chapter 4 thermochemistry
TRANSCRIPT
A. Energy Changes in Chemical Reactions
Thermochemistry
The study of changes in heat energy which take place during chemical reactions
Classify into:
• Exothermic reaction
• Endothermic reaction
TIPS: EX mean to go out/exit EN mean to come in/enter
Energy level diagram
The energy of the products is lower than
the total energy of the reactants
The energy of the products is higher than
the energy of the reactants
Energy
Reactants
Products
∆H = negative
Energy
Reactants
Products
∆H = positive
Exothermic reaction Endothermic reaction
Definition A chemical reaction that gives out heat to
the surroundings
A chemical reaction that absorbs heat
from the surroundings
What happen?
During exothermic reaction,
temperature of the surrounding
increases.
This is because heat given out from
the reaction is transferred to the
surroundings.
During endothermic reaction,
temperature of the surrounding
decreases.
This is because the reactants absorb
heat energy from the surroundings.
Heat of reaction, ∆H
The change in the amount
of heat in a chemical
reaction.
∆H negative: heat is given out ∆H positive: heat is absorbed
Step to construct energy level diagrams
Step 6 Label ∆H as positive or negative
Step 1 Identify whether the reaction is exothermic or
endothermic
Step 2 Draw and label the energy axis
Step 3 Draw the energy level for reactants and
products
Step 4 Draw an arrow from reactants level to the
products level
Step 5 Write the reactants and products based on the
balanced chemical equations
Construct energy level diagrams for the following thermochemical equations
• Zn + 2HCl → ZnCl2 + H2 ∆H = -152 kJ
• N2 + 2O2 → 2NO2 ∆H = +220 kJ
• KOH + HNO3 → KNO3 + H2O ∆H = -57 kJ
• C + 2S → CS2 ∆H = +220 kJ
• Ca(NO3)2 + K2CO3 → CaCO3 + 2KNO3 ∆H = +12 kJ
Energy change during formation and breaking of bonds
Usually a chemical reaction involves bond breaking and bond formation.
• Bond breaking : always requires energy
• Bond formation : always releases energy
Endothermic
Energy absorb for bond breaking is
more than energy released from bond
formation
∆H positive
Type of Reaction Energy Change Sign of ∆H
Exothermic
Energy absorb for bond breaking is
less than energy released from bond
formation
∆H negative
Application of exothermic and endothermic reaction
Cold packs
Contain chemicals (water & solid ammonium nitrate, NH4NO3) that react to absorb heat from surroundings.
• Help to reduce high temperature
• Help to reduced swelling
Application of exothermic and endothermic reaction
Hot packs
Contain chemicals (calcium chloride, CaCl2
or magnesium sulphate, MgSO4 and water)that react to release heat.
• Help to warm up something
• Help to lessen the pain of aching muscles
Application of exothermic and endothermic reaction
Reusable heat pack
Contain sodium acetate crystallization and re-solution system
Lye (drain cleaner)
Contain sodium hydroxide, NaOH
Heat of Reaction
• The change in the amount of heat in a chemical reaction.
• Symbol: ∆H
Different types of reactions
Combustion Heat of Combustion
Types of Reaction Heat of Reaction
Precipitation Heat of Precipitation
Displacement Heat of Displacement
Neutralization Heat of Neutralization
Heat of Combustion
The heat change when one mole of a substance is
completely burnt in oxygen under standard
conditions
Heat of reaction Definition
Heat of Precipitation The heat change when one mole of a precipitate is
formed from their ions in aqueous solution
Heat of Displacement
The heat change when one mole of a metal is
displaced from its salt solution by a more
electropositive metal
Heat of Neutralization The heat change when one mole of water is formed
from reaction between an acid and an alkali
Heat of Precipitation
Experiment to investigate the heat of precipitation between
silver nitrate solution and sodium chloride solution
Heat of Precipitation
Procedure:
1. Measure 25 cm3 of 0.5 mol dm-3 sodium chloride, NaCl solution using
a measuring cylinder and pour the solution into a polystyrene cup.
2. Measure and record the initial temperature of sodium chloride
solution.
3. Measure 25 cm3 of 0.5 mol dm-3 silver nitrate, AgNO3 solution using a
measuring cylinder and pour the solution into another polystyrene
cup.
4. Measure and record the initial temperature of silver nitrate solution.
5. Pour silver nitrate, AgNO3 solution into the polystyrene containing
sodium chloride, NaCl solution.
6. Stir the mixture and record the highest temperature, θ3
Tabulation of data:
Initial temperature silver nitrate solution :
Initial temperature sodium chloride solution :
Highest temperature :
Heat of Displacement
Experiment to investigate the heat of displacement of copper
by zinc
Copper(II) nitrate
0.5 g zinc powder
Polystyrene cup
Heat of Displacement
Procedure:
1. Measure 25 cm3 of 0.2 mol dm-3 copper(II) nitrate solution
using a measuring cylinder.
2. Pour the solution into a polystyrene cup.
3. Measure and record the initial temperature, θ1 of
copper(II) nitrate solution.
4. Measure 0.5 g of zinc powder and added into the
polystyrene quickly.
5. Stir the mixture and the highest temperature is recorded,
θ2
Heat of Neutralization
Describe an experiment to determine the heat of neutralization between dilute hydrochloric acid and sodium hydroxide solution.
100cm3 of 2 mol dm-3 dilute hydrochloric acid
100 cm3 of 2 mol dm-3 aqueous sodium hydroxide Plastic cup
Procedure:
1. 50 cm3 of 2.0 mol dm-3 sodium hydroxide solution is
measured using a measuring cylinder and poured into a
plastic cup.
2. The initial temperature of sodium hydroxide solution is
measured after a few minutes.
3. 50 cm3 of 2.0 mol dm-3 hydrochloric acid is measured using
another measuring cylinder and poured into a plastic cup.
4. The initial temperature of hydrochloric acid solution is
measured after a few minutes.
5. The hydrochloric acid is then poured quickly and carefully
into the sodium hydroxide solution.
6. The mixture is stirred using thermometer and the highest
temperature reached is recorded.
Heat of Neutralization
All neutralization process can be presented by the following ionic equation
H+ + OH- → H2O ∆H = -57 kJ mol-1
• Heat of neutralization for strong acid and strong alkali is same, that is -57 kJ mol-1
• Heat of neutralization for weak acid and strong alkali is less than -57 kJ mol-1
• Heat of neutralization for weak acid and weak alkali is much lesser
Strong acid & strong alkali
Explanation:
Strong acid ionise completely in water to produce high concentration of hydrogen ions
HCl → H+ + Cl-
1 mol of hydrogen ions reacts with 1 mol of hydroxide ions to form I mol of water to release 57 kJ of heat energy
KOH/NaOH with HCl/HNO3
Weak acid & strong alkali
Explanation:
Weak acid ionise partially in water to produce low concentration of hydrogen ions
CH3COOH + CH3COO- + H+
Most of the ethanoic acid still remains in the form of molecule
Thus the heat release is always less than 57 kJ
Weak acid & weak alkali
Explanation:
More heat energy is needed to dissociate both the weak acid and weak alkali completely to produce hydrogen ions.
Monoprotic acid & Diprotic acid
H2SO4 + 2NaOH → Na2SO4 + 2H2O ∆H = - 114.6 kJ HCl + NaOH → NaCl + H2O ∆H = - 57.3 kJ
Explain the differences of heat of reaction.
1. H2SO4 is a diprotic acid while HCl is monoprotic acid.
2. H2SO4 will produce 2 mole of H+ ions and HCl produce 1 mole of H+ ion
3. Neutralization of a diprotic acid will produce twice heat energy than monoprotic acid
Heat of Combustion
Describe an experiment to determine the heat of combustion of butanol in the laboratory.
In your description include a labelled diagram, procedure and tabulation of data
Heat of Combustion
Procedure:
1. 100 cm3 of water is measured using a measuring cylinder.
2. Poured into a copper tin.
3. The initial temperature of water is measured and
recorded, θ1 .
4. The spirit lamp is filled with butanol and weighed, x g
5. A spirit lamp is light and put under the copper tin.
Heat of Combustion
Procedure:
6. The water is stirred continuously with a thermometer
7. When the temperature of water increased by 30 °C, the
flame is put off
8. The spirit lamp is weighed again, y g
9. The highest temperature is recorded, θ2
Heat of Combustion
Results:
Mass of weight of spirit lamp + butanol / g x
Final mass of spirit lamp + butanol / g y
Mass of butanol used / g x-y = z
Highest temperature of water / °C θ2
Initial temperature of water / °C θ1
Increased in temperature / °C θ2 - θ1 = θ3
Heat of Combustion
When 2.7 g of glucose (C6H12O6) is burnt completely in excess oxygen, the heat released increases the temperature of 600 g of water by 12.5 °C. Calculate the heat of combustion of glucose.
[Specific heat capacity of water = 4.2 J g-1 °C-1, density of water = 1.0 g cm-3, RAM: H = 1, C = 12, O = 16]
When 1 mole of butanol is burnt in excess of oxygen, 2600 kJ of heat is produced. Calculate the mass of butanol needed to burn completely in oxygen in order to raise the temperature of 250 cm3 of water by 30 °C.
[Specific heat capacity of water = 4.2 J g-1 °C-1, density of water = 1.0 g cm-3, RAM: H = 1, C = 12, O = 16]
Comparing & contrasting heats of combustion
Draw a graph of the number of carbon atoms in alcohol
againts the magnitude of the heat of combustion.
Alcohol Molecular formula
Heat of combustion (kJ/mol)
Methanol -728
Ethanol -1376
Propanol -2016
Butanol -2678
Pentanol - 3332
Hexanol
From the graph:
1. State the relationship between the number of
carbon atoms in an alcohol and the heat of
combustion
2. Predict the heat of combustion of hexanol
Compare the heat of combustion between ethanol and butanol. Explain why there is a difference in the heat of combustion between ethanol and butanol.
1. The heat of combustion of butanol is higher than ethanol.
2. Butanol has higher number of carbons per molecules than ethanol
3. Butanol will produced more carbon dioxide and water than ethanol
4. The combustion of butanol produced more heat than ethanol.