# Hess's Law

> IB Chemistry SL · Reactivity 2: Energetics and Kinetics
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u5-hess-s-law/

This module covers Hess's Law of constant heat summation, how to construct enthalpy cycles, and calculate unknown enthalpy changes from formation and combustion data, a core exam skill for IB SL Chemistry energetics.

**Prerequisites:** [Enthalpy and enthalpy change](https://www.owlsprep.com/study/ib-chemistry-sl-u5-enthalpy-change/); [Standard enthalpies of formation and combustion](https://www.owlsprep.com/study/ib-chemistry-sl-u5-standard-enthalpies/)

## Learning objectives

- State Hess's Law and explain its theoretical basis in terms of enthalpy as a state function
- Calculate unknown enthalpy changes using Hess's Law cycles from multiple reaction steps
- Calculate reaction enthalpy from standard enthalpies of formation and combustion

## 1. Core Concept and Theoretical Basis

**Hess's Law** — The total enthalpy change for a chemical reaction is independent of the pathway taken between initial reactants and final products.

*Notation:* ΔH_{\text{total}} = ΔH_1 + ΔH_2 + ... + ΔH_n

*Example:* Converting graphite to diamond has the same enthalpy change whether done directly or via carbon dioxide as an intermediate.

Hess's Law is a consequence of the first law of thermodynamics (conservation of energy) and the fact that enthalpy is a **state function**. Only the initial and final states of the system matter, not how you get from one to the other.

> **info**
>
> This property lets us calculate enthalpy changes for reactions that cannot be measured directly, for example very slow reactions or incomplete reactions.

**Worked example:** Calculate the enthalpy change for $C_{(graphite)} \rightarrow C_{(diamond)}$ given: 1. $C_{(graphite)} + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -393.5 \text{ kJ mol}^{-1}$ 2. $C_{(diamond)} + O_2(g) \rightarrow CO_2(g) \quad \Delta H = -395.4 \text{ kJ mol}^{-1}$

1. Reverse the second equation to get $C_{(diamond)}$ as a product, flipping the sign of $\Delta H$:
2. $$CO_2(g) \rightarrow C_{(diamond)} + O_2(g) \quad \Delta H = +395.4 \text{ kJ mol}^{-1}$$
3. Add the two equations, cancel common species on both sides, and sum the enthalpy changes:
4. $$\begin{align*} C(\text{graphite}) + \cancel{O_2(g)} &\rightarrow \cancel{CO_2(g)} \quad \Delta H = -393.5 \\ \cancel{CO_2(g)} &\rightarrow C(\text{diamond}) + \cancel{O_2(g)} \quad \Delta H = +395.4 \\ \hline C(\text{graphite}) &\rightarrow C(\text{diamond}) \quad \Delta H = +1.9 \text{ kJ mol}^{-1} \end{align*}$$

## 2. Enthalpy Calculations from Standard Enthalpies of Formation

The most common exam application of Hess's Law uses tabulated standard enthalpies of formation ($\Delta H_f^\ominus$) to find the reaction enthalpy. The general formula comes from a Hess cycle that goes from reactants, back to their constituent elements in standard state, then to products.

$$\Delta H^\ominus_{\text{reaction}} = \sum \Delta H^\ominus_f(\text{products}) - \sum \Delta H^\ominus_f(\text{reactants})$$

**Worked example:** Calculate the standard enthalpy change for $2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l)$. Given: $\Delta H_f^\ominus(C_4H_{10}) = -126$ kJ mol⁻¹, $\Delta H_f^\ominus(CO_2) = -394$ kJ mol⁻¹, $\Delta H_f^\ominus(H_2O(l)) = -286$ kJ mol⁻¹, $\Delta H_f^\ominus(O_2(g)) = 0$ kJ mol⁻¹.

1. Enthalpy of formation of elements in their standard state is 0, so $O_2$ contributes nothing to the sum of reactants.
2. Calculate the sum for products, accounting for stoichiometry:
3. $$\sum \Delta H_f^\ominus (\text{products}) = 8(-394) + 10(-286) = -6012 \text{ kJ}$$
4. Calculate the sum for reactants:
5. $$\sum \Delta H_f^\ominus (\text{reactants}) = 2(-126) + 13(0) = -252 \text{ kJ}$$
6. Subtract reactants from products to get the final enthalpy change:
7. $$\Delta H^\ominus_{\text{rxn}} = -6012 - (-252) = -5760 \text{ kJ for 2 moles of } C_4H_{10}$$

> **Exam tip:** Always multiply each enthalpy value by the stoichiometric coefficient from your balanced equation, never just use the tabulated per-mole value directly.

## 3. Enthalpy Calculations from Standard Enthalpies of Combustion

When given enthalpies of combustion, the Hess cycle is constructed differently, leading to a reversed formula compared to formation data. The cycle goes from reactants to combustion products, then from combustion products back to the target products.

$$\Delta H^\ominus_{\text{reaction}} = \sum \Delta H^\ominus_c(\text{reactants}) - \sum \Delta H^\ominus_c(\text{products})$$

> **Formula Memory Hook**
>
> "Formation: Products minus Reactants; Combustion: Reactants minus Products"

**Worked example:** Calculate the enthalpy of formation of glucose $C_6H_{12}O_6(s)$ given: $\Delta H_c^\ominus(C_6H_{12}O_6) = -2800$ kJ mol⁻¹, $\Delta H_c^\ominus(C_{(graphite)}) = -394$ kJ mol⁻¹, $\Delta H_c^\ominus(H_2(g)) = -286$ kJ mol⁻¹.

1. Write the balanced target equation for glucose formation: $6C_{(graphite)} + 6H_2(g) + 3O_2(g) \rightarrow C_6H_{12}O_6(s)$. The enthalpy of this reaction is the enthalpy of formation we need.
2. Apply the combustion formula, remembering $\Delta H_c^\ominus(O_2) = 0$:
3. $$\begin{align*} \Delta H_f^\ominus &= \left[6(-394) + 6(-286)\right] - 1(-2800) \\ &= -4080 + 2800 = -1280 \text{ kJ mol}^{-1} \end{align*}$$

## Common pitfalls

- **Wrong:** Failing to flip the sign of $\Delta H$ when reversing a reaction step
  - Why it fails: Reversing a reaction swaps the direction of enthalpy flow: exothermic becomes endothermic and vice versa
  - Correct: Always flip the sign of $\Delta H$ whenever you reverse a chemical equation in your Hess cycle
- **Wrong:** Forgetting to scale $\Delta H$ by the reaction stoichiometry
  - Why it fails: Tabulated enthalpy values are always per mole, so multiple moles of a compound require scaling the enthalpy value
  - Correct: Multiply every enthalpy value by the coefficient of the compound in the balanced overall equation
- **Wrong:** Mixing up the order of products and reactants for combustion data
  - Why it fails: The formula for combustion is flipped compared to formation data, which is a common point of confusion
  - Correct: Draw the full Hess cycle from scratch if you cannot remember the formula, to confirm the order
- **Wrong:** Using a non-zero $\Delta H_f$ for an element in standard state
  - Why it fails: Students often incorrectly memorize or misread tabulated values for elemental species
  - Correct: Always remember $\Delta H_f$ of any element in its standard state is defined as 0, so it can be ignored in calculations
- **Wrong:** Incorrectly cancelling species when adding reaction steps
  - Why it fails: Rushing through the cycle leads to extra leftover species on one side of the equation
  - Correct: Cross out all species that appear on both sides of the summed equations, then check your overall equation matches the target reaction

## Cheatsheet

| Concept | Formula | Key Note |
| --- | --- | --- |
| General Hess Cycle | $\Delta H_{\text{total}} = \sum \Delta H_{\text{steps}}$ | Flip sign for reversed reactions, scale for stoichiometry |
| From Enthalpies of Formation | $\Delta H_{rxn}^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})$ | $\Delta H_f^\ominus$(element, standard state) = 0 |
| From Enthalpies of Combustion | $\Delta H_{rxn}^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})$ | Formula is flipped compared to formation |

## What's next

Hess's Law is the foundation for all further enthalpy and thermodynamics calculations in IB Chemistry, and is frequently combined with other topics like bond enthalpies, entropy, and Gibbs free energy in both multiple choice and extended response questions. Mastery of Hess cycle construction and enthalpy calculation is one of the highest-weight skills in the Energetics unit for SL. Beyond the IB, the concept that state functions are pathway independent is a core principle of all physical chemistry and chemical thermodynamics. Next, you will apply Hess's Law to calculate average bond enthalpies and connect enthalpy change to bond breaking and forming processes.

- [Collision theory and reaction rates](https://www.owlsprep.com/study/ib-chemistry-sl-u5-collision-theory-and-reaction-rates/)

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