Enthalpy change in solution

In a nutshell

The enthalpy change of solution is defined as the enthalpy change when water dissolves one mole of solute. This can be calculated with known enthalpy change of hydrations and lattice energy values. The route can be found by drawing energy cycles or energy level diagrams.

Equations

There are chemical equations you must be able to write for any species. This includes enthalpy change of hydration and enthalpy change of solution.

$one\>mole\>of\>gaseous\>ions\>→\>one\>mole\>of\>aqueous \>ions$

$one\>mole\>of\>solute\>→\>aqueous\>solute$

Dissolution

Bonds are broken and made when ionic lattices dissolve in water. First the bonds between the oppositely charged ions are broken. This is an endothermic reaction because energy is required to break these bonds. It's the opposite of lattice enthalpy.

Then new bonds between the separated ions and water molecules form. This is an exothermic reaction as energy is being released as the new bonds are made. The ions are getting hydrated. This enthalpy change is called the enthalpy change of hydration, $\triangle_{hyd}H$ . The enthalpy change of hydration is defined as the enthalpy change when water dissolves one mole of gaseous ions.

$one\>mole\>of\>gaseous\>ions\>→\>one\>mole\>of\>aqueous \>ions$

Example

$one\>mole\>of\>gaseous\>calcium\>ion\>→\>one\>mole\>of\>aqueous\>calcium\>ion$

$Ca^{2+}(g)→Ca^{2+}(aq)$

The overall effect of both of these reactions is the enthalpy change of solution, $\triangle_{sol}H$ . The enthalpy change of solution is defined as the enthalpy change when water dissolves one mole of solute.

$one\>mole\>of\>solute\>→\>aqueous\>solute$

Example

$one\>mole\>of\>sodium\>iodide\>→\>one\>mole\>of\>aqueous\>sodium\>iodide$

$NaI(s)→NaI(aq)$

The equation you should write in energy cycles is the direct route, shown below.

$one\>mole\>of\>sodium\>iodide\>→\>one\>mole\>of\>aqueous\>sodium\>ion+\>one\>mole\>of\>aqueous\>iodide\>ion$

$NaI(s)→Na^+(aq)+I^-(aq)$

Substances that are soluble tend to have an exothermic enthalpy change of solution. This allows the substance to be dissolved and hydrated in water.

Enthalpy change of solution

Enthalpy change of solution can be calculated by drawing energy cycles and energy level diagrams.

Energy cycles

The procedure to draw an energy cycle in order to calculate the enthalpy of solution is described below.

Procedure

1.	Write the chemical equation for the enthalpy of solution.
2.	Write the gaseous ions that make up the solute below it.
3.	Draw an arrow from the gaseous ions arrow to the solid solute and label it "lattice energy".
4.	Draw an arrow from the gaseous ions to the aqueous ions and label it "enthalpy of hydration".
5.	Calculate the enthalpy of solution by following a route with known values.

Example

Calculate the enthalpy change of solution, $\triangle_{sol}H$ , for sodium bromide. The lattice energy is $-736\> kJ\>mol^{-1}$ . The enthalpy of hydration for $Na^+(g)$ is $-406\> kJ\>mol^{-1}$ and for $Br^-(g)$ is $-348\> kJ\>mol^{-1}$ .

Write the direct chemical equation for the enthalpy change of solution for sodium bromide:

$NaBr(s)→Na^+(aq)+Br^-(aq)$

Below it, write the gaseous ions that form sodium bromide:

$Na^+(g)+Br^-(g)$

Draw the lattice energy arrows, label the arrows and write the known values:

Chemistry; Energetics II; KS5 Year 12; Enthalpy change in solution

Find the route that will give the enthalpy change of solution:

$\triangle_{sol}H=-lattice\>energy+enthalpy\>of\>hydrations \newline \triangle_{sol}H=-lattice\>energy+enthalpy\>of\>hydration\>of\>Na^+(g)+enthalpy\>of\>hydration\>of\>Br^-(g)$

Insert the known values to calculate the enthalpy change of solution:

$\triangle_{sol}H=-(-736)+(-406)+(-348)=-18\>kJ\>mol^{-1}$

The enthalpy change of solution for sodium bromide is $\underline{-18\>kJ\>mol^{-1}}$ .

It's exothermic. This means sodium bromide is soluble.

Energy level diagrams

Another way to calculate the enthalpy change of solution is by drawing energy level diagrams. Described below is the procedure to draw an energy level diagram for a solute to calculate the enthalpy change of solution.

Procedure

1.	Write the solid substance.
2.	On top, write the gaseous ions that form the solid substance.
3.	Draw a vertical arrow from the gaseous ions to the solid substance and label it lattice enthalpy.
4.	On the right, slightly above the solid substance, write the aqueous ions that form the solid substance.
5.	Draw a vertical arrow from the gaseous ions to the aqueous ions and label it enthalpy of hydration.
6.	Draw a vertical arrow from the solid substance to the aqueous ions and label it enthalpy change of solution.
7.	Calculate the enthalpy change of solution by following a route with known values.

Example

Calculate the enthalpy change of solution for potassium chloride. The lattice enthalpy is $-715\> kJ\>mol^{-1}$ . The enthalpy of hydration for $K^+(g)$ is $-322\> kJ\>mol^{-1}$ and for $Cl^-(g)$ is $-364\> kJ\>mol^{-1}$ .

Write the solid substance, potassium chloride:

$KCl(s)$

Write the gaseous ions above it:

$K^+(g)+Cl^-(g)$

Write the aqueous ions slightly above the solid potassium chloride on the right:

$K^+(aq)+Cl^-(aq)$

Draw the arrows and label them:

Chemistry; Energetics II; KS5 Year 12; Enthalpy change in solution

Find the route that will give the enthalpy change of solution:

$\triangle_{sol}H=-lattice\>enthalpy+enthalpy\>of\>hydrations \newline \triangle_{sol}H=-lattice\>enthalpy+enthalpy\>of\>hydration\>of\>K^+(g)+enthalpy\>of\>hydration\>of\>Cl^-(g)$

Insert known values to calculate the enthalpy change of solution:

$\triangle_{sol}H=-(-715)+(-322)+(-364)=29\>kJ\>mol^{-1}$

The enthalpy change of solution for potassium chloride is $\underline{29\>kJ\>mol^{-1}}$ .

It's slightly endothermic. However, other factors may result in potassium chloride being soluble in water.

Enthalpy of hydration factors

The enthalpy of hydration is the enthalpy when water dissolves one mole of gaseous ions. Factors that affect the enthalpy of hydration are the size and charge of the ions. The higher the charge, the more attraction between the ion and water. The smaller the ion, the higher the charge density. Therefore, water will feel a stronger attraction towards the ion. This will give a more exothermic value.