Hess's Law

IB Chemistry SL· 35 min read

1. 1. Core Concept and Theoretical Basis★★☆☆☆⏱ 10 min

📘 Definition

Hess's Law

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

The total enthalpy change for a chemical reaction is independent of the pathway taken between initial reactants and final products.

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.

📐 Worked Example

Calculate the enthalpy change for $C_{(g r a p hi t e)} \to C_{(d iam o n d)}$ given: 1. $C_{(g r a p hi t e)} + O_{2} (g) \to C O_{2} (g) Δ H = - 393.5 kJ mol^{- 1}$ 2. $C_{(d iam o n d)} + O_{2} (g) \to C O_{2} (g) Δ H = - 395.4 kJ mol^{- 1}$

1
Reverse the second equation to get $C_{(d iam o n d)}$ as a product, flipping the sign of $Δ H$ :
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. 2. Enthalpy Calculations from Standard Enthalpies of Formation★★★☆☆⏱ 15 min

The most common exam application of Hess's Law uses tabulated standard enthalpies of formation ( $Δ H_{f}^{⊖}$ ) 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 $2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{2} O (l)$ . Given: $Δ H_{f}^{⊖} (C_{4} H_{10}) = - 126$ kJ mol⁻¹, $Δ H_{f}^{⊖} (C O_{2}) = - 394$ kJ mol⁻¹, $Δ H_{f}^{⊖} (H_{2} O (l)) = - 286$ kJ mol⁻¹, $Δ H_{f}^{⊖} (O_{2} (g)) = 0$ 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. 3. Enthalpy Calculations from Standard Enthalpies of Combustion★★★☆☆⏱ 15 min

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})

📐 Worked Example

Calculate the enthalpy of formation of glucose $C_{6} H_{12} O_{6} (s)$ given: $Δ H_{c}^{⊖} (C_{6} H_{12} O_{6}) = - 2800$ kJ mol⁻¹, $Δ H_{c}^{⊖} (C_{(g r a p hi t e)}) = - 394$ kJ mol⁻¹, $Δ H_{c}^{⊖} (H_{2} (g)) = - 286$ kJ mol⁻¹.

1
Write the balanced target equation for glucose formation: $6 C_{(g r a p hi t e)} + 6 H_{2} (g) + 3 O_{2} (g) \to C_{6} H_{12} O_{6} (s)$ . The enthalpy of this reaction is the enthalpy of formation we need.
2
Apply the combustion formula, remembering $Δ H_{c}^{⊖} (O_{2}) = 0$ :
3
$\begin{align*} \Delta H_f^\ominus &= \left[6(-394) + 6(-286)\right] - 1(-2800) \\ &= -4080 + 2800 = -1280 \text{ kJ mol}^{-1} \end{align*}$

4. Common Pitfalls

Wrong move:

Failing to flip the sign of $Δ H$ when reversing a reaction step

Why:

Reversing a reaction swaps the direction of enthalpy flow: exothermic becomes endothermic and vice versa

Correct move:

Always flip the sign of $Δ H$ whenever you reverse a chemical equation in your Hess cycle

Wrong move:

Forgetting to scale $Δ H$ by the reaction stoichiometry

Why:

Tabulated enthalpy values are always per mole, so multiple moles of a compound require scaling the enthalpy value

Correct move:

Multiply every enthalpy value by the coefficient of the compound in the balanced overall equation

Wrong move:

Mixing up the order of products and reactants for combustion data

Why:

The formula for combustion is flipped compared to formation data, which is a common point of confusion

Correct move:

Draw the full Hess cycle from scratch if you cannot remember the formula, to confirm the order

Wrong move:

Using a non-zero $Δ H_{f}$ for an element in standard state

Why:

Students often incorrectly memorize or misread tabulated values for elemental species

Correct move:

Always remember $Δ H_{f}$ of any element in its standard state is defined as 0, so it can be ignored in calculations

Wrong move:

Incorrectly cancelling species when adding reaction steps

Why:

Rushing through the cycle leads to extra leftover species on one side of the equation

Correct move:

Cross out all species that appear on both sides of the summed equations, then check your overall equation matches the target reaction

5. Quick Reference Cheatsheet

Concept	Formula	Key Note
General Hess Cycle	$Δ H_{total} = \sum Δ H_{steps}$	Flip sign for reversed reactions, scale for stoichiometry
From Enthalpies of Formation	$Δ H_{r x n}^{⊖} = \sum Δ H_{f}^{⊖} (products) - \sum Δ H_{f}^{⊖} (reactants)$	$Δ H_{f}^{⊖}$ (element, standard state) = 0
From Enthalpies of Combustion	$Δ H_{r x n}^{⊖} = \sum Δ H_{c}^{⊖} (reactants) - \sum Δ H_{c}^{⊖} (products)$	Formula is flipped compared to formation

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · P1
Calculate enthalpy change from given data
2024 · P2
Construct Hess cycle from combustion data
2023 · P1
Identify correct enthalpy expression

Going deeper

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