WHAT IS ENTHALPY? - CHEMISTRY - 2026

The enthalpy is a measure of the amount of energy contained in a body (system) having a volume, is subjected to pressure and is interchangeable with its environment. It is represented by the letter H. The physical unit associated with it is the Joule (J = kgm2 / s2).

Mathematically it can be expressed as follows:

H = U + PV

Where:

H = Enthalpy

U = Internal energy of the system

P = Pressure

V = Volume

If both U and P and V are state functions, H will be as well. This is because at a given moment, some initial and final conditions can be given for the variable to be studied in the system.

What is the enthalpy of formation?

It is the heat absorbed or released by a system when 1 mole of a product of a substance is produced from its elements in their normal state of aggregation; solid, liquid, gaseous, solution or in its most stable allotropic state.

The most stable allotropic state of carbon is graphite, in addition to being at normal conditions of pressure 1 atmosphere and 25 ° C of temperature.

It is denoted as ΔH ° f. In this way:

ΔH ° f = H final - H initial

Δ: Greek letter that symbolizes the change or variation in the energy of a final state and an initial one. The subscript f signifies compound formation and the superscript or standard conditions.

Example

Considering the formation reaction of liquid water

H2 (g) + ½ O2 (g) H2O (l) ΔH ° f = -285.84 kJ / mol

Reagents: Hydrogen and Oxygen its natural state is gaseous.

Product: 1 mole of liquid water.

It should be noted that the enthalpies of formation according to the definition are for 1 mole of compound produced, so the reaction must be adjusted if possible with fractional coefficients, as seen in the previous example.

Exothermic and endothermic reactions

In a chemical process, the enthalpy of formation can be positive ΔHof> 0 if the reaction is endothermic, that is, it absorbs heat from the medium or negative ΔHof <0 if the reaction is exothermic with emission of heat from the system.

Exothermic reaction

Reactants have higher energy than products.

ΔH ° f <0

Endothermic reaction

The reactants have lower energy than the products.

ΔH ° f> 0

To correctly write a chemical equation, it must be molarly balanced. In order for the "Law of Conservation of Matter" to be fulfilled, it must also contain information on the physical state of the reactants and products, which is known as the state of aggregation.

It must also be taken into account that pure substances have a formation enthalpy of zero at standard conditions and in their most stable form.

In a chemical system where there are reactants and products, the enthalpy of reaction is equal to the enthalpy of formation under standard conditions.

ΔH ° rxn = ΔH ° f

Taking into account the above we have to:

ΔH ° rxn = ∑nproducts H ∑nreactive products Hreactive

Given the following fictitious reaction

aA + bB cC

Where a, b, c are the coefficients of the balanced chemical equation.

The expression for the enthalpy of reaction is:

ΔH ° rxn = c ΔH ° f C (a ΔH ° f A + b ΔH ° f B)

Assuming that: a = 2 mol, b = 1 mol, and c = 2 mol.

ΔH ° f (A) = 300 KJ / mol, ΔH ° f (B) = -100 KJ / mol, ΔH ° f (C) = -30 KJ. Calculate ΔH ° rxn

ΔH ° rxn = 2mol (-30KJ / mol) - (2mol (300KJ / mol + 1mol (-100KJ / mol) = -60KJ - (600KJ - 100KJ) = -560KJ

ΔH ° rxn = -560KJ.

It then corresponds to an exothermic reaction.

Enthalpy of formation values ​​of some inorganic and organic chemical compounds at 25 ° C and 1 atm of pressure

Exercises to calculate enthalpy

Exercise 1

Find the enthalpy of reaction of NO2 (g) according to the following reaction:

2NO (g) + O2 (g) 2NO2 (g)

Using the equation for the enthalpy of reaction we have:

ΔH ° rxn = ∑nproducts H ∑nreactive products Hreactive

ΔH ° rxn = 2mol (ΔH ° f NO2) - (2mol ΔH ° f NO + 1mol ΔH ° f O2)

In the table in the previous section we can see that the enthalpy of formation for oxygen is 0 KJ / mol, because oxygen is a pure compound.

ΔH ° rxn = 2mol (33.18KJ / mol) - (2mol 90.25 KJ / mol + 1mol 0)

ΔH ° rxn = -114.14 KJ

Another way to calculate the enthalpy of reaction in a chemical system is through the HESS LAW, proposed by the Swiss chemist Germain Henri Hess in 1840.

The law says: "The energy absorbed or emitted in a chemical process in which reactants are converted into products is the same whether it is carried out in one stage or in several".

Exercise 2

The addition of hydrogen to acetylene to form ethane can be accomplished in one step:

C2H2 (g) + 2H2 (g) H3CCH3 (g) ΔH ° f = - 311.42 KJ / mol

Or it can also happen in two stages:

C2H2 (g) + H2 (g) H2C = CH2 (g) ΔH ° f = - 174.47 KJ / mol

H2C = CH2 (g) + H2 (g) H3CCH3 (g) ΔH ° f = - 136.95 KJ / mol

Adding both equations algebraically we have:

C2H2 (g) + H2 (g) H2C = CH2 (g) ΔH ° f = - 174.47 KJ / mol

H2C = CH2 (g) + H2 (g) H3CCH3 (g) ΔH ° f = - 136.95 KJ / mol

C2H2 (g) + 2H2 (g) H3CCH3 (g) ΔH ° rxn = 311.42 KJ / mol

Exercise 3

(Taken from quimitube.com. Exercise 26. Hess's Law Thermodynamics)

As can be seen in the statement of the problem, only some numerical data appear, but the chemical reactions do not appear, therefore it is necessary to write them.

CH3CH2OH (l) + 3O2 (g) 2CO2 (g) +3 H2O (l) ΔH1 = -1380 KJ / mol.

The value of the negative enthalpy is written because the problem says that there is release of energy. We must also consider that they are 10 grams of ethanol, therefore we must calculate the energy for each mole of ethanol. For this the following is done:

The molar weight of ethanol is sought (sum of the atomic weights), a value equal to 46 g / mol.

ΔH1 = -300 KJ (46 g) ethanol = - 1380 KJ / mol

10 g ethanol 1mol ethanol

The same is done for acetic acid:

CH3COOH (l) + 2O2 (g) 2CO2 (g) + 2 H2O (l) ΔH2 = -840 KJ / mol

ΔH2 = -140 KJ (60 g acetic acid) = - 840 KJ / mol

10 g acetic acid 1 mole acetic acid.

In the previous reactions, the combustion of ethanol and acetic acid are described, so it is necessary to write the problem formula, which is the oxidation of ethanol to acetic acid with production of water.

This is the reaction the problem asks for. It is already balanced.

CH3CH2OH (l) + O2 (g) CH3COOH (l) + H2O (l) ΔH3 =?

Hess's law application

For this we multiply the thermodynamic equations by numerical coefficients to make them algebraic and to be able to correctly organize each equation. This is done when one or more reactants are not on the corresponding side of the equation.

The first equation remains the same because ethanol is on the reactant side as indicated by the problem equation.

The second equation must be multiplied by the coefficient -1 in such a way that the acetic acid that is as reactant can become the product

CH3CH2OH (l) + 3O2 (g) 2CO2 (g) + 3H2O (l) ΔH1 = -1380 KJ / mol.

- CH3COOH (l) - 2O2 (g) - 2CO2 (g) - 2H2O (l) ΔH2 = - (-840 KJ / mol)

CH3CH3OH + 3O2 -2O2 - CH3COOH 2CO2 + 3H2O -2CO2

-2H2O

They add algebraically and this is the result: the equation requested in the problem.

CH3CH3OH (l) + O2 (g) CH3COOH (l) + H2O (l)

Determine the enthalpy of the reaction.

In the same way as each reaction was multiplied by the numerical coefficient, the value of the enthalpies must also be multiplied

ΔH3 = 1x ΔH1 -1xΔH2 = 1x (-1380) -1x (-840)

ΔH3 = -1380 + 840 = - 540 KJ / mol

ΔH3 = - 540 KJ / mol.

In the previous exercise, ethanol has two reactions, combustion and oxidation.

In every combustion reaction there is formation of CO2 and H2O, while in the oxidation of a primary alcohol such as ethanol there is formation of acetic acid

References

1. Cedrón, Juan Carlos, Victoria Landa, Juana Robles (2011). General chemistry. Teaching material. Lima: Pontifical Catholic University of Peru.
2. Chemistry. Libretexts. Thermochemistry. Taken from hem.libretexts.org.
3. Levine, I. Physicochemistry. vol.2.