1. Spontaneity, Entropy, and Free Energy

2. Spontaneous Processes and Entropy

3. First Law of Thermodynamics
"Energy can neither be created nor destroyed", so while energy can be converted in a chemical process, the total energy remains constant
The energy of the universe is constant

4. Spontaneous Processes
Processes that occur without outside intervention
Spontaneous processes deal with the natural direction of a process and not with the speed at which it occurs-Spontaneous processes may be fast or slow
Many forms of combustion are fast
Conversion of diamond to graphite is slow

5. Entropy (S) A measure of the randomness or disorder
The driving force for a spontaneous process is an increase in the entropy of the universe
Entropy is a thermodynamic function describing the number of arrangements that are available to a system
Nature proceeds toward the states that have the highest probabilities of existing

6. Positional Entropy
The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved
Ssolid < Sliquid << Sgas

7. Entropy and the Second Law of Thermodynamics

8. Second Law of Thermodynamics
"In any spontaneous process there is always an increase in the entropy of the universe"
If a process is spontaneous in one direction then it can’t be spontaneous in the reverse direction
"The entropy of the universe is increasing"
For a given change to be spontaneous, DSuniv must be positive
We can predict whether a process is spontaneous by considering entropy changes that occur in the system and the surroundings
nSuniv = nSsys + nSsurr

10. State functions-all depend only on the change between the initial and final states of a system, not on the process by which the change occurs
For chemical reactions, this means that the thermodynamic state functions are independent of reaction pathways
Enthalpy change (nH), entropy change (nS), and free-energy change (nG) are state functions.

12. Standard state conditions
Superscript circle represents standard state conditions
All gases are at 1 atm pressure
All liquids are pure
All solids are pure
All solutions are at 1M concentration
Energy of formation of an element in its normal state is defined as zero
T used for standard state values is almost invariably room temperature, 25oC (298K), though standard state values can be calculated for other temperatures.

13. Enthalpy Review When bonds are formed, energy is released
In order to break bonds, energy must be absorbed
nH = Hproducts – Hreactants
if products have stronger bonds than reactants, then products have lower enthalpy than reactants and are more stable-energy is released, so exothermic
if products have weaker bonds than reactants, then products have higher enthalpy than reactants and are less stable-energy is absorbed, so endothermic
since substances like lowest possible energy state which gives them the greatest stability, exothermic reactions are more likely to occur spontaneously than endothermic reactions

14. Will a reaction be spontaneous?
Enthalpy change Entropy change Spontaneous?
Exothermic (nH < 0) Increase (nS > 0) Yes, always
Exothermic (nH < 0) Decrease (nS < 0) Only at lower Ts
Endothermic (nH > 0) Increase (nS > 0) Only at higher Ts
Endothermic (nH > 0) Decrease (nS < 0) No, never

15. The Effect of Temperature on Spontaneity

16. Direction of Heat Flow
Entropy changes in the surroundings are primarily determined by heat flow
Exothermic reactions in a system at constant temperature increase the entropy of surroundings
Endothermic reactions in a system at constant temperature decrease the entropy of surroundings
The impact of the transfer of a given quantity of energy as heat to or from the surroundings will be greater at lower temperatures

17. Sign of nSsurr depends on the direction of the heat flow
+ for exothermic process
– for endothermic process
Magnitude of nSsurr depends on both the quantity of energy that flows as heat and the temperature at which the energy is transferred
Entropy changes in the system are dominated by the change in the number of gaseous molecules
Fewer gaseous molecules on the product side means a decrease in entropy, and vs.
Molecular structure also plays a role-a complex molecule has a higher standard entropy than a simpler one

18. Free Energy (G), also called "Gibbs Free Energy"

19. Calculating Free Energy Change (constant temperature and pressure)
nG = nH - TnS
H is enthalpy
T is Kelvin temperature

20. Free Energy and Spontaneity
Reactions proceed in the direction that lowers their free energy (-nG)-process that occurs at constant temperature and pressure is spontaneous in the direction in which the free energy decreases
At low temperature enthalpy is dominant
At high temperature entropy is dominant

21. nH TnS nG Spontaneity
Spontaneous at all temperatures
Nonspontaneous at all temperatures
- - ??? Spontaneous if the absolute value of nH is greater than the absolute value of TnS (low temperature)
- Spontaneous at low temperatures
+ Nonspontanteous at high temperatures
+ + ??? Spontaneous if the absolute value of TnS is greater than the absolute value of nH (high temperature)
+ Nonspontaneous at low temperatures
- Spontaneous at high temperatures

22. Entropy Changes in Chemical Reactions

23. Constant Temperature and Pressure
Reactions involving gaseous molecules
The change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
9 molecules 10 molecules
nS increases

24. Third Law of Thermodynamics
"The entropy of a perfect crystal at O K is zero" (NO disorder, since everything is in perfect position)
Entropy-randomness or disorder of the system
Greater disorder, greater the entropy
Since zero entropy is defined to be a solid crystal at 0K, all substances have positive value for entropy
Liquids have higher entropy than solids
Gases have higher entropy than liquids
Particles in solution have higher entropy than solids
Two moles of a substance has a higher entropy than one mole

25. Calculating Entropy Change in a Reaction
Entropy is an extensive property (a function of the number of moles)
Generally, the more complex the molecule, the higher the standard entropy value
nS0 = Σ nS0products - Σ nS0reactants

26. Free Energy and Chemical Reactions

27. Standard Free Energy Change
nG0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states
nG0 cannot be measured directly
The more negative the value for nG0, the farther to the right the reaction will proceed in order to achieve equilibrium
Equilibrium is the lowest possible free energy position for a reaction
If nG is negative, the reaction is spontaneous
If nG is positive, the reaction is not spontaneous
If nG = 0, the reaction is at equilibrium

28. Calculating Free Energy Change
Method #1, for reactions at constant temperature:
nG0 = nH0 - TnS0
Method #2, an adaptation of Hess's Law:
Cdiamond(s) + O2(g)  CO2(g) nG0 = -397 kJ
Cgraphite(s) + O2(g)  CO2(g) nG0 = -394 kJ
Cdiamond(s)  Cgraphite(s) nG0 = -397 kJ - (-394 kJ) = -3kJ
(Complete example is on page 798)
Method #3, using standard free energy of formation (DGf0)
Standard Free Energy of Formation is the change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states
nGf0 of an element in its standard state is zero
nS0 = Σ nG0products - Σ nG0reactants

30. The Dependence of Free Energy on Pressure
Enthalpy, H
is not pressure dependent

31. Entropy, S entropy depends on volume, so it also depends on pressure
S large volume > S small volume
S low pressure > S high pressure
Standard free energy change gives spontaneity of reaction when all concentrations of reactants and products are in standard state concentrations (1M)
Free energy change, and thus spontaneity, of reaction will be different from standard free energy change if initial concentrations of reactants and products are not 1M

32. The standard free energy change can be related to energy change for other conditions by G = Go + RT ln(P)
G0 is the free energy of the gas at a pressure of 1 atm
G is the free energy of the gas at a pressure of P atm
R is the universal gas constant, T is Kelvin temperature
nG = nG0 + RT ln(Q)
Q is the reaction quotient (from the law of mass action, section 13.5)
R is the gas constant ( J/K ×mol)
nG0 is the free energy change for the reaction with all reactants and products at a pressure of 1 atm
nG is the free energy change for the reaction for the specified pressures of reactants and products

34. Free Energy and Equilibrium

35. Thermodynamic View of Equilibrium
Equilibrium point occurs at the lowest value of free energy available to the reaction system
At equilibrium, nG = 0 and Q = K
nG0 = -RT ln(K)
nG0 K
nG0 = 0 and K = 1, reactants and products equally favored at equilibrium
nG0 < 0 and K > 1, products favored at equilibrium
nG0 > 0 and K < 1, reactants are favored at equilibrium

36. Temperature Dependence of K
nG0 = -RT ln ( K) = nH0 -TnS0

38. Free Energy and Work
Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy
wmax = nG

39. Energy
Released in many chemical and physical processes that is available to do work

40. Free energy-energy available to do work
Spontaneous change-skier going downhill, waterfall, fuel fire at gasoline refinery, book slips from your hand and will start to fall, ice cubes in a cold drink will gradually melt.
Decrease in energy of system.
Occur by themselves without outside assistance.
Once conditions are right to begin, proceed on their own.
Once begins, tendency to continue until it is finished.
Some occur rapidly (pulling hand from hot stove), some slowly over many years (erosion of mountain), and some so slowly nothing appears to be happening (Gasoline-oxygen mixtures appear perfectly stable indefinitely at room T).
If heated, gasoline-oxygen mixture’s rate of reaction increases-can react explosively until all of one or the other is totally consumed.

41. Give substantial amounts of products at equilibrium.
Spontaneous reactions naturally favor formation of products at specified conditions
Give substantial amounts of products at equilibrium.
All release free energy (exergonic) which lowers the energy of the system (exothermic)
Exothermic changes have the tendency to proceed spontaneously
The word tendency should be emphasized.
Some exothermic reactions are not spontaneous, while not every nonspontaneous is endothermic.
Salts dissolving in water-spontaneous change that is endothermic.

42. Nonspontaneous change-pile of bricks in morning later becomes brick wall.
Continues only as long as receives some outside assistance.
Occurs only when some spontaneous event has occurred first.
Nonspontaneous reactions do not favor formation of products at specified conditions (there are more reactions formed over products).

43. Universal phenomenon that something that brings about randomness is more likely to occur than something that brings about order.
Your room does not spontaneously get more organized
It gets messy after it is cleaned
There is a tendency for things to become more disorganized and random.
Throw new deck of cards that was in order into air-land disordered.

44. Enthalpy
Enthalpy, H, (heat content of a substance): natural systems tend to go from a state of higher energy to a state of lower energy.
A ball rolls down a hill spontaneously, but not up.
The ball looses potential energy as it rolls downhill.
At the bottom of the hill it has zero potential energy.

45. Enthalpy is the heat content of a substance.
The change in enthalpy in a reaction is the difference between the heat contained in the products and the heat contained in the reactants.
In an endothermic reaction, heat is absorbed, which would create a positive enthalpy.
In an exothermic reaction, heat is given off, which means the reaction would have a negative enthalpy.

46. Entropy
Entropy, S, (disorder of a substance): natural processes tend to go from an orderly state to a disorderly state.
Place 3 layers of different colored marbles in a container and shake it
No longer ordered and no matter how much you shake box, chances getting it back to original pattern next to zero.
Entropy describes the degree of randomness in a system.
The larger the entropy value, the larger the degree of randomness of system.
Enthalpy/entropy state functions-depends on state of system.

47. Any event w/increase in entropy tends to be spontaneous
∆S = Sfinal -Sinitial or, for chemical systems as ∆S=Sproducts - Sreactants
If Sproducts > Sreactants is positive, there is an increase in entropy
Tends to be spontaneous.
Any event w/increase in entropy tends to be spontaneous

48. When nitroglycerine explodes, heat released and disorder increased
When nitroglycerine explodes, heat released and disorder increased. Both conditions are favorable.

49. Law of disorder-things move in direction of maximum disorder or randomness (chaos).
Entropy gas > entropy liquid > entropy solid.
Increases when substance divided into parts (dissolve S in L).
Tends to increase when # products molecules > # reactant molecules.

51. Physical State: Gas > Liquid > Solid
Entropy changes associated with changes in state can be predicted
Increasing the disorder increases entropy
Ssolids < Sliquids < Sgases
H2O(l)  H2O(g) ∆Ssystem > 0

52. Temperature:
An increase in the temperature of a substance is always accompanied by an increase in the random motion of its particles
Increased temperature means increased kinetic energy which means faster movement, more possible arrangements and increased disorder.
∆Ssystem > 0

54. Number of Molecules:
Entropy increases whenever reaction gives increase in number of gaseous particles.
Assuming no change in physical state, the entropy of a system usually increases when the number of gaseous product particles is greater than the number of gaseous reactant particles
Larger #number of gaseous particle, the more random arrangements are available
2SO3(g)  2SO2(g) + O2 (g) ∆Ssystem > 0

57. For same type of bonding, larger molecules possess higher entropy.
Molecular Size:
For same type of bonding, larger molecules possess higher entropy.
Relative Entropies:
CH4 < C2H6 < C3H8 < C4H10

58. Gas particles have more entropy when they can move freely:
Mixing:
Dissolving of gas in a solvent always results in a decrease in entropy
Gas particles have more entropy when they can move freely:
Sgas > Sdissolved gas
CO2(g)  CO2 (aq)
∆Ssystem < 0

59. Mixing:
With some exceptions, you can predict the change in entropy when a solid or a liquid dissolves to form a solution
Solute particles become dispersed throughout the solvent, increasing randomness and disorder of the particles
NaCl(s)  Na+(aq) + Cl-(aq) ∆Ssystem > 0

61. Expansion:

65. Think randomness:

68. Entropy, the universe and free energy
Spontaneous vs. Nonspontaneous Reactions

70. The law of disorder states that the entropy of the universe must increase as a result of a spontaneous reaction or process:
∆Suniverse > 0 for any spontaneous reaction.
Since the universe equals the system plus the surroundings, any change in the entropy of the universe is the sum of changes occurring in the system and surroundings.
∆Suniverse = ∆Ssystem + ∆Ssurroundings

71. Changes in the system’s enthalpy and entropy affects ∆Suniverse
The reaction or process is exothermic which means ∆Hsystem is negative.
Heat released raises T of surroundings so it increases the entropy of the surroundings, making ∆Ssurroundings positive.
The entropy of the system increases, so ∆Ssystem is positive.
Exothermic chemical reactions accompanied by an increase in entropy are all spontaneous .

72. Second Law of Thermodynamics:
Whenever a spontaneous event takes place in the universe, it is accompanied by an overall increase in entropy (in any isolated system, the degree of disorder can only increase).
Movement towards order requires E-creation of disorganization requires no E-spontaneous
Folding shirt takes more energy than throwing it on ground.
Entropy of substance varies with T-lower T, lower entropy.

73. In a spontaneous process the entropy of the universe must increase:
∆Ssystem + ∆Ssurroundings > 0
Universe becomes more disordered.

77. Spontaneous reactions:
Reaction is spontaneous if reaction is exothermic or entropy of system may increase, or both may occur (free energy is released)
Highly exergonic-
Exothermic-heat released-heat content  (favorable)
Entropy increases-more disorder (favorable)
Slightly exergonic-
exothermic (favorable) and offsets ii.
Entropy decreases-more order (unfavorable)
Exergonic-
Endothermic-heat content increases (unfavorable)
Entropy increases (favorable) and offsets i.

78. Nonspontaneous reactions:
Reaction is nonspontaneous if reaction is endothermic or entropy of system decreases, or both may occur
Highly endergonic
Endothermic-heat content increases (unfavorable)
Entropy decreases (unfavorable)
Slightly endergonic
Endothermic (unfavorable) and offsets ii.
Entropy increases (favorable)
Endergonic
Exothermic (favorable)
Entropy decreases (unfavorable) and offsets i.

85. (to the left)
(to the right)

91. Example:
A mixture of 2.00 mol of H2, 1.00 mol of N2, and 2.00 mol of NH3 is placed in a 1.00-L container at 472oC. Would the reaction N2(g)  +  3 H2(g) { 2 NH3(g)  Kc = proceed toward the right or the left?
Determine reaction quotient, Qc (expression having same form as Keq expression but whose concentration values are initial concentrations-given moles) and compare to Kc
For the reaction above Qc =
[NH3] [2.00]2
Qc = = = 0.500
[N2][H2] [1.00][2.00]3
Since Qc > Kc reaction proceeds left to reach equilibrium

92. Third Law of Thermodynamics:
At absolute zero, the entropy of a pure crystal is also zero.
P=1 atm/T>100oC, water highly disordered gas w/very high entropy.
If confined-molecules spread evenly throughout container and have constant motion.
When system cooled, water vapor condenses to form a liquid and the entropy is lowered.
Below 0oC, water molecules join together to form ice and are highly ordered, yet the order is not perfect
Water still has some entropy because there is enough thermal energy left to cause them to vibrate and rotate within general area of lattice sites.
Cool further, decrease thermal energy and entropy decrease
Absolute zero, ice in state of perfect order-entropy will be zero.

94. How to calculate standard entropy change:
1 mol substance at 298K (25oC) and P = 1 atm-standard entropy, So.
Units of J/Ko
So = (sum of So of products)- (sum of So of reactants)
So = Σ So products - Σ Soreactants
How to calculate standard entropy change:
Write the phase change as a balanced reaction.
Find standard entropies, So, from table.
Calculate the change in entropy by subtraction.

95. Example:
Urea (from urine) hydrolyses slowly in the presence of water to produce ammonia and carbon dioxide. What is the standard entropy change, in J/K, for this reaction when 1 mole of urea reacts with water?
CO(NH2)2(aq) + H2O(l)  CO2 (g) + 2NH3(g)
So = [CO2 (g) + (2)NH3 (g)] - [CO(NH2) 2(aq) + H2O(l)]      = [213.6 J/K + (2)192.5 J/K] - [173.8 J/K J/K]      = (598.6 J/K)-(243.8 J/K) = J/K
Since 2 molecules become 3 molecules, there is an increase in entropy, so the positive value.

96. Example:
Evaluate the entropy change for the reaction CO(g) + 3 H2(g)  CH4(g) + H2O(g).
So = (sum of So of products)-(sum of So of reactants)
∆S = [CH4(g) + H2O(g)] – [CO(g) + (3)H2(g)]
∆S = [ ] – [ (131)]
J/K = -216 J (K mol)-1
Since 4 mol of reactants form 2 mol of products (all in gaseous state), there is a decrease in entropy, so the large negative change for the reaction.

97. 3 mol gases on reactant side/1 mol of liquid on product side
And another:
CO(g) + 2 H2(g)  CH3OH(l)
3 mol gases on reactant side/1 mol of liquid on product side
Changing from more disordered to more ordered state.
S (Sreactants - Sproducts) should be negative since Sreactants > Sproducts.
So= [CH3OH(l) - CO(g)] – [(2)H2 (g)]
= J/Kmol J/Kmol - (2)130.7 J/Kmol
= J/K (negative as we had predicted).

98. Free Energy:
Reactions can do work/serve as pathways for conversion of heat into work.
Energy which is, or which can be, available to do useful work is called Gibbs free energy (G) or more often simply free energy.
It is the comparison of changes of enthalpy & entropy during chemical reaction.
It tells if there is energy available to do work.

99. G = H - T∆S where H is enthalpy, S is entropy, and T is absolute T
Free Energy:
G = H - T∆S where H is enthalpy, S is entropy, and T is absolute T
These factors can have either -/+ effect on whether or not a reaction will occur spontaneously or at all.
All spontaneous processes move toward equilibrium.
Reactions occur naturally to reach lower state of energy.

100. Change can only be spontaneous if it is accompanied by a decrease in free energy-G must be negative

102. H is negative/S is positive, G = H - T∆S = (-) - T(+)
Can predict at what temperature a nonspontaneous reaction will become spontaneous:
H is negative/S is positive, G = H - T∆S = (-) - T(+)
G will be negative regardless of value of the absolute temperature, T (can only be positive).
Change will occur spontaneously at all T
Increase in entropy-both factors favor spontaneity.
H is positive/S is negative, G = H - T∆S = (+) - T(-)
G will be positive at all temperatures and the change will always be nonspontaneous.
Decrease in entropy-both factors work against spontaneity
If H/S both positive, G = (+) - T(+)
When H/S have the same sign, temperature becomes critical in determining spontaneity of an event.
Only at relatively high T will value T∆S be larger than value of H so difference, G, is negative

109. Calculate free energy:
2 H2 + O2  2H2O
Go = Gof (products) - Gof (reactants)
Go = (2 x ) – (2 x ) = kJ/mol
Spontaneous because it is negative.

110. Compute Go for the hydrolysis of urea, CO(NH2)2,
Example:
Compute Go for the hydrolysis of urea, CO(NH2)2,
CO(NH2)2(aq) + H2O(l)  CO2(g) + 2 NH3(g)
Go = [CO2 + (2)NH3] - [CO(NH2)2 + H2O] = [ (2) ] - [ ] = ( kJ) - (-605.1) = kJ

111. Another example:
What is Go for the combustion of ethyl alcohol (C2H5OH) to give CO2(g) and H2O(g)?
Balance equation:C2H5OH(l) + 3O2(g) 2 CO2(g) + 3 H2O(g)
Go = [(2)CO2(g) + (3)H2O(g)] - [C2H5OH(l) + (3)O2(g)]
Go = [2 mol( kJ/mol) + 3 mol( kJ/mol)]
= [1 mol( kJ/mol) + 3 mol(0 kJ/mol)]
= ( kJ) - ( kJ)
= kJ

112. http://college. hmco. com/cgi-bin/SaCGI. cgi/ace1app. cgi

113. 2 NH4Cl(s) + CaO(s)  CaCl2(s) + H2O(l) + 2 NH3(g) -8.3 kJ/mole
Calculate Go in kJ for the following reactions, using the appropriate data tables.  Use Go = Gproducts - Greactants
SO3(g) + H2O(l)  H2SO4(l) 
-82.9 kJ/mole
2 NH4Cl(s) + CaO(s)  CaCl2(s) + H2O(l) + 2 NH3(g) 
-8.3 kJ/mole
CaSO4(s) + 2 HCl(g)  CaCl2(s) + H2SO4(l) 
+74.1 kJ/mole
C2H4(g) + H2O(l)  C2H5OH(l) 
-5.9 kJ/mole
Ca(s) + 2 H2SO4(l)  CaSO4(s) + SO2(g) + 2 H2O(l) 
kJ/mole

114. Is the following reaction spontaneous at 250C?
2H2(g) + O2(g)  2H2O(g)
Compute both ∆H and ∆ S
∆ H = (2* ∆ HH2O0) - (2* ∆ HH20 + 1* ∆ HO20)
∆ H = (2* ) - (2*0 + 1*0)
∆ H = kJ/mol
∆ S = (2* ∆ SH2O0) - (2* ∆ SH20 + 1* ∆ SO20)
∆ S = (2* 188.7) - (2* *205.0)
∆ S = J/mol*K = kJ/mol*K
∆ G = ∆ H - T* ∆ S
∆ G = kJ/mol - 298K*( kJ/mol*K)
∆ G = -457 kJ/mol
The reaction is spontaneous, since ∆ G < 0

115. Example:
Calculate the Go in kJ for the following reactions at 100oC using the appropriate data tables. 
5 SO3(g) + 2 NH3(g)  2 NO(g) + 5 SO2(g) + 3 H2O(g) 
delta H = kJ/mole
delta S = J/mole K
delta G = kJ/mole (spontaneous)
N2O(g) + NO2(g)  2 NO(g
delta H = kJ/mole
delta S =  J/mole K
delta G = kJ/mole (nonspontaneous)