#### Transcript Chapter 10 Energy

```Chapter 10
Energy
Energy: Types
Objectives:
1)To understand the general properties of
energy
2)To understand the concepts of
temperature and heat
3)To consider the direction of energy flow as
heat
Energy: Types
Energy: the ability to do work or produce heat
TWO TYPES
Potential energy (energy due to position or
composition) PE = mgh
Kinetic energy: energy is due to the motion of an
object KE=1/2 mv2
Law of Conservation of Energy: energy can be
converted from one form to another but can be
neither created nor destroyed.
Energy: Types
Work: force acting over a distance.
As energy is transferred some of it is
converted to heat (due to friction).
STATE FUNCTION: property of the system
that changes independently of its pathway.
Examples: ENERGY is a state function
work and heat are not.
Energy: Temperature and Heat
Temperature: measure
of the random motion
of the components of
a substance.
What will happen to the temperature
of the water on both sides?
Insulated Box and a thin metal sheet
Energy: Temperature and Heat
Heat: flow of energy
due to a temperature
difference.
T final = T hot (initital) + T cold (initial)
2
Exothermic and Endothermic
System: part of the universe on which we wish to
focus
Surroundings: include everything else in the
universe.
Exothermic: a process that results in the evolution
of heat. ENERGY FLOWS OUT
Endothermic: process that absorbs energy from
the surroundings ENERGY FLOWS IN.
10.4 Thermodynamics
• Objective: To understand how energy flow
affects internal energy
10.4 Thermodynamics
• Thermodynamics: study of energy
• First Law of Thermodynamics: The energy
of the universe is constant (same as Law
of Conservation of Energy)
• Internal energy (E)= sum of kinetic and
potential energy of the system
– This can be changed by a flow of work, heat
or both
10.4 Thermodynamics
• Change in Internal energy
DE=q+w
q= heat
w= work
Thermodynamic quantities consists of 2 Parts: a
number and a sign. The sign reflects the
system’s point of view.
10.4 Thermodynamics
• Change in Internal energy
DE=q+w
Surroundings
Surroundings
Energy
System
System
DE<0
DE>0
q is -, exothermic, heat out
q is +, endothermic, heat into system
w is -, system does work
on surroundings
w is +, if surroundings do work
on system
10.5 Energy Changes
Objective:
1) To understand how heat is measured
10.5 Energy Changes
Calorie: (metric system) the amount of
energy (heat) required to raise the
temperature of one gram of water by one
Celsius degree.
In food, kilocalorie. C
The joule (SI unit) can be most conveniently
defined in terms of the calorie.
1 calorie=4.184 joule
1 cal=4.184 J
10.5 Energy Changes
Express 60.1 cal of energy in units of joules.
60.1 cal x 4.184 J = 251 J
1 cal
Calculating energy requirements
Determine the amount of energy(heat) in joules
required to raise the temperature of 7.40 g
water from 29.0o to 46.0o
4.184 J x 7.40 g
goC
x 17.0 C = 526 J
10.5 Energy Changes
Another important factor is IDENTITY OF SUBSTANCE.
Specific heat capacity: amount of energy required to
change the temperature of one gram of a substance by
one Celsius degree.
Q=s x m x DT
Q= energy (heat) required
s=specific heat capacity
m= mass of the sample in grams
DT=change in temperature in Celsius degrees.
Table 10.1 p. 297
10.5 Energy Changes
A 1.6 g sample of a metal that has the
appearance of gold requires 5.8 J of
energy to change its temperature from 23
C to 41 C. Is the metal pure gold.
10.5 Energy Changes
5.8= s x 1.6 x (41-23)
5.8
=s
s=0.20 J/g C
1.6 x 18
pure gold s=0.13 J/g C
Section 10.6 Thermochemistry
(Enthalpy)
Objective: To consider the heat (enthalpy) of
chemical reactions.
Section 10.6 Thermochemistry
(Enthalpy)
Enthalpy: a special
energy function (DH)
Under constant
pressure, the change
in
enthalpy is equal to the
energy that flows as
heat.
DHp=heat
Section 10.6 Thermochemistry
(Enthalpy)
When 1 mole of
methane(CH4) is
burned at constant
pressure, 890kJ of
energy is released as
heat. Calculate DH
for a process in which
a 5.8 g sample of
methane is burned at
constant pressure.
qp=DH= -890kJ/mol CH4
5.8 g CH4 x 1mol=
16.0 g
Section 10.6 Thermochemistry
(Enthalpy)
When 1 mole of sulfur
dioxide reacts with
excess oxygen to form
sulfur trioxide at constant
pressure, 198.2kJ of
energy is released as
heat. Calculate DH for a
process in which a 12.8 g
sample of sulfur dioxide
reacts with oxygen at
constant pressure.
39.6kJ
Section 10.6 Thermochemistry
(Enthalpy)
Calorimeter is a device used to determine
the heat associated with a chemical
reaction.
Figure 10.6: A coffee-cup
calorimeter.
NCSSM Distance Learning T.I.G.E.R. - Chemistry page 4#thermo
10.7 Hess’s Law
• Objective: To understand Hess’s law
10.7 Hess’s Law
• Enthalpy is a state function
• Independent of path soooo
– The change in enthalpy is the same whether
the reaction takes place in one step or in a
series of steps. (Hess’s Law)
– Important: because it allows us to calculate
heats of reaction that might be difficult or
inconvenient to measure directly in a
calorimeter.
10.7 Hess’s Law
N2(g) + O2(g) 2NO2(g) DH=68kJ
N2(g) + O2(g) 2NO (g) DH2=180kJ
2NO (g) + O2(g) 2NO2 DH3= -112kJ
N2(g) + O2(g)
2NO2(g) DH=68kJ
10.7 Hess’s Law
Characteristics of Enthalpy Changes
1) If a reaction is reversed, the sign of DH is
also reversed.
2) The magnitude of DH 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 DH
is multiplied by the same integer.
10.7 Hess’s Law
Given the following data
4CuO(s)
2Cu2O(s) + O2(g) DH=288kJ
Cu2O(s)
Cu(s) + CuO(s)
DH=11 kJ
Calculate DH for the following reaction
2Cu(s) + O2(g)
2CuO(s)
Cu(s) + CuO(s)
2 Cu2O + O2
Cu2O DH= -(11kJ)
4CuO(s) DH=-(288kJ)
2(Cu(s) + CuO(s) Cu2O DH= -2(11kJ)
2 Cu2O + O2
4CuO(s) DH=-(288kJ)
2Cu(s) + O2(g)
2CuO(s)
DH= -310kJ
10.10 Energy as a Driving Force
Objective: To understand energy as a driving
force for natural processes.
10.10 Energy as a Driving Force
Energy spread: in a given process,
concentrated energy is dispersed widely.
This distributions happens every time an
exothermic process occurs.
Matter spread: molecules of a substance
spread out and occupy a larger volume.
10.10 Energy as a Driving Force
Entropy: natural tendency of the universe to
become disordered.
S (as things become more disordered), the
value of S increases
Second law of thermodynamics: The entropy
of the universe is always increasing.
Putting it all together!
Gibbs Free Energy
G=H-TDS
Temperature in Kelvin
Change in Free Energy
DG=DH-TDS (constant temperature)
Related to system
If DG is negative (spontaneous reaction), DS (+)
Calculations
• Let’s predict the spontaneity of the melting
of ice
H2O(s)
H2O(l)
DH =6.03 x 103 J/mol and DS=22.1 J/K mol
What is DG at -10o, 0o and 10oC?
Which is spontaneous?
DG=2.2x 102
DG=0
DG= -2.2 x 102
More calculations
Br2(l)
Br2(g)
DH=31.0 kJ/mol and DS=93.0 J/K mol
What is the normal boiling point?
DG= DH- T DS
0= DH- T DS
T= DH/D S=333K
Figure 10.7: Energy sources
used in the United States.
Figure 10.8: The earth’s
atmosphere.
Figure 10.9: The atmospheric CO2
concentration over the past 1000
years.
Figure 10.10: Comparing the
entropies of ice and steam.
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