The heat of solution or solution enthalpy is the heat that is absorbed or released during the dissolution process of a certain amount of solute in the solvent under the condition of constant pressure.
When a chemical reaction occurs, energy is needed to form and break bonds that allow new substances to form. The energy that flows for these processes to take place is heat, and thermochemistry is the branch of science responsible for studying them.
As for the term enthalpy, it is used to refer to the flow of heat when chemical processes occur under conditions of constant pressure. The creation of this term is attributed to the Dutch physicist Heike Kamerlingh Onnes (1853 – 1926), who discovered superconductivity.
How is it calculated?
To find the enthalpy, we must start from the first law of thermodynamics, which considers that the variation in the internal energy ΔU of a system is due to the absorbed heat Q and the work W performed by some external agent:
ΔU = Q + W
Where work is the negative integral over the entire volume of the product of pressure by the differential variation in volume. This definition is equivalent to the negative integral of the product of the scalar force and the displacement vector in mechanical work:
When the constant pressure condition mentioned above is applied, P may fall outside the integral; So the job is:
W = -P (V f -V o ) = -PΔ V
-Expression for enthalpy
If this result is substituted into Δ L, the following is obtained:
ΔU = Q – PΔ V
Q = ΔU + PΔV = U f – U or + P (V f –V o ) = U f + PV f – (U o + PV o )
The quantity U + PV is called enthalpy H , so that:
Q = H f – H o = Δ H
Enthalpy is measured in joules as it is energy.
Solution enthalpy
The initial components of a solution are solute and solvent and have an original enthalpy. When this dissolution takes place, it will have its own enthalpy.
In this case, the enthalpy variation in joules can be expressed as:
ΔH = solution H – reagents H
Or in the standard form of enthalpy ΔH or , when the result is in joule / mol
ΔH o = H or solution – H or reagents
If the reaction releases heat, the sign of ΔH is negative (exothermic process); if it absorbs heat (endothermic process), the signal will be positive. And, of course, the enthalpy value of the solution will depend on the concentration of the final solution.
applications
Many ionic compounds are soluble in polar solvents such as water. Solutions of salt (sodium chloride) in water or brine are in common use. However, the enthalpy of the solution can be considered as the contribution of two energies:
– One to break solute-solute and solvent-solvent bonds
– The other is required in the formation of new solvent-solvent bonds.
In the case of dissolution of an ionic salt in water, it is necessary to know the so-called reticular enthalpy of the solid and the enthalpy of hydration to form the solution, in the case of water. If it is not water, it is called the enthalpy of solvation .
The reticular enthalpy is the energy necessary for the rupture of the ionic network and formation of gaseous ions, a process that is always endothermic, since energy must be supplied to the solid to separate it into its constituent ions and bring them to a gaseous state.
On the other hand, hydration processes are always exothermic, as hydrated ions are more stable than gaseous ions.
In this way, solution creation can be exothermic or endothermic, depending on whether the disruption of the solute’s ionic network requires more or less energy than hydration provides.
Measurements with the calorimeter
In practice, it is possible to measure ΔH in a calorimeter, which basically consists of an insulated container equipped with a thermometer and a stirring rod.
As for the container, water is almost always spilled, which is the calorimetric liquid par excellence, as its properties are the universal reference for all liquids.
Old calorimeter used by Lavoisier. Source: Gustavocarra [CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0)].
Of course, the calorimeter materials also interfere with heat exchange, in addition to water. But the caloric capacity of the whole ensemble, called the calorimeter constant , can be determined separately from the reaction and then taken into account when it occurs.
The energy balance is as follows, remembering the condition that there is no energy leakage in the system:
Δ H solution + Δ H water + C calorimeter Δ t = 0
From where:
Δ H solution = – m water . c water . Δ T – C calorimeter Δ T = -Q water – Q calorimeter
And to get the default enthalpy:
– Solute mass: m s
– Molecular weight of solute: M s
– Water mass: m water
– Molecular weight of water: M water
– Molar heat capacity of water: water C ; m *
– Temperature change: ΔT
* CP , m of water is 75,291 J / mol. K
solved exercises
-Exercise 1
The enthalpy of formation of a solid KOH is potassium hydroxide Δ H or = 426 kJ / mol , which liquid water H 2 O is 285.9 kJ / mol .
It is also known that when metallic potassium hydroxide reacts with hydrogen and the resulting liquid water Δ H or = -2011 kJ / mol . With these data calculate the enthalpy of the KOH solution in water.
Solution
– The KOH is divided into its components:
KOH solid → K Solid + ½ O 2 + ½ H 2 ; Δ H or = – 426 kJ / mol
– Liquid water is formed:
½ O 2 + ½ H 2 → liquid H 2 O ; Δ H or = -285.9 kj / mol
– Now we have to shape the solution:
Solid K + H 2 O → ½ H 2 + aqueous KOH ; Δ H or = -2011 kJ / mol
Note that the KOH decay enthalpy sign has been reversed, which is due to Hess’ Law: when reactants become products, the enthalpy change does not depend on the steps taken and when the equation needs to be reversed, as in this case, enthalpy changes sign.
The energy balance is the algebraic sum of enthalpies:
– 426 kJ / K – 285.9 kJ / mol – 2011 kJ / mol = – 2722.9 kJ / mol
-Exercise 2
The enthalpy of the solution for the next reaction is determined in a constant pressure calorimeter and the calorimeter is known to be constant 342.5 J / K. When 1.423 g of sodium sulfate dissolved in Na 2 SO 4 , in 100.34 g of water, the temperature change is 0037 K. Calculates the standard enthalpy of solution for Na 2 SO 4 from these data.
Solution
The standard enthalpy of the solution is eliminated from the equation given above:
For sodium sulfate: M s = 142.04 g/mol; m s = 1,423 g
E for water: m water = 100.34 g; Water M = 18.02 g/mol; C water; m = 75.291 J / K mol
ΔT = 0.037 K
Calorimeter C = 342.5 J / K
