for education

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
• 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
A + B C + D + Energy

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
• 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
Suniv = Ssys + Ssurr

9. Entropy and Exothermic Processes
• Entropy and a process being exothermic are
related, but a process doesn’t have to be both
• Exothermic process
– Ssurr = positive, exothermic = more disorder in
surroundings
• Endothermic process
– Ssurr = negative, endothermic = less disorder in
surroudings
– ssystem 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
• H = -125 kJ, 25 oC + 273 = 298 K
• Ssurr = -125 kJ/298 K = 419 J/K
• Remember 125 kJ = 125000 J
• ssurr is positive, calculation refers
to system!

13. Example
• 2nd reaction
• H = 778 kJ, 25 oC + 273 = 298 K
• S = 778 kJ/298 K = -2.61 kJ/K
• Remember 778 kJ = 778000 J

14. Free Energy, G
• Typically defined as the energy available to
perform work
• G determines whether a process is spontaneous
or not (G = 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
G = H - TS

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

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

17. Dependence on Temperature
• 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.
G = H - TS

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

20. Example of Gibbs Energy
• For the reaction at 298 K, the values of H and
S are 58.03 kJ and 176.6 J/K, respectively.
What is the value of G at 298 K?

21. Example of Gibbs Energy
• For the reaction at 298 K, the values of H and
S are 58.03 kJ and 176.6 J/K, respectively.
What is the value of G at 298 K?
G = H - TS
G = 58.03 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:
So = 198.7 J/K
2NH3(g)
N2(g) + 3H2(g) Ho = 92.2 kJ

23. 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:
So = 198.7 J/K
2NH3(g)
N2(g) + 3H2(g) Ho = 92.2 kJ
G = H  TS
G = 92.2 kJ  (298 K)(-198.7 J/K)(1 kJ/1000 J)
= -33.0 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 = +98.8 kJ; S = +141.5 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)(0.1415 kJ/K)
• = (98.8 kJ)  (42.2 kJ)
• = 56.6 k
• Reaction is NOT spontaneous!

26. Third Example
• 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!
At what temperature does this become spontaenous?

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 (S = 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
• Typically @ 25 oC (298 K)
• Designated by o sign (similar to degree sign)

30. Calculating Entropy Change in a
Reaction
0 0 o
reaction products reactants
p r
S n S n S
  
 
 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!
0 0 o
reaction products reactants
p r
S n S n S
  
 

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 198 J/Kmol
H2 131 J/Kmol
CH4 186 J/Kmol
H2O 189 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 198 J/Kmol
H2 131 J/Kmol
CH4 186 J/Kmol
H2O 189 J/Kmol
s
oreaction = [186 + 189] – [198 + 3(131)] = 216 J/Kmol

35. Standard Free Energy Change
• G0 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
– G0 cannot be measured directly
– The more negative the value for G0, 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 (Gf
0):
0 0 0
(products) (reactants)
p f r f
G n G n G
    
 
Gf
0 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) = - 1378 kJ
0 0 0
(products) (reactants)
p f r f
G n G n G
    
 

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) 2 Fe (s) + 3 CO2(g)
Go = -742.2 -137.2 0 -394.4

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) 2 Fe (s) + 3 CO2(g)
Go (kJ/mol)= -742.2 -137.2 0 -394.4
Go = [0 + (3 x -394.4)] – [-742.2 + (3 x -137.2)] =
-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 Ho and
So values are given --- you find Ho
reaction and
So
reaction, then plug in to get Go
reaction

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
• R = Universal gas constant = 8.314 J/K mol
• T = Temperature in Kelvin
• K = Equilibrium constant = [Pproducts]/[Preactants]
• System is at equilbrium (no net movement)
when Go = 0 (K = 1)
• If the system is NOT at equilibrium, use Q
instead of K
G = Go + RTln(Q)
Go = -RTln(K) at equilibrium

44. Example
• For the synthesis of ammonia where Go =
-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 = 0.020 atm

45. Example
• For the synthesis of ammonia where Go =
-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 = 0.020 atm
G = Go + RTln(Q)
G = -33.3 kJ/mole + (8.314 J/K mol)(298 K) ln( [0.02 atm]2 )
___________________
[1.0 atm][3.0 atm]3
G = -33.3 kJ/mole + (8.314 J/K mol)(298 K) ln(1.5 x 10-5)
G = -60.5 kJ/mol

46. Example
• The synthesis of methanol from carbon
dioxide and hydrogen has a Go = -25.1
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 Go = -25.1
kJ/mol. What is the value of the equilibrium
constant at 25 oC?
G = Go + RTln(K)
Go = - 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, Go = -5.40 kJ/mole. Give the
equilibrium K for this process.

49. Example
• At 298 K, Go = -5.40 kJ/mole. Give the
equilibrium constant K for this process.
• Go = -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 Ssurr 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