EXOTHERMIC REACTION
-Gives out heat to the surroundings
-Tempareture of the surroundings increases
-Reactants have more energy compared to the products
-Bond formation releases more energy than the energy that required in the bond breaking
ENDOTHERMIC REACTION
-Absorb heat from the surroundings
-Temperature of the surroundings decreases
-Reactants have less energy than the products.
-Bond formation releases less energy than the energy that required in the bond breaking.
Heat of reaction
-is the change in the amount of heat in a chemical reaction
- symbol: Δ H
-unit: kJ mol-1
-Standard conditions:
- Temperature: 25 ̊C / 298 K
- Pressure: 1 atm
- Concentration of solution: 1.0 mol dm-3
In an exothermic reaction:
Heat released during the reaction=Heat absorbed by the solution
In an endothermic reaction:
Heat absorbed
during the reaction =Heat lost by the solution
For calculation:
Heat change = Heat absorbed or given out
by the aqueous solution
= mcθ
Where m= mass of the solution (in g)
°C= specific heat capacity of the solution
( in J g-1̊°C-1)
θ= temperature change in the solution ( in °C)
The following are the assumptions made during the calculation of the heat of reaction:
(a)The solution is dilute. It has the same density as water,which is 1 g cm-3
(b)The solution has same specific heat capacity as water,which is 4.2 J g-1°C-1
(c)No heat is lost to or absorbed by the surroundings
e.g.
When 30cm3 of potassium hydroxide solution is added to 25cm3 nitric acid,the results are as follows.
Initial temperature of potassium hydroside solution =29 ̊°C
Initial temperature of nitric acid= 27 °̊C
Highest temperature of the mixture= 38 °̊C
Calculate the heat change in this reaction.
(density of solution:1 g cm-3. Specific heat capacity of solution: 4.2 J g-1 ̊C-1)
Answer :
Average initial temperature of the solutions= 27+29
Average initial temperature of the solutions= 27+29
2
= 28 ̊C
ΔH= mcθ
= (30 + 25) cm3 (4.2)J g -1 ̊C-1 (38-28) ̊C
= 2310 J
=2.31 kJ Heat of Precipitation
Heat change when 1 mole of a precipitate is formed from its ions in aqueous solution
Example of precipitation reaction:
Pb(NO3)2 + K2SO4 ---------------> PbSO4 + 2KNO3
ΔH = -50kJ mol-1 Heat of Displacement
Heat change when one mole of a metal is displaced from its salt solution by a more electropositive metal
Example of displacement reaction:
Zn + CuSO4 ---------------------> ZnSO4 + Cu
Δ H= -210kJ mol-1
Heat of neutralization
Heat produced when 1 mole of water is formed from the reaction between an acid and an alkali
** Complete neutralization of a strong diprotic acid with an alkali produces
double amount of heat as compared to a strong monoprotic acid.
e.g. of monoprotic acid: HNO3, HCl
diprotic acid: H2SO4
This is because a diprotic acid produces 2 moles of hydrogen ions when it dissociates in water.
H2SO4 -----------> 2H+ + SO42-
2 moles of hydrogen ions produces 2 moles of water when reacted with
hydroxide ion from an alkali.
2H+ + 2OH- ------------> 2H2O
E.g of neutralization reaction
HCl + NaOH -----------------> NaCl + H2O
Δ H = -57.3 kJ mol-1
H2SO4 + 2NaOH -----------> Na2SO4 + 2H2O
Δ H= -114.6 kJ mol-1
Heat of combustion
Heat produced when one mole of a substance is completely combusted in excess oxygen under standard condition.
E.g CH4 + 2O2 -----------> CO2 + 2H2O
Δ H=-890 kJ mol-1
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