## Finding Heat of Reaction

\[\begin{gathered}

{\text{Heat of Reaction, }}\Delta {\text{H}} \hfill \\

{\text{ = }}\frac{{{\text{Change of Energy}}}}{{{\text{Number of Mole of Reactant/Product}}}} \hfill \\

\end{gathered} \]

Q = mcθ

m = mass of the substance that absorbs the heat

c = specific heat capacity of the substance that absorbs the heat

θ = change of temperature

**Notes**:

- We always use the specific heat capacity of water as an approximation of the specific heat capacity for all kinds of aqueous solution.
- We also use the density of water as an approximation of the density of all aqueous solution. Since the density of water is 1g/cm³, therefore every single cm³ of solution/water is equal to 1 g. For example 50cm³ ≡ 50g and 200cm³ ≡ 200g.

**Example 1**:

When 50 cm³ of sulphuric acid is added to 40 cm³ of potassium hydroxide solution, the temperature of the solution increases from 29.0°C to 42.0°C. Calculate the change of heat energy in the reaction. (Specific heat capacity of the solution = 4.2 Jg^{-1}°C^{-1})

**Answer**:

m = 50 cm³ + 40 cm³ = 90 cm³

c = 4.2 Jg^{-1}°C^{-1}

θ = 42 – 29 = 13°C

Change of heat energy

= mcθ

= (90)(4.2)(13) = 4914J

**Example 2**:

When 20 cm³ solution of potassium carbonate is added into 20 cm³ of calcium nitrate solution, and the mixture is stirred immediately. The temperature of the solution change from 29.5°C to 28.0°C. Calculate the change of heat energy in this reaction. (Specific heat capacity of the solution = 4.2 Jg^{-1}°C^{-1})

**Answer**:

m = 20 cm³ + 20 cm³ = 40 cm³

c = 4.2 Jg^{-1}°C^{-1}

θ = 29.5 – 28 = 1.5°C

Change of heat energy

= mcθ

= (40)(4.2)(1.5) = 252J

**Example 3**:

[Specific heat capacity of the solution = = 4.2 Jg^{-1}°C^{-1}, density of the solution =1 g/cm³ ]

**Answer**:

The chemical equation of the reaction,

_{2}O

Number of mole of hydrochloric acid,

\[\begin{gathered}

n = \frac{{MV}}{{1000}} \hfill \\

n = \frac{{(2.0)(100)}}{{1000}} \hfill \\

n = 0.2mol \hfill \\

\end{gathered} \]

Number of mole of sodium hydroxide,

\[\begin{gathered}

n = \frac{{MV}}{{1000}} \hfill \\

n = \frac{{(2.0)(100)}}{{1000}} \hfill \\

n = 0.2mol \hfill \\

\end{gathered} \]

Number of mole of water produced = 0.2 mol

Change of heat energy,

= mcθ

= (100+100)(4.2)(42.0-28.5) = 11340J

The heat of neutralisation,

\[\begin{gathered}

\Delta {\text{H = }}\frac{{{\text{Heat Change}}}}{{{\text{Number of Mole of Water Produced}}}} \hfill \\

\Delta {\text{H = }}\frac{{11340}}{{0.2}} \hfill \\

\Delta {\text{H = 56700Jmo}}{{\text{l}}^{ – 1}} \hfill \\

\end{gathered} \]