Transcript Plan for Wed, 15 Oct 08

Plan for Wed, 15 Oct 08
• Lecture
– The nature of energy (6.1)
– Enthalpy and calorimetry (6.2)
Thermodynamics
• The study of energy and its interconversion.
– Energy is the capacity to do work or to produce heat
– Energy is neither created nor destroyed…it is only converted from
one form to another.
• Some examples of energy interconversion...
– sunlight: radiant (electromagnetic) energy...plants harvest sunlight
and convert it to sugars (chemical energy)
– heat: EM radiation is converted to kinetic energy when it “excites”
the motion of molecules
– food: the sugars (chemical energy content) from plants are
metabolized by animals (different chemical energy content),
allowing them to move around and stay alive (kinetic energy)
– gas: long-dead plants and animals (chemical energy) are distilled
and refined and delivered to your car where they are burned up
and converted to kinetic energy and heat.
Energy: Kinetic vs. Potential
•
Potential Energy (PE) – the energy
an object has by virtue of its
placement in a field of force, like
gravity.
–
•
PE = mgh, where m = mass, g =
acceleration due to gravity, h = height
Ball A has higher PE than ball B.
Kinetic Energy (KE) – the energy an
object has by virtue of its motion
–
KE = ½ mv2, where m = mass, v =
velocity
ball A’s PE is converted to KE when
it rolls down the hill...
ball A transfers its KE to ball B by
doing work
it also loses some of its PE to
frictional heating of the hill.
Energy = KE + PE
• Total Energy is the sum of kinetic
energy and potential energy.
• Energy is readily converted
between these two forms.
• If the system of interest is isolated
(no exchange with surroundings),
then total energy of the system is
constant.
Example: Mass on a Spring
• Initial PE = ½ kx2
• At x = 0:
PE
– PE = 0
– KE = ½ mv2 = ½ kx2
x
• Units of Energy
Joule = kg.m2/s2
• Example:
x=0
– Init. PE = 10 J
– M = 10 kg
– Vmax = [2(PE)/M]1/2 = 1.4m/s
Molecular Potential Energy
Molecules are subject to more forces than just gravity.
(and Chemical Bonds)
Petrucci, Fig. 7.9
Molecular Kinetic Energy
How do molecules
move around?
Translation: motion through space
Rotation: motion about the
center of mass
Vibration: motion directed
through chemical bonds
Petrucci, Fig. 7.9
Internal Energy: E = Ek + Ep
Molecular Kinetic Energy Molecular Potential Energy
(and chemical bonds)
Chemical reactions and phase changes involve
changes in both the kinetic and potential energy
of molecules.
Petrucci, Fig. 7.9
Conservation of Energy
First Law of Thermodynamics:
Total energy of the universe is
constant.
• To help keep track of energy
flow let’s define...
• System: that one part of the
universe you are interested in
(i.e., you define it).
• Surroundings: everything
else in the universe.
Surroundings
System
q transfer
w transfer
• Energy gained by the
system must be lost by the
surroundings.
• Energy exchange can be
in the form of heat (q),
work (w), or both.
Heat vs. Work
• Heat: a transfer of energy between two objects due to a temperature
gradient (heat is measured in J).
• Work: a force acting over a distance to move an object (work is
measured in J).
What do these have to do with molecular KE and PE?
• We can think of KE and PE as an accounting of the energy that
exists in a system…
• Heat and work are the modes by which a system can exchange
energy with its surroundings.
By measuring the heat absorbed or evolved, and the
work performed on or by the system, we can obtain
information about how the internal energy of the
system changes during a process.
Heat Exchange: Exothermic
Energy
Water @ 80° C
Einitial
Water @ 20° C
q
• Exothermic Process:
evolution of the system that
results in heat transfer to
the surroundings.
Efinal
Efinal < Einitial
• q < 0 (heat is lost)
Another Example of Exothermic
Heat Exchange: Endothermic
Energy
Water @ 80° C
Efinal
Water @ 20° C
• Endothermic Process:
evolution of the system
that results in heat
transfer to the system.
q
Einitial
Efinal > Einitial
• q > 0 (heat is gained)
Another Example of Endothermic
Defining Work
• In chemical processes, a
common type of work is the
expansion of gas against a
constant external pressure
• Let’s isolate this system; no
heat exchange with the
surroundings:
q=0
• System does “work” on the
surroundings by pushing back
the atmosphere:
w<0
Defining Work (cont)
Work: energy transferred as a
force applied over a distance.
• How much work does the
system do?
• Pext = force/area
• |w| = force x distance
= Pext x A x Dh
= Pext DV
• w = - Pext DV (note sign)
Heat and Work
q = heat > 0
w = work < 0
System gains energy as
heat (q), causing the gas
to expand.
The expanding gas
exerts force on the
piston, causing it to
move, doing work (w)
on the surroundings.
System loses energy as
work done on surr.
w=Fxd
= -PDV
P = Psurr
DV of sys!!!
system
Heat and Work in a Chemical Reaction
Zn (s) + 2HCl (aq)  Zn2+(aq) + H2 (g) + 2H2O + Cl-(aq)
System loses
energy as heat.
Production of
H2(g) causes
piston to move...
system loses
energy as work
done on
surroundings.
Conservation of Energy Revisited
DE = q + w
• If system gives heat • If system does work
q < 0 (q is negative)
w < 0 (w is negative)
• If system gets heat
• If work done on system
q > 0 (q is positive)
w > 0 (w is positive)
Zumdahl, Ex 6.3
• A hot-air balloon is inflated
from 4 x 106 L to 4.5 x 106
L by the addition of 1.3 x
108 J of heat. If the
balloon expands against
an external pressure of 1
atm, what is DE for this
process?
• First, define the system:
the balloon.
DE = q + w
= (1.3 x 108 J) + (-PDV)
= (1.3 x 108 J) + (-1 atm (Vfinal - Vinit))
= (1.3 x 108 J) + (-0.5 x 106 L.atm)
Conversion: 101.3 J per L.atm
 (-0.5 x 106 L.atm) x (101.3 J/L.atm) = -5.1 x 107 J
DE = (1.3 x 108 J) + (-5.1 x 107 J)
= 8 x 107 J
In English: the system gained more energy through
heat than it lost doing work. Therefore, the overall
energy of the system has increased.
Energy and Sign Convention

Energy
Einitial
Eout
Efinal
DE < 0
Energy
Efinal
Einitial
DE > 0
Ein
• If system loses
energy:
Efinal < Einitial
Efinal-Einitial = DE < 0.
• If system gains
energy:
Efinal > Einitial
Efinal-Einitial = DE > 0.
Suppose that you have two identical 500-mL bottles
of water sitting on your desk at equilibrium.
Bottle 1 has been kept at 25°C since bottling.
Bottle 2 came from the same spring, but has been
frozen, thawed, transported by air in an
unpressurized compartment, and allowed to fluctuate
wildly in temperature before being at equilibrium on
your desk.
Which bottle has the higher internal energy?
State Functions
• A State Function is a
function in which the
value only depends on
the initial and final
state….NOT on the
pathway taken.
Los Alamos
• In this example, start in
Seattle, end in Denver,
but you take different
paths to get from one
place to the other.
U as a state fxn
DU = q + w
Combustion of gas in a bomb
calorimeter:
DV = 0  w = -PDV = 0
All the energy produced in
the combustion rxn is
evolved as heat.
Combustion of gas in your
car:
DV > 0  w = -PDV < 0
Some of the energy
produced is evolved as heat,
and some is evolved as work
done on the surroundings.
BUT IN BOTH CASES DU IS
THE SAME.
Measuring DE
• On an earlier slide, I mentioned that by
measuring:
– the heat (q) exchanged between a system and its
surroundings
– the work (w) performed on or by a system
we could determine how the internal energy of a
system changes during a process.
• How exactly do we go about measuring q and w?
• It is kind of a pain, actually, to measure w, so let’s
concentrate on ways to measure q…
Enthalpy
Enthalpy is defined as:
H  E  PV
Consider a process carried out at constant P (w = PDV):
DE  qP  w  qP - PDV
 qP  DE  PDV
Note similarity to our definition of enthalpy:
 qP  DE  PDV  DH
q p  DH
What is this telling us?
We can track the heat flow
in a process occurring at
constant P and that will give
us direct info about DE.
Heat Capacity
Heat Capacity (C): the energy required
to raise the temp. of a sample of a
substance by 1oC; the ability of a
substance to absorb heat.
Specific heat (s): the energy required
to raise the temperature of 1 gram of
a substance by 1oC.
If we know the specific heat of a
substance, the mass, and the
temperature change, we can
determine the heat flow.
qrxn  s  m  DT
Constant P Calorimetry
NaOH + HCl  H2O + NaCl; DH = -58 kJ
The heat evolved in this reaction is trapped
in the water…this heat increases the
average Ek of the water molecules, leading
to an increase in the temperature of the
water.
qsystem = -qsurroundings
Formally, q is the amount of heat that must be
exchanged with the surroundings to return the
system to its original temperature.
In calorimetry we don’t let it escape…q goes
into raising the temperature of the water.