2. SCechtaiopnte6r.61: Thermochemistry
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Define energy, work, potential energy, kinetic energy, system and
surroundings, endothermic and exothermic.
Be able to identify each of the above.
Understand E = q + w and use the equation to find the sign of E.
Know H = H p
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s H reactants.
Use calorimetry to solve for a variable or estimate a temperature of a
system.
Use Hesss law to calculate enthalpy of a reaction.
Use Standard enthalpy of formation to solve for enthalpy of reaction: H
= 裡nHre
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3. SCechtaiopnte6r.61: Thermochemistry
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Unit 6: Thermodynamics
Energy
Types of Energy
Kinetic Energy
Potential Energy
Chemical Energy
Energy Changes in Chemical Reactions
Thermodynamics
First Law of Thermodynamics
Heat vs. Work
Enthalpy
Exothermic and Endothermic Processes
Change in Enthalpy (H)
Enthalpy of Reactions (Hrxn)
System vs. Surroundings in a Chemical Reaction
Calorimetry
Specific Heat and Heat Capacity
Calorimeters
Bomb Calorimeters
Standard Enthalpy of Formation and Enthalpy of Reactions
Standard States and Standard Enthalpy Changes
Calculating H Directly
Hess Law
Heat of Solution
Energy Use and the Environment
Present Sources of Energy
Energy Consumption
Environmental Problems Attributed to Fuel Consumption
4. Section 6.1
The Nature of Energy
Capacity to do work or to produce heat.
Law of conservation of energy energy can
be converted from one form to another but
can be neither created nor destroyed.
The total energy content of the universe is
constant.
Energy
5. Section 6.1
The Nature of Energy
Potential energy energy due to position or
composition.
Kinetic energy energy due to motion of the
object and depends on the mass of the object
and its velocity.
Energy
6. Section 6.1
The Nature of Energy
In the initial position, ball A has a higher
potential energy than ball B.
Initial Position
7. Section 6.1
The Nature of Energy
After A has rolled down the hill, the potential energy
lost by A has been converted to random motions of
the components of the hill (frictional heating) and to
the increase in the potential energy of B.
Final Position
8. Section 6.1
The Nature of Energy
Heat involves the transfer of energy between
two objects due to a temperature difference.
Work force acting over a distance.
Energy is a state function; work and heat are not
State Function property that does not
depend in any way on the systems past or
future (only depends on present state).
Energy
9. Section 6.1
The Nature of Energy
System part of the universe on which we
wish to focus attention.
Surroundings include everything else in the
universe.
Chemical Energy
10. Section 6.1
The Nature of Energy
Endothermic Reaction:
Heat flow is into a system.
Absorb energy from the surroundings.
Exothermic Reaction:
Energy flows out of the system.
Energy gained by the surroundings must be
equal to the energy lost by the system.
Chemical Energy
11. Section 6.1
The Nature of Energy
exothermic process? Explain.
CONCEPT CHECK!
Is the freezing of water an endothermic or
12. Section 6.1
The Nature of Energy
a) Your hand gets cold when you touch
ice.
b) The ice gets warmer when you touch it.
c) Water boils in a kettle being heated on a
stove.
d) Water vapor condenses on a cold pipe.
e) Ice cream melts.
Exo
Endo
Endo
Exo
Endo
CONCEPT CHECK!
Classify each process as exothermic or
endothermic. Explain. The system is
underlined in each example.
13. Section 6.1
The Nature of Energy
CONCEPT CHECK!
For each of the following, define a system and
its surroundings and give the direction of energy
transfer.
a) Methane is burning in a Bunsen burner
in a laboratory.
b) Water drops, sitting on your skin after
swimming, evaporate.
14. Section 6.1
The Nature of Energy
Which is lower in energy: a mixture of
hydrogen and oxygen gases, or water?
CONCEPT CHECK!
Hydrogen gas and oxygen gas react violently to
form water. Explain.
15. Section 6.1
The Nature of Energy
Thermodynamics
The study of energy and its interconversions is
called thermodynamics.
Law of conservation of energy is often called the
first law of thermodynamics.
16. Section 6.1
The Nature of Energy
Internal Energy
Internal energy E of a system is the sum of the
kinetic and potential energies of all the
particles in the system.
To change the internal energy of a system:
E = q + w
q represents heat
w represents work
17. Section 6.1
The Nature of Energy
Work vs Energy Flow
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18. Section 6.1
The Nature of Energy
Thermodynamic quantities consist of two parts:
Number gives the magnitude of the change.
Sign indicates the direction of the flow.
Internal Energy
19. Section 6.1
The Nature of Energy
Internal Energy
Sign reflects the systems point of view.
Endothermic Process:
q is positive
Exothermic Process:
q is negative
20. Section 6.1
The Nature of Energy
Sign reflects the systems point of view.
System does work on surroundings:
w is negative
Surroundings do work on the system:
w is positive
Internal Energy
21. Section 6.1
The Nature of Energy
Work = P A h = PV
P is pressure.
A is area.
h is the piston moving
a distance.
V is the change in
volume.
Work
22. Section 6.1
The Nature of Energy
Work
For an expanding gas, V is a positive quantity
because the volume is increasing. Thus V and
w must have opposite signs:
w = PV
To convert between L揃atm and Joules, use 1
L揃atm = 101.3 J.
23. Section 6.1
The Nature of Energy
EXERCISE!
Which of the following performs more work?
a) A gas expanding against a pressure
of 2 atm from 1.0 L to 4.0 L.
b) A gas expanding against a pressure
of 3 atm from 1.0 L to 3.0 L.
They perform the same amount of work.
24. Section 6.1
The Nature of Energy
|work| > |heat|
|work| < |heat|
b) Work is done on a gas and the process is
exothermic.
|work| > |heat|
|work| < |heat|
E = negative
E = positive
E = positive
E = negative
CONCEPT CHECK!
Determine the sign of E for each of the
following with the listed conditions:
a) An endothermic process that performs work.
25. Section 6.2
Enthalpy and Calorimetry
Change in Enthalpy
State function
H = q at constant pressure
H = Hproducts Hreactants
26. Section 6.2
Enthalpy and Calorimetry
EXERCISE!
Consider the combustion of propane:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
H = 2221 kJ
Assume that all of the heat comes from the combustion of
propane. Calculate H in which 5.00 g of propane is burned in
excess oxygen at constant pressure.
252 kJ
27. Section 6.2
Enthalpy and Calorimetry
Calorimetry
Science of measuring heat
Specific heat capacity:
The energy required to raise the temperature of one
gram of a substance by one degree Celsius.
Molar heat capacity:
The energy required to raise the temperature of one
mole of substance by one degree Celsius.
28. Section 6.2
Enthalpy and Calorimetry
Calorimetry
If two reactants at the same temperature are mixed and
the resulting solution gets warmer, this means the
reaction taking place is exothermic.
An endothermic reaction cools the solution.
30. Section 6.2
Enthalpy and Calorimetry
Calorimetry
Energy released (heat) = m s T
m = mass of solution (g)
s = specific heat capacity (J/属C揃g)
T = change in temperature (属C)
31. Section 6.2
Enthalpy and Calorimetry
b) 50属C
c) Between 10属C and 50属C
CONCEPT CHECK!
A 100.0 g sample of water at 90属C is added to a 100.0
g sample of water at 10属C.
The final temperature of the water is:
a) Between 50属C and 90属C
32. Section 6.2
Enthalpy and Calorimetry
c) Between 10属C and 50属C
Calculate the final temperature of the water.
23属C
CONCEPT CHECK!
A 100.0 g sample of water at 90.属C is added to a 500.0 g
sample of water at 10.属C.
The final temperature of the water is:
a) Between 50属C and 90属C
b) 50属C
33. Section 6.2
Enthalpy and Calorimetry
c) Between 10属C and 50属C
Calculate the final temperature of the water.
18属C
CONCEPT CHECK!
You have a Styrofoam cup with 50.0 g of water at 10.属C. You
add a 50.0 g iron ball at 90. 属C to the water. (sH2O= 4.18 J/属C揃g
and sFe= 0.45 J/属C揃g)
The final temperature of the water is:
a) Between 50属C and 90属C
b) 50属C
34. Section 6.3
Hesss Law
In going from a particular set of reactants to a particular
set of products, the change in enthalpy is the same
whether the reaction takes place in one step or in a
series of steps.
35. Section 6.3
Hesss Law
This reaction also can be carried out in two distinct steps,
with enthalpy changes designated by H2 and H3.
N2(g) + O2(g) 2NO(g) H2= 180 kJ
2NO(g) + O2(g) 2NO2(g) H3= 112 kJ
N2(g) + 2O2(g) 2NO2(g) H1 = 68 kJ
N2(g) + 2O2(g) 2NO2(g) H2 + H3 = 68 kJ
H1 = H2 + H3 = 68 kJ
37. Section 6.3
Hesss Law
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38. Section 6.3
Hesss Law
Characteristics of Enthalpy Changes
If a reaction is reversed, the sign of H is also reversed.
The magnitude of H is directly proportional to the
quantities of reactants and products in a reaction. If the
coefficients in a balanced reaction are multiplied by an
integer, the value of H is multiplied by the same
integer.
39. Section 6.3
Hesss Law
Example
Consider the following data:
Calculate H for the reaction
3 2 2
2 H2(g) O2(g) 常
2 H2O(g)
2 2
NH (g) 常
1
N (g) 3
H (g) H = 46 kJ
H = 484 kJ
2 N2(g) 6 H2O(g) 常
3 O2(g) 4 NH3(g)
40. Section 6.3
Hesss Law
Problem-Solving Strategy
Work backward from the required reaction, using
the reactants and products to decide how to
manipulate the other given reactions at your
disposal.
Reverse any reactions as needed to give the
required reactants and products.
Multiply reactions to give the correct numbers of
reactants and products.
41. Section 6.3
Hesss Law
Example
Reverse the two reactions:
Desired reaction:
2 N2(g) 6 H2O(g) 常
3 O2(g) 4 NH3(g)
2 2 3
2 H2O(g) 常
2 H2(g) O2(g)
2 2
1
N (g) 3
H (g) 常
NH (g) H = 46 kJ
H = +484 kJ
42. Section 6.3
Hesss Law
Example
Multiply reactions to give the correct numbers of
reactants and products:
4( ) 4(H = 46 kJ )
Desired reaction:
2 N2(g) 6 H2O(g) 常
3 O2(g) 4 NH3(g)
2 2 3
1 3
2 2
N (g) H (g) 常
NH (g)
3(2 H2O(g) 常常2 H2(g) O2(g) ) 3(H = +484 kJ )
43. Section 6.3
Hesss Law
Example
Final reactions:
2 N2(g) 6 H2(g) 常
4 NH3(g)
6 H2O(g) 常
6 H2(g) 3 O2(g)
Desired reaction:
2 N2(g) 6 H2O(g) 常
3 O2(g) 4 NH3(g)
H = +1268 kJ
H = 184 kJ
H = +1452 kJ
44. Section 6.4
Standard Enthalpies of Formation
Standard Enthalpy of Formation (Hf属)
Change in enthalpy that accompanies the formation of
one mole of a compound from its elements with all
substances in their standard states.
45. Section 6.4
Standard Enthalpies of Formation
Conventional Definitions of Standard States
For a Compound
For a gas, pressure is exactly 1 atm.
For a solution, concentration is exactly 1 M.
Pure substance (liquid or solid)
For an Element
The form [N2(g), K(s)] in which it exists at 1 atm and
25属C.
46. Section 6.4
Standard Enthalpies of Formation
A Schematic Diagram of the Energy Changes for the Reaction
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
H属reaction= (75 kJ) + 0 + (394 kJ) + (572 kJ) = 891 kJ
47. Section 6.4
Standard Enthalpies of Formation
Problem-Solving Strategy: Enthalpy Calculations
1. When a reaction is reversed, the magnitude of H
remains the same, but its sign changes.
2. When the balanced equation for a reaction is multiplied
by an integer, the value of H for that reaction must be
multiplied by the same integer.
48. Section 6.4
Standard Enthalpies of Formation
Problem-Solving Strategy: Enthalpy Calculations
3. The change in enthalpy for a given reaction can be
calculated from the enthalpies of formation of the
reactants and products:
H属rxn = 裡npHf(products) - 裡nrHf(reactants)
4. Elements in their standard states are not included in the
Hreactioncalculations because Hf属 for an element in its
standard state is zero.
49. Section 6.4
Standard Enthalpies of Formation
Na(s)
H2O(l)
NaOH(aq)
H2(g)
H属 = 368 kJ
0
286
470
0
EXERCISE!
Calculate H属 for the following reaction: 2Na(s) + 2H2O(l)
2NaOH(aq) + H2(g)
Given the following information:
Hf属 (kJ/mol)
52. Section 6.5
Present Sources of Energy
The Earths Atmosphere
Transparent to visible light from the sun.
Visible light strikes the Earth, and part of it is changed to
infrared radiation.
Infrared radiation from Earths surface is strongly
absorbed by CO2, H2O, and other molecules present in
smaller amounts in atmosphere.
Atmosphere traps some of the energy and keeps the
Earth warmer than it would otherwise be.
![Section 6.4
Standard Enthalpies of Formation
Conventional Definitions of Standard States
For a Compound
For a gas, pressure is exactly 1 atm.
For a solution, concentration is exactly 1 M.
Pure substance (liquid or solid)
For an Element
The form [N2(g), K(s)] in which it exists at 1 atm and
25属C.](https://image.slidesharecdn.com/thermochemistry-240426094945-0ba68f7e/85/thermochemistry-pptxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx-45-320.jpg)
