Transcript Thermodynamics

Thermodynamics
• First Law – energy can
never be created nor
destroyed. (from CH 9)
• We want to be able to
predict spontaneity:
under given conditions,
occurs without outside
intervention
• Different than fast or
slow!
Enthalpy to Predict Spontaneity?
•
•
•
•
•
•
C (s) + O2 (g)  CO2 (g) ΔHo = -393.5 kJ
This process is exothermic and spontaneous.
Most exothermic processes are spontaneous.
H2O (s)  H2O (l)
ΔHo = 6.01 kJ
This process is endothermic and spontaneous.
Dissolving most ionic compounds in water is
endothermic and spontaneous.
• However, many (most?) endothermic
processes are not spontaneous.
• Enthalpy alone will not predict spontaneity.
Spontaneous
Non-spontaneous
DH = 0
Entropy
• Entropy – S – How energy is distributed
among energy levels of particles in universe
• disorder of a system
• Consider placing cards in order
• If you have 2 cards, there is a 1 in 2 chance
they will be in order. (2!)
• If you have 13 cards (1 suit), there is a 1 in
6 x 109 chance they will be in order. (13!)
• If you have 52 cards, there is a 1 in 8 x 1067
chance they will be in order. (52!)
Probability and Microstates
Room of unnamed
S117 student
More than disorder: Which would require more
energy starting with a empty room: to put everything
in order or to make it look exactly like this?
Probability and Microstates
• Positional
Probability
– Gas expansion
– Phase change
– solubility
Entropy and Positional Probability
• ΔS = Sf – Si
• Sgas >> Sliquid > Ssolid
• S increases when a solute (solid or liquid) is
dissolved.
• S decreases when a solute (gas) is dissolved.
• S increases as temperature increases.
Processes Which Increase Entropy
Change in Entropy
• For these systems, predict the sign of the
change in entropy (based positional prob.)
– N2 (g) + 3H2 (g)  2NH3 (g)
– AgCl (s)  Ag+ (aq) + Cl- (aq)
– Unfolded protein  folded protein
Isothermal Expansion of Ideal Gas
• DE = 0
• q = -w
• Work and heat are
path dependent
– One step: free
expansion
– One step: against
mass of 1/4m
– Two step, etc
Work obtained from Expansion
What is work for 1 step expansion against m = 1/4mi?
Answer: -3/4 PDV
2 steps
6 steps
W = -PDV
W = -1.4PDV
𝑞𝑟𝑒𝑣
𝑉2
= −𝑤𝑟𝑒𝑣 = 𝑛𝑅𝑇 𝑙𝑛
𝑉1
Work Required for Compression
Draw PV diagram for 1 step against M, then 2
step against M/2 followed by M
P
P
V
V
W = 3PDV
W = 2PDV
W = 1.4PDV
Reversible Process
• Reversible process:
infinite number of steps
(100 % efficient)
• One step: 3PV - .75PV to
get back to same place
– Surroundings gain
2.25PV of work in
form of heat
– Ordered energy into
disordered energy
Real Process
• More work into
system than you
Expansion:
can get out
• But no change in
universe, so that
Compression
means more heat
into surroundings
• Cycle: increase
entropy of universe
heat
heat
Entropy
Entropy is statistical
DS = k ln
𝜔2
𝜔1
Entropy is macroscopic
DS = nR ln
DS =
k = R/Na
𝑞𝑟𝑒𝑣
𝑇
From 𝑞𝑟𝑒𝑣 = 𝑛𝑅𝑇 𝑙𝑛
𝑉2
𝑉1
𝑉2
𝑉1
Change of State:
Equilibrium Process
DS =
DS =
DS =
𝑞𝑟𝑒𝑣
𝑇
𝐻𝑓𝑢𝑠𝑖𝑜𝑛
𝑇𝑚𝑝
𝐻𝑣𝑎𝑝𝑜𝑟
𝑇𝑏𝑝
Second Law
• Second Law – In any spontaneous process,
the entropy of the universe increases.
• ΔSuniv = ΔSsystem + ΔSsurroundings > 0
• ΔSuniv > 0 , forward reaction is spontaneous
• ΔSuniv < 0 , reverse reaction is spontaneous
• ΔSuniv = 0 , equilibrium
Temperature Dependence
• ΔSuniv = ΔSsystem + ΔSsurroundings
• Entropy change in the surroundings
primarily due to heat flow
• ΔSuniv = ΔSsystem + ΔHsurroundings/T
Exothermicity is more
important at low
temperature. Explain this
conceptually.
Exothermicity is a
major driving force
Spontaneity of Ice Melting: An Endothermic Process
J W Gibbs
• Gibbs thought it would be easier to examine
only the properties of the system to predict
spontaneity. (Rather than worrying about
the entire universe!)
• He rewrote the 2nd Law
ΔSsurr = qsurr / T
At constant T and P, qsurr = -ΔHsys
ΔSsurr = -ΔHsys / T
Gibbs
called
this term
“free
energy”
ΔSuniv = ΔSsys + ΔSsurr
ΔSuniv = ΔSsys +[ -ΔHsys / T]
-TΔSuniv = ΔHsys - TΔSsys
ΔG = ΔHsys - TΔSsys
ΔG = ΔHsys - TΔSsys
ΔG = -TΔSuniv
ΔSuniv ΔG
+
0
< 0 Forward Spontaneous
> 0 Reverse Spontaneous
= 0 Equilibrium
ΔG < 0
Moves Right
ΔG > 0
0
Moves Left
Third Law
• Unlike H or E, the absolute entropy of a
compound can be measured.
• Third Law – the entropy of a pure crystalline
substance @ 0 K is 0. (all motion stops)
• S = q / T (J/K)
• Appendix 3 lists So
• These can be used to calculate the ΔSo for a
reaction.
• ΔSorxn = Σ ΔSo (products) - Σ ΔSo (reactants)
Conceptual Understanding of
absolute entropy
• What is the cause of the
trends you see?
• Why isn’t entropy of
aluminum = zero, like
heat of formation?
Substance
So (J/mol K)
Al2O3 (s)
51
H2 (g)
131
Al (s)
28
H2O (g)
189
Standard Free Energy of
Formation
• Standard Free Energy of Formation – ΔGof
– The free energy change for the formation
of 1 mole of a substance in the standard
state.
• Can calculate ΔGof
ΔGo = ΣΔGof (products) - ΣΔGof (reactants)
• Can calculate from ΔHo and ΔSo
ΔGorxn= ΔHorxn – TΔSorxn
Can calculate from Hess’s Law
Example Problem
•HCl (g) + NH3 (g)  NH4Cl (s)
•Find ΔGo @ 298 K
ΔHof (kJ/mol)
So (J/K mol) ΔGof (kJ/mol)
HCl (g)
-92.06
186.9
-95.3
NH3 (g)
-46.11
192.5
-16.5
NH4Cl (s)
-314.43
94.6
-202.9
Example Problem
• 2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s)
• Find ΔGo at 298K
ΔHof (kJ/mol)
Sof (J/K)
Al (s)
0
28.33
Fe2O3 (s)
-824.2
87.40
Al2O3 (s)
-1675.7
50.92
Fe (s)
0
27.28
Example Problem
• 2 SO2 (g) + O2 (g)  2 SO3 (g)
• Find ΔGo at 298 K and 1500 K
ΔHof (kJ/mol)
Sof (J/K)
O2 (g)
0
205.03
SO2 (g)
-296.83
248.11
SO3 (g)
-395.72
256.65
What assumption is made in the second problem?
Free Energy and Keq
• ΔGo is at standard state (P = 1 atm, conc = 1
M, 298 K)
• At non-standard conditions,
• ΔG = ΔGo + RT lnQ
• R = 8.314 J/(molK) = 8.314 x 10-3 kJ/(molK)
Example Problem
• 2 SO2 (g) + O2 (g)  2 SO3 (g)
• ΔGo at 298 K is –141.77 kJ
• What is ΔG at 298 K if PSO2 = 0.50 atm, PO2
= 1.0 atm and PSO3 = 0.10 atm ?
Relationship of Free Energy and
Equilibrium
• ΔG = ΔGo + RT lnQ
• At equilibrium
– Q = Keq
– ΔG = 0
• 0 = ΔGo + RT lnKeq
• ΔGo = - RT lnKeq
o /RT
-ΔG
• Keq = e
Determination of Equilibrium
Constant
• 2 SO2 (g) + O2 (g)  2 SO3 (g)
• ΔGo at 298 K is –141.77 kJ
• What is Keq at 298K for this
reaction?
Answer: Keq = 7.1 x 1024
Rule of Thumb:
What change in
free energy will
lead to a 10-fold
change in
equilibrium?
Free Energy Diagrams
• “Uphill” and “downhill”
• Condition dependent
• Hydrolysis of ATP has
DGo = -30.5kJ/mol
• Does the reaction have
more potential for energy
release at phyisological
conditions of
[ATP]=3.5mM, [ADP] =
1.8 mM, [Pi] = 5.0mM
Application
• Glutamine is an important carrier of
nitrogen in metabolism. It can be made
from glutamate and ammonia, a reaction
with a standard free energy of 14 kJ at 25
oC. What is the equilibrium constant for
this reaction? What is the equilibrium
constant for this reaction if it is coupled to
ATP hydrolysis?