Finding Heat of Reaction


Heat of Reaction, ΔH= Change of Energy Mole of reactant/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:
  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 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 + 2H2O

Number of mole of hydrochloric acid,
n= MV 1000 = (2.0)(100) 1000 =0.2mol
Number of mole of sodium hydroxide,
n= MV 1000 = (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,
ΔH= Heat Change Number of mole of water produced ΔH= 11340 0.2 =56,700Jmo l 1 =26.7kJmo l 1


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