1. Why All Things Happen … Zumdahl’s Chapter 16
Spontaneity, Entropy, Free Energy, and
Why All Things Happen …
“The Universe Becomes Less Predictable”
Zumdahl’s Chapter 16

2. Spontaneous Processes and Entropy
One of the main objectives in studying thermodynamics, as far as chemists are concerned, is to be able to predict whether or not a reaction will occur when reactants are brought together under a special set of conditions (for example, at a certain temperature, pressure, and concentration).

3. Entropy is Probability
S = k ln W in modern symbolism.
W is an actual count of how many different ways the Universe could be arranged without being detectably different macroscopically.
And it is usually enormous!
For example, how many different poker hands might be in some player’s possession?
W  (52)(51)(50)(49)(48) / 5! or 2,598,960.
For 4 players, that’s ~1.481024 different games.
Over twice Avogadro’s Number!

4. Poker Microstates
One microstate in poker might be a flush; all cards of the same suit.
Wflush = 4(13)(12)(11)(10)(9) / 5! = 5148 as the number of ways to get a flush on the deal.
But Wflush / Wtotal gives ~505:1 odds against.
So flushes-on-the-deal are fairly ignorable.

5. Chemical Microstates Positional
In a solid, molecules are frozen in position.
But a liquid can swap molecular positions without macroscopic consequence: Sliq > Ssolid
A gas is far more chaotic: Sgas >> Sliquid!

6. 2nd Law of Thermodynamics
“In any spontaneous process, the entropy of the Universe increases.”
We must include consideration of a system’s environment to apply this law.
For example, condensing a gas implies a large decrease in the system’s entropy! Ssys << 0
Fortunately, the (latent) heat of vaporization gets released to force the surroundings to occupy higher energy levels, so Ssurr >> 0 and Suniv > 0!
Suniv = Ssys + Ssurr  0

7. haos Norse Mythology Valhalla is the abode of the Norse gods.
But, contrary to many other mythologies, Norse gods are not immortal.
Valhalla is held up by a giant tree, the roots of which are being gnawed by a serpent.
The serpent will succeed, and when it does, Valhalla and the Universe will fall.
The serpent’s name is
haos

8. Universal Chaos, Suniv The Norsemen were right!
There is Chaos growing in the Universe all the time at the expense of Order. It is now a fundamental principle of Science.
It’s called “entropy,” S, and is a state function that must always increase for the Universe as a whole, but some System’s S may decrease.

9. Entropy (S) is a measure of the randomness or disorder of a system.
DS = Sf - Si
If the change from initial to final results in an increase in randomness
Sf > Si
DS > 0
+DS; favors a spontaneous reaction for a chemical reaction
-DS; favors a non-spontaneous reaction for a chemical reaction or does not happen
18.2

10. Spontaneous Physical and Chemical Processes
A waterfall runs downhill
A lump of sugar dissolves in a cup of coffee
At 1 atm, water freezes below 0 0C and ice melts above 0 0C
Heat flows from a hotter object to a colder object
A gas expands in an evacuated bulb
Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2

11. spontaneous
nonspontaneous
18.2

12. Enthalpy is a factor in whether a reaction is spontaneous; but is not the only factor.
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = kJ
H+ (aq) + OH- (aq) H2O (l) DH0 = kJ
H2O (s) H2O (l) DH0 = 6.01 kJ
NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ
H2O
18.2

13. An exothermic reaction (-DH) favors a spontaneous reaction since products are at a lower potential energy But does not Guarantee it.
Reactants
-DH
Potential Energy
Products

14. Guidelines for Determining if DS is + or -.
1. For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s) H2O (l)
DS > 0

15. Guidelines for Determining if DS is + or -.
Entropy often increases when one material dissolves in another.
Entropy increases as Temperature increases.

16. Guidelines for Determining if DS is + or -.
4. When gases are produced (or consumed)
If a reaction produces more gas molecules than it consumes, DS0 > 0.
If the total number of gas molecules diminishes, DS0 < 0.
If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number.
What is the sign of the entropy change for the following reaction?
2Zn (s) + O2 (g) ZnO (s)
The total number of gas molecules goes down, DS is negative.
18.3

17. Guidelines for Determining if DS is + or -.
5. Entropy increases as number of moles of gases increases during chemical reaction.
N2 (g) + 3 H2 (g) -> 2 NH3 (g)
4 moles gas
2 moles gas
- DS
6. Entropy increases as the complexity of the molecule increases.
Higher entropy
C4H10 or
CH4

18. How does the entropy of a system change for each of the following processes?
(a) Condensing water vapor
Randomness decreases
Entropy decreases (DS < 0)
(b) Forming sucrose crystals from a supersaturated solution
Randomness decreases
Entropy decreases (DS < 0)
(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (DS > 0)
(d) Subliming dry ice
Randomness increases
Entropy increases (DS > 0)
18.2

20. Standard Entropy Values; S°
240

21. Thermodynamics
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
energy, enthalpy, pressure, volume, temperature
, entropy
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
6.7

22. Entropy Changes in the System (DSsys)
The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C.
rxn
aA + bB cC + dD
DS0
rxn
dS0(D)
cS0(C) = [ + ] -
bS0(B)
aS0(A)
DS0
rxn
nS0(products) = S
nS0(reactants) -
What is the standard entropy change for the following reaction at 250C?
2CO (g) + O2 (g) CO2 (g)
S0(CO) = J/K•mol
S0(O2) = J/K•mol
S0(CO2) = J/K•mol
DS0
rxn
= 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
= – [ ] = J/K•mol

23. 3 Factors Involved in Determining Whether a Reaction is Spontaneous or Not
DH; a negative value (exothermic reactions) favors a spontaneous reaction
DS; a positive value (disorder increases) favors a spontaneous reaction
Temperature; If the above factors conflict; temperature determines if reaction is spontaneous or not

24. Why do Chemical Reactions Occur? Free Energy

25. For a constant-temperature process:
Gibbs Free Energy
For a constant-temperature process:
Gibbs free energy (G)
DG = DHsys -TDSsys
DG < The reaction is spontaneous in the forward direction.
DG > The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
DG = The reaction is at equilibrium.

26. DG=DH-TDS DS +DH -DH +DS +DS Spontaneous at all Temperatures
High Temperatures DH
+DH
-DS
-DH
-DS
Spontaneous at
Low Temperatures
NonSpontaneous at
all Temperatures

27. Why do Chemical Reactions Occur? Free Energy
(-) favorable
(+) favorable
(-) spontaneous always
(+) unfavorable
(-) unfavorable
(+) nonspontaneous always
(-) Low T
(+) High T
(+) Low T
(-) High T

28. DG; CHANGE IN GIBB’S FREE ENERGY
DG = DHsys -TDSsys
Determine if reaction is spontaneous or not.
wmax; free energy available to do useful work. DG = wmaximum
DHsys; represents energy available
for useful work
-TDSsys; represents energy that can’t
Be harnessed to do useful work

29. Ways to Obtain DGreaction
From DGfo Values
From Gibb’s Free Energy Equation
DG = DHsys -TDSsys

30. The standard free-energy of reaction (DG0 rxn ) is the free-energy change for a reaction when it occurs under standard-state conditions.
aA + bB cC + dD
DG0
rxn
dDG0 (D) f
cDG0 (C) = [ + ] -
bDG0 (B)
aDG0 (A)
DG0
rxn
nDG0 (products) f = S
nDG0 (reactants) -
Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f
DG0 of any element in its stable form is zero. f
18.4

31. What is the standard free-energy change for the following reaction at 25 0C?
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DG0
rxn
nDG0 (products) f = S
mDG0 (reactants) -
DG0
rxn
6DG0 (H2O) f
12DG0 (CO2) = [ + ] -
2DG0 (C6H6)
DG0
rxn =
[ 12 x – x –237.2 ] – [ 2 x ] = kJ
Is the reaction spontaneous at 25 0C?
DG0 = kJ
< 0
spontaneous
18.4

32. Temperature and Spontaneity of Chemical Reactions
CaCO3 (s) CaO (s) + CO2 (g)
DH0 = kJ; Does not favor spontaneous reaction
DS0 = J/K; Favors spontaneous reaction
DG0 = DH0 – TDS0
Determine the temperature range where the
above reaction will be spontaneous.
0 = DH0 – TDS0
-DG; spontaneous
DG=0;equilibrium
+DG;nonspontaneous
DH0 = TDS0
T=DH0/ DS0
T = kJ/ kJ/K =1108K

33. Gibbs Free Energy and Phase Transitions
Ice melting;
Liquid vaporizing
Solid subliming
H2O (l) H2O (g)
Phase changes happen at equilibrium
DG0 = 0
DG0 = DH0 – TDS0
T=DH0/ DS0
Determine the boiling point of water given that
DSvap = J/mole K and DHvap = kJ/mole.
T = 40.6 kJ/mole = 373 K = 100 °C
kJ/mole K
18.4

34. Gibbs Free Energy and Chemical Equilibrium
-DG; spontaneous
DG=0;equilibrium
+DG;nonspontaneous
Q<K; forward rxn spontaneous
Q=K ;equilibrium
Q>K;forward rxn nonspontaneous
DG = DG0 + RT lnQ(K)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
(K)Q is the reaction quotient = (products/reactants)
DGo is change in Gibbs Free Energy
under standard state conditions
DG is change in Gibbs Free Energy
under experimental conditions

35. Chemical equilibrium
Chemical Equilibrium predicts how reaction will proceed and how far can it go.
Chemical equilibrium does not indicate rates of reactions, just extent of reaction at equilibrium and how far the reaction is from equilibrium
Dynamic nature of chemical reaction
aA + bB <------> cC + dD
A, B decrease with time, C, D increase with time until no change.
Equilibrium: rxn proceeds to a point where reactant to product balanced the reverse rxn, product to reactant.

36. Equilibrium Equations and Equilibrium Constants
Consider the following general equilibrium reaction:
aA + bB + …  mM + nN + …
Where A, B, … are the reactants;
M, N, …. are the products;
a, b, ….m, n, …. are coefficients in the balanced equation.
At equilibrium, the composition of the reaction mixture obeys an equilibrium equation.

37. Equilibrium Equations and Equilibrium Constants
The value of K varies with temperature.

38. The Law of Mass Action K depends on temperature ONLY
K = equilibrium constant = kforward/kreverse for single-step reactions
K depends on temperature ONLY
Don’t use solids or pure liquids (i.e. water) for the Law of Mass Action only gases or aqueous solutions
Gases have partial pressure equilibrium constants

39. What about non-equilibrium reactions?
Use the reaction quotient equation!
Can use Q to figure out which direction the reaction will go:
Q = K reaction is at equilibrium
Q > K reverse reaction rate is greater than forward reaction rate (leftward shift)
Q < K forward reaction rate is greater than reverse reaction rate (rightward shift)

40. 4 rules to determine what value to use for the concentrations in Q
Rule 1: Solvents (e.g. water): [l] is equal to the mole fraction of the solvent. Thus, [H2O] always equals 1. or (l)=1 so ignore
Rule 2: Pure solids in equilibrium with a solution (e.g. CaCO3(s), Fe(OH)3(s)): [s] always equals 1.
Rule 3:Gases in equilibrium with a solution (e.g. CO2(g), O2(g)): [i] equals the partial pressure of the gas (atm)
Rule 4: Compounds dissolved in water: [i] is always reported in units of moles/liter (molar).

41. Homework # 58
Using dat from Appendix 4, calculate ∆G for the reaction:
2H2S (g) + SO2 (g)  3S (s) + 2 H2O (g)
For the following conditions at 25°C:
PH2S = 1.0 x 10-4 atm
PSO2 = 1.0 x 10-2 atm
PH2O = 3.0 x 10-2 atm

42. Answer 58. ΔG° = 3(0) + 2(-229) - [2(-34) + 1(-300.)] = -90. kJ
ΔG = ΔG° + RT ln = -90. kJ +
ΔG = -90. kJ kJ = -50. kJ

43. 66. Consider the following reaction at 298K
66. Consider the following reaction at 298K 2 SO2(g) + O2(g) → 2 SO3(g); An equilibrium mixture contains O2(g) and SO3 (g) at partial pressures of .50 atm and 2.0 atm respectively. Using data from Appendix 4 determine the equilibrium partial pressure of SO2 in the mixture. Will this reaction be most favored at a high or low temp assuming standard conditions?
ΔG°= final -initial
ΔG° = 2(−371 kJ) − [2(−300. kJ)] = −142 kJ
ΔG° = −RT ln K,
ln K = =
K = e = 7.76 × 1024
K = 7.76 × 1024 = = 1.0 × atm
From the negative value of ΔG°, this reaction is spontaneous at standard conditions.
There are more molecules of reactant gases than product gases, so ΔS° will be negative (unfavorable).
Therefore, this reaction must be exothermic (ΔH° < 0). When ΔH° and ΔS° are both negative, the reaction will be spontaneous at relatively low temperatures where the favorable ΔH° term dominates.