1. Chapter 16 Spontaneity, Entropy and Free Energy
DE Chemistry
Dr. Walker

2. Laws Of Thermodynamics
The first law of thermodynamics is a statement of the law of conservation of energy:
Energy can be neither created nor destroyed. In other words, the energy of the universe is constant.

3. Energy in Chemical Reactions
A + B C + D + Energy
The potential energy is broken in chemical bonds in compounds A and B
The potential energy is the chemical bonds of C and D is lower
The excess has been given off (energy) as thermal energy, or heat, which is kinetic energy transferred to the surroundings

4. Spontaneity Spontaneous Processes
Processes that occur without outside intervention
Spontaneous processes may be fast or slow
Many forms of combustion are fast
Conversion of graphite to diamond is slow
Kinetics is concerned with speed, thermodynamics with the initial and final state

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 (one of the laws of thermodynamics!)
The universe favors chaos!!
Entropy is a thermodynamic function describing the number of arrangements that are available to a system
Nature proceeds towards the states that have the highest probabilities of existing

6. Second Law of Thermodynamics
"In any spontaneous process there is always an increase in the entropy of the universe"
"The entropy of the universe is increasing"
For a given change to be spontaneous, Suniverse must be positive
Suniv = Ssys + Ssurr

7. 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

8. Entropy and Temperature
Suniv = Ssys + Ssurr
Entropy changes in the surrounding are primarily determined by heat flow
The magnitude of entropy is dependent on the temperature
The lower the temperature, the higher the impact to the surroundings of the transfer of energy
Remember the books’ $50 example

9. Entropy and Exothermic Processes
Entropy and a process being exothermic are related, but a process doesn’t have to be both
Exothermic process
DSsurr = positive, exothermic = more disorder in surroundings
Endothermic process
DSsurr = negative, endothermic = less disorder in surroudings
Dssystem must increease to obey 2nd Law of Thermodynamics

10. More on Entropy Why is the sign different?
The enthalpy (H) concerns the system
Our entropy here concerns the surroundings
As usual, temperatures must be in Kelvin

11. Example

12. Example 1st reaction DH = -125 kJ, 25 oC + 273 = 298 K
DSsurr = -125 kJ/298 K = 419 J/K
Remember 125 kJ = J
Dssurr is positive, calculation refers to system!

13. Example 2nd reaction DH = 778 kJ, 25 oC + 273 = 298 K
DS = 778 kJ/298 K = kJ/K
Remember 778 kJ = J

14. Free Energy, G G = H - TS
Typically defined as the energy available to perform work
G determines whether a process is spontaneous or not (DG = negative = spontaneous)
Takes into account enthalpy, entropy, and temperature
Symbol honors Josiah Gibbs (sometimes called Gibbs Free Energy or Gibbs energy), a physics professor at Yale during the late 1800’s who was important in developing much of modern thermodynamics

15. Energy Diagrams Left – A spontaneous reaction
The products have a lower free energy (G) than the reactants (DG < 0)
Right – A nonspontaneous reaction
The reactants have a higher free energy than the products (DG > 0)

16. What If? DG = 0? As a result The reaction is at equilibrium
DG < 0 The reaction is spontaneous.
DG > 0 The reaction is nonspontaneous.
DG = 0 The reaction mixture is at equilibrium.
-It’s not moving forward or reverse overall – multiple
processes, same speed

17. Dependence on Temperature
DG = DH - TDS
Notice there are two terms in the equation
Spontaneity can change when the temperature changes.
Temperature and spontaneity are not necessarily correlated.
A reaction with a negative entropy (loss of randomness) would be LESS spontaneous at higher temperatures – it doesn’t want to happen, but is pushed by the extra heat.

18. Reaction Rates Free energy (G) values tell us if reactions will occur
They do not tell us how fast reactions will occur.
During a reaction
Reactant particles must physically collide
They must collide with enough energy to break the bonds in the reactant
Some reactions require the addition of heat energy. This gives the reactants the extra energy needed (more collisions, harder collisions) for this process to occur.

19. H, S, G and Spontaneity
G = H - TS
H is enthalpy, T is Kelvin temperature
Value of H
Value of TS
Value of G
Spontaneity
Negative
Positive
Spontaneous
Nonspontaneous
???
Spontaneous if the absolute value of H is greater than the absolute value of TS (low temperature)
Spontaneous if the absolute value of TS is greater than the absolute value of H (high temperature)

20. Example of Gibbs Energy
For the reaction at 298 K, the values of DH and DS are kJ and J/K, respectively. What is the value of DG at 298 K?

21. Example of Gibbs Energy
For the reaction at 298 K, the values of DH and DS are kJ and J/K, respectively. What is the value of DG at 298 K?
G = H - TS
G = kJ – (298 K)(176.6 J/K)
G = 5.40 kJ Not spontaneous at this temperature

22. Second Example
Calculate the standard free-energy change at 25 oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes:
2NH3(g)
N2(g) + 3H2(g)
DHo = 92.2 kJ
DSo = 198.7 J/K

23. Second Example DG = DH  TDS
Calculate the standard free-energy change at 25 oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes:
2NH3(g)
N2(g) + 3H2(g)
DHo = 92.2 kJ
DSo = 198.7 J/K
DG = DH  TDS
DG = 92.2 kJ  (298 K)( J/K)(1 kJ/1000 J)
= kJ

24. Third Example
Iron metal can be produced by reducing iron(III) oxide with hydrogen:
Fe2O3(s) + 3 H2(g)  2 Fe(s) + 3 H2O(g) H = kJ; S = J/K
(a) Is this reaction spontaneous under standard-state conditions at 25 °C?
(b) At what temperature will the reaction become spontaneous?

25. Third Example
To determine whether the reaction is spontaneous at 25 °C, we need to determine the sign of G = H  TS. At 25 C (298 K), G for the reaction is
G = H  TS = (98.8 kJ)  (298 K)( kJ/K)
= (98.8 kJ)  (42.2 kJ)
= 56.6 k
Reaction is NOT spontaneous!

26. Third Example
At what temperature does this become spontaenous?
At temperatures above 698 K, the TS term becomes larger than H, making the Gibbs energy negative and the process spontaneous
Why is this postive this time?
Remember, here we’re dealing with the system, not the surroundings!

27. Entropy Changes in Chemical Reactions
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
Typically, more moles of gas, more entropy

28. Third Law of Thermodynamics
"The entropy of a perfect crystal at O K is zero" (NO disorder, since everything is in perfect position)
No movement = 0 K
No disorder = no entropy (DS = 0)

29. Standard State Conditions
We’ve seen that quantities such as entropy (S), enthalpy (H), and free energy (G) are dependent upon the conditions present
Standard State
One set of conditions at which quantities can be compared
Pure solids/liquids/gases at 1 atm pressure
Solutes at 1 M concentration
25 oC (298 K)
Designated by o sign (similar to degree sign)

30. Calculating Entropy Change in a Reaction
Calculates standard entropy of a reaction, uses standard entropies of compounds
Entropy is an extensive property (a function of the number of moles)
Generally, the more complex the molecule, the higher the standard entropy value

31. Calculating Standard Entropy - Example

32. Calculating Standard Entropy - Example
2 (28 J/Kmol) + 3 (189 J/Kmol) – 51 J/K mole – 3 (131 J/Kmol) = 179 J/K
System gained entropy – water more complex than hydrogen!

33. Second Example Evaluate the entropy change for the reaction:
CO + 3 H2 -> CH4 + H2O
in which all reactants and products are gaseous.
So values: CO J/Kmol
H J/Kmol
CH J/Kmol
H2O J/Kmol

34. Second Example Evaluate the entropy change for the reaction:
CO + 3 H2 -> CH4 + H2O
in which all reactants and products are gaseous.
So values: CO J/Kmol
H J/Kmol
CH J/Kmol
H2O J/Kmol
Dsoreaction = [ ] – [ (131)] = 216 J/Kmol

35. Standard Free Energy Change
DG0 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
DG0 cannot be measured directly
The more negative the value for DG0, 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

36. Calculating Free Energy of Formation
Using standard free energy of formation (Gf0):
Gf0 of an element in its standard state is zero!

37. Calculating Free Energy of Formation - Example

38. Calculating Free Energy of Formation - Example
2 (-394kJ/mol) + 4 (-229 kJ/mol) – 2 (-163 kJ/mol) – 3 (0) = kJ

39. Second Example
Calculate the standard free energy for the reaction below and determine whether it is spontaneous at standard conditions.
Fe2O3(s) + 3 CO (g) Fe (s) + 3 CO2(g)
DGo =

40. Second Example
Calculate the standard free energy for the reaction below and determine whether it is spontaneous at standard conditions.
Fe2O3(s) + 3 CO (g) Fe (s) + 3 CO2(g)
DGo (kJ/mol)=
DGo = [0 + (3 x )] – [ (3 x )] =
-29.4 kJ/mol

41. Additional Note on Energies of Formation
Examples on enthalpy are not shown, but could be calculated the same way.
There are numerous other possibilities for calculating these
Another type of problem could be where DHo and DSo values are given --- you find DHoreaction and DSoreaction, then plug in to get DGoreaction

42. The Dependence of Free Energy on Pressure
Enthalpy, H, is not pressure dependent
Entropy, S
entropy depends on volume, so it also depends on pressure
Slarge volume > Ssmall volume
Slow pressure > Shigh pressure

43. Free Energy and Equilibrium
DG = DGo + RTln(Q)
DGo = -RTln(K) at equilibrium
R = Universal gas constant = J/K mol
T = Temperature in Kelvin
K = Equilibrium constant = [Pproducts]/[Preactants]
System is at equilbrium (no net movement) when DGo = 0 (K = 1)
If the system is NOT at equilibrium, use Q instead of K

44. Example For the synthesis of ammonia where DGo =
-33.3 kJ/mole, determine whether the reaction is spontaneous at 25 C with the following pressures:
PN2 = 1.0 atm, PH2 = 3.0 atm, PNH3 = atm

45. Example For the synthesis of ammonia where DGo =
-33.3 kJ/mole, determine whether the reaction is spontaneous at 25 C with the following pressures:
PN2 = 1.0 atm, PH2 = 3.0 atm, PNH3 = atm
DG = DGo + RTln(Q)
DG = kJ/mole + (8.314 J/K mol)(298 K) ln( [0.02 atm] )
___________________
[1.0 atm][3.0 atm]3
DG = kJ/mole + (8.314 J/K mol)(298 K) ln(1.5 x 10-5)
DG = kJ/mol

46. Example
The synthesis of methanol from carbon dioxide and hydrogen has a DGo = kJ/mol. What is the value of the equilibrium constant at 25 oC?

47. Example
The synthesis of methanol from carbon dioxide and hydrogen has a DGo = kJ/mol. What is the value of the equilibrium constant at 25 oC?
DG = DGo + RTln(K)
DGo = - RTln(K)
-25.1 kJ/mol = -(8.314 J/molK)(298)lnK
Solve for ln K = 10.1, K = 2 x 104

48. Example
At 298 K, DGo = kJ/mole. Give the equilibrium K for this process.

49. Example
At 298 K, DGo = kJ/mole. Give the equilibrium constant K for this process.
DGo = -RTln(K)
-5.40 kJ/mole = -(8.31 J/K mol)(298 K)ln K
2.18 = ln K
8.85 = K

50. Free Energy and Work
The maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy
The amount of work obtained is always less the maximum
Work is changed to heat in surroudings – You lose efficiency, but the DSsurr increases (favorable!)
Henry’s Bent’s First Two Laws of Thermodynamics
1st Law: You can’t win, you can only break even
2nd Law: You can’t break even