1. Chapter 17 Thermochemistry

2. Section 17.1: The flow of energy
Thermochemistry:
Study of energy changes that occur during chemical reactions and changes in state

3. Section 17.1: The flow of energy
energy changes can either occur through heat transfer or work
heat (q), energy is transferred from a warmer object to a cooler object (always)
Adding heat increases temperature

4. 17.1: Endothermic & Exothermic processes
Kinetic energy vs. potential energy
potential energy is determined by the strength of repulsive
and attractive forces between atoms
In a chemical reaction, atoms are recombined into new arrangements that have different potential energies
change in potential energy: due to absorption and release of energy to and from the surroundings

5. 17.1: Endothermic & Exothermic processes
Two parameters crucial in Thermochemistry:
a) system--part of the universe attention is focused
b) surroundings--everything else in the universe
System + surrounding = universe
Fundamental goal of Thermochem.: study the heat flow between the system and its surroundings

6. 17.1: Endothermic & Exothermic processes
If System (energy) Surrounding (energy) by the same amount
the total energy of the universe does not change
Law of conservation of energy

7. 17.1: Endothermic & Exothermic processes
Direction of heat flow is given from the point of view of the system
So, endothermic process: system absorbs heat from the surroundings (system heats up)
Heat flowing into the system = +q
Exothermic process: heat is released into the surroundings (system cools down, -q)

8. 17.1: Endothermic & Exothermic processes
Example 1:
A container of melted paraffin wax is allowed to stand at room temperature (r.t.) until the wax solidifies. What is the direction of heat flow as the liquid wax solidifies? Is the process exothermic or endothermic?
Answer: Heat flows from the system (paraffin) to the surroundings (air)
Process: exothermic

9. 17.1: Endothermic & Exothermic processes
Example 2:
When solid Ba(OH)2▪8H2O is mixed in a beaker with solid NH4SCN, a reaction occurs. The beaker quickly becomes very cold. Is the reaction exothermic or endothermic?
Answer: Endothermic
surrounding = beaker and air
System = chemicals within beaker

10. 17.1: Units of heat flow Two units used:
a) calorie (cal)—amount of heat required to raise the temperature of 1g of pure water by 1oC
b) joule (j)—1 joule of heat raises the temperature of 1 g of pure water oC
Joule = SI unit of energy

11. 17.1 Heat capacity & specific heat
Heat capacity is the quantity of heat needed to raise the temperature of an object exactly 1oC
Heat capacity depends on:
a) mass
b) chemical composition
So, the greater the mass the greater the heat capacity
eg.: cup of water vs. a drop of water
(cup of water = greater heat capacity)

12. 17.1: Heat capacity & specific heat
Specific heat: amount of heat required to raise the temperature of 1g of a substance by 1oC
Table 17.1 (p.508): List of specific heats of substances

13. Specific heat calculation
C = q = heat (joules/calories)
m * ΔT mass (g) * ΔTemp. (oC)
ΔT = Tf –Ti (Tf = final temperature)
(Ti = initial temperature)
C = j or cal
(g * oC) (g * oC)

14. Example 1
The temperature of a 95.4 g piece of copper increases from 25.0 oC to 48.0 oC when the copper absorbs 849 j of heat. What is the specific heat of copper?
unknown: Ccu
Know:
mass copper = 95.4 g
ΔT = Tf – Ti = (48.0 oC – 25.0 oC)
= 23.0 oC
q = 849 j

15. Example1 C = q m * ΔT C = 849 j 95.4 g * 23.0 oC C = 0.387 j/g * oC
Sample problem 17.1, page 510

16. Example 2
How much heat is required to raise the temperature of g of mercury (Hg) 52 oC?
unknown: q
Know:
mass Hg = g
ΔT = 52.0 oC
CHg = 0.14 j/(g * oC)

17. Example 2 Problem #4 page 510 C = q m * ΔT q = CHg * m * ΔT
q = (j/g * oC) * g * 52 oC
q = 1.8 x 103 j (1.8 kj)
Problem #4 page 510

18. Measuring Enthalpy Changes
Section 17.2
Measuring Enthalpy Changes

19. 17.2: Enthalpy (measuring heat flow)
Calorimetry:
accurate measurement of heat flow
into or out of a system in chemical
and physical processes
In calorimetry, heat released by a system is equal to the heat absorbed by its surroundings and vice versa
Instrument used to measure absorption or release of heat is a calorimeter

20. 17.2: Enthalpy (measuring heat flow)
Two types of calorimeters:
a) Constant-Pressure calorimeter (eg. foam cups)
As most reactions occur at constant pressure we can say that:
A change in enthalpy (ΔH) = heat supplied (q)
So, a release of heat (exothermic) corresponds to a decrease in enthalpy (at constant pressure)
An absorption of heat (endothermic) corresponds to an increase in enthalpy (constant pressure)

21. 17.2: Enthalpy (measuring heat flow)
b) Constant-Volume Calorimeters (eg. bomb calorimeters)
Substance is burned (in the presence of O2) inside a chamber surrounded by water (high pressure)
Heat released warms the water
Figure 17.6 (p. 512) Bomb calorimeter.

22. 17.2: Thermochemical equations
A chemical equation that includes the enthalpy change is called
a thermochemical equation
Reactants and products at their usual physical state (at 25 oC) given at standard pressure (101.3 kPa)
So, the heat of reaction (or ΔH) for the above equation is kJ

23. 17.2: Thermochemical equations
So, rewrite the equation as follows:
Other reactions absorb heat from the surroundings, eg.:
Rewrite to show heat of reaction

24. 17.2: Thermochemical equations
Amount of heat released/absorbed during a reaction depends on the number of moles of reactants involved
eg.:

25. Enthalpy Diagrams Enthalpy of reactants greater than of products
Diagram A:
Enthalpy of reactants greater than of products
Diagram B:
Enthalpy of reactant less than of products

26. 17.2: Thermochemical equations
Physical states of reactants and products must be stated:
Vaporization of H2O(l) requires more heat (44.0 kJ)

27. Example1
1. Calculate the amount of heat in (kJ) required to decompose 2.24 mol
NaHCO3(S)
Known:
Unknown:
ΔH = ?
2.24 mol NaHCO3 decomposes
ΔH = 129 kJ (2 mol NaHCO3)
Solve:
129 kJ = ΔH
2 mol NaHCO3(s) mol NaHCO3(s)
ΔH = (129 kJ) * 2.24 mol NaHCO3(s)
2 mol NaHCO3(s)
ΔH = kJ
Sample problem 17.3; p. 516

28. Example 2
2. When carbon disulfide is formed from its elements, heat is absorbed. Calculate the amount of heat in (kJ) absorbed when 5.66 g of carbon disulfide is formed.
Known:
Unknown:
ΔH = ?
5.66 g CS2 is formed
ΔH = 89.3 kJ (1 mol CS2(l))
Molar mass: CS2(l): C = 12.0 g/mol
2 *S = 32.1 g/mol = 64.2 g/mol
76.2 g/mol

29. Example 2
Solve:
1. Moles CS2(l) = g CS = mol CS2(l) g/mol CS2(l)
kJ = ΔH
1 mol CS2(l) mol CS2(l)
ΔH = (89.3 kJ) * mol CS2(l)
1 mol CS2(l)
ΔH = kJ

30. 17.3: Heat in changes of state
Objective:
-Heats of Fusion and Solidification
-Heats of Vaporization and Condensation
-Heat of solution

31. 17.3: Heat of fusion and solidification
The temperature remains constant when a change of state occurs via a gain/loss of energy
Heat absorbed by 1 mole of a solid during melting (constant temperature) is the molar heat of fusion (ΔHfus)
Molar heat of solidification (ΔHsolid) is the heat lost by 1 mole of liquid as it solidifies (constant temperature)
So, ΔHfus = ΔHsolid

32. 17.3: Heat of fusion and solidification
Figure 17.9: Enthalpy changes and changes of state

33. 17.3: Heat of fusion and solidification

34. 17.3: Heats of Vaporization and Condensation
Molar heat of vaporization (ΔHvap): Amount of heat required to vaporize one mole of a liquid at the liquid’s normal boiling point
Molar heat of condensation (ΔHcond): heat released
when 1 mole of vapor condenses
So, ΔHvap = -ΔHcond

35. 17.3: Heat of vaporization and condensation
Figure 17.10: Heating curve of water

36. 17.3: Heat of solution
There is heat released/gained when a solute dissolves in a solvent
The enthalpy change due to 1 mole of a substance dissolving: molar heat of solution (ΔHsoln)

37. 17.3: Heat of solution Applications: hot/cold packs Hot pack: