & & \mathrm{C(s, graphite)} + \dfrac{1}{2}~\mathrm{O_2(g)} &\longrightarrow& \mathrm{CO(g)} & & \Delta_{\mathrm{f}} H^{\circ} = -110.5~\mathrm{kJ~mol^{-1}} \\[0.5ex] \text{Step 2:} & & \mathrm{CO(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} &\longrightarrow& \mathrm{CO_2(g)} & & \Delta_{\mathrm{r}} H^{\circ} = -283.0~\mathrm{kJ~mol^{-1}} \\[0.2ex] \hline \\[-1ex] \text{Overall:} & & \mathrm{C(s, graphite)} + \mathrm{O_2(g)} &\longrightarrow& \mathrm{CO_2(g)} & & \Delta_{\mathrm{f}} H^{\circ} = -393.5~\mathrm{kJ~mol^{-1}} \end{array} \]

C(s) + O₂(g) CO(g) + ½ O₂(g) CO₂(g) Overall: −393.5 kJ mol−1 Step 1: −110.5 kJ mol−1 Step 2: −283.0 kJ mol−1 0 −110.5 −393.5 kJ mol−1 Enthalpy, H
Energy level diagram for the formation of CO2 via Hess's Law.

The intermediate level shows CO(g) + ½ O2(g) because the diagram tracks all species through the pathway. The unused ½ O2 from the starting materials “carries through” until it reacts in Step 2.

Therefore, known enthalpies of reactions can be used to determine the enthalpies of other reactions that cannot be measured directly. The intermediate steps do not have to occur in reality; they are simply a mathematical tool for calculating the overall enthalpy change.

TipProblem-Solving Strategy for Hess’s Law

When using Hess’s law to determine an unknown ΔH, follow these steps:

  1. Identify the target reaction you want to find ΔH for
  2. Write out all given reactions with their known ΔH values
  3. Apply transformation rules to manipulate the given reactions:
    • Reverse reactions if a species appears on the wrong side
    • Multiply or divide reactions to get correct stoichiometric coefficients
  4. Add the transformed reactions algebraically, canceling species that appear on both sides
  5. Add the corresponding ΔH values to get the final answer

For formation reactions, the target compound should be the only product, formed from elements in their standard states.

ImportantSignificant Figures: Scaling vs. Summing in Hess’s law

In Hess’s law calculations, two different operations require different significant figure rules:

Scaling applies when you multiply or divide a single reaction. If you double a reaction with ΔH = +50.1 kJ:

  • Calculation: 2 × 50.1 = 100.2
  • The coefficient 2 is exact (from stoichiometry)
  • Apply multiplication rule → round to 3 sig figs: 1.00 × 102 kJ

Summing applies when you add multiple reactions together. If you add two separate reactions that each have ΔH = +50.1 kJ:

  • Calculation: 50.1 + 50.1 = 100.2
  • Apply addition rule → round to tenths place: 100.2 kJ

Same arithmetic, different answers. When you multiply a reaction by 2, you’re scaling one measurement by an exact count. You’re not adding two separate measurements together, even though the math looks identical.

See Scaling vs. Summing for the complete explanation.

Example: Find the enthalpy of formation for acetylene (C2H2), given the following data.

\[ \begin{array}{lcrllll} \textrm{Reaction 1:} & & \mathrm{H_2(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} &\longrightarrow& \mathrm{H_2O(l)} & & \Delta_{\mathrm{f}} H^{\circ} = -285.83~\mathrm{kJ~mol^{-1}} \\[0.5ex] \textrm{Reaction 2:} & & \mathrm{C(s,graphite)} + \mathrm{O_2(g)} &\longrightarrow& \mathrm{CO_2(g)} & & \Delta_{\mathrm{f}} H^{\circ} = -393.5~\mathrm{kJ~mol^{-1}} \\[0.5ex] \textrm{Reaction 3:} & & 2~\mathrm{C_2H_2(g)} + 5~\mathrm{O_2(g)} &\longrightarrow& 4~\mathrm{CO_2(g)} + 2~\mathrm{H_2O(l)} & & \Delta_{\mathrm{r}} H^{\circ} = -2598~\mathrm{kJ~mol^{-1}} \\[0.2ex] \end{array} \]

Recall the definition of a formation reaction which is “one mole of a product is formed from its constituent elements in their standard states”. For acetylene, the target reaction is:

\[ \mathrm{2~C(s,graphite)} + \mathrm{H_2(g)} \longrightarrow \mathrm{C_2H_2(g)} \]

Solution

Looking at the given reactions, we need C2H2(g) as a product, but Reaction 3 has it as a reactant. We must reverse Reaction 3. We also need 2 moles of C(s, graphite), so Reaction 2 must be multiplied by 2. Reaction 1 produces H2O(l), which we need because the reversed Reaction 3 will have H2O(l) as a reactant that we want to cancel. However, Reaction 3 produces 2 moles of C2H2(g), but we only need 1 mole for the formation reaction, so we must divide the reversed reaction by 2.

Transform the reactions.

  1. Leave Reaction 1 alone
  2. Multiply Reaction 2 by 2
  3. Reverse Reaction 3 and divide by 2

Write out the reaction transformations and state function transformations and then combine.

\[ \begin{array}{lcrllll} \textrm{Reaction 1:} & & \mathrm{H_2(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} &\longrightarrow& \mathrm{H_2O(l)} & & \Delta_{\mathrm{f}} H^{\circ} = -285.83~\mathrm{kJ~mol^{-1}} \\[0.5ex] \textrm{Reaction 2:} & & 2~\mathrm{C(s,graphite)} + 2~\mathrm{O_2(g)} &\longrightarrow& 2~\mathrm{CO_2(g)} & & \Delta_{\mathrm{r}} H^{\circ} = -787.0~\mathrm{kJ~mol^{-1}} \\[0.5ex] \textrm{Reaction 3:} & & 2~\mathrm{CO_2(g)} + \mathrm{H_2O(l)} &\longrightarrow& \mathrm{C_2H_2(g)} + \dfrac{5}{2}~\mathrm{O_2(g)} & & \Delta_{\mathrm{r}} H^{\circ} = 1299~\mathrm{kJ~mol^{-1}} \\[0.2ex] \hline \\[-1ex] \textrm{Overall:} & & 2~\mathrm{C(s, graphite)} + \mathrm{H_2(g)} &\longrightarrow& \mathrm{C_2H_2(g)} & & \Delta_{\mathrm{f}}H^{\circ} = 226.\bar{1}7~\mathrm{kJ~mol^{-1}} \end{array} \]

C₂H₂(g) 2 C(s) + H₂(g) + 5/2 O₂(g) 2 C(s) + H₂O(l) + 2 O₂(g) 2 CO₂(g) + H₂O(l) Overall: +226.2 kJ mol−1 Rxn 1: −285.8 kJ mol−1 Rxn 2: −787.0 kJ mol−1 Rxn 3: +1299 kJ mol−1 +226.2 0 −285.8 −1072.8 kJ mol−1 Enthalpy, H
Energy level diagram for the formation of C2H2 via Hess's Law.
WarningCommon Mistakes with Hess’s Law

Forgetting to change the sign when reversing a reaction

When you flip a reaction to get a species on the correct side, you must reverse the sign of ΔH. If the forward reaction is exothermic (ΔH < 0), the reverse is endothermic (ΔH > 0).

Forgetting to scale ΔH when multiplying/dividing reactions

If you multiply an entire reaction by 2 to get correct stoichiometry, you must also multiply ΔH by 2. Similarly, dividing a reaction by 2 requires dividing ΔH by 2.

Not identifying the target reaction correctly

For a formation reaction, you need exactly 1 mole of product formed from elements in their standard states. If your final equation has 2 moles of product, you have not found the standard enthalpy of formation.

Incorrectly canceling species

Only species that appear on opposite sides with identical states and coefficients cancel completely. Be careful with states of matter: H2O(l) does not cancel with H2O(g).

Arithmetic errors when adding ΔH values

Keep track of positive and negative signs carefully. Reversed reactions contribute positive values if the original was negative, and vice versa.

Practice Problems

Practice

Calculate the enthalpy change for the reaction:

\[ \mathrm{N_2(g)} + 2~\mathrm{O_2(g)} \longrightarrow 2~\mathrm{NO_2(g)} \]

Given the following reactions and enthalpy changes:

\[ \begin{alignat*}{3} &\textrm{Reaction 1:} \quad && \dfrac{1}{2}~\mathrm{N_2(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} &&\longrightarrow \mathrm{NO(g)} &\quad& \Delta_{\mathrm{f}} H^{\circ} &&= 90.3~\mathrm{kJ~mol^{-1}} \\[1ex] &\textrm{Reaction 2:} \quad && \mathrm{NO(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} &&\longrightarrow \mathrm{NO_2(g)} &\quad& \Delta_{\mathrm{r}} H^{\circ} &&= -57.1~\mathrm{kJ~mol^{-1}} \end{alignat*} \]

Solution

To form 2 NO2(g), we need to combine the two given reactions appropriately.

Multiply Reaction 1 by 2:

\[ \mathrm{N_2(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{NO(g)} \quad \Delta_{\mathrm{r}} H^{\circ} = 180.6~\mathrm{kJ~mol^{-1}} \]

Multiply Reaction 2 by 2:

\[ 2~\mathrm{NO(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{NO_2(g)} \qquad \Delta_{\mathrm{r}} H^{\circ} = -114.2~\mathrm{kJ~mol^{-1}} \]

Add the transformed reactions:

\[ \begin{array}{lcllcll} & \quad & \mathrm{N_2(g)} + \mathrm{O_2(g)} &\longrightarrow& 2~\mathrm{NO(g)} && \Delta_{\mathrm{r}} H^{\circ} = 180.6~\mathrm{kJ~mol^{-1}} \\[0.5ex] & \quad & 2~\mathrm{NO(g)} + \mathrm{O_2(g)} &\longrightarrow& 2~\mathrm{NO_2(g)} && \Delta_{\mathrm{r}} H^{\circ} = -114.2~\mathrm{kJ~mol^{-1}} \\[0.2ex] \hline \\[-1ex] & \quad & \mathrm{N_2(g)} + 2~\mathrm{O_2(g)} &\longrightarrow& 2~\mathrm{NO_2(g)} && \Delta_{\mathrm{r}} H^{\circ} = 66.4~\mathrm{kJ~mol^{-1}} \end{array} \]

Answer: ΔrH° = 66.4 kJ mol−1

N₂(g) + 2 O₂(g) 2 NO(g) + O₂(g) 2 NO₂(g) Step 1: +180.6 kJ mol−1 Step 2: −114.2 kJ mol−1 Overall: +66.4 kJ mol−1 +180.6 +66.4 0 kJ mol−1 Enthalpy, H
Energy level diagram for the formation of 2 NO2 via Hess's Law.


Practice

The combustion of methane can be represented as a two-step process. Calculate ΔrH° for the second step given the following:

\[ \begin{alignat*}{3} &\textrm{Step 1:} \quad && \mathrm{CH_4(g)} + \dfrac{3}{2}~\mathrm{O_2(g)} &&\longrightarrow \mathrm{CO(g)} + 2~\mathrm{H_2O(g)} &\quad& \Delta_{\mathrm{r}} H^{\circ} &&= -519~\mathrm{kJ~mol^{-1}} \\[1ex] &\textrm{Overall:} \quad && \mathrm{CH_4(g)} + 2~\mathrm{O_2(g)} &&\longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{H_2O(g)} &\quad& \Delta_{\mathrm{c}} H^{\circ} &&= -802~\mathrm{kJ~mol^{-1}} \end{alignat*} \]

What is ΔrH° for Step 2?

\[ \mathrm{CO(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} \]

Solution

According to Hess’s law, the overall enthalpy change equals the sum of the individual steps:

\[ \Delta_{\mathrm{r}} H_{\mathrm{overall}}^{\circ} = \Delta_{\mathrm{r}} H_{\mathrm{step~1}}^{\circ} + \Delta_{\mathrm{r}} H_{\mathrm{step~2}}^{\circ} \]

Rearranging:

\[ \begin{align*} \Delta_{\mathrm{r}} H_{\mathrm{step~2}}^{\circ} &= \Delta_{\mathrm{r}} H^{\circ}_{\mathrm{overall}} - \Delta_{\mathrm{r}} H_{\mathrm{step~1}}^{\circ} \\[1.5ex] &= -802~\mathrm{kJ~mol^{-1}} - (-519~\mathrm{kJ~mol^{-1}}) \\[1.5ex] &= -802~\mathrm{kJ~mol^{-1}} + 519~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -283~\mathrm{kJ~mol^{-1}} \end{align*} \]

Answer: ΔrH° = −283 kJ mol−1 for Step 2

CH₄(g) + 2 O₂(g) CO(g) + 2 H₂O(g) + ½ O₂(g) CO₂(g) + 2 H₂O(g) Step 1: −519 kJ mol−1 Step 2: −283 kJ mol−1 Overall: −802 kJ mol−1 0 −519 −802 kJ mol−1 Enthalpy, H
Energy level diagram for the combustion of CH4 via Hess's Law.


Practice

Calculate ΔfH° for the reaction:

\[ \mathrm{2~C(s,graphite)} + \mathrm{H_2(g)} \longrightarrow \mathrm{C_2H_2(g)} \]

Given:

\[ \begin{alignat*}{3} &\textrm{Reaction 1:} \quad && \mathrm{C_2H_2(g)} + \dfrac{5}{2}~\mathrm{O_2(g)} &&\longrightarrow 2~\mathrm{CO_2(g)} + \mathrm{H_2O(l)} &\quad& \Delta_{\mathrm{c}} H^{\circ} &&= -1299.6~\mathrm{kJ~mol^{-1}} \\[1ex] &\textrm{Reaction 2:} \quad && \mathrm{C(s,graphite)} + \mathrm{O_2(g)} &&\longrightarrow \mathrm{CO_2(g)} &\quad& \Delta_{\mathrm{c}} H^{\circ} &&= -393.5~\mathrm{kJ~mol^{-1}} \\[1ex] &\textrm{Reaction 3:} \quad && \mathrm{H_2(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} &&\longrightarrow \mathrm{H_2O(l)} &\quad& \Delta_{\mathrm{c}} H^{\circ} &&= -285.8~\mathrm{kJ~mol^{-1}} \end{alignat*} \]

Note: Reactions 2 and 3 are labeled with ΔcH° (combustion enthalpy) because these values are typically determined from combustion experiments. For elements that combust to form a single product, the combustion enthalpy of the element equals the formation enthalpy of the product: ΔcH°[C(s)] = ΔfH°[CO2(g)] and ΔcH°[H2(g)] = ΔfH°[H2O(l)].

Solution
  1. Need C2H2(g) as a product, so reverse Reaction 1.
  2. Need 2 moles of C(s, graphite) as reactants, so multiply Reaction 2 by 2.
  3. Need 1 mole of H2(g) as a reactant, so leave Reaction 3 alone.

Transformations:

\[ \begin{array}{lclllll} \textrm{Reaction 1 (reversed):} & & 2~\mathrm{CO_2(g)} + \mathrm{H_2O(l)} &\longrightarrow& \mathrm{C_2H_2(g)} + \dfrac{5}{2}~\mathrm{O_2(g)} & & \Delta_{\mathrm{r}} H^{\circ} = 1299.6~\mathrm{kJ~mol^{-1}} \\[0.5ex] \textrm{Reaction 2 (×2):} & & 2~\mathrm{C(s,graphite)} + 2~\mathrm{O_2(g)} &\longrightarrow& 2~\mathrm{CO_2(g)} & & \Delta_{\mathrm{r}} H^{\circ} = -787.0~\mathrm{kJ~mol^{-1}} \\[0.5ex] \textrm{Reaction 3:} & & \mathrm{H_2(g)} + \dfrac{1}{2}~\mathrm{O_2(g)} &\longrightarrow& \mathrm{H_2O(l)} & & \Delta_{\mathrm{c}} H^{\circ} = -285.8~\mathrm{kJ~mol^{-1}} \\[0.2ex] \hline \\[-1ex] \textrm{Overall:} & & 2~\mathrm{C(s, graphite)} + \mathrm{H_2(g)} &\longrightarrow& \mathrm{C_2H_2(g)} & & \Delta_{\mathrm{f}}H^{\circ} = 226.8~\mathrm{kJ~mol^{-1}} \end{array} \]

Answer: ΔfH° = 226.8 kJ mol−1 (the standard enthalpy of formation of acetylene)

Note: The reaction numbers (1, 2, 3) refer to the given reactions in the problem, not the order in which they’re applied. The pathway shown in the diagram below follows Rxn 3 → Rxn 2 (×2) → Rxn 1 (reversed).

C₂H₂(g) 2 C(s) + H₂(g) + 5/2 O₂(g) 2 C(s) + H₂O(l) + 2 O₂(g) 2 CO₂(g) + H₂O(l) Overall: +226.8 kJ mol−1 Rxn 3: −285.8 kJ mol−1 Rxn 2 (×2): −787.0 kJ mol−1 Rxn 1 (rev): +1299.6 kJ mol−1 +226.8 0 −285.8 −1072.8 kJ mol−1 Enthalpy, H
Energy level diagram for the formation of C2H2 via Hess's Law.


Summary

Hess’s Law

Hess’s law states that the total enthalpy change for a reaction is independent of the pathway taken. This is a consequence of enthalpy being a state function.

Transformation Rules

  1. Reversing a reaction changes the sign of ΔH
  2. Scaling a reaction by a factor n multiplies ΔH by n
  3. Adding reactions adds their ΔH values

Problem-Solving Approach

  1. Identify the target reaction
  2. Transform given reactions (reverse, scale) to align species correctly
  3. Add transformed reactions, canceling intermediates
  4. Sum the transformed ΔH values