# Investigation into how much heat is produced in a neutralisation reaction.

Imthiar Khan                Chemistry Coursework

Investigation into how much heat is produced in a neutralisation reaction.

Background Information.

All neutralisation reactions give out more heat than is taken in. This type of reaction is called an exothermic reaction.

The basic reaction will be H   with OH ions. The H  will always be the acid and the OH  will always be the alkali.

When;

H+    +    OH-            H2O    +    Salt

1mol              1mol               1mol               1mol

= 58 000 jmol of energy is given off (data book value)

This is regardless of which acid and which alkali are used for the reaction, but the type of salt does depend on the type of acid and alkali used.

The equation used to find the ΔHr is

ΔHr = m    x    4.20    x    ΔT

m = mass of solution in beaker

4.2 = specific heat capacity of water

ΔT = Heat change measured on graph

The ΔT value will change according to the concentration of acid that is being used.

Why does concentration of acid affect the ΔHr value?

1. The higher the concentration the more particles
2. The more particles the more chance of collision
3. The more chance of collision means there is more chance of a successful collision.
4. This means the temperature will increase more rapidly
5. This will make the ΔT value larger
6. Which in turn makes the ΔHr value larger.

However, the ΔHr value should always be in proportion to the energy given out by 1mol of acid. You can gain this value by dividing or multiplying where necessary to gain the same concentration.

Preliminary Results

To gain some further knowledge about the reaction and what to expect, I did a preliminary experiment. The results were as follows;

The heat increase was 12.5ºC. When looking at the graph the delta T value is 12.5ºC. This value when substituted into the formula

ΔHr = m x 4.2 x ΔT

Should give results approximate to 58000Jmol.

= 50 x 4.2 x 12.5

= 2625J

However, this is when 0.05 moles of acid reacts with 0.05 moles of alkali. To find out for 1 mole we must multiply from 0.05 to 1. To do this we multiply by 20.

1 mole = 2625 x 20

= 52500Jmol

This result shows that the degree of accuracy is ...