Hess' Law

CIE A-Level Chemistry· 25 min read

1. Definition and Core Principle of Hess' Law★★☆☆☆⏱ 7 min

📘 Definition

Hess' Law

Hess' law states that the total enthalpy change for a chemical reaction is independent of the path taken between the initial reactants and final products. This is a consequence of the law of conservation of energy.

Example:

If reaction $A \to C$ occurs directly or via $A \to B \to C$ , then $Δ H (A \to C) = Δ H (A \to B) + Δ H (B \to C)$ .

Hess' law is critical because many enthalpy changes cannot be measured directly in the lab. For example, the enthalpy of formation of methane cannot be measured directly, as carbon and hydrogen do not react spontaneously to form methane under standard conditions. Hess' law lets us calculate this value indirectly using measurable data.

📐 Worked Example

Given: $Δ H_{1}$ for $C (s) + O_{2} (g) \to C O_{2} (g) = - 393$ kJ mol⁻¹, $Δ H_{2}$ for $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g) = - 283$ kJ mol⁻¹. Calculate $Δ H$ for $C (s) + \frac{1}{2} O_{2} (g) \to C O (g)$ .

1
Construct the enthalpy cycle: the target reaction forms CO, which can then combust to form $C O_{2}$ . The alternative path is direct combustion of C to $C O_{2}$ . By Hess' law: $Δ H_{target} + Δ H_{2} = Δ H_{1}$
2
Rearrange to solve for $Δ H_{target}$ :
3
$\Delta H_{\text{target}} = \Delta H_1 - \Delta H_2$
4
Substitute the given values:
5
$\Delta H_{\text{target}} = (-393) - (-283) = -110 \text{ kJ mol}^{-1}$

Exam tip:

Always label all arrows in your enthalpy cycle clearly to avoid sign errors

2. Enthalpy Cycles for Formation and Combustion★★★☆☆⏱ 8 min

CIE exams most commonly ask for enthalpy changes of reaction using two types of tabulated data: standard enthalpies of formation, and standard enthalpies of combustion. Each has a standard formula derived directly from Hess' law:

From enthalpies of formation: $Δ H_{r} = \sum Δ H_{f} (products) - \sum Δ H_{f} (reactants)$
From enthalpies of combustion: $Δ H_{r} = \sum Δ H_{c} (reactants) - \sum Δ H_{c} (products)$

📐 Worked Example

Calculate the enthalpy change of combustion of propene: $C_{3} H_{6} (g) + \frac{9}{2} O_{2} (g) \to 3 C O_{2} (g) + 3 H_{2} O (l)$ . Use these $Δ H_{f}$ values (kJ mol⁻¹): $C_{3} H_{6} (g) = + 20$ , $C O_{2} (g) = - 393$ , $H_{2} O (l) = - 286$ , $O_{2} (g) = 0$ .

1
Apply the formula for enthalpy of reaction from formation values:
2
$\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$
3
Calculate the sum of product enthalpies:
4
$3 \times (-393) + 3 \times (-286) = -2037 \text{ kJ mol}^{-1}$
5
Calculate the sum of reactant enthalpies:
6
$\Delta H_f(C_3H_6) + \frac{9}{2} \times \Delta H_f(O_2) = 20 + 0 = 20 \text{ kJ mol}^{-1}$
7
Substitute into the formula:
8
$\Delta H_r = -2037 - 20 = -2057 \text{ kJ mol}^{-1}$

3. Multi-Step Reaction Enthalpy Calculations★★★★☆⏱ 10 min

When you are given multiple reaction equations and asked to find the enthalpy change for a target reaction, you can rearrange and add the given equations following these steps:

Write down the target reaction with correct stoichiometry
Adjust each given reaction: reverse the reaction if needed, and flip the sign of $Δ H$
Scale coefficients to match the target, and scale $Δ H$ by the same factor
Add all adjusted reactions and their $Δ H$ values to get the final result

📐 Worked Example

Given: 1) $2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g) Δ H = - 196$ kJ mol⁻¹ 2) $2 S O_{2} (g) + O_{2} (g) + 2 H_{2} O (l) \to 2 H_{2} S O_{4} (a q) Δ H = - 544$ kJ mol⁻¹ Calculate $Δ H$ for $S O_{3} (g) + H_{2} O (l) \to H_{2} S O_{4} (a q)$ .

1
Reverse equation 1 and divide all coefficients by 2, flip the sign and halve $Δ H$ :
2
$SO_3(g) \rightarrow SO_2(g) + \frac{1}{2}O_2(g) \quad \Delta H_1 = +98 \text{ kJ mol}^{-1}$
3
Divide equation 2 by 2, keep direction the same, halve $Δ H$ :
4
$SO_2(g) + \frac{1}{2}O_2(g) + H_2O(l) \rightarrow H_2SO_4(aq) \quad \Delta H_2 = -272 \text{ kJ mol}^{-1}$
5
Add the two adjusted equations, cancel common species on opposite sides:
6
$SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)$
7
Add the enthalpy values to get the final result:
8
$\Delta H = +98 + (-272) = -174 \text{ kJ mol}^{-1}$

Exam tip:

Always check that the final equation matches the target exactly after cancelling species

4. Common Pitfalls

Wrong move:

Using $Δ H_{r} = \sum Δ H_{c} (products) - \sum Δ H_{c} (reactants)$

Why:

The formula for combustion is reversed compared to formation because of how the enthalpy cycle is constructed

Correct move:

Always use $Δ H_{r} = \sum Δ H_{c} (reactants) - \sum Δ H_{c} (products)$ for combustion calculations

Wrong move:

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

Why:

Reversing a reaction swaps reactants and products, so the direction of the enthalpy change flips

Correct move:

Always change the sign of $Δ H$ when you reverse a reaction in Hess' law calculations

Wrong move:

Not scaling $Δ H$ when adjusting reaction stoichiometry

Why:

Enthalpy change is an extensive property proportional to the amount of substance reacting

Correct move:

If you multiply reaction coefficients by $n$ , multiply $Δ H$ by $n$ , and vice versa for dividing

Wrong move:

Assuming all elements have $Δ H_{f} = 0$ regardless of state

Why:

Only elements in their standard (most stable) state have $Δ H_{f} = 0$

Correct move:

For example, $Δ H_{f}$ of $C l (g)$ is not zero, only $C l_{2} (g)$ has $Δ H_{f} = 0$

Wrong move:

Drawing arrows in the wrong direction in enthalpy cycles

Why:

Arrow direction determines whether you add or subtract $Δ H$ values, leading to sign errors

Correct move:

Always draw arrows from reactants to their constituent elements, or from all combustion products to the reaction species

5. Quick Reference Cheatsheet

Calculation Type	Formula
Enthalpy from $Δ H_{f}$	$Δ H_{r} = \sum Δ H_{f} (products) - \sum Δ H_{f} (reactants)$
Enthalpy from $Δ H_{c}$	$Δ H_{r} = \sum Δ H_{c} (reactants) - \sum Δ H_{c} (products)$
Reverse a reaction	$Δ H_{new} = - Δ H_{old}$
Scale reaction by $n$	$Δ H_{new} = n \times Δ H_{old}$
Multiple reaction steps	$Δ H_{target} = \sum (adjusted Δ H values)$

What's Next

Hess' law is the foundation for all further enthalpy and energy calculations in CIE A-Level Chemistry, and links directly to upcoming topics including bond enthalpies, Born-Haber cycles for lattice enthalpy, and Gibbs free energy calculations. Mastery of Hess' law sign conventions and cycle construction is critical to avoid losing simple marks in both multiple-choice and structured questions. The core principle that state function changes are independent of path is also applied in electrochemistry for calculating cell potentials from half-cell values, making this a transferable skill across the syllabus. Build on your knowledge with the following topics: