# Heat of Reaction

1. The Heat of Reaction is the heat absorbed in a reaction at standard state condition between the numbers of moles of reactants shown in the equation for the reaction.
2. The Heat of reaction is represented by the symbol ∆H.
3. The unit of ∆H is kJmol-1
4. If the reaction is exothernic, ∆H shows a value of negative. If the reaction is endothernic, ∆H shows a value of positive.
5. The 4 heat of reaction that you need to know in the syllabus
1. Heat of Precipitation
The heat of precipitation of a substance is the heat change when 1 mole of precipitate formed from its solution of ion under standard state condition.
2. Heat of Displacement
The Heat of Displacement of an element is the heat change when 1 mole of the element was displaced from its compound under standard state condition.
3. Heat of Neutralisation
Heat of neutralisation is the heat change when 1 mol of water is formed by the neutralisation of hydrogen ions by hydroxide ions with measurements made under standard conditions.
4. Heat of Combustion
The Heat of Combustion of a substance is the heat energy evolved when 1 mole of the substance is completely burnt in oxygen.

## Calculating Heat Change

Heat change in a chemical reaction is directly proportional to the number of mole of reactant that takes part in a reaction or number of mole of product been produced.

Example 1:

N(g) + ½ O2(g) → NO2(g) ∆H = +66 kj mol-1

Calculate the heat change when 0.1 mole of nitrogen dioxide is formed in the reaction that shown above.

Heat change = 0.1 x 66kJ = 6.6 kJ

Example 2:

C(s) + O2(g) → CO2 (g)   ∆H=  393.5kJ mol-1

How much energy is released when 4g of carbon combust completely in excess oxygen. (Relative atomic mass of carbon = 12 )

Number of mole of carbon = 4g 12gmo l −1 = 1 3 mol

Total heat been released = 1 3 ×393.5kJ=131.2kJ

Example 3:

CH3OH(l) +  O2 (g) → CO2 (g) + 2H2O (l)     ∆H =  560 kJ mol-1

Find the mass of alcohol that need to be combusted, in excess of oxygen, to release 140 kJ of heat energy.
[Relative atomic mass: H=1; C=12; 0=16]

Number of mole of alcohol = 140kJ 560kJmo l −1 =0.25mol

Relative molecular mass of CH3OH = 12 + 3(1) + 16 + 1 = 32

Mass of alcohol = 0.25mol x 32g/mol = 8g

Example 4:

Calculate the heat change when excess zinc powder is added into 50cm³ of copper(II) sulphate solution 0.2 mol dm-3
CuSO4(ak) + Zn(p) → ZnSO4(ak) + Cu(p)
∆H =  190 kJ mol-1

Number of mole of copper(II) sulphate solution
n= MV 1000 n= (0.2)(50) 1000 =0.01mol

Heat change = 0.01 x 190kJ = 1.9kJ

Example 5:

The heat of combustion of carbon to CO2 is -393.5kJ/mol. Calculate the heat released upon formation of 35.2g of CO2 from carbon and oxygen gas.

Relative molecular mass of carbon dioxide = 12 + 2(16) = 44

Number of mole of carbon dioxide = 35.2/44 = 0.8 mol

C + O2  CO2

Number of mole of carbon = 0.8 mol

Heat released = 0.8 x 393.5kJ = 314.8 kJ

## 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:

1. We always use the specific heat capacity of water as an approximation of the specific heat capacity for all kinds of aqueous solution.
2. 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)

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)

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 ]

The chemical equation of the reaction,

HCl + NaOH → KCl + 2H2O

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,
$\begin{gathered} \Delta {\text{H = }}\frac{{{\text{Heat Change}}}}{{{\text{Number of Mole of Water Produced}}}} \hfill \\ \Delta {\text{H = }}\frac{{11340}}{{0.2}} \hfill \\ \Delta {\text{H = 56700Jmo}}{{\text{l}}^{ – 1}} \hfill \\ \end{gathered}$