- The heat of neutralisation is the heat energy evolved when an acid reacts with a base, per mole of the acid or base. All measurements are made under standard state conditions.
- An example of the enthalpy change of neutralisation is the heat change obtained in the reaction between sodium hydroxide and hydrochloric acid. The equation for the reaction is
∆H = -57 kJ mol-1
Examples of Neutralization Reaction
Equation | Half Equation |
| HCl + NaOH → NaCl + H2O | H+(aq) + OH–(aq) → H2O(l) |
| H2SO4 + 2KOH → K2SO4 + 2H2O | H+(aq) + OH–(aq) → H2O(l) |
| CH3COOH + NaOH → CH3COONa +H2O | H+(aq) + OH–(aq) → H2O(l) |
| 2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O | H+(aq) + OH–(aq) → H2O(l) |
- The heat of neutralisation of a strong acid with a strong alkali is almost the same for all acids and alkalis. This is because the same reaction always takes places. The reaction is H+(aq) + OH–(aq) → H2O(l)
- Heat change of neutralization reaction is affected by 3 factors:
- Quantity of acid and alkali
- Basicity of the acid and alkali
- Strength of acid and alkali
Example:
An experiment is carried out by adding 25cm³ of sodium hydroxide 0.5 mol/dm³ into 25 cm³ of dilute nitric acid 0.5 mol/dm³. Calculate the temperature change of the mixture. [Specific heat capacity of the solution = = 4.2 Jg-1°C-1, density of the solution =1 g/cm³ ]
Answer:
Number of mole of NaOH,
\[\begin{gathered}
n = \frac{{MV}}{{1000}} \hfill \\
n = \frac{{(0.5)(25)}}{{1000}} \hfill \\
n = 0.125mol \hfill \\
\end{gathered} \]
Number of mole of HNO3,
\[\begin{gathered}
n = \frac{{MV}}{{1000}} \hfill \\
n = \frac{{(0.5)(25)}}{{1000}} \hfill \\
n = 0.125mol \hfill \\
\end{gathered} \]
Number of mole of water produced = 0.0125mol
Amount of heat released, Q = 0.0125 x 57,000J = 712.5J
Mass of the solution, m = 25 + 25 = 50 cm³
Specific heat capacity of the solution = 4.2 Jg-1°C-1
Q = mcθ
712.5 = 50(4.2)θ
θ = 3.4°
Neutralisation Between “Strong Acid and Strong Alkali” vs “Weak Acid and Strong Alkali”
- Below is the comparison of the heat of neutralisation of the reaction between “strong acid and strong alkali” and “weak acid and strong alkali”.
Strong Acid and Strong Alkali
HCl + NaOH → NaCl + H2O
∆H = -57kJ mol-1
H2SO4 + 2KOH → K2SO4 + 2H2O
∆H = -57kJ mol-1
2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O
∆H = -57kJ mol-1
Weak Acid and Strong Alkali
CH3COOH + NaOH → CH3COONa +H2O
∆H = -55kJ mol-1
HCN + KOH → KCN + H2O
∆H = -12kJ mol-1
- The heat of neutralisation between strong acid and strong alkali is almost the same (-57kJ mol-1 ) for all acid and all alkali.
- Nevertheless, the heat of neutralisation between weak acid and strong alkali is always less than -57kJ mol-1.
- This is because weak acids are partially decomposed.
- During the reaction of neutralisation, the acid molecules will be decomposed to form hydrogen ions and react with the hydroxide ions from alkali.
- Part of the heat released during the formation of water molecule is used to decompose the acid to form hydrogen ion.
Temperature Change in the Reaction Involving Change of Volume or Concentration
Altering the Volume of the Solution Without Altering the Concentration
If an experiment is repeated by altering the volume without altering the concentration, the temperature change will remain the same.
Example 1:
When 50 cm³ of dilute hydrochloric acid 2 mol/dm³ is added into 50 cm³ of potassium hydroxide 2 mol/dm³, the temperature increase 13°C. What is the temperature increase if 300 cm³ of dilute hydrochloric acid 2 mol/dm³ is added into 300 cm³ of potassium hydroxide solution 2 mol/ dm³?
Answer:
The volume of the reactants increases by 6 times (300/50) whereas the concentration of the solution remain unchanged. Therefore the change of the temperature remains the same.
Temperature increase = 13°C
Altering the Concentration of the Solution Without Altering the Volume
If an experiment is repeated by altering the concentration by n time without altering the volume of the solution, the temperature change will be n time of the initial temperature change as well.
Example 2:
In an experiment, 50 cm³ of lead(II) nitrate solution 0.2 mol/dm³ is added into 50 cm³ of sodium carbonate solution 0.2 mol/dm³, the increase of temperature is 2.4°C. what is the increase in temperature if 50 cm³ of lead(II) nitrate solution 0.6 mol/dm³ is added into 50 cm³ of sodium carbonate solution 0.6 mol/dm³?
Answer:
The volume of the reactants remains unchanged whereas the concentration of the solution increases by 3 times (0.6/0.2). Therefore the change of the temperature
θ = 2.4°C x 3 = 7.2°C