bab 3 termokimia kssm

1. Chapter 3:
Thermochemistry

2. 3.1 Exothermic & Endothermic Reactions
Endothermic Exothermic
Thermochemistry

3. Exothermic reaction Exothermic reaction
• released heat to the environment
• surrounding temperature
• Examples: respiration, oxidation,
dissolving NaOH in showing hands
feeling cold.
•absorb heat from the environment
•surrounding temperature
•Example: Photosynthesis, dissolving
ammonium salts in water.
Situation:
Situation: hands feel hot when the detergent powder
on the hands is dipped into water.
when a disinfectant liquid containing,
alcohol evaporates while sanitizing hands.

4. Exothermic Energy level diagram Endothermic
reactant energy > product energy
ΔH = negative
reactant energy>product energy
ΔH = positive
Exothermic Endothermic
Mg+ H2SO4 MgSO4+ H2 ΔH = 467 kJ/mol N2+ O2 -------> 2NO ΔH = +180 kJmol-1
1) 1 mole of N2 gas reacts with 1 mole of O2 gas to
produce 2 moles of NO.
2)A total of 180 kJof heat energy is ABSORBED from
the surrounding.
3)During the reaction, the temperature of the
surrounding decreases.
1) 1 mole of Mg reacts with 1 mole of H2SO4 produces 1 mole of
MgSO4 and 1 mole of H2 gas.
2) A total of 467 kJof heat energy is RELEASED to the surrounding.
3)During the reaction, the temperature of the surrounding increases.

5. Energy changes during bond formation
Exothermic Reactions
Heat energy released during bondformation in the reaction product is LARGERthan the
heat energy absorbed to break the bond in the reaction material.
Endothermic Reactions
Heat energy absorbed to breakthebond in the reaction material is LARGERthan the heat
energy released during bond formation in the reaction product.

6. Exercise 1

7. 3.2 Heat of Reaction
Precipitation Displacement Neutralization Combustion
• Heat change when
1 mole of
precipitate is
formed from their
ions in an aqueous
solution.
• Heat change
when one mole
metal is
displaced from
its salt solution
by a more
electropositive
metal.
• heat change
when one mole
of water is
formed from
neutralizatiof acids
and alkalis.
• Heat of combustion is
the heat that
released when one
mole of substance is
completely burnt in
excess
O2.

8. 3.2 Reaction Heat
1) Precipitation
2) Displacement
3) Neutralization
4) Combustion

9. Find the calculation of the heat of reaction, ∆H
Step 1: Write a balanced equation
Step 2: Find number of mole
Step 3: Calculate the heat change, Q =mcθ
Step 4: Find the heat of precipitation,∆H
Q = total heat released in the reaction only
∆H= total heat released for 1 mole

10. 1) Heat of Precipitation
Question 1:
100 cm3 solution of lead (II) nitrate, Pb (NO3)2, 1.0 mol/dm3mixed with 100 cm3sodium sulfate solution,
Na2SO4 .1.0 mol/dm3. The temperature of the reaction mixture increased from 30.0 ° Cto 33.0 ° C.
Calculate the heat of precipitation of lead (II) sulfate,PbSO4.
[Specific heat capacity of solution, c = 4.2 Jg−1° C−1; density of solution = 1 g cm−3]
Step 1: Write a balanced equation Step 3: Calculate the heat change, Q =mcθ
Q = mcθ
=200x 4.2 x3
= 2520J
= 2.52 kJ
Step 4: Find the heat of precipitation,∆H
Step 2: Find the PbSO mole4
n Mol Pb (NO3)2= n Mol PbSO4
n=MV
1000
= (1) (100) = 0.1 mol Pb (NO3)2
1000
0.1 mol Pb (NO3) 2 = 0.1 mol PbSO4
Pb(NO3)2 + Na2SO4 ------> PbSO4 + 2 NaNO3
= Q
n
∆H
= 2.52
0.1
= 25.2 kJ
Ans: ∆H= - 25.2 kJ

11. Find the change in temperature, θ
AgNO3+ KCl →AgCl + KNO3 ∆H= - 65.5 kJ/mol
If 20 cm3 of argentum nitrate solution, AgNO30.5 mol/ dm3 mixed with 20 cm3potassium chloride
solution, KCl 0.5 mol/dm3, calculate the rise in the temperature of mixture.
[Given: The specific heat capacity of the solution, c = 4.2 Jg−1° C−1, density of solution = 1 g cm−3]

12. 2) Heat of Displacement
Question 1:
Magnesium powder, excess Mg is added to 50 cm3 solution of iron (II) sulfate, FeSO4
0.25 mol/dm3. The temperature of the reaction mixture increased by 4.0° C. Calculate the heat of
displacement of iron, Fe from its salt solution.
[Given: The specific heat capacity of the solution, c = 4.2 Jg−1° C−1, density of solution = 1 g cm−3]
Step 1:Write a balanced equation
Mg+ FeSO4 →Fe + MgSO4
Step 3: Calculate the heat change, Q =mcθ
Q = mcθ
= 50 x4.2 x 4
= 840 J
= 0.84 kJ
Step 2:Find the mole of Fe
n Mol Fe = n Mol FeSO4
n = MV
1000
(0.25) (50)
1000
= 0.0125 mol Fe
Step 4: Find the heat of displacement,

13. 3) Heat of Neutralization (Exothermic)
Question 1:
60 cm3 sodium hydroxide solution, NaOH 2.0 mol/dm3 mixed together with 60 cm3ethanoic acid solution,
CH3COOH 2.0 mol/dm3, the highest temperature of the mixed solution is 40.5 ° C.The initial temperature of
sodium hydroxide solution, NaOH is 28.0 ° C and ethanoic acid solution, CH3COOH is 28.0 °C. Calculate the heat of
neutralization.
[Specific heat capacity of solution, c = 4.2 J
g−1° C−1;density of solution = 1 g cm−3]
Step 1: Write a balanced equation NaOH +
CH3COOH -------> CH3COONa + H20
Step 2: Find the H20 mole
n Mol NaOH = n Mol H20 n
=MV
1000
(2) (60)
1000
= 0.12 mol NaOH = 0.12 mol H20
Step 3: Calculate the heat change, Q =mcθ
Q = mCθ
= (60 +60) x4.2 x (40.5-28) =
120x4.2x12.5
= 6.3 kJ
Step 4: Find the heat of neutralization,

14. 4) Heat of Combustion