Hess cycles

What you'll learn

What Hess’s law says and why it works.
How to draw and use Hess cycles for formation and combustion data.
How to calculate an unknown enthalpy change using a cycle.
How to avoid the common sign, state-symbol and coefficient mistakes.

The energy idea you already need

In Energetics, we often measure or calculate an enthalpy change, written $ΔH\Delta H$ .

Definition

Enthalpy change

An enthalpy change, $ΔH\Delta H$ , is the heat energy change for a reaction at constant pressure, usually measured in kJ mol⁻¹.

If $ΔH\Delta H$ is negative, the reaction is exothermic: energy is transferred to the surroundings. If $ΔH\Delta H$ is positive, the reaction is endothermic: energy is taken in from the surroundings.

For Hess cycles, the key point is that an enthalpy change is linked to the initial and final chemical states, not the detailed path between them.

Standard enthalpy changes

Many Hess cycle questions use standard enthalpy changes, shown using the symbol $ΔH∘\Delta H^\circ$ .

Definition

Standard conditions

For Edexcel A-Level Chemistry, standard conditions usually mean 298 K and 100 kPa, with substances in their standard states. For solutions, concentration is usually 1 mol dm⁻³.

A standard state is the physical state of a substance under standard conditions. For example, oxygen is O₂(g), carbon is C(s, graphite), and water is H₂O(l) at 298 K and 100 kPa.

Common Mistake

Ignoring state symbols

H₂O(l) and H₂O(g) have different enthalpy values. In Hess cycle calculations, always use the data that matches the state symbols in the equation.

Hess’s law

Definition

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 works because enthalpy is a state function. A state function depends only on the current state of the system, not on how the system got there.

So if a reaction can go directly from reactants to products, or indirectly through another set of substances, the total $ΔH\Delta H$ must be the same for both routes.

Key Idea

The big idea

In a Hess cycle, two different routes connect the same starting and finishing substances. The enthalpy changes around the cycle must add up consistently.

What a Hess cycle is

A Hess cycle is a diagram that uses arrows to compare a direct reaction route with an indirect route using known enthalpy changes.

The two most common A-Level cycles use enthalpies of formation or enthalpies of combustion.

Formation and combustion Hess cycle templates

The arrows matter. If you go with an arrow, use the enthalpy value as written. If you go against an arrow, change the sign.

Tip

Arrow rule

Following an arrow keeps the sign of $ΔH\Delta H$ . Going against an arrow reverses the sign of $ΔH\Delta H$ .

Enthalpy of formation cycles

Definition

Standard enthalpy of formation

The standard enthalpy of formation, $ΔHf∘\Delta H_f^\circ$ , is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

For example:

graphite)+2H2(g)→CH4(g)\text{C(s, graphite)} + 2\text{H}_2\text{(g)} \to \text{CH}_4\text{(g)}

This represents the formation of one mole of methane, so its enthalpy change is $ΔHf∘\Delta H_f^\circ$ for CH₄(g).

Elements in their standard states have $ΔHf∘=0\Delta H_f^\circ = 0$ kJ mol⁻¹. For example, O₂(g), H₂(g), N₂(g), Cl₂(g), Br₂(l), I₂(s), and C(s, graphite) all have zero standard enthalpy of formation.

The formation formula

For a reaction:

reactants→products\text{reactants} \to \text{products}

the standard enthalpy change of reaction can be found using:

ΔHr∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_r^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)}

Remember to multiply each $ΔHf∘\Delta H_f^\circ$ value by the balancing coefficient in the equation.

Example

Using enthalpies of formation

Calculate the standard enthalpy change for:

CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \to \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}

Use the following standard enthalpies of formation:

CH₄(g): −74.8 kJ mol⁻¹
O₂(g): 0 kJ mol⁻¹
CO₂(g): −393.5 kJ mol⁻¹
H₂O(l): −285.8 kJ mol⁻¹

Calculate the total formation enthalpy of the products, including coefficients:
$mol−1\sum \Delta H_f^\circ \text{(products)} = -393.5 + 2(-285.8) = -965.1 \text{ kJ mol}^{-1}$
Calculate the total formation enthalpy of the reactants:
$mol−1\sum \Delta H_f^\circ \text{(reactants)} = -74.8 + 2(0) = -74.8 \text{ kJ mol}^{-1}$
Apply products minus reactants:
$mol−1\Delta H_r^\circ = -965.1 - (-74.8) = -890.3 \text{ kJ mol}^{-1}$

So the reaction is exothermic, with $ΔHr∘=−890.3\Delta H_r^\circ = -890.3$ kJ mol⁻¹.

Common Mistake

Forgetting the coefficient

If the equation contains 2H₂O(l), you must use 2 times the enthalpy value for H₂O(l). The data value is for one mole, but the equation may contain more than one mole.

Enthalpy of combustion cycles

Definition

Standard enthalpy of combustion

The standard enthalpy of combustion, $ΔHc∘\Delta H_c^\circ$ , is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, with all substances in their standard states.

Combustion cycles are useful when you are given combustion data for the reactants and products of a reaction. In these cycles, both sides are burned to the same final substances, often CO₂(g) and H₂O(l).

For combustion cycles, the formula is:

ΔHr∘=∑ΔHc∘(reactants)−∑ΔHc∘(products)\Delta H_r^\circ = \sum \Delta H_c^\circ \text{(reactants)} - \sum \Delta H_c^\circ \text{(products)}

Notice this is the opposite order from the formation formula.

Key Idea

Formation vs combustion

Formation cycles use products minus reactants. Combustion cycles use reactants minus products.

Example

Using enthalpies of combustion

Calculate the enthalpy change for the hydrogenation of ethene:

C2H4(g)+H2(g)→C2H6(g)\text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{(g)} \to \text{C}_2\text{H}_6\text{(g)}

Use the following standard enthalpies of combustion:

C₂H₄(g): −1411 kJ mol⁻¹
H₂(g): −286 kJ mol⁻¹
C₂H₆(g): −1560 kJ mol⁻¹

Add the combustion enthalpies of the reactants:
$mol−1\sum \Delta H_c^\circ \text{(reactants)} = -1411 + (-286) = -1697 \text{ kJ mol}^{-1}$
Add the combustion enthalpies of the products:
$mol−1\sum \Delta H_c^\circ \text{(products)} = -1560 \text{ kJ mol}^{-1}$
Use reactants minus products:
$mol−1\Delta H_r^\circ = -1697 - (-1560) = -137 \text{ kJ mol}^{-1}$

So the hydrogenation of ethene has $ΔHr∘=−137\Delta H_r^\circ = -137$ kJ mol⁻¹.

Common Mistake

Using the formation formula for combustion data

With combustion data, do not do products minus reactants. The cycle arrows go down from both reactants and products to the same combustion products, so the correct shortcut is reactants minus products.

Rearranging Hess cycles to find an unknown

Sometimes the unknown is not $ΔHr∘\Delta H_r^\circ$ . You may be asked to find an unknown enthalpy of formation or combustion.

The safest method is to write the normal Hess formula first, then substitute the unknown as a symbol.

Example

Finding an unknown enthalpy of formation

The standard enthalpy of combustion of propane is −2220 kJ mol⁻¹:

C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \to 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}

Find $ΔHf∘\Delta H_f^\circ$ for C₃H₈(g), using:

CO₂(g): $ΔHf∘=−393.5\Delta H_f^\circ = -393.5$ kJ mol⁻¹
H₂O(l): $ΔHf∘=−285.8\Delta H_f^\circ = -285.8$ kJ mol⁻¹
O₂(g): $ΔHf∘=0\Delta H_f^\circ = 0$ kJ mol⁻¹

Let the unknown formation enthalpy of propane be $x$ kJ mol⁻¹.
Use the formation equation for the combustion reaction:
$ΔHr∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_r^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)}$
Substitute the values:
$\left[3(-393.5) + 4(-285.8)\right] - \left[x + 5(0)\right]$
Calculate and solve:
$−2220=−2323.7−xx=−103.7\begin{aligned} -2220 &= -2323.7 - x \\ x &= -103.7 \end{aligned}$

So $ΔHf∘\Delta H_f^\circ$ for C₃H₈(g) is −103.7 kJ mol⁻¹.

Using equations instead of diagrams

You can also use Hess’s law by manipulating thermochemical equations.

A thermochemical equation is a balanced chemical equation with an enthalpy change attached to it.

The rules are:

If you reverse an equation, change the sign of $ΔH\Delta H$ .
If you multiply an equation by a number, multiply $ΔH\Delta H$ by the same number.
If you add equations, add their enthalpy changes.

This is the same chemistry as a Hess cycle, just written algebraically.

Tip

Sanity check

Most combustion enthalpies are negative because combustion is exothermic. If your calculated combustion value comes out positive, check your signs and arrows carefully.

Why Hess cycles are useful

Some enthalpy changes are hard to measure directly. For example, a reaction may be too slow, incomplete, unsafe, or may produce several products.

Hess’s law lets you calculate the desired enthalpy change using reactions that are easier to measure, such as combustion reactions or formation reactions.

In practical energetics, measured values often come from calorimetry using:

$mc\Delta T$
$n = m / M$ or $n = c V$
$ΔH=−q/n\Delta H = -q/n$

The negative sign appears because the heat change of the reaction is opposite to the heat change of the surroundings.

Common Mistake

Calorimetry values are often less exothermic

Experimental combustion enthalpies are often less negative than data book values because of heat loss, incomplete combustion, evaporation of fuel, or heating the apparatus as well as the water.

A quick method for any Hess calculation

When you meet a Hess cycle calculation, slow down and be systematic.

Balance the target equation.
Identify whether the data are formation, combustion, or other thermochemical equations.
Multiply each enthalpy value by the balancing coefficient.
Choose the correct sign using either the cycle arrows or the formula.
Give the final answer in kJ mol⁻¹ for the reaction as written.

Exam technique

In the exam

Check whether the data are $ΔHf∘\Delta H_f^\circ$ or $ΔHc∘\Delta H_c^\circ$ before choosing the formula.
Include coefficients in your sums; enthalpy values are per mole of substance, not per equation automatically.
Watch state symbols, especially H₂O(l) versus H₂O(g), and quote your final answer with a sign and units.

Self review

Check yourself

Why is $ΔHf∘\Delta H_f^\circ$ for O₂(g) equal to zero, but not necessarily for O₃(g)?
In a combustion Hess cycle, why is the shortcut reactants minus products?
What happens to $ΔH\Delta H$ if you reverse a thermochemical equation?

What you'll learn

What Hess’s law says and why it works.
How to draw and use Hess cycles for formation and combustion data.
How to calculate an unknown enthalpy change using a cycle.
How to avoid the common sign, state-symbol and coefficient mistakes.

The energy idea you already need

In Energetics, we often measure or calculate an enthalpy change, written $ΔH\Delta H$ .

Definition

Enthalpy change

An enthalpy change, $ΔH\Delta H$ , is the heat energy change for a reaction at constant pressure, usually measured in kJ mol⁻¹.

For Hess cycles, the key point is that an enthalpy change is linked to the initial and final chemical states, not the detailed path between them.

Standard enthalpy changes

Many Hess cycle questions use standard enthalpy changes, shown using the symbol $ΔH∘\Delta H^\circ$ .

Definition

Standard conditions

For Edexcel A-Level Chemistry, standard conditions usually mean 298 K and 100 kPa, with substances in their standard states. For solutions, concentration is usually 1 mol dm⁻³.

A standard state is the physical state of a substance under standard conditions. For example, oxygen is O₂(g), carbon is C(s, graphite), and water is H₂O(l) at 298 K and 100 kPa.

Common Mistake

Ignoring state symbols

H₂O(l) and H₂O(g) have different enthalpy values. In Hess cycle calculations, always use the data that matches the state symbols in the equation.

Hess’s law

Definition

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 works because enthalpy is a state function. A state function depends only on the current state of the system, not on how the system got there.

So if a reaction can go directly from reactants to products, or indirectly through another set of substances, the total $ΔH\Delta H$ must be the same for both routes.

Key Idea

The big idea

In a Hess cycle, two different routes connect the same starting and finishing substances. The enthalpy changes around the cycle must add up consistently.

What a Hess cycle is

A Hess cycle is a diagram that uses arrows to compare a direct reaction route with an indirect route using known enthalpy changes.

The two most common A-Level cycles use enthalpies of formation or enthalpies of combustion.

Formation and combustion Hess cycle templates

The arrows matter. If you go with an arrow, use the enthalpy value as written. If you go against an arrow, change the sign.

Tip

Arrow rule

Following an arrow keeps the sign of $ΔH\Delta H$ . Going against an arrow reverses the sign of $ΔH\Delta H$ .

Enthalpy of formation cycles

Definition

Standard enthalpy of formation

The standard enthalpy of formation, $ΔHf∘\Delta H_f^\circ$ , is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

For example:

graphite)+2H2(g)→CH4(g)\text{C(s, graphite)} + 2\text{H}_2\text{(g)} \to \text{CH}_4\text{(g)}

This represents the formation of one mole of methane, so its enthalpy change is $ΔHf∘\Delta H_f^\circ$ for CH₄(g).

The formation formula

For a reaction:

reactants→products\text{reactants} \to \text{products}

the standard enthalpy change of reaction can be found using:

ΔHr∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_r^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)}

Remember to multiply each $ΔHf∘\Delta H_f^\circ$ value by the balancing coefficient in the equation.

Example

Using enthalpies of formation

Calculate the standard enthalpy change for:

CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \to \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}

Use the following standard enthalpies of formation:

CH₄(g): −74.8 kJ mol⁻¹
O₂(g): 0 kJ mol⁻¹
CO₂(g): −393.5 kJ mol⁻¹
H₂O(l): −285.8 kJ mol⁻¹

Calculate the total formation enthalpy of the products, including coefficients:
$mol−1\sum \Delta H_f^\circ \text{(products)} = -393.5 + 2(-285.8) = -965.1 \text{ kJ mol}^{-1}$
Calculate the total formation enthalpy of the reactants:
$mol−1\sum \Delta H_f^\circ \text{(reactants)} = -74.8 + 2(0) = -74.8 \text{ kJ mol}^{-1}$
Apply products minus reactants:
$mol−1\Delta H_r^\circ = -965.1 - (-74.8) = -890.3 \text{ kJ mol}^{-1}$

So the reaction is exothermic, with $ΔHr∘=−890.3\Delta H_r^\circ = -890.3$ kJ mol⁻¹.

Common Mistake

Forgetting the coefficient

If the equation contains 2H₂O(l), you must use 2 times the enthalpy value for H₂O(l). The data value is for one mole, but the equation may contain more than one mole.

Enthalpy of combustion cycles

Definition

Standard enthalpy of combustion

For combustion cycles, the formula is:

ΔHr∘=∑ΔHc∘(reactants)−∑ΔHc∘(products)\Delta H_r^\circ = \sum \Delta H_c^\circ \text{(reactants)} - \sum \Delta H_c^\circ \text{(products)}

Notice this is the opposite order from the formation formula.

Key Idea

Formation vs combustion

Formation cycles use products minus reactants. Combustion cycles use reactants minus products.

Example

Using enthalpies of combustion

Calculate the enthalpy change for the hydrogenation of ethene:

C2H4(g)+H2(g)→C2H6(g)\text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{(g)} \to \text{C}_2\text{H}_6\text{(g)}

Use the following standard enthalpies of combustion:

C₂H₄(g): −1411 kJ mol⁻¹
H₂(g): −286 kJ mol⁻¹
C₂H₆(g): −1560 kJ mol⁻¹

Add the combustion enthalpies of the reactants:
$mol−1\sum \Delta H_c^\circ \text{(reactants)} = -1411 + (-286) = -1697 \text{ kJ mol}^{-1}$
Add the combustion enthalpies of the products:
$mol−1\sum \Delta H_c^\circ \text{(products)} = -1560 \text{ kJ mol}^{-1}$
Use reactants minus products:
$mol−1\Delta H_r^\circ = -1697 - (-1560) = -137 \text{ kJ mol}^{-1}$

So the hydrogenation of ethene has $ΔHr∘=−137\Delta H_r^\circ = -137$ kJ mol⁻¹.

Common Mistake

Using the formation formula for combustion data

Rearranging Hess cycles to find an unknown

Sometimes the unknown is not $ΔHr∘\Delta H_r^\circ$ . You may be asked to find an unknown enthalpy of formation or combustion.

The safest method is to write the normal Hess formula first, then substitute the unknown as a symbol.

Example

Finding an unknown enthalpy of formation

The standard enthalpy of combustion of propane is −2220 kJ mol⁻¹:

C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \to 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}

Find $ΔHf∘\Delta H_f^\circ$ for C₃H₈(g), using:

CO₂(g): $ΔHf∘=−393.5\Delta H_f^\circ = -393.5$ kJ mol⁻¹
H₂O(l): $ΔHf∘=−285.8\Delta H_f^\circ = -285.8$ kJ mol⁻¹
O₂(g): $ΔHf∘=0\Delta H_f^\circ = 0$ kJ mol⁻¹

Let the unknown formation enthalpy of propane be $x$ kJ mol⁻¹.
Use the formation equation for the combustion reaction:
$ΔHr∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_r^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)}$
Substitute the values:
$\left[3(-393.5) + 4(-285.8)\right] - \left[x + 5(0)\right]$
Calculate and solve:
$−2220=−2323.7−xx=−103.7\begin{aligned} -2220 &= -2323.7 - x \\ x &= -103.7 \end{aligned}$

So $ΔHf∘\Delta H_f^\circ$ for C₃H₈(g) is −103.7 kJ mol⁻¹.

Using equations instead of diagrams

You can also use Hess’s law by manipulating thermochemical equations.

A thermochemical equation is a balanced chemical equation with an enthalpy change attached to it.

The rules are:

If you reverse an equation, change the sign of $ΔH\Delta H$ .
If you multiply an equation by a number, multiply $ΔH\Delta H$ by the same number.
If you add equations, add their enthalpy changes.

This is the same chemistry as a Hess cycle, just written algebraically.

Tip

Sanity check

Most combustion enthalpies are negative because combustion is exothermic. If your calculated combustion value comes out positive, check your signs and arrows carefully.

Why Hess cycles are useful

Some enthalpy changes are hard to measure directly. For example, a reaction may be too slow, incomplete, unsafe, or may produce several products.

Hess’s law lets you calculate the desired enthalpy change using reactions that are easier to measure, such as combustion reactions or formation reactions.

In practical energetics, measured values often come from calorimetry using:

$mc\Delta T$
$n = m / M$ or $n = c V$
$ΔH=−q/n\Delta H = -q/n$

The negative sign appears because the heat change of the reaction is opposite to the heat change of the surroundings.

Common Mistake

Calorimetry values are often less exothermic

Experimental combustion enthalpies are often less negative than data book values because of heat loss, incomplete combustion, evaporation of fuel, or heating the apparatus as well as the water.

A quick method for any Hess calculation

When you meet a Hess cycle calculation, slow down and be systematic.

Balance the target equation.
Identify whether the data are formation, combustion, or other thermochemical equations.
Multiply each enthalpy value by the balancing coefficient.
Choose the correct sign using either the cycle arrows or the formula.
Give the final answer in kJ mol⁻¹ for the reaction as written.

Exam technique

In the exam

Check whether the data are $ΔHf∘\Delta H_f^\circ$ or $ΔHc∘\Delta H_c^\circ$ before choosing the formula.
Include coefficients in your sums; enthalpy values are per mole of substance, not per equation automatically.
Watch state symbols, especially H₂O(l) versus H₂O(g), and quote your final answer with a sign and units.

Self review

Check yourself

Why is $ΔHf∘\Delta H_f^\circ$ for O₂(g) equal to zero, but not necessarily for O₃(g)?
In a combustion Hess cycle, why is the shortcut reactants minus products?
What happens to $ΔH\Delta H$ if you reverse a thermochemical equation?

What you'll learn

The energy idea you already need

Standard enthalpy changes

Hess’s law

What a Hess cycle is

Enthalpy of formation cycles

The formation formula

Enthalpy of combustion cycles

Rearranging Hess cycles to find an unknown

Using equations instead of diagrams

Why Hess cycles are useful

A quick method for any Hess calculation

Review Hess cycles by teaching Genie

Remember key concepts with flashcards

Hess cycles

What you'll learn

The energy idea you already need

Standard enthalpy changes

Hess’s law

What a Hess cycle is

Enthalpy of formation cycles

The formation formula

Enthalpy of combustion cycles

Rearranging Hess cycles to find an unknown

Using equations instead of diagrams

Why Hess cycles are useful

A quick method for any Hess calculation

Review Hess cycles by teaching Genie

Remember key concepts with flashcards

Hess cycles

What you'll learn

The energy idea you already need

Standard enthalpy changes

Hess’s law

What a Hess cycle is

Enthalpy of formation cycles

The formation formula

Enthalpy of combustion cycles

Rearranging Hess cycles to find an unknown

Using equations instead of diagrams

Why Hess cycles are useful

A quick method for any Hess calculation

Review Hess cycles by teaching Genie

Remember key concepts with flashcards

Hess cycles Revision Guide

Hess cycles

What you'll learn

The energy idea you already need

Standard enthalpy changes

Hess’s law

What a Hess cycle is

Enthalpy of formation cycles

The formation formula

Enthalpy of combustion cycles

Rearranging Hess cycles to find an unknown

Using equations instead of diagrams

Why Hess cycles are useful

A quick method for any Hess calculation

Review Hess cycles by teaching Genie

Remember key concepts with flashcards

Hess cycles Revision Guide