Download Introduction to Thermodynamics I. Conservation of Energy

Introduction to Thermodynamics
I.
Conservation of Energy
A.
First Law of Thermodynamics = energy cannot be created or destroyed, only
converted between different forms
1. Example: CH4(g) + 2O2(g)
CO2(g) + 2H2O + energy
a) Reaction gives off energy as heat
b) Potential energy stored in chemical bonds is lowered
c) Total energy is unchanged
2.
Uses and Shortcomings
a) Lets us keep track of energy flow in processes
b) Does not tell us if or why a given process occurs
c) Does not tell us direction of a chemical reaction
II.
Entropy
A.
The “Heat Tax”
1. Conversion of energy between forms is inefficient:
2. Usually, some energy is lost as heat
3. The fewer energy conversion, the better
B. Direction of Processes and Reactions
1) Examples:
a) Ball at the top of a hill
Ball at the bottom of a hill
b) Steel + H2O + O2
Rust
c) Gas in one part of a container
Gas filling a container
d) Ice at 5 oC
Water at 5 oC
2) What is in common?
a) Exothermic? Not ice melting or gas expanding
b) Increased Disorder = Increased Entropy = +DS
3) Entropy = S = driving force of spontaneous reactions = disorder or random
a) Probability (likelihood): there are many ways for objects/molecules to
be disordered, but only a few to be ordered
b) Nature proceeds towards the most likely state
= state with greatest number of energetically
equivalent arrangements
4) Expansion of a Gas
a) Possible arrangements of 4 gas molecules in a 2-bulb system
b) Microstates = possible configurations of a particular arrangement
Microstates
ABCD
---
B
ABC
ABD
ACD
BCD
D
C
B
A
AB
CD
AC
BD
AD
BC
CD
AB
BD
AC
BC
AD
Order
1 microstate
Ordered
4 microstates
Somewhat
Disordered
6 microstates
Fully disordered
c) Entropy selects most likely arrangement = 2 molecules in each bulb
5) Probabilities
Probability of all molecules in left bulb
Number of Molecules
Probability
1
½
2
½x½=¼
3
½ x ½ x ½ = 1/8
5
1/25 = 1/32
10
1/210 = 1/1024
n
1/2n
6.022 x 1023
1 x 10-23
6) Positional Entropy = entropy depending on configuration in space
a) Changes in state depend on positional entropy
b) Ssolid < Sliquid << Sgas
c) Larger volume allows many more available positions for particles
d) Dissolving a solid provides more volume for particles to occupy
e) Example: What has highest positional entropy
i) 1 mole solid CO2 or 1 mole of gaseous CO2?
ii) 1 mole N2 at 1 atm or 1 mole of N2 at 0.01 atm?
f) Example: Predict the sign of the entropy change for
i) Dissolving solid sugar into water
ii) Iodine vapor condensing to crystals on a surface
7) Second Law of Thermodynamics = in any spontaneous process, there is
always an increase in the entropy of the universe
a) Energy is conserved = constant
b) Entropy is always increasing
c) DSuniverse = DSsystem + DSsurroundings
d) For a given process: if DSuniverse = + the process is spontaneous
if DSuniverse = - the process is not spontaneous
e) Life = constant battle against entropy
i) Large molecules are assembled from smaller ones
ii) Organizing a cell is DSsystem = - the process is not spontaneous
iii) Fortunately, it is DSuniverse that must be positive in a process
iii) DSsurroundings = large + for life to occur
III. Temperature and Spontaneity
A.
Change in state: 1 mol = 18 ml H2O(l)
1 mol = 31 L H2O(g)
1) DSsurrounding depends on flow of heat into or out of the system
a) Heat increases the motion (randomness) of particles
b) Exothermic reactions release heat to surroundings DSsurrounding = +
c) Endothermic reactions absorb heat from surroundings DSsurrounding = d) Vaporization of water is endothermic DSsurrounding = 2)
B.
DSuniverse = DSsystem + DSsurrounding = (+) + (-) = +/- ?
a) Depends on the temperature
b) If T > 100 oC, DSuniverse = +
If T < 100 oC, DSuniverse = -
Temperature Effects
1) DSsurroundings depends on heat flow
a) Exothermic reactions usually favors spontaneity
b) Spontaneity usually lowers the energy of the starting material as it
becomes product
c) The difference of these energies = heat released to surroundings
2) Importance of Exothermicity of DSuniverse depends on Temperature
a) Adding heat to hot surroundings has little effect
b) Adding heat to cold surroundings has a large effect
c) Heat transfer is more important at low temperatures
3) In Summary
a) Sign of DSsurr depends on direction of heat transfer
b) Magnitude of DSsurr depends on T
c)
ΔSsurr
Heat(J)
ΔH


Temperatur e(K)
T
d) The (-) is there because DH is for the system, which is opposite of
DH of the surroundings
4) Example: Find DSsurr at 25 oC
a) Sb2S3(s) + 3Fe(s)
2Sb(s) + 3FeS(s) DH = -125 kJ/mol
b) Sb4O6(s) + 6C(s)
4Sb(s) + 6CO(g) DH = +778 kJ/mol
IV. Free Energy
A.
Free Energy = G = H – TS
1) DGprocess = DH – TDS
ΔG
2) Divide by –T
3)
ΔG
DSuniv  
T
ΔH


 ΔS  ΔSsurr  ΔS  ΔSuniv
T
T
4) A process is spontaneous if DG = 5) Chemists use DG rather than DS because we only need to know system
B. Example: Predicting Spontaneity using DG
1) H2O(s)
H2O(l) DHo = 6030 J/mol, DSo = 22.1 J/K mol
T
T
DHo
DSo
DSsurr
oC
K
J/mol
J/K•mol
-10
263
6030
22.1
-22.9
0
273
6030
22.1
10
283
6030
22.1
DSuniv
TDSo
DGo
J/mol
J/mol
-0.8
5810
+220
-22.1
0.0
6030
0
-21.3
0.8
6250
-220
J/K•mol J/K • mol
2) Classifying Processes/Reactions based on DH and DS
ΔG  ΔH - TDS
3) Example: At what T is Br2(l)
Br2(g) spontaneous (1 atm) given that
DH = 31.0 kJ/mol and DS = 93.0 J/Kmol
a) Spontaneous when DG = b) Set DG = 0 and solve for T
0  DH - TDS 
DH 31000J/mol
T

 333K
DS
93J/Kmol
c) When T > 333K, TDS > DH and DG = - (Entropy controlled)
d) When T < 333K, TDS < DH and DG = + (Enthalpy controlled)
e) 333K is the boiling point of Br2(l)