What is enthalpy variation?

enthalpy variation is a physical quantity represented by the abbreviation ΔH (the Greek letter Δ means variation and the letter H represents enthalpy) which indicates the amount of energy absorbed or released by a chemical reaction.

• Endothermic reaction: when energy is absorbed;

• Exothermic reaction: when energy is released.

Expression for calculating enthalpy change

As with all variations worked on in Physics, the enthalpy variation is determined from the subtraction between the final result and the initial one. As the end of a reaction corresponds to the products and the beginning corresponds to the reactants, therefore:

ΔH = Products - Reagents

As every chemical substance has a specific amount of energy (enthalpy), to calculate the enthalpy variation, it is necessary to take into account the energy amount of each member of the reaction, as follows:

• Reagents: if the reaction has more than one product, their enthalpies must also be added;

If a reaction presents the following equation:

A + 2 B →

The enthalpy of the reactants will be given by adding the enthalpy of reactant A to the sum of twice the enthalpy of B (this is because there are 2 moles of B in the equation).

Hr = HA + 2.HB

• Products: if the reaction has more than one product, their enthalpies must be added together;

If a reaction presents the following equation:

A + 2 B → C + 3D

The enthalpy of the reactants will be given by the sum of the enthalpy of the product C with the sum of the triple enthalpy of D (this is because there are 3 moles of D in the equation).

Hp = HC + 3.HD

Thus, the expression to calculate the enthalpy variation is given by the subtraction between the enthalpy of the products and the enthalpy of the reactants of a chemical reaction.

ΔH = Hp - Hr

Interpreting the result of enthalpy variation

As it is a subtraction between the enthalpies of products and reagents, the result of the enthalpy variation can be positive or negative.

• Negative: when the enthalpy of the reactants is greater than that of the products.

Hr > Hp

• Positive: when the enthalpy of the products is greater than that of the reactants.

Hp > Hr

From the result of the enthalpy variation, we can classify the reaction into endothermic or exothermic, according to the following criterion:

• Endothermic: when the result is positive.

ΔH > 0

• Exothermic: when the result is negative.

ΔH < 0

Examples of how to determine the enthalpy change of a chemical reaction

1st Example (UFF-RJ): Consider the default enthalpy values ​​of formation (ΔH^O_f) in KJ.mol^-1 at 25 °C, of ​​the following substances:

• CH_4(g) = -74,8

• CHCl₃₍₁₎ = - 134,5

• HCl_(g) = - 92,3

For the following reaction:

CH_4(g) + 3Cl_2(g) → CHCl₃₍₁₎ + 3HCl_(g)

The ΔH^O_f it will be:

a) - 151.9 KJ.mol^-1

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b) 168.3 KJ.mol^-1

c) - 673.2 KJ.mol^-1

d) - 336.6 KJ.mol^-1

e) 841.5 KJ.mol^-1

To determine the enthalpy variation of this reaction, we must perform the following steps:

• 1^O Step: Check if the equation is balanced. If not, it must be balanced.

The equation provided by the exercise is balanced, so we have 1 mole of CH₄ and 3 mol of Cl₂ in the reagents, and 1 mol of CHCl₃ and 3 mol of HCl in the products.

• 2^O Step: Calculate the amount of enthalpy of the products (Hp), multiplying the amount in mol by the enthalpy supplied and then adding the results:

Hp = CHCl product₃ + HCl product

Hp = 1.(-134.5) + 3.(-92.3)

Hp = -134.5 + (-276.9)

Hp = -134.5 - 276.9

Hp = - 411.4 KJ.mol^-1

• 3^O Step: Calculate the amount of enthalpy of the reactants (Hr), multiplying the amount in mol by the enthalpy given and then adding the results:

Note: As the Cl₂ it is a simple substance, so its enthalpy is 0.

Hr = CH reagent₄ + Cl reagent₂

Hr = 1.(-74.8) + 3.(0)

Hr = -74.8 + 0

Hr = -74.8 KJ.mol^-1

• 4^O Step: Calculate the enthalpy change, using the values ​​found in the previous steps in the following expression:

H^O_f = Hp - Hr

H^O_f = - 411,4 - (-74,8)

H^O_f = -411,4 + 74,8

H^O_f = -336.6 KJ.mol^-1

2nd Example (UEM-PR): It is possible to prepare oxygen gas in the laboratory by carefully heating potassium chlorate, according to the reaction:

2KClO_3(s) → 2KCl_(s) + 3O_2(g) ΔH= +812 KJ/mol

Assuming that the enthalpy of KCl_(s) is worth +486 KJ/Mol and considering the system at 25 °C and 1 atm, what is the default enthalpy value of KClO_3(s) in KJ/mol?

To determine the enthalpy of one of the components of the equation from the knowledge of the enthalpy variation, we must perform the following steps:

• 1^O Step: Check if the equation is balanced. If not, it must be balanced.

The equation provided by the exercise is balanced, so we have 2 mol of KclO₃ in the reagent, and 2 mol of KCl and 3 mol of O₂ on the products.

• 2^O Step: Calculate the amount of enthalpy of the products (Hp), multiplying the amount in mol by the enthalpy supplied and then adding the results:

Note: As the O₂ it is a simple substance, so its enthalpy is 0.

Hp = product KCl+ product O₂

Hp = 2.(+486) + 3.(0)

Hp = +972 + 0

Hp = 972 KJ.mol^-1

• 3^O Step: Calculate the enthalpy of the KClO reagent₃, using the enthalpy variation, provided in the statement, and the products enthalpy calculated in the previous step, in the following expression:

H^O_f = Hp - Hr

812= 972 - 2.(Hr)

2Hr = 972-812

2Hr = 160

Hr = 160
2

Hr = 80 KJ/mol

Would you like to reference this text in a school or academic work? Look:

DAYS, Diogo Lopes. "What is enthalpy variation?"; Brazil School. Available in: https://brasilescola.uol.com.br/o-que-e/quimica/o-que-e-variacao-entalpia.htm. Accessed on July 27, 2021.