# Heat of Solution

When a solute dissolves in a solvent, an enthalpy change occurs. The heat of solution is the difference in enthalpy between the dissolving solute and a solvent under constant pressure, resulting in infinite dilution. It is also known as the enthalpy of solution and is measured using a calorimeter. It can either be positive (endothermic) or negative (exothermic).

Hot and cold packs work due to the heat of solution of the chemicals inside these packs. These packs have an inner pouch that bursts when squeezed, allowing the chemical to dissolve in water. Heat is then released in the hot pack and absorbed in the cold pack.

The molar heat (enthalpy) of solution is the energy released or absorbed per mole of solute dissolved in a solvent.

## Formula

The equation for the heat of solution is:

ΔH_{soln} = m x c_{p} x ΔT

ΔH_{soln}: Heat of solution

m: Mass of the solution

c_{p}: Specific heat of the solvent

ΔT: Temperature difference before and after the dissolution

## How to Calculate Heat of Solution

To calculate the heat of solution, let us understand the steps involved in the dissolution process.

**Step 1**: Breaking of Solute

The intermolecular forces holding the solute molecules disappear, separating the molecules. As a result, the solute breaks up. It is an endothermic process since heat is supplied to break the bonds. Let ΔH_{1} be the enthalpy change and ΔH_{1} > 0.

**Step 2**: Breaking of Solvent

This process is similar to the breaking of solute. Heat must be supplied to break apart the solvent molecules. Let ΔH_{2} be the enthalpy change in this process and, just like before, ΔH_{2} > 0.

**Step 3**: Combination of Solute and Solvent Molecules

The third and final step is for the two types of molecules to combine and form a solution. Let ΔH_{3} be the enthalpy of combination and, since heat is released when two molecules combine to form a chemical bond, ΔH_{3} < 0. Therefore, the process is exothermic.

The sum of the enthalpies in each step gives the heat of solution.

ΔH_{soln} = ΔH_{1} + ΔH_{2} + ΔH_{3}

The heat of solution can be greater than zero (ΔH_{soln} > 0) or less than zero (ΔH_{soln} < 0) depending upon the difference between ΔH_{1} + ΔH_{2} and ΔH_{3}. If the heat of solution is equal to zero (ΔH_{soln} = 0), the solution is called an ideal solution. Otherwise, the solution is called a non-ideal solution.

For an ideal solution

ΔH_{soln} = ΔH_{1} + ΔH_{2} + ΔH_{3} = 0

=> ΔH_{1} + ΔH_{2} = – ΔH_{3}

## Example Problems

**Problem 1. **Calculate the heat of solution when 50 g of hydrated salt is dissolved in 100 g of water, resulting in a temperature increase of 20 °C. The specific heat of water is 4.184 J/g-°C.

**Solution.**

Given m = 50 g + 100 g = 150 g c_{p} = 4.184 J/g-°C ΔT = 20 °C

The heat of solution is

ΔH_{soln} = m x c_{p} x ΔT

=> ΔH_{soln} = 150 g x 4.184 J/g-°C x 20 °C

=> ΔH_{soln} = 12552 J or 12.6 kJ

**Problem 2.** 120 g of water experienced an increased temperature of 2 °C when 0.24 g of NaOH was dissolved. Calculate the enthalpy of solution in kJ/mol of NaOH. (Molar mass of NaOH is 40 g/mol, and the specific heat of water is 4.18 J/g-°C)

**Solution**

Given m_{NaoH} = 0.24 g m_{water} = 120 g c_{p} = 4.184 J/g-°C ΔT = 2 °C

The heat absorbed by the solution is

q = m_{water} x c_{p} x ΔT

=> q = 120 g x 4.184 J/g-°C x 2 °C

=> q = 1004 J

The dissolution of NaOH gave 1004 J of energy. Therefore, ΔH_{soln} = -1004 J

Number of moles of NaOH: n = m_{NaOH} /M_{NaOH} = 0.24 g /40 g-mol^{-1} = 0.006 mol

The enthalpy of solution of 0.006 mol of NaOH is 1004 J

Therefore, the enthalpy of solution of 1 mol of NaOH is = 1004 J/0.006 mol = 167333 J/mol or 167.3 kJ/mol.

**References**