D3.4 State and explain Hess’s law | Ontario Grade 12 Chemistry (SCH4U)

Ontario Grade 12 Chemistry

Energy Changes and Rates of Reaction

Finding an enthalpy change by combining chemical equations

Some reactions release heat, but measuring their enthalpy change directly may be difficult. Hess’s law lets us determine that change using other reactions with known enthalpy changes. The key is to combine balanced chemical equations so their overall reaction matches the reaction of interest.

What you will learn

State Hess’s law in words.
Explain why enthalpy changes can be added when chemical equations are combined.
Use equation reversal and multiplication correctly when applying Hess’s law.

1. A prerequisite: reaction equations and enthalpy change

A balanced chemical equation shows the relative amounts of reactants and products. It conserves each type of atom. A reaction’s enthalpy change, written as

\Delta H

, describes the heat transferred at constant pressure for the amounts shown in the equation. It is usually reported in kilojoules, or kilojoules per reaction as written.

A negative

\Delta H

means the reaction releases heat to its surroundings. A positive

\Delta H

means it absorbs heat. For example, the sign of an enthalpy change tells us the direction of heat transfer; it does not tell us how quickly the reaction occurs.

A chemical equation and its enthalpy change are linked. If we reverse the equation, the heat transfer reverses too, so the sign of

\Delta H

changes. If we multiply every coefficient in an equation by the same number, we multiply its enthalpy change by that number.

\Delta H_{\mathrm{reversed}}=-\Delta H_{\mathrm{forward}}

The equation’s coefficients specify the reaction amounts associated with its $\Delta H$ .
Reversing a reaction changes the sign of $\Delta H$ .
Scaling a reaction scales its $\Delta H$ by the same factor.

2. State and explain Hess’s law

Hess’s law states that the enthalpy change for a reaction is the same whether the reaction occurs in one step or in several steps, provided the overall reactants and products are the same.

Consider an observable result: burning a sample can warm its surroundings. The measured temperature change reflects heat transferred during that process. At the particle level, atoms in the reactants are rearranged into products. If an overall change is described by a sequence of reactions, the intermediate substances are produced in one step and used in another. They cancel when the equations are added.

The final combined equation therefore describes the same overall change in reactants and products. Since the enthalpy change depends on that overall change, not on the chosen route, the enthalpy changes for the steps add to give the enthalpy change for the combined reaction. This is the basis of Hess’s law.

In practice, arrange known equations so that, after any needed reversals or scaling, their sum is the target equation. Add their enthalpy changes using the same reversals and scaling. Cancel substances that appear on both sides. The remaining equation must match the target, including its coefficients.

\Delta H_{\mathrm{overall}}=\sum_i \Delta H_i

Hess’s law: the enthalpy change depends on the overall reaction, not the route taken.
When equations are added, substances appearing on both sides cancel.
The enthalpy changes are added using the same operations as the equations.

3. Rules for combining equations

First, compare the target equation with the available equations. A substance needed as a reactant in the target must appear on the reactant side after the equations are adjusted. A substance needed as a product must appear on the product side.

If an available equation has a substance on the wrong side, reverse the whole equation and change the sign of its enthalpy change. If the target requires twice the amount shown, multiply every coefficient by two and multiply

\Delta H

by two as well. A coefficient applies to every substance in the equation, not just the substance being matched.

After combining the adjusted equations, cancel identical substances on opposite sides. Check that atoms are conserved and that the result is exactly the target equation. Then add the adjusted enthalpy values, keeping their signs and units.

The value of

\Delta H

is tied to the reaction as written. It is not a fixed number independent of the equation’s coefficients. Keep enough digits during addition, then round the final result to a precision supported by the given data.

\Delta H_{\mathrm{combined}}=\sum_i n_i\Delta H_i

Reverse an equation: reverse the sign of its enthalpy change.
Multiply an equation: multiply its enthalpy change by the same factor.
Cancel intermediates and verify the final balanced equation.

4. Interpreting the result

Once the equations have been combined, the sign of the calculated value describes the overall heat transfer for the target reaction as written. A negative result indicates heat is released; a positive result indicates heat is absorbed.

Units must follow the data through the calculation. If the supplied values are in kilojoules for the reactions as written, their adjusted sum is also in kilojoules for the combined reaction as written. If the target equation is subsequently multiplied or reversed, its enthalpy change must be adjusted too.

Hess’s law does not require the target reaction to occur by the particular steps used in the calculation. Those equations are a way to determine the overall enthalpy change. The essential checks are that the adjusted equations add to the target and that their adjusted enthalpy changes have been combined consistently.

Read the sign only after confirming the target equation and enthalpy arithmetic.
Report the enthalpy change for the equation and amounts actually written.
The steps used for calculation need not be the physical route of the target reaction.

Worked example

Finding the enthalpy change for carbon dioxide formation

Use the two reactions below to find

\Delta H

for the formation of carbon dioxide from graphite and oxygen. Treat the enthalpy values as data for the equations as written.

Identify the target
The target is carbon dioxide formation from graphite. The first given reaction forms carbon monoxide, so carbon monoxide must cancel as an intermediate.
$\mathrm{C(s, graphite)}+\mathrm{O_2(g)}\rightarrow\mathrm{CO_2(g)}$
Prepare the given equations
The first equation forms one mole of carbon monoxide using one-half mole of oxygen. The second consumes that carbon monoxide and forms carbon dioxide. Their equations already place carbon monoxide on opposite sides, so it will cancel when they are added.
$\begin{aligned}\mathrm{C(s, graphite)}+\frac{1}{2}\mathrm{O_2(g)}&\rightarrow\mathrm{CO(g)} &&\Delta H_1=-110.5\ \mathrm{kJ}\\\mathrm{CO(g)}+\frac{1}{2}\mathrm{O_2(g)}&\rightarrow\mathrm{CO_2(g)} &&\Delta H_2=-283.0\ \mathrm{kJ}\end{aligned}$
Add the equations and enthalpy changes
Add the equations term by term. Carbon monoxide appears once as a product and once as a reactant, so it cancels. The two half-moles of oxygen combine to one mole. Add the two negative enthalpy changes because neither equation was reversed or scaled.
$\Delta H=(-110.5\ \mathrm{kJ})+(-283.0\ \mathrm{kJ})=-393.5\ \mathrm{kJ}$
Check and report
The remaining balanced equation matches the target. The negative sign means the reaction releases heat. The data are given to one decimal place, so report the result to one decimal place.
$\mathrm{C(s, graphite)}+\mathrm{O_2(g)}\rightarrow\mathrm{CO_2(g)}\qquad\Delta H=-393.5\ \mathrm{kJ}$

Answer: The enthalpy change for the target reaction is

-393.5\ \mathrm{kJ}

as written.

Check: Carbon, oxygen, and the reaction amounts balance in the final equation. The enthalpy values were added without changing their signs because both equations were used as written.

Common mistakes and how to avoid them

Reversing an equation but keeping its original enthalpy sign.

Correction: Change the sign of

\Delta H

whenever the equation is reversed.

Multiplying an equation’s coefficients but leaving its enthalpy change unchanged.

Correction: Multiply

\Delta H

by the same factor as every equation coefficient.

Adding enthalpy values before checking whether the equations produce the target reaction.

Correction: Adjust and add the equations first, cancel intermediates, and verify the target. Then add the matching enthalpy changes.

Treating an intermediate that cancels as part of the overall reaction.

Correction: Cancel a substance when the same amount appears on both sides of the combined equation.

Lesson summary

Hess’s law says that a reaction’s enthalpy change is the same for any route with the same overall reactants and products.
Add balanced equations to obtain the target equation, cancelling substances that appear on both sides.
Reverse or scale enthalpy changes whenever the corresponding equations are reversed or scaled.
Check the final equation, units, sign, and reported precision.

Check your understanding

Question 1

A reaction has

\Delta H=-45\ \mathrm{kJ}

as written. What is

\Delta H

when the equation is reversed?

$-45\ \mathrm{kJ}$
$+45\ \mathrm{kJ}$
$-90\ \mathrm{kJ}$
$+22.5\ \mathrm{kJ}$

Show answer and explanation

+45\ \mathrm{kJ}

Reversing the reaction reverses the heat transfer, so the enthalpy change changes sign but keeps the same magnitude.

Question 2

An equation with

\Delta H=+30\ \mathrm{kJ}

is multiplied by three. What enthalpy change belongs to the multiplied equation?

$+10\ \mathrm{kJ}$
$+30\ \mathrm{kJ}$
$+90\ \mathrm{kJ}$
$-90\ \mathrm{kJ}$

Show answer and explanation

+90\ \mathrm{kJ}

Every coefficient is tripled, so the enthalpy change is also tripled:

3(+30\ \mathrm{kJ})=+90\ \mathrm{kJ}

.

Key terms

Enthalpy change: The heat transferred at constant pressure for a reaction as written, represented by $\Delta H$ .
Intermediate: A substance formed in one step and used in another that cancels from the overall equation.
Hess’s law: The principle that an overall reaction has the same enthalpy change regardless of the route taken between the same reactants and products.

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D3.4 · State and explain Hess’s law