1. Chapter 16 Spontaneity, Entropy, and Free Energy
Why do some things happen and others don’t?

2. 16.1 Spontaneous Processes and Entropy
Entropy and the Second Law of Thermodynamics
The Effect of Temperature on Spontaneity
16.4 Free Energy
16.5 Entropy Changes in Chemical Reactions
16.6 Free Energy and Chemical Reactions
16.7 The Dependence of Free Energy on Pressure
16.8 Free Energy and Equilibrium
16.9 Free Energy and Work
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3. To compare kinetics with thermodynamics To define spontaneity
Objectives
To compare kinetics with thermodynamics
To define spontaneity
To answer why some things happen and others don’t.
To define Entropy
To assign positive or negative sign depending on the process
To understand why a diverging diamond interchange is safer than other designs
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4. Thermodynamics vs. Kinetics
Domain of Kinetics
Rate of a reaction depends on the pathway from reactants to products.
Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products.
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5. Spontaneous Processes and Entropy
Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process.
A spontaneous process is one that occurs without outside intervention.
Examples??
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7. What should happen to the gas when you open the valve?
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.
What should happen to the gas when you open the valve?
The gas should spread evenly throughout the two bulbs.
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8. The Expansion of An Ideal Gas Into an Evacuated Bulb
What is the driving force for the process? Why does it happen?
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9. Entropy (S) is a measure of molecular randomness or disorder.
The driving force for a spontaneous process is an increase in the entropy (S) of the universe.
Entropy (S) is a measure of molecular randomness or disorder.
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10. Is a deck of cards usually ordered or disordered?
Entropy
Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state.
Nature spontaneously proceeds toward the states that have the highest probabilities of existing.
Is a deck of cards usually ordered or disordered?
There are so many ways for a deck of cards to be disordered, but
Only one way for it to be ordered…
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11. The Microstates That Give a Particular Arrangement (State)
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12. Therefore: Ssolid < Sliquid << Sgas Why do solutions form?
Positional Entropy
A gas expands into a vacuum because the expanded state has the highest positional probability of states available to the system.
Therefore: Ssolid < Sliquid << Sgas
Why do solutions form?
Why do ice cubes get smaller, even though they are in the freezer?
If a process moves in a direction of greater disorder or higher entropy, the sign is positive.
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13. Predict the sign of S for each of the following, and explain:
Concept Check
Predict the sign of S for each of the following, and explain:
The evaporation of alcohol
The freezing of water
Compressing an ideal gas at constant temperature
Heating an ideal gas at constant pressure
Dissolving NaCl in water + –
a) + (a liquid is turning into a gas)
b) - (more order in a solid than a liquid)
c) - (the volume of the container is decreasing)
d) + (the volume of the container is increasing)
e) + (there is less order as the salt dissociates and spreads throughout the water)
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14. To compare kinetics with thermodynamics To define spontaneity
Objectives Review
To compare kinetics with thermodynamics
To define spontaneity
To answer why some things happen and others don’t.
To define Entropy
To assign positive or negative sign depending on the process
Why is a diverging diamond safer? #12
Work Session Page 782 # 1, 19, 23 (think IGL and volume), 25, 33, 35
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15. To define the Second Law of Thermodynamics
Objectives
To define the Second Law of Thermodynamics
Wait, what was the first? I’m constantly forgetting it…
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16. Second Law of Thermodynamics
In any spontaneous process there is always an increase in the entropy of the universe.
What is the sign of a spontaneous process?
The entropy of the universe is increasing.
The total energy of the universe is constant, but the entropy is increasing.
Suniverse = ΔSsystem + ΔSsurroundings
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17. Second Law of Thermodynamics
Suniverse = ΔSsystem + ΔSsurroundings
How can life exist if it requires cells to take small molecules to make bigger ones?
Is this a spontaneous process?
Does S increase?
Is this consistent with the 2nd Law of Thermodynamics?
For a spontaneous process, how are ΔSsys & ΔSsurr related?
For a spontaneous process, how are ΔSsys & ΔSsurr related?
They have to combine such that entropy of the universe is increased.
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18. To define the Second Law of Thermodynamics 1st Law of Thermodynamics:
Objectives Review
To define the Second Law of Thermodynamics
1st Law of Thermodynamics:
The amount of energy in the Universe is constant
Work Session #8
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19. To predict spontaneity of a process based on ΔS and ΔH
Objectives
To predict spontaneity of a process based on ΔS and ΔH
To understand the effect of temperature on Spontaneity
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20. Which side is favored by Entropy?
H2O (l) H2O (g) water is the system…
18 ml L @1atm, 100C
Positional Probability?
What is the sign of ΔSsystem?
What direction is the heat flow? ΔHsystem?
…what about the surroundings?
What is the sign of ΔSsurr? Heat flow? ΔHsurr?
What is the overall ΔS of the universe?
ΔSuniv = ΔSsys + ΔSsurr
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21. Concept Check
For the process A(l) A(s), which direction involves an increase in energy randomness(-ΔH)? Positional randomness(+ΔS)? Explain your answer.
As temperature increases/decreases , does energy or positional randomness take precedence? Why?
At what temperature is there a balance between energy randomness and positional randomness?
Since energy is required to melt a solid, the reaction as written is exothermic. Thus, energy randomness favors the right (product; solid). Since a liquid has less order than a solid, positional randomness favors the left (reactant; liquid).
As temperature increases, positional randomness is favored (at higher temperatures the fact that energy is released becomes less important). As temperature decreases, energy randomness is favored.
There is a balance at the melting point.
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22. The magnitude of ΔSsurr depends on the temperature.
ΔSsurr = - ΔH T
The sign of ΔSsurr depends on the direction of the heat flow. –ΔH, +ΔSsurr
The magnitude of ΔSsurr depends on the temperature.
Why does ice freeze, but only at lower temperatures? Balance between energy randomness and positional randomness.
Liquid  Solid
At high temps, ΔSsurr = - ΔH is very small…as T gets smaller, exothermicity becomes a stronger driving force T
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23. ΔSsurr = - ΔH enthalpy determines the sign T
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24. ΔSsurr = - ΔH temperature determines the magnitude T
Heat flow (constant P) = change in enthalpy = ΔH
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25. Interplay of Ssys and Ssurr in Determining the Sign of Suniv ΔSuniv = ΔSsys + ΔSsurr
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26. To predict spontaneity of a process based on ΔS and ΔH
Objectives Review
To predict spontaneity of a process based on ΔS and ΔH
To understand the effect of temperature on Spontaneity
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27. To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH
Objectives
To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH
To predict spontaneity based on Gibbs Free Energy
To conceptually and mathematically use Gibbs Equation
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28. ΔG = ΔH – TΔS (at constant T and P) Negative ΔG means positive ΔSuniv
Free Energy (G)
ΔG = ΔH – TΔS (at constant T and P)
Negative ΔG means positive ΔSuniv
positive ΔSuniv means Spontaneous!
Negative ΔG means Spontaneous!
ΔG < 0 means spontaneous
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29. How to determine spontaneity based on ΔS and ΔH …
For a process to be considered spontaneous, it must be both :
exothermic (favor energy randomness (-ΔH) and
Favor positional randomness (entropy (+ΔS))
ΔG = ΔH – TΔS
If the opposite of the above conditions exist, we can conclude that the process is not spontaneous.
Since energy is required to melt a solid, the reaction as written is exothermic. Thus, energy randomness favors the right (product; solid). Since a liquid has less order than a solid, positional randomness favors the left (reactant; liquid).
As temperature increases, positional randomness is favored (at higher temperatures the fact that energy is released becomes less important). As temperature decreases, energy randomness is favored.
There is a balance at the melting point.
Spontaneous
Not Spontaneous
Don’t Know ΔH - + ΔS
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30. Describe the following as
spontaneous/non-spontaneous/cannot tell, and explain.
A reaction that is:
Exothermic and becomes more positionally random
Spontaneous
Exothermic and becomes less positionally random
Cannot tell
Endothermic and becomes more positionally random
Endothermic and becomes less positionally random
Not spontaneous
Explain how temperature affects your answers.
a) Spontaneous (both driving forces are favorable). An example is the combustion of a hydrocarbon.
b) Cannot tell (exothermic is favorable, positional randomness is not). An example is the freezing of water, which becomes spontaneous as the temperature of water is decreased.
c) Cannot tell (positional randomness is favorable, endothermic is not). An example is the vaporization of water, which becomes spontaneous as the temperature of water is increased..
d) Not spontaneous (both driving forces are unfavorable).
Questions "a" and "d" are not affected by temperature. Choices "b" and "c" are explained above.
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31. Effect of ΔH and Δ S on Spontaneity ΔG = ΔH – TΔS
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32. A liquid is vaporized at its boiling point. Predict the signs of: w q
Concept Check
A liquid is vaporized at its boiling point. Predict the signs of: w q H S
Ssurr G
Explain your answers. – +
As a liquid goes to vapor, it does work on the surroundings (expansion occurs). Heat is required for this process. Thus, w = negative; q = H = positive. S = positive (a gas is more disordered than a liquid), and Ssurr = negative (heat comes from the surroundings to the system); G = 0 because the system is at its boiling point and therefore at equilibrium.
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33. Predict the spontaneity of: H2O(s)  H2O(l)
Exercise
Predict the spontaneity of: H2O(s)  H2O(l)
Given: ΔH = 6.03 X 103 J/mol ΔS = 22.1 J/K*mol
5-Step with Gibbs!
ΔG = ΔH – TΔS
Try at -10 oC, 0 oC, and 10 oC
-10 oC, ΔG = +2.2 X 102 J/mol
0 oC, ΔG = 0 J/mol
10 oC, ΔG = -2.2 X 102 J/mol
S = 132 J/K·mol
Ssurr = -132 J/k·mol
G = 0 kJ/mol
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34. Exercise
The value of Hvaporization of substance X is 45.7 J/mol, and its normal boiling point is 72.5°C.
Calculate S, Ssurr, and G for the vaporization of one mole of this substance at 72.5°C and 1 atm.
ΔSsurr = - ΔH T
S = 132 J/K·mol
Ssurr = -132 J/K·mol
G = 0 kJ/mol
S = 132 J/K·mol @ point, no other S is occuring
Ssurr = -132 J/k·mol
G = 0 kJ/mol
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35. Find the normal boiling point of liquid Br2. Br2(l)  Br2(g)
Exercise
Find the normal boiling point of liquid Br2.
Br2(l)  Br2(g)
ΔH = 31.0 KJ/mol ΔS = 93.0 J/K*mol
5-Step with Gibbs!
ΔG = ΔH – TΔS
S = 132 J/K·mol
Ssurr = -132 J/k·mol
G = 0 kJ/mol
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36. Spontaneous Reactions
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37. To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH
Objectives Review
To understand Gibbs Free Energy (ΔG) in relation to ΔS and ΔH
To predict spontaneity based on Gibbs Free Energy
To conceptually and mathematically use Gibbs Equation
Work Session: 27, 29, 30, 31
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38. To predict the sign of ΔS for a given chemical reaction.
Objectives
To predict the sign of ΔS for a given chemical reaction.
To calculate ΔS°reaction given ΔS° values for products and reactants
To calculate ΔS°reaction , ΔH°reaction , ΔG°reaction
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39. ΔS changes in Chemical Reactions
Entropy changes in chemical reactions are determined by positional probability.
This means that the more molecules, the more possible configurations, the more disorder.
Predict the sign of ΔS for the following reaction:
H2  2H
N2(g) + 3H2(g)  2NH3(g)
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
CaCO3(s)  CaO(s) + CO2(g)
2SOS(g) + O2(g)  2SO3(g)
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40. Concept Check
Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The potential energy of AB is much greater than that of either reactant.
Predict the signs of:
H Ssurr Suniv
Explain.
Since the POTENTIAL energy of the products is greater than the average bond energies of the reactants, the reaction is endothermic as written.
Thus, the sign of H is positive; Ssurr = -H/T, so S is negative, but close to zero (cannot tell for sure); and Suniv is positive.
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41. Third Law of Thermodynamics…there is a 3rd Law??
So far, we’ve discussed the relative change in entropy, ΔS. Can we assign an absolute S?
The entropy of a perfect crystal at 0 K is zero.
A perfect crystal represents the lowest possible entropy.
Or in other words, everything is perfectly ordered, in its place, just plain ducky….
Therefore, from 0 K, as the temperature increases, so does the randomness of vibrational motion and, therefore, the entropy of the substance.
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42. Third Law of Thermodynamics…there is a 3rd Law??
Oh, yeah, …
Since a perfect crystal at absolute zero is an unattainable ideal, we take the Third Law as a standard although we have never actually observed it…….
You are welcome….
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43. Standard Enthalpy Values (H°)
Represent the increase in enthalpy of formation of a compound.
ΔH°reaction = ΣnpH°products – ΣnrH°reactants
n = moles (balanced coefficient)
H° = standard enthalpy (from a given list in problem # or on pages A20-A21)
See formula on AP reference sheet.
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44. Standard Entropy Values (S°)
Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure.
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
n = moles (balanced coefficient)
S° = standard entropy (from a given list in problem # or on pages A20-A21)
See formula on AP reference sheet.
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45. Calculate Δ S° for the following reaction:
Exercise
Calculate Δ S° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information:
S° (J/K·mol)
Na(s)
H2O(l)
NaOH(aq)
H2(g)
Δ S°= –11 J/K
[2(50) + 131] – [2(51) + 2(70)] = –11 J/K
ΔS°= –11 J/K
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46. Calculate Δ S° for the following reaction:
Exercise
Calculate Δ S° for the following reaction:
2NiS(s) + 3O2(g) → 2SO2(g) + 2NiO(s)
Given the following information:
S° (J/K·mol)
SO2(g) 248
NiO(s) 38
O2(g)
NiS(s) 53
Δ S°= –149 J/K
[2(50) + 131] – [2(51) + 2(70)] = –11 J/K
ΔS°= –11 J/K
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47. Calculate Δ S° for the following reaction:
Exercise
Calculate Δ S° for the following reaction:
Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g)
Given the following information:
S° (J/K·mol)
Al2O3(s) 51
H2(g) 131
Al(s) 28
H2O(g) 189
Δ S°= 179 J/K
[2(50) + 131] – [2(51) + 2(70)] = –11 J/K
ΔS°= –11 J/K
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48. Standard Free Energy Change (ΔG°)
The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.
ΔG° = ΔH° – TΔS°
ΔG°reaction = ΣnpG°products – ΣnrG°reactants
ΔH°reaction = ΣnpH°products – ΣnrH°reactants
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
A20-A21 for ΔS°, ΔH°, ΔG° data…
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49. A stable diatomic molecule spontaneously forms from its atoms.
Concept Check
A stable diatomic molecule spontaneously forms from its atoms.
Predict the signs of:
H° S° G°
Explain.
The reaction is exothermic, more ordered, and spontaneous.
Thus, the sign of H is negative; S is negative; and G is negative.
– – –
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50. H° S° G° Predict the sign of: Explain
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51. To predict the sign of ΔS for a given chemical reaction.
Objectives
To predict the sign of ΔS for a given chemical reaction.
To calculate ΔS°reaction given ΔS°values for products and reactants
To calculate ΔS°reaction , ΔH°reaction , ΔG°reaction
Work Session # 33, 34, 37, 39, 40, 41, 43, 45, 47, 51 (like Hess’s Law!) A20-A21 for ΔS°, ΔH°, ΔG° data…
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52. Consider the following system at equilibrium at 25°C.
Concept Check
Consider the following system at equilibrium at 25°C.
PCl3(g) + Cl2(g) PCl5(g)
G° = −92.50 kJ
What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain.
The ratio will decrease.
S is negative (unfavorable) yet the reaction is spontaneous (G is negative). Thus, H must be negative (exothermic, favorable). Thus, as the temperature is increased, the reaction proceeds to the left, decreasing the ratio of partial pressure of PCl5 to the partial pressure of PCl3.
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53. Free Energy and Pressure
G = G° + RT ln(P) or
ΔG = ΔG° + RT ln(Q)
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54. ln(K) vs. 1/T (for both endothermic and exothermic cases)
Concept Check
Sketch graphs of:
G vs. P
H vs. P
ln(K) vs. 1/T (for both endothermic and exothermic cases)
G vs. P: A natural log graph (levels off as P increases).
H vs. P: no relationship (slope of zero).
lnK vs 1/T: endothermic - straight line, negative slope
exothermic - straight line, positive slope
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55. The Meaning of ΔG for a Chemical Reaction
A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.
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56. The equilibrium point occurs at the lowest value of free energy available to the reaction system.
ΔG = 0 = ΔG° + RT ln(K)
ΔG° = –RT ln(K)
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57. Change in Free Energy to Reach Equilibrium
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58. Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction
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59. Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy.
wmax = ΔG
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60. All real processes are irreversible.
Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.
All real processes are irreversible.
First law: You can’t win, you can only break even.
Second law: You can’t break even.
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61. Calculate H, E, q, and w for the process you described above.
Concept Check
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.
Calculate H, E, q, and w for the process you described above.
All are equal to zero.
All are equal to zero. Since it is a constant temperature process, H = 0 and E = 0. The gas is working against zero pressure (evacuated bulb) so w = 0. E = q + w, so q = 0.
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62. Concept Check
Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant.
b) Given your answer to part a, what is the driving force for the process?
Entropy
Some students are probably aware of the concept of entropy. This is a good introduction to the concept.
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63. The Microstates That Give a Particular Arrangement (State)
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64. Negative ΔG means positive ΔSuniv.
Free Energy (G)
A process (at constant T and P) is spontaneous in the direction in which the free energy decreases.
Negative ΔG means positive ΔSuniv.
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