Transcript H - Bruder Chemistry

Chapter 6
Energy
Thermodynamics
1
Energy is...
The ability to do work.
 Conserved.
 made of heat and work.
 a state function.
 independent of the path, or how you get
from point A to B.
 Work is a force acting over a distance.
 Heat is energy transferred between
objects because of temperature difference.

2
Potential Energy
Potential energy is energy an object possesses
by virtue of its position or chemical
composition.
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3
Kinetic Energy
Kinetic energy is energy an object possesses
by virtue of its motion.
1
KE =  mv2
2
4
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Conversion of Energy


5
Energy can be converted from one type to
another.
For example, the cyclist above has potential
energy as she sits on top of the hill.
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Conversion of Energy
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6
As she coasts down the hill, her potential
energy is converted to kinetic energy.
At the bottom, all the potential energy she had
at the top of the hill is now kinetic energy.
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Units of Energy
The SI unit of energy is the joule (J).
kg m2
1 J = 1 
s2
 An older, non-SI unit is still in
widespread use: the calorie (cal).
1 cal = 4.184 J
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The universe
is divided into two halves.
 the system and the surroundings.
 The system is the part you are
concerned with.
 The surroundings are the rest.
 Exothermic reactions release energy to
the surroundings.
 Endo thermic reactions absorb energy
from the surroundings.
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Definitions:
System and Surroundings
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The system includes the
molecules we want to
study (here, the hydrogen
and oxygen molecules).
The surroundings are
everything else (here, the
cylinder and piston).
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Definitions: Work

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10
Energy used to
move an object over
some distance is
work.
w=Fd
where w is work, F
is the force, and d is
the distance over
which the force is
exerted.
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Heat

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11
Energy can also be
transferred as heat.
Heat flows from
warmer objects to
cooler objects.
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Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
12
CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
N 2 + O 2 + heat  2NO
Potential energy
2NO
13
Heat
N2 + O2
Same rules for heat and work
Heat given off is negative.
 Heat absorbed is positive.
 Work done by system on surroundings
is negative.
 Work done on system by surroundings
is positive.
 Thermodynamics- The study of energy
and the changes it undergoes.

14
What is work?
Work is a force acting over a distance.
 w= F x Dd
 P = F/ area
 d = V/area
 w= (P x area) x D (V/area)= PDV
 Work can be calculated by multiplying
pressure by the change in volume at
constant pressure.
 units of liter - atm L-atm
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15
Work needs a sign
If the volume of a gas increases, the
system has done work on the
surroundings.
 work is negative
 w = - PDV
 Expanding work is negative.
 Contracting, surroundings do work on
the system w is positive.
 1 L atm = 101.3 J
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Examples
What amount of work is done when 15
L of gas is expanded to 25 L at 2.4 atm
pressure?
 If 2.36 J of heat are absorbed by the gas
above. what is the change in energy?
 How much heat would it take to change
the gas without changing the internal
energy of the gas?
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Surroundings
System
Energy
DE <0
18
Surroundings
System
Energy
DE >0
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Direction
Every energy measurement has three
parts.
1. A unit ( Joules of calories).
2. A number how many.
3. and a sign to tell direction.
 negative - exothermic
 positive- endothermic
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First Law of Thermodynamics

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Energy is neither created nor destroyed.
In other words, the total energy of the universe is
a constant; if the system loses energy, it must be
gained by the surroundings, and vice versa.
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First Law of Thermodynamics
The energy of the universe is constant.
 Law of conservation of energy.
 q = heat
 w = work
 DE = q + w
 Take the systems point of view to
decide signs.
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Internal Energy
The internal energy of a system is the sum of all
kinetic and potential energies of all components
of the system; we call it E.
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Internal Energy
By definition, the change in internal energy, DE,
is the final energy of the system minus the initial
energy of the system:
DE = Efinal − Einitial
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Changes in Internal Energy

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When energy is
exchanged between
the system and the
surroundings, it is
exchanged as either
heat (q) or work (w).
That is, DE = q + w.
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DE, q, w, and Their Signs
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Exchange of Heat between
System and Surroundings
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When heat is absorbed by the system from
the surroundings, the process is endothermic.
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Exchange of Heat between
System and Surroundings
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When heat is absorbed by the system from
the surroundings, the process is endothermic.
When heat is released by the system into the
surroundings, the process is exothermic.
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State Functions

However, we do know that the internal energy
of a system is independent of the path by
which the system achieved that state.
– In the system below, the water could have reached room
temperature from either direction.
29
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State Functions
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Therefore, internal energy is a state function.
It depends only on the present state of the
system, not on the path by which the system
arrived at that state.
And so, DE depends only on Einitial and Efinal.
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State Functions


However, q and w are
not state functions.
Whether the battery is
shorted out or is
discharged by running
the fan, its DE is the
same.
– But q and w are different
in the two cases.
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Enthalpy
Symbol is H
 Change in enthalpy is DH
 delta H
 If heat is released the heat content of the
products is lower
 DH is negative (exothermic)
 If heat is absorbed the heat content of
the products is higher
 DH is positive (endothermic)
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Enthalpy
If a process takes place at constant
pressure (as the majority of processes we
study do) and the only work done is this
pressure-volume work, we can account
for heat flow during the process by
measuring the enthalpy of the system.
 Enthalpy is the internal energy plus the
product of pressure and volume:

H = E + PV
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Enthalpy
abbreviated H
 H = E + PV (that’s the definition)
 at constant pressure.
 DH = DE + PDV


the heat at constant pressure qp can be
calculated from
34

DE = qp + w = qp - PDV

qp = DE + P DV = DH
Enthalpy
Since DE = q + w and w = -PDV, we can
substitute these into the enthalpy
expression:
DH = DE + PDV
DH = (q+w) − w
DH = q
 So, at constant pressure, the change in
enthalpy is the heat gained or lost.
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Endothermicity and
Exothermicity
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A process is
endothermic when
DH is positive.
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Endothermicity and
Exothermicity
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A process is
endothermic when
DH is positive.
A process is
exothermic when
DH is negative.
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Enthalpy of Reaction
The change in
enthalpy, DH, is the
enthalpy of the
products minus the
enthalpy of the
reactants:
DH = Hproducts − Hreactants
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Enthalpy of Reaction
This quantity, DH, is called the enthalpy of
reaction, or the heat of reaction.
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The Truth about Enthalpy
1.
2.
3.
40
Enthalpy is an extensive property.
DH for a reaction in the forward
direction is equal in size, but opposite
in sign, to DH for the reverse reaction.
DH for a reaction depends on the state
of the products and the state of the
reactants.
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Calorimetry
Measuring heat.
 Use a calorimeter.
 Two kinds
 Constant pressure calorimeter (called a
coffee cup calorimeter)
 heat capacity for a material, C is
calculated
 C= heat absorbed/ DT = DH/ DT
 specific heat capacity = C/mass

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Calorimetry
molar heat capacity = C/moles
 heat = specific heat x m x DT
 heat = molar heat x moles x DT
 Make the units work and you’ve done
the problem right.
 A coffee cup calorimeter measures DH.
 An insulated cup, full of water.
 The specific heat of water is 1 cal/gºC
 Heat of reaction= DH = sh x mass x DT
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Heat Capacity and Specific Heat
The amount of energy required to raise the
temperature of a substance by 1 K (1C) is its
heat capacity.
43
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Heat Capacity and Specific Heat
We define specific heat capacity (or simply
specific heat) as the amount of energy
required to raise the temperature of 1 g of a
substance by 1 K.
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Heat Capacity and Specific Heat
Specific heat, then, is
Specific heat =
s=
45
heat transferred
mass  temperature change
q
m  DT
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Examples
The specific heat of graphite is 0.71
J/gºC. Calculate the energy needed to
raise the temperature of 75 kg of
graphite from 294 K to 348 K.
 A 46.2 g sample of copper is heated to
95.4ºC and then placed in a calorimeter
containing 75.0 g of water at 19.6ºC. The
final temperature of both the water and
the copper is 21.8ºC. What is the specific
heat of copper?
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Calorimetry
Constant volume calorimeter is called a
bomb calorimeter.
 Material is put in a container with pure
oxygen. Wires are used to start the
combustion. The container is put into a
container of water.
 The heat capacity of the calorimeter is
known and tested.
 Since DV = 0, PDV = 0, DE = q
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Constant Pressure Calorimetry
By carrying out a
reaction in aqueous
solution in a simple
calorimeter such as this
one, one can indirectly
measure the heat change
for the system by
measuring the heat
change for the water in
the calorimeter.
48
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Constant Pressure Calorimetry
Because the specific heat
for water is well known
(4.184 J/g-K), we can
measure DH for the
reaction with this
equation:
q = m  s  DT
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Bomb Calorimeter
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
thermometer

stirrer

full of water

ignition wire

Steel bomb

sample
Properties
intensive properties not related to the
amount of substance.
 density, specific heat, temperature.
 Extensive property - does depend on
the amount of stuff.
 Heat capacity, mass, heat from a
reaction.

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Hess’s Law
Enthalpy is a state function.
 It is independent of the path.
 We can add equations to to come up
with the desired final product, and add
the DH
 Two rules
 If the reaction is reversed the sign of DH
is changed
 If the reaction is multiplied, so is DH
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Hess’s Law
Hess’s law states that
“[i]f a reaction is
carried out in a series
of steps, DH for the
overall reaction will
be equal to the sum of
the enthalpy changes
for the individual
steps.”
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Hess’s Law
Because DH is a state
function, the total
enthalpy change
depends only on the
initial state of the
reactants and the final
state of the products.
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H (kJ)
O2 NO2
-112 kJ
180 kJ
N2 2O2
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NO2
68 kJ
Standard Enthalpy
The enthalpy change for a reaction at
standard conditions (25ºC, 1 atm , 1 M
solutions)
 Symbol DHº
 When using Hess’s Law, work by
adding the equations up to make it look
like the answer.
 The other parts will cancel out.
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Standard Enthalpies of Formation
Standard enthalpies of formation, DHf°, are
measured under standard conditions (25 °C
and 1.00 atm pressure).
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Calculation of DH
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
DH = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ
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Example
Given
5
C 2 H 2 (g) + O 2 (g)  2CO 2 (g) + H 2 O( l)
2
DHº= -1300. kJ
C(s) + O 2 (g)  CO 2 (g)
DHº= -394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l)
2
DHº= -286 kJ

calculate DHº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
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Example
Given
O 2 (g) + H 2 (g)  2OH(g) DHº= +77.9kJ
O 2 (g)  2O(g) DHº= +495 kJ
H 2 (g)  2H(g) DHº= +435.9kJ
Calculate DHº for this reaction
O(g) + H(g)  OH(g)
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Standard Enthalpies of Formation
Hess’s Law is much more useful if you
know lots of reactions.
 Made a table of standard heats of
formation. The amount of heat needed
to for 1 mole of a compound from its
elements in their standard states.
 Standard states are 1 atm, 1M and 25ºC
 For an element it is 0
 There is a table in Appendix 4 (pg A22)

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Standard Enthalpies of Formation
Need to be able to write the equations.
 What is the equation for the formation
of NO2 ?
 ½N2 (g) + O2 (g)  NO2 (g)
 Have to make one mole to meet the
definition.
 Write the equation for the formation of
methanol CH3OH.
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Since we can manipulate the
equations
We can use heats of formation to figure
out the heat of reaction.
 Lets do it with this equation.
 C2H5OH +3O2(g)  2CO2 + 3H2O
 which leads us to this rule.

63
Since we can manipulate the
equations
We can use heats of formation to figure
out the heat of reaction.
 Lets do it with this equation.
 C2H5OH +3O2(g)  2CO2 + 3H2O
 which leads us to this rule.

( DH of products) - ( DH of reactants) = DH o
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