Enthalpy (H) is the total energy of a system, i.e. the sum of all the types of energy it contains at constant pressure. It is expressed in joules (J) or kilojoules (kJ).
Any substance involved in a reaction contains a certain amount of internal energy. When a particle of matter, whether an atom or a molecule, is formed, a certain amount of energy is accumulated. This energy can be found in the form:
- kinetic energy linked to the movement of electrons around the nucleus and to the movement of molecules and atoms (vibration, rotation and translation);
- potential energy from the forces of attraction between nucleons, between nuclei and electrons, in chemical bonds between atoms, and in molecular interactions.
The sum of all these energies is the enthalpy of the substance.
However, it is difficult to determine the internal energy of a substance experimentally. It is easier to measure the heat absorbed or released during a reaction by :
The variation in enthalpy (ΔH) corresponds to the energy absorbed or released during a reaction at constant pressure and temperature. This energy is also known as the ‘heat of reaction’.
It is also expressed in joules (J) or kilojoules (kJ).
This heat of reaction, known as the ‘enthalpy variation’, corresponds to the variation in the total energy of the system during a physical or chemical transformation at constant pressure.
To calculate the change in enthalpy, we need to differentiate between the enthalpy of the products and that of the reactants, which is given by the following formula:
ΔH = Hproducts − Hreactants
The variation in energy during a reaction can be visualised using an enthalpy diagram. Such a graph shows the relative enthalpy of reactants and products by means of horizontal steps located at different levels.
The enthalpy variation corresponds to the difference in height between the levels and its sign indicates whether the reaction is endothermic or exothermic.
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Endothermic reaction |
Exothermic reaction |
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ΔH positive
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ΔH negative
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- The variation in molar enthalpy (ΔH ) corresponds to the variation in enthalpy associated with the transformation of one mole of a substance under given conditions. It is measured in kJ/mol.
- The variation in standard enthalpy (ΔH°), or standard molar enthalpy, corresponds to the variation in enthalpy associated with the transformation of one mole of a particular substance at TPA (temperature of 25°C and pressure of 100 kPa). It is also measured in kJ/mol.
When the variation in molar enthalpy is determined under standard conditions (in this case, TPA), we refer to the variation in standard molar enthalpy (ΔH°). This corresponds to the enthalpy variation associated with the transformation of one mole of a particular substance at TPA. The unit used here is the kilojoule per mole (kJ/mol).
The standard molar enthalpy variation has the advantage that it can be determined for a large number of transformations, since it relates to a precise quantity of matter. It is used in stoichiometric calculations to determine the value of an enthalpy variation for a quantity of substance other than one mole.
A change in enthalpy can be measured using a calorimeter. It can also be determined by observing an energy diagram or by calculating the energy balance of a reaction. A stoichiometric calculation can also be used to find the value of the change in enthalpy.
How much energy is released in the following reaction if a mass of |5{,}50| of hydrogen (|H_2|) is consumed by enough fluorine (|F_2|)?
|H_{2\text{(g)}}+F_{2\text{(g)}} \rightarrow 2\ HF_{\text{(g)}} + 536{,}6\ \text{kJ}|
Solution
We need to find out how many moles of hydrogen (|H_2|) are equivalent to |5{,}50\ \text{g}| by cross-referencing this with the molar mass of |H_2|, which is |2{,}02\ \text{g/mol}|.
|\dfrac{2{,}02\ \text{g}}{1\ \text{mol}}=\dfrac{5{,}50\ \text{g}}{?\ \text{mol}}|
|? = 2{,}72\ \text{mol}|
We then need to cross-produce to determine the amount of energy released for |2{,}72\ \text{mol}| of |H_2|, knowing that, for |1\ \text{mole}| of |H_2|, this amount is equal to |536{,}6\ \text{kJ}|.
|\dfrac{536{,}6\ \text{kJ}}{1\ \text{mol}}=\dfrac{?}{2{,}72\ \text{mol}}|
|? = 1\ 460\ \text{kJ}|
So, if the reaction is carried out with |5{,}50\ \text{g}| of hydrogen (|H_2|), a quantity of energy of |1\ 460\ \text{kJ}| would be released.
|\Delta H = -1\ 460\ \text{kJ}|