$\text{HeatofReaction,}\Delta \text{H=}\frac{\text{ChangeofEnergy}}{\text{Moleofreactant/product}}$

Change of energy, Q = mcθ

m = mass of the substance that absorb the heat

c = specific heat capacity of the substance that absorb 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 into 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 carbonatye is added into 20 cm³ of calcium nitrate solution, and the mixture is stirred immediately. The temperature of the solution chage 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**:

When 100 cm³ of hydrochloric acid, 2 mol/dm³ is added into l00 cm³ of sodium hydroxide, 2 mol/dm³, the temperature of the mixture incerases from 28.5°C to 42.0°C. Find the heat of neutralisation of the reaction.

[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,

HCl + NaOH → KCl + 2H

_{2}ONumber of mole of hydrochloric acid,

$$n=\frac{MV}{1000}=\frac{(2.0)(100)}{1000}=0.2mol$$

Number of mole of sodium hydroxide,

$$n=\frac{MV}{1000}=\frac{(2.0)(100)}{1000}=0.2mol$$

Number of mole of water produced = 0.2 mol

Change of heat energy,

= mcθ

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

Heat of neutralisation,

$$\begin{array}{l}\Delta H=\frac{\text{HeatChange}}{\text{Numberofmoleofwaterproduced}}\\ \Delta H=\frac{11340}{0.2}=56,700Jmo{l}^{-1}=26.7kJmo{l}^{-1}\end{array}$$

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