Transcript No Slide Title

Changes of Energy - 1
• Change of energy depends only on initial and final states
– In general:
E  E1  E2  q  w
– Constant volume:
E  qv
• Pressure-volume work only
• No electrical work, etc.
T2
– From definition of heat capacity:

dq 
• C 
dT 

E  q v  Cv dT
T1
Energy vs. Enthalpy
• Define a new state function, enthalpy:
– Why?
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H  E  PV
H  E  PV 
Think about doing an experiment in which you measure heat capacity
It is convenient to do this at constant pressure (bench top experiment)
You add certain amount of heat, qp, and measure T
For solids and liquids, your measurement of Cp is similar to what you measure
when the same experiment is carried out at constant volume (Cv)
For solids and liquids, H = E and Cp = Cv
HOWEVER, when working with a gas, you get different results.
You add some heat, qp, raise the temperature, T
But the gas expands, V = nRT/P
E  q p
The system did work on the surroundings: w = -P V
Temperature does not go up as much as we expected, lost energy through work
We just add this lost work to the energy to get the enthalpy


Energy vs. Enthalpy
– At constant pressure
– At constant volume
H  C p T  q p
– What is the difference
E  H  w p
E  CV T  qV
  PV


 Cv  C p T
 qv  qp
Measuring Enthalpy
Remember H E + PV
Since both E and PV are state properties, H must be a state
property. It is therefore an exact differential!
Thus basic calculus applies:
dH= dE + d(PV)= dE + PdV + V dP
Since for reversible pressure-volume systems,
dE= dq + dw= dq - PdV
dH= dq - VdP
So for constant volume pressure processes:
H= qP
Another Example
This summation property allows us to use information seemingly only
tangentially related to our reaction of interest to get to an end point.
We want to know:
3 O2(g,1atm) + 2 glycine (s)  1 urea (s) + 3 CO2 (g, 1atm) + 3 H2O (l)
But we only have:
3 O2(g,1atm) + 2 glycine (s)  4 CO2(g,1atm) + 2 H2O (l) + 2 NH3(g)
H= -1163.5 kJ/mol
and
H2O (l) + urea (s)  CO2 (g) + 2 NH3 (g)
H= 133 kJ/mol
Substracting these last two equations gives the desired result:
3 O2(g,1atm) + 2 glycine (s)  4 CO2(g,1atm) + 2 H2O (l) + 2 NH3(g)
H2O (l) + urea (s)  CO2 (g) + 2 NH3 (g)
=
3 O2(g,1atm) + 2 glycine (s)  1 urea (s) + 3 CO2 (g, 1atm) + 3 H2O (l)
H= -1163.5 kJ/mol - 133 kJ/mol = -1296.8 kJ/mol
mol= mole of reaction!
Note: To get a biologically reasonable reaction:
glycine(s) + H2O (l)  glycine (aq)
Standard Enthalpies
• We are only ever concerned with H
• But we use H to calculate, what is the value of H
– “Actual energy” from E = mc2
• Impractical and unnecessary
– Arbitrarily agree on a zero
– The convention is to assign zero enthalpy to all elements in their
most stable states at 1atm pressure.
– The standard enthalpy of a compound is defined to be the
enthalpy of formation of 1 mol of compund at 1atm pressure from
its elements in their standard states.
– Although temperature is not specified in the defenition of standard
enthalpy, usually use 25 °C = 298 K
Standard Enthalpies
Enthalpy of reactants and products can be calculated from their heats
of formation.
2 C(s) + 3 H2(g) + 1/2 O2 (g)  C2H5OH (l)
H°298=0 is the most stable state at standard T and P.
From the table: H°298,f= -276.98 kJ/mol
But this is not a clean reaction in practice. How do we measure it?
Clean Reactions
We sum up the enthalpies of “clean” reactions.
C(graphite)
H2(g)
+
O2(g)
+ 1/2 O2(g)
2 CO2(g)
+ 3


H2O(l) 
CO2
H2O
C2H5OH(l)
+ 3O2 (g)
Bond Energies
So we can calculate the heats of
formation for many compounds and
the reaction enthalpies for many
transformations.
From these we can “back-out”
average values for how much energy
it takes to break a bond.
HH
| |
H-C-C-O-H
| |
H H
Bond
C-C
C=C
CC
C-H
C-N
C-O
C=0
C-S
N-H
O-O
O-H
S-H
H2
N2
O2
C(graphite)
D(kJ/mol)
344
615
812
415
292
350
725
259
391
143
463
339
436
945.4
498.3
716.7
Errors with Bond Energies
C(graphite) + H2(g) + O2(g) 
C(graphite)

H2(g)

O2(g)

C(g)+ 2H(g) + 2O(g)
C=O + C-O + C-H + O-H
Calculated Hf
Measured Hf
OH
|
H-C=O
C(g)
2H(g)
2O(g)
716.7 kJ/mol
436.0 kJ/mol
498.3 kJ/mol
1651 kJ/mol
1953 kJ/mol
-302 kJ/mol
-423.76 kJ/mol
Complex Systems
100W ~ 9000 kJ/day
1 g protein/carbohydrate=
15kJ= 3.6 Cal
1g fat= 35kJ= 8.4 Cal
System
Surroundings
Carnot Cycle
State 1
Thot
q=0,
w=CVT
w= -nRThotln(V2/V1)
q=-w
isothermal
adiabatic
adiabatic
State 4
Tcold
State 2
Thot
isothermal
w= -nRTcoldln(V4/V3)
q=-w
q=0,
w=-CVT
State 3
Tcold
Carnot Cycle, Schematic View
Thot
E
Tcold
The engine operates between two reservoirs to and from which heat can be
transferred.
We put heat into the system from the hot reservoir and heat is expelled into the cold
reservoir.
Questions about Thermodynamic Cycles
How much of the heat put in
at high temperature can be
converted to work?
Can two engines with the
same temperature difference
drive one another?
What does entropy have to
do with it?
Clausius
b. Jan. 2, 1822, Prussia
d. Aug. 24, 1888, Bonn
"Heat cannot of itself
pass from a colder to a
hotter body."
Carnot Cycle: Step 1
State 1
Thot
isothermal
adiabatic
adiabatic
isothermal
State 4
Tcold
State 2
Thot
State 3
Tcold
Thot
Heat converted to work!
Isothermal reversible Expansion:
E=0
Energy of an ideal gas depends only on temperature
V2

w   PdV  nRT ln(
V1
V2
)
V1
q= -w
Carnot Cycle: Step 2
State 1
Thot
isothermal
adiabatic
adiabatic
isothermal
State 4
Tcold
State 2
Thot
State 3
Tcold
Energy lost to expansion
Isothermal reversible Expansion:
q=0
No heat transferred in adiabatic process
w  E  CV T  PV
CV dT   PdV  
Tcol d

Cv
Thot
V3

nRT
dV
V
dT
dV
 nR
T
V
V2
Cv ln(
Tcold
V
)  nRln 3
Thot
V2
Carnot Cycle: Steps 3 & 4
State 1
Thot
isothermal
State 2
Thot
adiabatic
adiabatic
isothermal
State 4
Tcold
State 3
Tcold
Energy lost to expansion
Isothermal reversible Expansion:
q=0
No heat transferred in adiabatic process
w  E  CV T  PV
CV dT   PdV  
Tcol d

Cv
Thot
V3

nRT
dV
V
dT
dV
 nR
T
V
V2
Cv ln(
Tcold
V
)  nRln 3
Thot
V2
Carnot Cycle: Summary
Step1
Step2
Step3
Step4
w
-nRThotln(V2/V1)
-CV T
-nRTcoldln(V4/V3)
CV T
q
-w
0
-w
0
E
0
w
0
w
Tcold
V3
Cv ln(
)  nR ln
Thot
V2
w1 + w2 + w3 + w4= -nRThotln(V2/V1) - nRTcoldln(V4/V3)
q1 + q2 + q3 + q4= nRThotln(V2/V1) + nRTcoldln(V4/V3)
q = -w
Carnot Cycle: PV Diagram
-nRThotln(V2/V1)
Pressure
1
4 CV T
2
3
-nRTcoldln(V4/V3)
Volume
-CV T