Heat of solution how to calculate it, applications and exercises

The heat of solution or enthalpy of solution is the heat that is absorbed or given off during the dissolution process of a certain amount of the solute in the solvent, under the condition of constant pressure.

When a chemical reaction takes place, energy is required both to form and to break bonds that allow the formation of new substances. The energy that flows for these processes to take place is heat, and thermochemistry is the branch of science that is responsible for studying them..

Regarding the term enthalpy, this It is used to denote 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), the same one who discovered superconductivity.

Article index

• 1 How is it calculated?
  • 1.1 -The expression for enthalpy
• 2 Applications
  • 2.1 Measurements with the calorimeter
• 3 Solved exercises
  • 3.1 -Exercise 1
  • 3.2 -Exercise 2
• 4 References

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 done on it by some external agent:

ΔU = Q + W

Where work is the negative integral over the entire volume of the product of pressure times the differential change in volume. This definition is equivalent to the negative integral of the scalar product of the force and the displacement vector in mechanical work:

When the constant pressure condition mentioned above is applied, P can go out of the integral; therefore the job is:

W = -P (V_F -V_or) = -PΔV

-The expression for enthalpy

If this result is substituted in ΔOR is obtained:

ΔU = Q - PΔV

Q = ΔU + PΔV = U_F - OR_or + P (V_F -V_or) = U_F + PV_F - ( OR_or + PV_or )

The amount U + PV is called enthalpy H, so that:

Q = H_F - H_or = ΔH

Enthalpy is measured in joules, since it is energy.

Solution enthalpy

The initial components of a solution are solute and solvent, and they have an original enthalpy. When this dissolution takes place, it will have its own enthalpy.

In this case, the enthalpy change in joules can be expressed as:

ΔH = H_solution - H_reagents

Either in standard enthalpy form ΔH^or, where the result is in joule / mol

ΔH^or = H^or _solution - H^or_reagents

If the reaction gives off heat, the sign of ΔH it is negative (exothermic process), if it absorbs heat (endothermic process) the sign will be positive. And naturally, the value of the solution enthalpy 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. Now, 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 that required in the formation of new solute-solvent bonds..

In the case of the dissolution of an ionic salt in water, it is required to know the so-called lattice enthalpy of solid and the enthalpy of hydration to form the solution, in the case of water. If it is not water, then it is called enthalpy of solvation.

The lattice enthalpy is the energy required to break up the ionic network and form 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 the gaseous state.

On the other hand, hydration processes are always exothermic, since hydrated ions are more stable than ions in the gaseous state..

In this way, the creation of the solution can be exothermic or endothermic, depending on whether the breakdown of the ionic lattice of the solute 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 fitted with a thermometer and a stirring rod.

As for the container, water is almost always poured into it, which is the calorimetric liquid par excellence, since its properties are the universal reference for all liquids..

Of course the materials of the calorimeter are also involved in heat exchange, in addition to water. But the caloric capacity of the whole set, called constant calorimeter, 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 are no energy leaks in the system:

ΔH _solution + ΔH _Water + C _calorimeter ΔT = 0

From where:

ΔH _solution = - m _Water . c _Water . ΔT - C _calorimeter ΔT = -Q _Water - Q _calorimeter

And to get the standard enthalpy:

- Solute mass: m_s

- Solute molecular weight: M_s

- Mass of water: m_Water

- Molecular weight of water: M_Water

- Molar heat capacity of water: C_{Water ; m}*

- Temperature change: ΔT

* C_P.m of water is 75.291 J / mol. K

Solved exercises

-Exercise 1

The enthalpy of formation of solid potassium hydroxide KOH is ΔH^or = +426 KJ / mol, that of liquid water H_twoOr is it 285.9 kJ / mol.

Furthermore, it is known that when metallic potassium hydroxide reacts with liquid water, hydrogen and ΔH^or = -2011 kJ / mol. With these data, calculate the enthalpy of solution of KOH in water.

Solution

- KOH breaks down into its components:

KOH_solid → K_solid + ½ O_two + ½ H_two;  ΔH^or = - 426 kJ / mol

- Liquid water is formed:

½ O_two + ½ H_two → H_twoOR_liquid;  ΔH^or = -285.9 kJ / mol

- Now you have to form the solution:

K_solid + H_twoO → ½ H_two + KOH_aqueous ; ΔH^or = -2011 kJ / mol

Note that the sign of the enthalpy of disintegration of KOH has been inverted, which is due to Hess's Law: when the reactants are converted into products, the enthalpy change does not depend on the steps followed and when the equation needs to be inverted , as in this case, the enthalpy changes sign.

The energy balance is the algebraic sum of the enthalpies:

- 426 kJ / K - 285.9 kJ / mol - 2011 kJ / mol = -2722.9 kJ / mol

-Exercise 2

The enthalpy of solution for the next reaction is determined in a constant pressure calorimeter and the calorimeter constant is known to be 342.5 J / K. When 1,423 g of sodium sulfate Na is dissolved_twoSW₄ in 100.34 g of water, the temperature change is 0.037 K. Calculate the standard enthalpy of solution for Na_twoSW₄ from these data.

Solution

The standard enthalpy of the solution is solved from the equation given above:

For sodium sulfate: M_s = 142.04 g / mol; m_s = 1,423 g

And for the water: m_Water = 100.34 g; M_Water = 18.02 g / mol; C_{water; m} = 75.291 J / K mol

ΔT = 0.037 K

C _calorimeter = 342.5 J / K

References

1. Cengel, Y. 2012. Thermodynamics. 7th Ed. Mc.Graw Hill. 782 - 790
2. Engel, T. 2007. Introduction to Physicochemistry: Thermodynamics. Pearson Education. 63-78.
3. Giancoli, D. 2006. Physics: Principles with Applications. 6th… Ed Prentice Hall. 384-391.
4. Maron, S. 2002. Fundamentals of Physicochemistry. Limusa. 152-155.
5. Serway, R., Jewett, J. (2008). Physics for Science and Engineering. Volume 1. 7th. Ed. Cengage Learning. 553-567.