Hess's Law

Hess's law is one of the most powerful tools in energetics. It allows us to calculate enthalpy changes for reactions that are difficult or impossible to measure directly. Once you understand how to construct enthalpy cycles, you can solve a huge range of problems. This lesson covers the law itself, both formation and combustion routes, multi-step manipulation problems, and a visual Hess's law cycle diagram.

---

Statement of Hess's Law

Hess's law states that the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.

This is a direct consequence of the law of conservation of energy. If you could get more (or less) energy by taking a different route from the same reactants to the same products, you would be creating (or destroying) energy, which violates the first law of thermodynamics.

In practical terms, this means that if a direct reaction is hard to measure, we can calculate its enthalpy change by finding an alternative route through intermediate steps whose enthalpy changes are known.

---

Visualising Hess's Law Cycles

A Hess's law cycle is a diagram that shows two different routes from the same starting point to the same finishing point. The enthalpy change along Route 1 (the direct route) must equal the total enthalpy change along Route 2 (the indirect route).

graph TD
    A["Reactants"] -->|"ΔH (direct, Route 1)"| B["Products"]
    A -->|"ΔH₁ (step 1)"| C["Intermediate"]
    C -->|"ΔH₂ (step 2)"| B

By Hess's law: ΔH = ΔH₁ + ΔH₂

This principle applies regardless of how many intermediate steps are involved.

---

Enthalpy Cycles Using Enthalpies of Formation

One of the most common applications is calculating the enthalpy of a reaction using standard enthalpies of formation.

The General Formula

For any reaction:

ΔH°reaction = Σ ΔH°f (products) − Σ ΔH°f (reactants)

This works because we can imagine decomposing all reactants into their elements (the reverse of formation) and then forming all products from those elements.

graph TD
    A["Reactants"] -->|"ΔH°rxn (Route 1)"| B["Products"]
    A -->|"−Σ ΔH°f(reactants)"| C["Elements in standard states"]
    C -->|"+Σ ΔH°f(products)"| B

Worked Example 1: Combustion of Methane via Formation Data

Calculate the standard enthalpy change for:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Given:

Substance	ΔH°f / kJ mol⁻¹
CH₄(g)	−75
CO₂(g)	−394
H₂O(l)	−286
O₂(g)	0

Solution:

ΔH°rxn = [ΔH°f(CO₂) + 2 × ΔH°f(H₂O)] − [ΔH°f(CH₄) + 2 × ΔH°f(O₂)]

ΔH°rxn = [(−394) + 2(−286)] − [(−75) + 2(0)]

ΔH°rxn = [−394 − 572] − [−75]

ΔH°rxn = −966 + 75 = −891 kJ mol⁻¹

Sign check: This is the combustion of methane — it must be exothermic (negative). Our answer is −891 kJ mol⁻¹. Correct.

---

Enthalpy Cycles Using Enthalpies of Combustion

When you are given combustion data rather than formation data, a different formula applies:

ΔH°reaction = Σ ΔH°c (reactants) − Σ ΔH°c (products)

Notice the subtraction is reversed compared to the formation formula. This is because combustion takes us from the substances down to combustion products, so the cycle arrows point differently.

graph TD
    A["Reactants"] -->|"ΔH°rxn (Route 1)"| B["Products"]
    A -->|"+Σ ΔH°c(reactants)"| C["Combustion products: CO₂ + H₂O"]
    B -->|"+Σ ΔH°c(products)"| C

Worked Example 2: Hydrogenation of Ethene

Calculate ΔH for: C₂H₄(g) + H₂(g) → C₂H₆(g)

Given:

Substance	ΔH°c / kJ mol⁻¹
C₂H₄(g)	−1411
H₂(g)	−286
C₂H₆(g)	−1560

Solution:

ΔH°rxn = Σ ΔH°c(reactants) − Σ ΔH°c(products)

ΔH°rxn = [(−1411) + (−286)] − [(−1560)]

ΔH°rxn = −1697 + 1560 = −137 kJ mol⁻¹

---

Multi-Step Hess's Law Problems

Some exam questions give you several thermochemical equations and ask you to calculate ΔH for a target reaction. The systematic approach is:

1. Write the target equation clearly.
2. Examine each given equation — decide if it needs to be reversed, multiplied, or used as-is.
3. Reverse equations as needed (and change the sign of ΔH).
4. Multiply equations as needed (and multiply ΔH by the same factor).
5. Add the manipulated equations — intermediate species should cancel.
6. Add the ΔH values to get the answer.

Worked Example 3: Hydrogenation of Ethyne

Calculate ΔH for: C₂H₂(g) + 2H₂(g) → C₂H₆(g)

Given:

• (1) C₂H₂(g) + 5/2 O₂(g) → 2CO₂(g) + H₂O(l), ΔH₁ = −1300 kJ mol⁻¹
• (2) C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH₂ = −1560 kJ mol⁻¹
• (3) H₂(g) + ½O₂(g) → H₂O(l), ΔH₃ = −286 kJ mol⁻¹

Solution:

We need C₂H₂ on the left → use equation (1) as-is. We need C₂H₆ on the right → reverse equation (2): ΔH = +1560. We need 2H₂ on the left → multiply equation (3) by 2: ΔH = 2 × (−286) = −572.

Adding the equations and cancelling CO₂, H₂O, and O₂ on both sides gives:

C₂H₂(g) + 2H₂(g) → C₂H₆(g)

ΔH = −1300 + 1560 + (−572) = −312 kJ mol⁻¹

Worked Example 4: Formation of Ethanol from Combustion Data

Calculate ΔH°f for C₂H₅OH(l) given:

• ΔH°c [C(s, graphite)] = −394 kJ mol⁻¹
• ΔH°c [H₂(g)] = −286 kJ mol⁻¹
• ΔH°c [C₂H₅OH(l)] = −1367 kJ mol⁻¹

The formation equation is: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)

Using the combustion cycle:

ΔH°f = Σ ΔH°c(reactants) − Σ ΔH°c(products)

Only carbon and hydrogen combust (not O₂); ethanol is the product and its combustion enthalpy is used on the product side:

ΔH°f = [2(−394) + 3(−286)] − [(−1367)]

ΔH°f = [−788 − 858] − [−1367]

ΔH°f = −1646 + 1367 = −279 kJ mol⁻¹

The data book value is −277 kJ mol⁻¹ — very close, confirming our calculation.

---

Common Sign Errors in Hess's Law

Error	What goes wrong	How to avoid it
Forgetting to reverse the sign when reversing an equation	The final answer has the wrong sign or wrong magnitude	Always write the new ΔH next to the reversed equation immediately
Subtracting reactants from products with formation data	Gets the signs backwards	Remember: ΔH = Σf(products) − Σf(reactants)
Subtracting reactants from products with combustion data	Gets the signs backwards	Remember: ΔH = Σc(reactants) − Σc(products) — opposite order
Forgetting to multiply ΔH when multiplying an equation	Intermediate species don't cancel properly	Always multiply ΔH by the same factor as the equation
Including ΔH°f of elements (e.g. O₂) as non-zero	Adds a false contribution	ΔH°f of any element in its standard state = 0

---

Real-World Application: Industrial Ammonia Production

The Haber process produces ammonia: N₂(g) + 3H₂(g) → 2NH₃(g)

This reaction is difficult to study directly at standard conditions because it requires high temperature and pressure. However, using Hess's law with formation data (ΔH°f [NH₃] = −46 kJ mol⁻¹):

ΔH°rxn = 2(−46) − [0 + 0] = −92 kJ mol⁻¹

Knowing this value is negative tells engineers that the forward reaction is exothermic, which is why low temperatures favour a higher yield (Le Chatelier's principle), but the rate is too slow without a catalyst and elevated temperature.

---

Summary

• Hess's law: ΔH is independent of route.
• Formation route: ΔH = ΣΔH°f(products) − ΣΔH°f(reactants).
• Combustion route: ΔH = ΣΔH°c(reactants) − ΣΔH°c(products).
• Multi-step: reverse, multiply, add — always adjust ΔH when manipulating equations.
• Always perform a sign check at the end.

---

A-Level Deep Dive: Hess's Law

Spec mapping

Edexcel 9CH0 specification Topic 8.2 requires application of Hess's law to calculate enthalpy changes that cannot be measured directly, using ΔH°_f, ΔH°_c, or any combination of routes between the same initial and final states (refer to the official specification document for exact wording). Hess cycles are the foundation for every Born–Haber calculation in Topic 13 and underpin the indirect determination of enthalpies of hydration, solution, and lattice energy. Hess questions appear on Paper 1 as multi-mark calculations and on Paper 3 as synoptic items combining Hess with other topics. Core Practical 9 (CP9) uses a Hess cycle to determine the enthalpy of formation of MgO indirectly via the reactions of Mg and MgO with HCl, since direct combustion of magnesium under controlled conditions is impractical.

---

Worked example with full mark scheme

Question (8 marks):

The enthalpy of formation of CuSO₄·5H₂O cannot be measured directly. A student determined it using two enthalpies of solution:

CuSO₄(s) + aq → CuSO₄(aq) ΔH₁ = −66 kJ mol⁻¹ CuSO₄·5H₂O(s) + aq → CuSO₄(aq) ΔH₂ = +11 kJ mol⁻¹

(a) Construct a Hess cycle linking CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s) to the two measured enthalpies. (3)

(b) Calculate the enthalpy of hydration of anhydrous CuSO₄ to form the pentahydrate. (3)

(c) State two reasons why this indirect method is more accurate than direct measurement. (2)

Solution with mark scheme:

(a) Cycle: CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s); both routes lead to CuSO₄(aq) by addition of (excess) water.

CuSO₄(s) ──ΔH₁──→ CuSO₄(aq) CuSO₄·5H₂O(s) ──ΔH₂──→ CuSO₄(aq) Required: CuSO₄(s) + 5H₂O(l) ──ΔH_hyd──→ CuSO₄·5H₂O(s)

By Hess: ΔH_hyd + ΔH₂ = ΔH₁ ⟹ ΔH_hyd = ΔH₁ − ΔH₂.

M1 — both arrows drawn correctly with correct directions. A1 — common destination CuSO₄(aq) identified. B1 — Hess relationship written algebraically.

(b) ΔH_hyd = (−66) − (+11) = −77 kJ mol⁻¹.

M1 — substitution. M1 — correct subtraction including signs. A1 — correct numerical answer with sign.

(c) Direct hydration is slow and incomplete in practice; the indirect method uses two fast, complete dissolutions. B1. The two enthalpies of solution can be measured at the same temperature in the same calorimeter, minimising systematic error. B1.

Total: 8 marks (M3 A3 B2).

---

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks):

Use the standard enthalpies of formation below to calculate the standard enthalpy change for the thermite reaction:

2Al(s) + Fe₂O₃(s) → 2Fe(s) + Al₂O₃(s)

ΔH°_f[Fe₂O₃(s)] = −824 kJ mol⁻¹; ΔH°_f[Al₂O₃(s)] = −1676 kJ mol⁻¹.

(a) Calculate ΔH°_r. (3) (b) State and explain whether the reaction would be expected to require initiation. (3)

Mark scheme decomposition by AO: