2. LEARNING OUTCOME
• Describe the basic concept of endothermic and exothermic
• Solve numerical problems involving enthalpy change.
• Explains the exothermic and endothermic reactions with the aids of
simple energy level diagram.
• Explains the energy change during formation and breaking of bonds.
• Solve numerical problems involving enthalpy change during a simple
chemical reaction
4. INTRODUCTION
THE LAW OF CONVERSION states that
energy may neither be created nor destroyed.
However, energy can be changed from one form
to another.
5. CHEMICAL ENERGY
• All chemical substance contain chemical energy
• During chemical reactions, the chemical energy in the
reactants will be changed into other form of energy;
heat or thermal energy, mechanical energy, electrical
energy, light energy and sound energy.
• Unit for measuring energy is Joule (J).
1kJ = 1000J
6. THERMOCHEMISTRY
•Thermochemistry is the study of changes in heat
energy during chemical reactions.
•There are two types of chemical reactions:
a) Exothermic reaction
b) Endothermic reaction
7. a) Exothermic Reaction
• Exothermic reaction is a chemical
reaction that gives out heat to the
surroundings
• In exothermic reaction,
i. Chemical energy in reactant is
change into heat energy
ii. The heat energy is transferred to the
surroundings
iii. Temperature of the surroundings
increases
8. b) Endothermic Reaction
• Endothermic reaction is a chemical
reaction that absorbs heat from
the surroundings.
• In an endothermic reaction,
i. Heat energy is absorbed from the
surroundings
ii. The heat energy is changed into
chemical energy
iii. Temperature of the surroundings
decreases
9. ENTHALPY, H
• Every chemical substance has a certain amount of chemical energy. This
energy is known as enthalpy or heat energy content with symbol of H
and is different for different substances.
• Heat of reaction or enthalpy is the amount of heat energy released or
absorbed. It is given the symbol ΔH and is usually measured in
kilojoules (kJ).
• Only ΔH can be measured, not H for the initial or final state of a system
ΔH = total energy content of products – total energy content of reactant
ΔH = Hproducts - Hreactant
10. •In exothermic reaction,
the product formed has
less energy than the
reactant, Hproduct <
Hreactants. Therefore, the
ΔH is negative.
11. •In endothermic
reaction, the products
formed have more
energy than reactants,
Hproducts> Hreactants.
Therefore, ΔH is
positive.
13. Factors affecting heat quantities
• The amount if heat contained by an object depends
primarily on three factors:
i. The mass of material
ii. The temperature
iii. The kind of material and its ability to absorb or
retain heat.
14. Calculating enthalpy changes
ΔH = total energy content of products – total energy content of
reactant
ΔH = Hproducts - Hreactant
Q = mcθ
15. Specific Heat
• Specific heat: the amount of heat energy required to raise the
temperature of 1g of a substance of 1oC.
• Units are: J g-1 oC-1
Specific heat of water (or aqueous solutions) =
4.184 J = 1 cal
16. Heat Quantities
• The heat required to raise the temperature of 1.00 g of water 1 oC is known as
a calorie
• Calorie (with a capital “C”): dietary measurement of heat. Food has
Food has potential energy stored in the chemical bonds of food.1 Cal = 1 kcal
= 1000 cal. The SI unit for heat is the joule. It is based on the mechanical
energy requirements.1.00 calorie = Joules
17. Q = ΔH = heat energy change (J or calories)
m = mass (g)
c = specific heat (J g1 oC-1)
Specific heat of water = 4.184 J g1 oC-1
θ = change in temperature
Q = mcθ
18. Example 1
How much heat in joules will be absorbed when 32.0g of water is
heated from 25.0 oC to 80.0oC?
Q= mcθ
Q= ?
M=32.0g
C=4.18 J g-1 oC-1
Θ= 80-25=55
Q= 32 X 4.18 X 55
=7360 J
19. Example 2
When 435 J of heat is added to 3.4g
of olive oil at 21oC, the temperature
increases to 85oC. What is the
specific heat of olive oil?
Q= mcθ
Q= 435 J
M=3.4g
C= ?
θ = 85-21 = 64
435 = 3.4 X c X 64
C=(3.4 X 64)/435
=0.5 J g-1 OC-1
20. Enthalpy Change During Breaking and
Formation of Bonds
•Chemical reaction involves bond breaking and
bond formation
•Bond breaking need energy endothermic
•Bond formation releases energy exothermic
•The balance of heat release to the surroundings.
21. • In an exothermic reaction, the heat energy released during the
formation of the bonds in the products is greater than the heat
energy absorbed to break the bonds in the reactants.
Bond enthalpy reactants< bond enthalpy products
• In an endothermic reaction, the heat energy absorbed to
break the bonds in the reactants is greater than the heat
energy released during the formation of the bonds in the
products.
Bond enthalpy reactants> bond enthalpy products
22. Enthalpy Change of a Reaction
Using Thermochemical Equations
Mg(s) + ½ O2(g) MgO(s) + heat Equation 1
Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Equation A ∆H = -A
MgO(s) + 2HCl(aq) MgCl2(aq) +H2O(l) Equation B ∆H = -B
H2(g) + ½ O2(g) H2O(l) Equation C ∆H = +C
The final ΔH = -A + (-B) + C
23. Example:
Using Thermochemical Equations
Calcium oxide combines with water to produce calcium hydroxide and heat
(exothermic reaction).
CaO(s) + H2O(l) Ca(OH)2(s) + 65.2 kJ OR
CaO(s) + H2O(l) Ca(OH)2(s) ΔH = -65.2 kJ mol-1
These Δ H values assume 1 mole of each compound (based on the coefficients)
How many kJ of heat are produced when 7.23 g of CaO react?
1. Write out and balance equation: already balanced
2. Determine the number of moles: 7.23 g/56.01 gmol-1 = 0.129mol
3. Multiply: 0.129 mol CaO x 65.2 kJ mol-1
* Notice that mol / mol cancel out and you’re left with kJ
4. Solve = 8.41 kJ
24. LETS TRY!
Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes
to sodium carbonate, water, and carbon dioxide.
How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)?
1. Write out and balance the equation
2. Determine the number of moles of NaHCO3(s)
3. Set up the ratio
4. Solve for x
5. State, with justification, whether the reaction is endothermic or
exothermic.
25. Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to
sodium carbonate, water, and carbon dioxide.
2NaHCO3(s) + 129 kJ Na2CO3(s) + H2O(g) + CO2(g) OR
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) ΔH = 129 kJ
How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)?
1. Balanced equation
2. Moles NaHCO3(s) = 2.24 mol
3. Ratio: x kJ / 2.24 NaHCO3(s) = 129 kJ / 2mol
4. Solve for x = 144 kJ
5. The reaction is endothermic because energy is being absorbed / the
ΔH is positive.
27. TYPES OF ENTHALPY
TYPE OF REACTION HEAT OF REACTION
Precipitation Heat of precipitation
Displacement Heat of displacement
Neutralization Heat of neutralization
Combustion Heat of combustion
30. Since most of the chemical reactions carried out to determine the heat
of reaction involve aqueous solutions. Some assumptions are made:
• Density of any aqueous solution is equal to density of water, 1g cm-3
• The specific heat of any aqueous solution is equal to water, 4.2 J g-1
oC-1
31. HEAT OF PRECIPITATION
Heat change when one mole of the
precipitation is formed from their ions in
aqueous solution
•Unit: kJ mol-1
32. Example:
• 100 cm3 of 1.0 mol dm−3 of lead(II) nitrate, Pb(NO3)2
solution is added to 100 cm3 of 1.0 mol dm−3 of
sodium sulphate, Na2SO4 solution. The temperature of
the mixture rises from 30.0 °C to 33.0 °C. Calculate the
heat of precipitation of lead(II) sulphate, PbSO4
• [Specific heat capacity of solution, c = 4.2 J g−1 °C−1;
density of solution = 1 g cm−3]
33.
34.
35. Lets try!
The following equation shows the precipitation of silver chloride, AgCl
If 20cm3 of 0.5 mol dm-3 of silver nitrate, AgNO3 solution is added to
20cm3 of 0.5 mol dm-3 of potassium chloride, KCl solution, calculate the
rise in temperature of the mixture.
[Specific heat capacity of solution, c = 4.2 J g−1 °C−1; density of solution
= 1 g cm−3]
Answer: 3.9 °C
36.
37. HEAT OF NEUTRALIZATION
Heat of neutralization is defined as the change of heat
or enthalpy during the formation of one mole of water
by the neutralisation reaction of required amounts of
acid and a base.
HCl + NaOH NaCl + H2O ΔH = -57.34 kJ mol-1
38. Example:
60 cm3 of 2.0 mol dm−3 of sodium hydroxide, NaOH solution is added
into 60 cm3 of 2.0 mol dm−3 of ethanoic acid,CH3COOH.The highest
temperature of the mixture is 40.5 °C. The initial temperature of
sodium hydroxide, NaOH solution is 28.0 °C, and ethanoic acid,
CH3COOH solution is 28.0 °C. Calculate the heat of neutralisation.
[Specific heat capacity of solution, c = 4.2 J g−1 °C−1; density of solution
= 1 g cm−3]
39.
40.
41. HEAT OF DISPLACEMENT
• Heat change when one mole of a metal is displaced from
its salt solution by a more electropositive metal.
• Unit kJ mol-1
Example: Zn + CuSO4 ZnSO4 + Cu
• Zinc is more electropositive than copper
• Zinc can replace copper from the solution
42. Example:
An excess of magnesium powder, Mg is added into 50 cm3 of 0.25 mol
dm−3of iron(II)sulphate, FeSO4 solution. The temperature of the mixture
increases by 4.0 °C. Calculate the heat of displacement of iron, Fe from
its salt solution.
[Specific heat capacity of solution, c = 4.2 J g−1 °C−1; density of solution
= 1 g cm−3]
43.
44.
45. HEAT OF COMBUSTION
•Reaction between a substance and oxygen to form
carbon dioxide and water
Example: CH4 + 2O2 CO2 + 2H2O
The heat is given out (exothermic reaction)
•The heat of combustion is the heat released when 1
mole of a substance is completely burnt in excess
oxygen, O2.
•ΔH = negative (exothermic)
•Unit = kJ mol-1
46. Example:
The thermochemical equation for the complete combustion of
ethanol, C2H5OH is shown below.
Calculate the mass of ethanol, C2H5OH needed to burn
completely in excess oxygen, O2 in order to raise the
temperature of 200 cm3 of water, H2O by 50.0 °C. (Assume
that no heat is lost to the surroundings)
