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) |
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°
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
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