# Hess’s law: explanation, applications, examples

In this way, the “unknown” ΔH of the reaction can be calculated. For example, the diagram above helps to understand this idea.

There are four reactions: AD, AB, BC, and CD. The reaction AD is the one with the highest ΔH, since it produces, so to speak, the most notorious chemical changes. The other reactions, meanwhile, have smaller ΔH, since they consist of alternative steps or alternate reactions to arrive at the same products D.

Therefore, the sum of ΔH _{1}, ΔH _{2,} and ΔH_{3}, will be equal to ΔH. Knowing this, ΔH, or any other enthalpy, can be calculated by applying simple clearings. This is Hess’s law.

__Explanation of Hess’s law__

__Explanation of Hess’s law__

**Calculation of the enthalpy of a reaction**

The reason why it is true that:

ΔH = ΔH _{1} + ΔH _{2} + ΔH _{3}

It’s because enthalpy, or enthalpy change, is a function of the state. This means that their values are not modified by the number of intermediate steps, stages, or reactions. For example, ΔH will remain the same even if there is a ΔH _{10} that must be considered in the algebraic sum. Enthalpies could be visualized as the tops of mountains.

No matter how the steps are directed towards them, the height that separates them from the sea level will never change. And it is precisely the heights that are added or subtracted to determine the unknown height of a nearby mountain.

Returning to the previous diagram: A is a mountain, D another mountain, and so on. The difference in height between the tops of A and D is the greatest. Adding the heights between AB, BC and CD will equal the height of AD. This is what is done with the enthalpies of their respective chemical reactions.

**Enthalpy of a reaction from enthalpies of formations**

One of the most valuable consequences of Hess’s law is that it allows the determination of the enthalpies of any chemical reaction, taking the enthalpies of formation of all the compounds (reactants and products) involved.

How do you read the diagram this time?

One trick is to always add the enthalpies when going in the direction of the arrow, and subtract them when the arrow is in the opposite direction.

Thus, ΔHºrxn, standard reaction enthalpy, is added to ΔHºf (reactants), standard enthalpy of formation of the reactants, and ΔHºf (products), standard enthalpy of product formation are subtracted from them.

**Reaction sums**

Diagrams can get very difficult to interpret, especially when there are many reactions that need to be considered. Therefore, the sum of the reactions is used. This method greatly facilitates the calculation of enthalpies and speeds up the understanding of Hess’s law.

Suppose for example that we want to determine the ΔHºrxn of the following reaction:

A + B + C → ABC

And we also have two other reactions:

A + B → AB (ΔHº _{1} )

AB + C → ABC (ΔHº _{2} )

It is interesting that A + B + C is on the left side (reactants) and that ABC is on the right side (products). Then, we simply proceed to add the last two reactions:

A + B → AB

__AB + C → ABC__

A + AB + C → ABC + AB

Since AB is on both sides, it is eliminated. And so we have:

A + B + C → ABC

ΔHºrxn = ΔHº _{1} + ΔHº _{2}

Adding the reactions, the use of any diagram is omitted.

__Hess’s law applications__

__Hess’s law applications__

Hess’s law allows us to obtain, without the need for experiments, the enthalpies for different reactions or chemical phenomena. Some of them are listed below:

- Enthalpies of formation for unstable compounds or intermediates, as there are usually in organic chemistry.
- Enthalpies of phase transitions, in the study of crystalline solids.
- Enthalpies of allotropic transitions, such as the one that occurs between graphite and diamond.
- Likewise, Hess’s law is used to determine the lattice energy of a solid and the electronic affinities of some atoms.

**Examples: solved exercises**

**– Example 1**

Calculate the ΔHrxn of the following reaction:

2HCl (g) + F _{2} (g) → 2HF (l) + Cl _{2} (g)

If you have the following reactions and their respective enthalpies on hand:

4HCl (g) + O _{2} (g) → 2H _{2} O (l) + 2Cl _{2} (g) (ΔH = -202.4 kJ / mol)

1 / 2H _{2} (g) + 1 / 2F _{2} (g) → HF (l) (ΔH = -600.0 kJ / mol)

H _{2} (g) + 1 / 2O _{2} (g) → H _{2} O (l) (ΔH = -285.8 kJ / mol)

**Ordering the equations**

To begin, we must add the reactions such that HCl and F _{2} are on the left side, and HF and Cl _{2 are} on the right side. But more importantly, it is to note that both H _{2} and H _{2} O are not in the equation of interest. Therefore, we must cancel them in the sum, and match the stoichiometric coefficients by multiplication:

2HCl (g) + 1 / 2O _{2} (g) → H _{2} O (l) + Cl _{2} (g) (ΔH = -202.4 kJ / mol) * (1/2)

This equation was multiplied by 1/2 to have 2HCl instead of 4HCl

H _{2} (g) + F _{2} (g) → 2HF (l) (ΔH = -600.0 kJ / mol) * (2)

This equation was multiplied by 2 to have F _{2} and not 1 / 2F _{2}

H _{2} O (l) → H _{2} (g) + 1 / 2O _{2} (g) (ΔH = -285.8 kJ / mol) * (-1)

Meanwhile, the latter was multiplied by -1 to be able to “flip” it. Thus, we have the equations ordered and ready to add.

**Sum of equations**

Adding everything up gives us:

2HCl (g) + 1 / 2O _{2} (g) → H _{2} O (l) + Cl _{2} (g) (ΔH = -101.2 kJ / mol)

H _{2} (g) + F _{2} (g) → 2HF (l) ΔH = -1200.0 kJ / mol)

__H _{2} O (l) → __

__H__

_{2}(g) + 1 / 2O_{2}(g) (ΔH = 285.8 kJ / mol)2HCl (g) + F _{2} (g) → 2HF (l) + Cl _{2} (g)

Notice that the terms 1 / 2O _{2} , H _{2} O, and H _{2} cancel out because they are on both sides of the arrow. The enthalpies also add up, giving:

ΔHrx = 285.8 kJ / mol + (-101.2 kJ / mol) + (-1200.0 kJ / mol)

This expression is the same as the one in the beginning:

ΔH = ΔH _{1} + ΔH _{2} + ΔH _{3}

And so, we have that ΔHrx is equal to:

ΔHrx = -1015.4 kJ / mol

**– Example 2**

In this example, it will be seen that the enthalpies of the alternating reactions correspond to the enthalpies of formation of the compounds of interest.

We want to determine ΔHrxn for the following decomposition reaction:

2SO _{3} (g) → 2SO _{2} (g) + O _{2} (g)

And the enthalpies of formation of the compounds SO _{3} and SO _{2} are counted at hand :

S (s) + O _{2} (g) → SO _{2} (g) (ΔH = -296.8 kJ / mol)

S (s) + 3 / 2O _{2} (g) → SO _{3} (g) (ΔH = -395.6 kJ / mol)

**Ordering the equations**

We will proceed to solve this exercise in the same way as the previous example. It is convenient that SO _{3} is to the right, and that it be multiplied by 2. For that, we must “flip” the equation for the formation of SO _{3} by multiplying it by -1, and then by multiplying it by 2:

2SO _{3} (g) → 2S (s) + 3O _{2} (g) (ΔH = -395.6 kJ / mol) * (-2)

We also multiply the SO _{2} formation equation by 2:

2S (s) + 2O _{2} (g) → 2SO _{2} (g) (ΔH = -296.8 kJ / mol) * (2)

**Sum of equations**

Now we proceed to add the equations:

2SO _{3} (g) → 2S (s) + 3O _{2} (g) (ΔH = 791.2 kJ / mol)

__2S (s) + 2O _{2} (g) → 2SO _{2} (g) ( __

__Δ__

__H = -593.6 kJ / mol)__

2SO _{3} (g) → 2SO _{2} (g) + O _{2} (g)

Note that the term 2S is removed because it is on both sides of the arrow. Likewise, 3O _{2} is subtracted 2O _{2,} giving an O _{2} . The sum of the enthalpies, and therefore, the value of ΔHrxn will be:

ΔH = ΔHrxn = 791.2 kJ / mol + (-593.6 kJ / mol)

= 197.6 kJ / mol

Like any decomposition reaction, it is endothermic, so its enthalpy is positive.

**Alternative method**

There is a method to achieve this same result more directly and easily. This was mentioned in a previous section.

2SO _{3} (g) → 2SO _{2} (g) + O _{2} (g)

To determine the ΔHrxn for this reaction, we must calculate ΔHºf (reactants) and ΔHºf (products). The ΔHºf (SO _{3} ) is equal to -395.6 kJ / mol, while the ΔHºf (SO _{2} ) is equal to -296.8 kJ / mol. The ΔHºf (O _{2} ) is equal to 0, since elemental oxygen exists as an O _{2} molecule and not as free O atoms.

So, we have:

ΔHrxn = ΔHºf (products) – ΔHºf (reactants)

= [ΔHºf (SO _{2} ) + ΔHºf (O _{2} )] – ΔHºf (SO _{3} )

= ΔHºf (SO _{2} ) – ΔHºf (SO _{3} )

However, we must multiply both enthalpies of formation by 2, in order to equalize the stoichiometric coefficients with respect to the chemical equation:

ΔHrxn = 2ΔHºf (SO _{2} ) – 2ΔHºf (SO _{3} )

And calculating we have:

ΔHrxn = 2 (-296.8 kJ / mol) – 2 (-395.6 kJ / mol)

= 197.6 kJ / mol

In fact, this is the way in which it is usually preferred to solve all exercises in which Hess’s law is applied.

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