# Enthalpy of Neutralisation Calculations

This is part of the HSC Chemistry course under the topic Properties of Acids and Bases.

### HSC Chemistry Syllabus

• Conduct a practical investigation to measure the enthalpy of neutralisation (ACSCH093)

### Enthalpy of Neutralisation Calculation Examples

This video walks through examples involving enthalpy of neutralisation calculations for HSC Chemistry.

### Enthalpy of Neutralisation – The Basics

The enthalpy change of neutralisation is the amount of energy when 1 mole of water is formed from the reaction of an acid and a base.

Since neutralisation reactions are exothermic, the final solution will increase in temperature.

By measuring the maximum change in temperature, the amount of energy produced by neutralisation and absorbed by the solution can be calculated:

$$q = mc \Delta T$$

Where:

• q = quantity of heat (amount of energy absorbed by the solution in J)
• m = mass of the final solution (unit depends on c - usually kg)
• c = the specific capacity of the final solution
• ΔT = the change in temperature given in Kelvin or degree Celsius

### What is Specific Heat Capacity?

The specific heat capacity is the amount of energy required to raise the temperature of a substance by 1 Kelvin (or ºC) per unit mass of that substance.

For example, pure water has a specific heat capacity of 4.18 x 103 J kg-1 K-1

This means 4.18 x 10J of energy is required to raise the temperature of 1 kg of water by 1 Kelvin or 1 degree Celsius.

### Calculating Molar Enthalpy Change (∆H)

The enthalpy change of neutralisation is given by the equation:

$$\Delta H = -\frac{q}{n}$$

where:

• ΔH is the enthalpy change of neutralisation (exothermic)
• n is the number of moles of water formed
• q is the energy absorbed by the solution

There is a negative sign in the calculation of ∆because q represents the energy absorbed (+) by the final solution while ∆H represents the energy released (–) by neutralisation by mole of water formed.

Since the value of q will always be inaccurate due to unaccounted energy loss, there will be a discrepancy between the theoretical value of ∆H of neutralisation and the experimental value.