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# Hess' Law

Sometimes, you may be asked to find out an enthalpy change that you can't measure directly. In these cases, you can use something called Hess' law. This involves taking enthalpy changes you do know and using them to work out the unknown value you don't know.

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Hess' law is a relationship used in physical chemistry. It states that the enthalpy change of a reaction is independent of the route taken. It is also known as Hess' law of constant heat summation.

• We will be looking at Hess' law.
• We'll explore how to represent Hess' law using energy diagrams and Hess' cycles.
• You'll learn how to calculate unknown enthalpy changes using enthalpy changes of combustion and formation.
• You'll be able to practice your skills using worked examples.

## What is Hess' law?

A reaction can take many different routes. Let's say you start with reactant A and want to get to product B. You can go directly from A to B, or you could go backwards all the way through the alphabet, passing through Z, X, Y, until you reach B. But according to Hess' law, it doesn't matter how many steps you take - provided you start and end with the same species, the enthalpy change will be the same. Hess' law reiterates the first law of thermodynamics, which says that energy is always conserved in a chemical reaction.

This is useful because it allows us to work out the unknown enthalpy change of one particular route, using the known enthalpy changes of another route. We'll look at some examples in just a second. But first, let's look at how we can represent Hess' law.

## How do we show Hess' law?

To show Hess' law in action, we need to consider the different routes of a reaction. Like in the introduction above, let's say that we start with reactant A. It reacts directly to form product B. We can call this reaction route 1. However, A can also react to form intermediate Z, which reacts to form intermediate Y. Y then forms X and X in turn reacts to form B. We can call this indirect route, from A, all the way through Z, Y and X, and finally ending at B, route 2. In both cases, we started and ended with the same species. Hess' law tells us that the enthalpy changes of the two routes will be the same.

We can display the different routes a reaction can take in two ways:

• Using an energy diagram
• Using a Hess' cycle

### Energy diagrams

Energy diagrams show the energy level of species at different points in a reaction. The difference in energy levels between products and reactants gives us the enthalpy change of the reaction. Here, you can see that both routes start with A and end up at B. The overall enthalpy change is the same, even though route 1 went directly from A to B, whereas route 2 went through Z, X and Y.

Fig. 1 - An energy diagram showing Hess' law

### Hess' cycle

Hess' cycles are another simple way of showing the different routes of a reaction. You don't need to show the energy levels of the different species involved. Instead, you simply map out the two different routes like a flow chart.

Fig. 2 - Hess' cycle

Remember that routes 1 and 2 both start with the same reactants and end with the same products. This means that they have the same overall enthalpy change. By writing in some of the enthalpy changes that we do know, we can calculate an enthalpy change that we don't know.

What if we want to find out the enthalpy change of route 1? Well, if we know the enthalpy changes of all the reactions in route 2, we can calculate the enthalpy change of route 1.

Fig. 3 - Hess' cycle example

The enthalpy change of route 1 equals all of the enthalpy changes of route 2 added together. Here,

$$\Delta H_1$$ = $$\Delta H_2+ \Delta H_3+\Delta H_4+\Delta H_5$$

## Examples of Hess' law

Now that we know how Hess' law works, let's apply it to some real-life calculations.

### Hess' law and enthalpy of formation

The first type of calculation we are going to look at involves using enthalpies of formation to calculate the enthalpy change of a reaction.

The enthalpy of formation, $$\Delta H_f^\Theta$$ , is the enthalpy change when one mole of a substance is formed from its elements, with all species in their standard states and under standard conditions.

The reactants and products in a chemical reaction are always made up of the same elements. This means that we can create a Hess' cycle with an indirect route going via these elements, using enthalpies of formation to help us. Here's an example.

$$C_3H_6(g) + H_2(g)\rightarrow C_3H_8(g)$$

 Species ${\mathrm{C}}_{3}{\mathrm{H}}_{6}\left(\mathrm{g}\right)$ (Propene) ${\mathrm{H}}_{2}\left(\mathrm{g}\right)$ (Hydrogen) ${\mathrm{C}}_{3}{\mathrm{H}}_{8}\left(\mathrm{g}\right)$ (Propane) ${{\mathrm{\Delta H}}_{\mathrm{f}}}^{\mathrm{\theta }}\left(\mathrm{kJ}{\mathrm{mol}}^{-1}\right)$ +20.4 +0.0 -103.8

Remember that the enthalpy of formation for any elemental molecule is always 0.

For this reaction, the direct route goes straight from our reactants, propene and hydrogen, to our product, propane. We'll call this route 1. We don't know the enthalpy change of this route. However, we do know the enthalpies of formation for each of the species involved. We can use them to make up an indirect route, route 2, that goes from reactants to elements to products, and work out the enthalpy change of that instead. Let's draw a Hess' cycle and write in the enthalpies of formation that we know. We can ignore the enthalpy of formation of hydrogen, because it is 0.

Fig. 4 - Hess' cycle for the formation of propane

Make sure you draw all your arrows going in the right direction. Enthalpies of formation always go from the elements to the compounds.

First we go from propene and hydrogen to their elements. This is the reverse of one of the enthalpy changes that we do know - the enthalpy of formation of propene. We know it is the reverse because the black arrow showing the reaction's enthalpy change goes in the opposite direction. We therefore need to take the negative of this enthalpy change. So far, route 2's enthalpy change is $-\left(20.4\right)=-20.4\mathrm{kJ}{\mathrm{mol}}^{-1}$.

We now need to go from the elements to propane. This is one of the enthalpy changes that we do know - the enthalpy of formation of propane. This time, we are following the black arrow in the right direction, so we add this enthalpy value to route 2's current enthalpy change. It now looks like this: $-20.4+\left(-103.8\right)=-124.2\mathrm{kJ}{\mathrm{mol}}^{-1}$

We've made it! We've successfully gone from our reactants, propene and hydrogen, to our product, propane, via the indirect route, and have calculated an enthalpy change: $-124.2\mathrm{kJ}{\mathrm{mol}}^{-1}$. Hess' law tells us that the enthalpy change of a reaction is always the same, regardless of the route taken. Therefore, the enthalpy change of the direct route, route 1, is also $-124.2\mathrm{kJ}{\mathrm{mol}}^{-1}$.

### Hess' law and enthalpy of combustion

Both sides of a chemical reaction always combust to produce the same products. This means that instead of making up an indirect route that goes via elements, we can go via their combustion products instead. This involves using enthalpies of combustion.

Enthalpy of combustion, ${{\mathrm{\Delta H}}_{\mathrm{c}}}^{\mathrm{\theta }}$, is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions.

$$C_3H_6(g)+H_2(g)\rightarrow C_3H_8(g)$$

 Name ${\mathrm{C}}_{3}{\mathrm{H}}_{6}\left(\mathrm{g}\right)$ ${\mathrm{H}}_{2}\left(\mathrm{g}\right)$ ${\mathrm{C}}_{3}{\mathrm{H}}_{8}\left(\mathrm{g}\right)$ ${{\mathrm{\Delta H}}_{\mathrm{c}}}^{\mathrm{\theta }}\left(\mathrm{kJ}{\mathrm{mol}}^{-1}\right)$ -2058 -286 -2469

To start with, let's draw a Hess' cycle. The direct route goes directly from the reactants, propene and hydrogen, to the product, propane. The indirect route goes via their combustion products, which are carbon dioxide and water. We can fill in the enthalpies of combustion:

Fig. 5 - Hess' cycle for the reaction between propene and hydrogen gas to form propane

In order to fully balance the three equations, we'd need to add oxygen to the reactants and products. However, the enthalpies of combustion of these oxygen molecules would cancel out, so we don't have to worry about including them.

Route 2 first takes us from propene and hydrogen to their combustion products. Notice how we are going in the direction of two black arrows - the enthalpy change for this is equal to the enthalpies of combustion of propene and hydrogen. So far, our route's overall enthalpy change looks like this: $\left(-286\right)+\left(-2058\right)=-2344\mathrm{kJ}{\mathrm{mol}}^{-1}$.

We now need to go from the combustion products to the overall product of the reaction, propane. This is the reverse of the enthalpy of combustion of propane - we are going in the opposite direction to the black arrow. We therefore need to take away propane's enthalpy of combustion from our route's overall enthalpy change: $-2344-\left(-2469\right)=125\mathrm{kJ}{\mathrm{mol}}^{-1}$.

Once again, we've successfully completed route 2, making it from our reactants to our products. The overall enthalpy change is $125\mathrm{kJ}{\mathrm{mol}}^{-1}$ . Hess' law states that the enthalpy change of a reaction is the same regardless of the route taken, and so we know that the enthalpy change for route 1 must also be $125\mathrm{kJ}{\mathrm{mol}}^{-1}$. This is our final answer.

You probably noticed that we used the same reaction for both examples, and yet produced slightly different answers. This is because measuring enthalpy changes, be it enthalpy of formation or enthalpy of combustion, is extremely tricky to do accurately. Values vary slightly. However, we reckon that 0.8 of a kilojoule per mole isn't that big of a difference!

Let's look at one more example. This time, we're going to use enthalpies of combustion to calculate the enthalpy of formation of ethane, C2H6.

$$2C(s)+3H_2(g)\rightarrow C_2H_6(g)$$

 Species $\mathrm{C}\left(\mathrm{s}\right)$ ${\mathrm{H}}_{2}\left(\mathrm{g}\right)$ ${\mathrm{C}}_{2}{\mathrm{H}}_{6}\left(\mathrm{g}\right)$ ${{\mathrm{\Delta H}}_{\mathrm{c}}}^{\mathrm{\theta }}\left(\mathrm{kJ}{\mathrm{mol}}^{-1}\right)$ -394 -286 -1560

First, draw a Hess' cycle. The direct route goes straight from our reactants, carbon and hydrogen, to our product, ethane. The indirect route goes via the combustion products. These are the same for both sides of the reaction, so we don't need to write them out - we can just put 'Combustion products'. However, notice that in our equation, we have two moles of carbon and three moles of hydrogen. Our indirect route therefore needs to feature two lots of the enthalpy of combustion of carbon, and three lots of the enthalpy of combustion of hydrogen.

Fig. 6 - Hess' cycle for the formation of ethane

Following route 2 from carbon and hydrogen to the combustion products, we get an enthalpy change of $2\left(-394\right)+3\left(-286\right)=-1646\mathrm{kJ}{\mathrm{mol}}^{-1}$.

To get from the combustion products to ethane, we need the reverse of the enthalpy of combustion of ethane. Route 2's overall enthalpy change is therefore $-1646-\left(-1560\right)=-86\mathrm{kJ}{\mathrm{mol}}^{-1}$. This means that route 1's enthalpy change, which is the enthalpy of formation of ethane, is also $-86\mathrm{kJ}{\mathrm{mol}}^{-1}$.

## The applications of Hess' law

As we explored above, some enthalpy changes are hard to measure. We can use Hess' law to calculate these instead. Examples include:

• Calculating lattice enthalpies.
• Calculating the enthalpy of formation of benzene, ${\mathrm{C}}_{6}{\mathrm{H}}_{6}$.
• Calculating the enthalpy change when graphite turns into diamond.

## Hess' Law - Key takeaways

• Hess' law states that the enthalpy change of a reaction is independent of the route taken. It is also known as Hess' law of constant heat summation.
• We can use Hess' law to calculate unknown enthalpy changes using enthalpy changes that we do know.
• We can represent Hess' law using an energy diagram or a Hess' cycle.
• Hess' law calculations often involve enthalpies of formation and enthalpies of combustion.

#### Flashcards in Hess' Law 18

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What is Hess' law?

Hess' law is a relationship used in physical chemistry. It states that the enthalpy change of a reaction is independent of the route taken.

How do you do Hess' law calculations?

To do Hess' law calculations, you draw a Hess' cycle. This shows a direct route, going from reactants to products, and an indirect route, going via intermediates. You write in the enthalpy changes of the indirect route involving the products, reactants, and intermediates, and use them to calculate the enthalpy change of the direct route.

How is Hess' law applied in calculating enthalpy?

Hess' law states that the enthalpy change of a reaction is independent of the route taken. This means that we can calculate the enthalpy change of the direct route of a reaction using the enthalpy changes of an indirect route. For example, say you want to calculate the enthalpy change of the reaction between ethene and hydrogen to form ethane. You can work this out using the enthalpy changes of formation of the three species involved.

Can Hess' law be applied to Gibbs free energy?

Hess' law also extends to Gibbs free energy. The overall change in Gibbs free energy in a reaction is always the same, regardless of the route taken.

Take the direct route of the reaction A + B → D. Let's say that the indirect route consists of the reactions A + B → 2C and 2C → D. The changes in Gibbs free energy for these two reactions are ΔGoand ΔGo2 respectively. The overall change in Gibbs free energy of the direct route is therefore ΔGo1 + ΔGo2.

How do you use Hess' law?

Hess' law states that the enthalpy change of a reaction is independent of the route taken. This means that we can use it to calculate an enthalpy change that we don't know, using enthalpy changes that we do know. For example, you might want to calculate the enthalpy change of a reaction. You can do this using the enthalpy changes of formation of all of the reactants and products involved.

## Test your knowledge with multiple choice flashcards

What is the value of the enthalpy of formation for an uncombined element, in kJ mol-1?

Which of the following are true about Hess' cycles involving standard enthalpy of formation?

Which of the following are true about Hess' cycles involving standard enthalpy of combustion?

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