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Introduction to Entropy
by Mike Roller
Entropy (S)= a measure of randomness or disorder
MATTER IS ENERGY.
ENERGY IS INFORMATION.
EVERYTHING IS INFORMATION.
PHYSICS SAYS THAT
STRUCTURES...
BUILDINGS, SOCIETIES,
IDEOLOGIES... WILL SEEK
THEIR POINT OF LEAST
ENERGY.
THIS MEANS THAT
THINGS FALL.
THEY FALL FROM HEIGHTS
OF ENERGY AND
STRUCTURED INFORMATION
INTO MEANINGLESS,
POWERLESS DISORDER.
THIS IS CALLED
ENTROPY.
Entropy: Time’s Arrow
In any spontaneous process, the entropy of the universe increases.
ΔSuniverse > 0
Another version of the 2nd Law:Energy spontaneously spreads out if it has no outside resistance
Entropy measures the spontaneous dispersal of energy as a function of temperature
How much energy is spread out How widely spread out it becomesEntropy change = “energy dispersed”/T
Second Law of Thermodynamicsoccurs without outside intervention
Entropy of the Universe
ΔSuniverse = ΔSsystem + ΔSsurroundings
Positional disorder Energetic disorder
ΔSuniverse > 0 spontaneous process
Both ΔSsys and ΔSsurr positive
Both ΔSsys and ΔSsurr negative
ΔSsys negative, ΔSsurr positive
ΔSsys positive, ΔSsurr negative
spontaneous process.
nonspontaneous process.
depends
depends
Entropy of the Surroundings(Energetic Disorder)
SystemHeat Entropy
Surroundings
SystemHeat Entropy
Surroundings
T
ΔHΔS sys
surr Low T large entropy change (surroundings)
High T small entropy change (surroundings)
ΔHsys < 0
ΔHsys > 0
ΔSsurr > 0
ΔSsurr < 0
Positional Disorder and Probability
Probability of 1 particle in left bulb = ½
" 2 particles both in left bulb = (½)(½) = ¼
" 3 particles all in left bulb = (½)(½)(½) = 1/8
" 4 " all " = (½)(½)(½)(½) = 1/16
" 10 " all " = (½)10 = 1/1024
" 20 " all " = (½)20 = 1/1048576
" a mole of " all " = (½)6.021023
The arrangement with the greatest entropy is the one with the highest probability (most “spread out”).
Ssolid < Sliquid << Sgas
Entropy of the System: Positional Disorder
Ludwig Boltzmann Orderedstates
Disorderedstates
Low probability(few ways)
High probability(many ways)
Low S
High S
Ssystem Positional disorder
S increases with increasing # of possible positions
Ludwig Boltzmann
The Third Law:
The entropy of a perfect crystal at 0 K is zero.
Everything in its place
No molecular motion
The Third Law of Thermodynamics
Entropy Curve
Solid GasLiquid
S(qrev/T)(J/K)
Temperature (K)0
0
fusion
vaporization
S° (absolute entropy) can be calculated for any substance
Entropy Increases with...• Melting (fusion) Sliquid > Ssolid
ΔHfusion/Tfusion = ΔSfusion
• Vaporization Sgas > Sliquid
ΔHvaporization/Tvaporization = ΔSvaporization
• Increasing ngas in a reaction
• Heating ST2 > ST1 if T2 > T1
• Dissolving (usually) Ssolution > (Ssolvent + Ssolute)
• Molecular complexity more bonds, more entropy
• Atomic complexity more e-, protons, neutrons
Recap: Characteristics of Entropy
• S is a state function
• S is extensive (more stuff, more entropy)
• At 0 K, S = 0 (we can know absolute entropy)
• S > 0 for elements and compounds in their standard states
• ΔS°rxn = nS°products - nS°reactants
• Raise T increase S
• Increase ngas increase S
• More complex systems larger S
Entropy and Gibbs Free Energy
by Mike Roller
Entropy (S) Review• ΔSuniverse > 0 for spontaneous processes
• ΔSuniverse = ΔSsystem + ΔSsurroundings
positional
energetic
• We can know the absolute entropy value for a substance
• S° values for elements & compounds in their standard states are tabulated (Appendix C, p. 1019)
• For any chemical reaction, we can calculate ΔS°rxn:
• ΔS°rxn = S°(products) - S°(reactants)
ΔSuniverse and Chemical Reactions
ΔSuniverse = ΔSsystem + ΔSsurroundings
For a system of reactants and products,
ΔSuniverse = ΔSrxn – ΔHrxn/T
• If ΔSuniverse > 0, the reaction is spontaneous
• If ΔSuniverse < 0, the reaction is not spontaneous
– The reverse reaction is spontaneous
• If ΔSuniverse = 0, the reaction is at equilibrium
– Neither the forward nor the reverse reaction is favored
C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)
CompoundC6H12O6(s)
O2(g)
CO2(g)
H2O(g)
ΔH°f (kJ/mol)-1275
0-393.5-242
S° (J/mol K)212205214189
ΔSuniverse = ΔSrxn – ΔHrxn/TΔS°rxn = S°(products) - S°(reactants)
= [6 S°(CO2(g)) + 6 S°(H2O(g))] – [S°(C6H12O6(s)) + 6 S°(O2(g))]
= [6(214) + 6(189)] – [(212) + 6(205)] J/K
ΔS°rxn = 976 J/K
ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)
= [6 ΔH°f(CO2(g)) + 6 ΔH°f(H2O(g))] – [ΔH°f(C6H12O6(s)) + 6 ΔH°f(O2(g))]
= [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ
ΔH°rxn = -2538 kJ
C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)
CompoundC6H12O6(s)
O2(g)
CO2(g)
H2O(g)
ΔH°f (kJ/mol)-1275
0-393.5-242
S° (J/mol K)212205214189
ΔSuniverse = ΔSrxn – ΔHrxn/T
ΔS°rxn = 976 J/K (per mole of glucose)
ΔH°rxn = -2538 kJ (per mole of glucose)
At 298 K,
ΔS°universe = 0.976 kJ/K – (-2538 kJ/298 K)
ΔS°universe = 9.5 kJ/K
–ΔG means +ΔSuniv
A process (at constant T, P) is spontaneous if free energy decreases
Gibbs Free Energy (G)
Josiah Gibbs
G = H – TSAt constant temperature,
ΔG = ΔH – TΔS(system’s point of view)
ΔG = ΔH – TΔS
Divide both sides by –T
-ΔG/T = -ΔH/T + ΔS
ΔSuniverse = ΔS – ΔH/T
ΔG and Chemical Reactions
ΔG = ΔH – TΔS• If ΔG < 0, the reaction is spontaneous
• If ΔG > 0, the reaction is not spontaneous
– The reverse reaction is spontaneous
• If ΔG = 0, the reaction is at equilibrium
– Neither the forward nor the reverse reaction is favored
• ΔG is an extensive state function
Ba(OH)2(s) + 2NH4Cl(s) BaCl2(s) + 2NH3(g) + 2 H2O(l)
ΔH°rxn = 50.0 kJ (per mole Ba(OH)2)
ΔS°rxn = 328 J/K (per mole Ba(OH)2)
ΔG = ΔH - TΔS
ΔG° = 50.0 kJ – 298 K(0.328 kJ/K)
ΔG° = – 47.7 kJ Spontaneous
At what T does the reaction stop being spontaneous?
The T where ΔG = 0.
ΔG = 0 = 50.0 kJ – T(0.328 J/K)
50.0 kJ = T(0.328 J/K)
T = 152 K not spontaneous below 152 K
Effect of ΔH and ΔS on Spontaneity
ΔH
–
+
–
+
ΔS
+
+
–
–
Spontaneous?
Spontaneous at all temps
Spontaneous at high temps• Reverse reaction spontaneous at low temps
Spontaneous at low temps• Reverse reaction spontaneous at high temps
Not spontaneous at any temp
ΔG = ΔH – TΔSΔG negative spontaneous reaction
1. ΔG° = ΔG°f(products) - ΔG°f(reactants)
• ΔG°f = free energy change when forming 1 mole of
compound from elements in their standard states
2. ΔG° = ΔH° - TΔS°
3. ΔG° can be calculated by combining ΔG° values for several reactions
• Just like with ΔH° and Hess’s Law
Ways to Calculate ΔG°rxn
2H2(g) + O2(g) 2 H2O(g)
1. ΔG° = ΔG°f(products) - ΔG°f(reactants)
ΔG°f(O2(g)) = 0
ΔG°f(H2(g)) = 0
ΔG°f(H2O(g)) = -229 kJ/mol
ΔG° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ
2. ΔG° = ΔH° - TΔS°ΔH° = -484 kJ
ΔS° = -89 J/K
ΔG° = -484 kJ – 298 K(-0.089 kJ/K) = -457 kJ
2H2(g) + O2(g) 2 H2O(g)
3. ΔG° = combination of ΔG° from other reactions (like Hess’s Law)
2H2O(l) 2H2(g) + O2(g) ΔG°1 = 475 kJ
H2O(l) H2O(g) ΔG°2 = 8 kJ
ΔG° = - ΔG°1 + 2(ΔG°2)
ΔG° = -475 kJ + 16 kJ = -459 kJ
Method 1: -458 kJ
Method 2: -457 kJ
Method 3: -459 kJ
What is Free Energy, Really?
• NOT just “another form of energy”
• Free Energy is the energy available to do useful work
• If ΔG is negative, the system can do work (wmax = ΔG)
• If ΔG is positive, then ΔG is the work required to make the process happen
– Example: Photosynthesis
– 6 CO2 + 6 H2O C6H12O6 + 6 O2
– ΔG = 2870 kJ/mol of glucose at 25°C
– 2870 kJ of work is required to photosynthesize 1 mole of glucose