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ENTROPY & GIBBS FREE
ENERGY
ENTROPY AND GIBBS FREE ENERGY
entropy (∆S)- a measure of disorder or randomness
entropy =  disorder
The units for entropy are J/mol·K or J/K
ENTROPY AND GIBBS FREE ENERGY
ENTROPY AND GIBBS FREE ENERGY
SECOND LAW OF
THERMODYNAMICS
the entropy of the
universe is constantly
increasing
∆S solid < ∆S liquid < ∆S gas
ENTROPY AND GIBBS FREE ENERGY
ΔS = Sproducts – Sreactants
When Sproducts>Sreactants, ΔS>0
ENTROPY AND GIBBS FREE ENERGY
ΔS = Sproducts – Sreactants
When Sproducts<Sreactants, ΔS<0
ENTROPY AND GIBBS FREE ENERGY
ENTROPY AND GIBBS FREE ENERGY
Predict the sign of ∆S for the following reactions:
H2(g)  2 H(g) +∆S
2 H2(g) + O2(g)  2 H2O(l) - ∆S
2 NO2(g)  N2O4(g) - ∆S
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) +∆S
condensation of steam to liquid - ∆S
ENTROPY AND GIBBS FREE ENERGY
Predict the sign of ∆S for the following reactions:
sublimation of dry ice + ∆S
2 NH3(g)  N2(g) + 3 H2(g)+ ∆S
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g) + ∆S
2 H2O(l)  2 H2(g) + O2(g) + ∆S
NaCl(s)  NaCl(aq)+ ∆S
ENTROPY AND GIBBS FREE ENERGY
To solve for the entropy change of a reaction
at standard conditions (SATP):
ΔS°= ∑ΔS°(products) - ∑ΔS°(reactants)
Don’t forget that the equations should be
multiplied by the appropriate factor, as
necessary.
ENTROPY AND GIBBS FREE ENERGY
Example 1: Calculate the standard entropy of the following
reaction:
C2H4(g) + H2(g)  C2H6(g)
SºC2H6(g)
= 229.5J/mol·K
SºC2H4(g)
= 219.8J/mol·K
SºH2(g)
= 130.6J/mol·K
ΔS° = ∑ΔS°(products) - ∑ΔS°(reactants)
= [1 mol x 229.5J/mol·K] – [(1 mol x 219.8J/mol·K)
+ (1 mol x 130.6J/mol·K)]
= (229.5J/K) – (350.4J/K)
= -120.9J/K
Therefore the standard entropy is -120.9J/K
Do you think this reaction will spontaneously occur
at room temperature?
ENTROPY AND GIBBS FREE ENERGY
Example 1: Calculate the standard entropy of the
following reaction:
C2H4(g) + H2(g)  C2H6(g)
Do you think this reaction will spontaneously occur
at room temperature?
Since molecules tend to
become more and more
disordered, and this reaction
results in more organization
(fewer moles of gas), then
this reaction is most likely
not spontaneous at room
temperature
ENTROPY AND GIBBS FREE ENERGY
How are entropy and enthalpy related?
ΔGº = ΔHº - TΔSº
Gibbs free energy
(J or kJ)
enthalpy Temperature (K)
entropy
Gibbs free energy is the energy that is available to do useful work.
A reaction will spontaneously occur if ΔG<0 (exergonic reaction)
A reaction will NOT spontaneously occur if ΔG>0 (endergonic reaction)
ENTROPY AND GIBBS FREE ENERGY
A reaction will spontaneously occur if ΔG<0 (exergonic reaction)
A reaction will NOT spontaneously occur if ΔG>0 (endergonic reaction)
ENTROPY AND GIBBS FREE ENERGY
Reactions with a negative ΔH and positive ΔS all have a
negative ΔG
ΔGº = ΔHº - TΔSº
(-) - T(+)
Exergonic & spontaneous at all temperatures
Reactions with a positive ΔH and negative ΔS all have a
positive ΔG
ΔGº = ΔHº - TΔSº
(+) - T(-)
Endergonic & nonspontaneous at all temperatures and will only occur
with the continuous input of energy
ENTROPY AND GIBBS FREE ENERGY
Reactions with a negative ΔH and negative ΔS all have a negative ΔG
ΔGº = ΔHº - TΔSº
(-) - T(-)
ΔG will be negative in temperature conditions where the value of TΔS is lower than the value of ΔH. Thus the
reaction is spontaneous only at lower temperatures.
Example: 2SO2(g) + O2(g)  2SO3(g)
This reaction is spontaneous at temperatures below 786ºC (1059K) and nonspontaneous above this temperature
ENTROPY AND GIBBS FREE ENERGY
Reactions with a positive ΔH and positive ΔS all have a negative ΔG
ΔGº = ΔHº - TΔSº
(+) - T(+)
ΔG will be negative in temperature conditions where the value of TΔS is higher than the value of ΔH. Thus the
reaction is spontaneous only at higher temperatures.
Example: H2O(s)  H2O(l)
This reaction is spontaneous at temperatures above 0ºC (273K) and nonspontaneous below this temperature
ENTROPY AND GIBBS FREE ENERGY
ΔH>0 ΔH<0
ΔS>0 Spontaneity depends
on T
(spontaneous at higher
temperatures)
Spontaneous at all
temperatures
ΔS<0 Nonspontaneous
(proceeds only with a
continuous input of
energy)
Spontaneity depends
on T
(spontaneous at lower
temperatures)
Summary: Spontaneous and Non-Spontaneous Reactions
ENTROPY AND GIBBS FREE ENERGY
Example 2: Is the following reaction spontaneous at
SATP?
CO(NH2)2(aq) + H2O(l)  CO2(g) + 2NH3(g)
ΔHº =119kJ
ΔSº = 354.8J/K = 0.3548KJ/K
T = 25ºC = 298K
ΔGº = ΔHº - TΔSº
= (119kJ) – (298K)(0.3548kJ/K)
= (119kJ) – (105.7kJ)
= 13kJ
ΔGº>0, so this reaction is not spontaneous
Therefore this reaction is not spontaneous at SATP
ENTROPY AND GIBBS FREE ENERGY

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Tang 05 entropy and gibb's free energy

  • 1. ENTROPY & GIBBS FREE ENERGY
  • 2. ENTROPY AND GIBBS FREE ENERGY
  • 3. entropy (∆S)- a measure of disorder or randomness entropy =  disorder The units for entropy are J/mol·K or J/K ENTROPY AND GIBBS FREE ENERGY
  • 4. ENTROPY AND GIBBS FREE ENERGY SECOND LAW OF THERMODYNAMICS the entropy of the universe is constantly increasing
  • 5. ∆S solid < ∆S liquid < ∆S gas ENTROPY AND GIBBS FREE ENERGY
  • 6. ΔS = Sproducts – Sreactants When Sproducts>Sreactants, ΔS>0 ENTROPY AND GIBBS FREE ENERGY
  • 7. ΔS = Sproducts – Sreactants When Sproducts<Sreactants, ΔS<0 ENTROPY AND GIBBS FREE ENERGY
  • 8. ENTROPY AND GIBBS FREE ENERGY Predict the sign of ∆S for the following reactions: H2(g)  2 H(g) +∆S 2 H2(g) + O2(g)  2 H2O(l) - ∆S 2 NO2(g)  N2O4(g) - ∆S C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) +∆S condensation of steam to liquid - ∆S
  • 9. ENTROPY AND GIBBS FREE ENERGY Predict the sign of ∆S for the following reactions: sublimation of dry ice + ∆S 2 NH3(g)  N2(g) + 3 H2(g)+ ∆S C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g) + ∆S 2 H2O(l)  2 H2(g) + O2(g) + ∆S NaCl(s)  NaCl(aq)+ ∆S
  • 10. ENTROPY AND GIBBS FREE ENERGY To solve for the entropy change of a reaction at standard conditions (SATP): ΔS°= ∑ΔS°(products) - ∑ΔS°(reactants) Don’t forget that the equations should be multiplied by the appropriate factor, as necessary.
  • 11. ENTROPY AND GIBBS FREE ENERGY Example 1: Calculate the standard entropy of the following reaction: C2H4(g) + H2(g)  C2H6(g) SºC2H6(g) = 229.5J/mol·K SºC2H4(g) = 219.8J/mol·K SºH2(g) = 130.6J/mol·K ΔS° = ∑ΔS°(products) - ∑ΔS°(reactants) = [1 mol x 229.5J/mol·K] – [(1 mol x 219.8J/mol·K) + (1 mol x 130.6J/mol·K)] = (229.5J/K) – (350.4J/K) = -120.9J/K Therefore the standard entropy is -120.9J/K Do you think this reaction will spontaneously occur at room temperature?
  • 12. ENTROPY AND GIBBS FREE ENERGY Example 1: Calculate the standard entropy of the following reaction: C2H4(g) + H2(g)  C2H6(g) Do you think this reaction will spontaneously occur at room temperature? Since molecules tend to become more and more disordered, and this reaction results in more organization (fewer moles of gas), then this reaction is most likely not spontaneous at room temperature
  • 13. ENTROPY AND GIBBS FREE ENERGY How are entropy and enthalpy related? ΔGº = ΔHº - TΔSº Gibbs free energy (J or kJ) enthalpy Temperature (K) entropy Gibbs free energy is the energy that is available to do useful work. A reaction will spontaneously occur if ΔG<0 (exergonic reaction) A reaction will NOT spontaneously occur if ΔG>0 (endergonic reaction)
  • 14. ENTROPY AND GIBBS FREE ENERGY A reaction will spontaneously occur if ΔG<0 (exergonic reaction) A reaction will NOT spontaneously occur if ΔG>0 (endergonic reaction)
  • 15. ENTROPY AND GIBBS FREE ENERGY Reactions with a negative ΔH and positive ΔS all have a negative ΔG ΔGº = ΔHº - TΔSº (-) - T(+) Exergonic & spontaneous at all temperatures Reactions with a positive ΔH and negative ΔS all have a positive ΔG ΔGº = ΔHº - TΔSº (+) - T(-) Endergonic & nonspontaneous at all temperatures and will only occur with the continuous input of energy
  • 16. ENTROPY AND GIBBS FREE ENERGY Reactions with a negative ΔH and negative ΔS all have a negative ΔG ΔGº = ΔHº - TΔSº (-) - T(-) ΔG will be negative in temperature conditions where the value of TΔS is lower than the value of ΔH. Thus the reaction is spontaneous only at lower temperatures. Example: 2SO2(g) + O2(g)  2SO3(g) This reaction is spontaneous at temperatures below 786ºC (1059K) and nonspontaneous above this temperature
  • 17. ENTROPY AND GIBBS FREE ENERGY Reactions with a positive ΔH and positive ΔS all have a negative ΔG ΔGº = ΔHº - TΔSº (+) - T(+) ΔG will be negative in temperature conditions where the value of TΔS is higher than the value of ΔH. Thus the reaction is spontaneous only at higher temperatures. Example: H2O(s)  H2O(l) This reaction is spontaneous at temperatures above 0ºC (273K) and nonspontaneous below this temperature
  • 18. ENTROPY AND GIBBS FREE ENERGY ΔH>0 ΔH<0 ΔS>0 Spontaneity depends on T (spontaneous at higher temperatures) Spontaneous at all temperatures ΔS<0 Nonspontaneous (proceeds only with a continuous input of energy) Spontaneity depends on T (spontaneous at lower temperatures) Summary: Spontaneous and Non-Spontaneous Reactions
  • 19. ENTROPY AND GIBBS FREE ENERGY Example 2: Is the following reaction spontaneous at SATP? CO(NH2)2(aq) + H2O(l)  CO2(g) + 2NH3(g) ΔHº =119kJ ΔSº = 354.8J/K = 0.3548KJ/K T = 25ºC = 298K ΔGº = ΔHº - TΔSº = (119kJ) – (298K)(0.3548kJ/K) = (119kJ) – (105.7kJ) = 13kJ ΔGº>0, so this reaction is not spontaneous Therefore this reaction is not spontaneous at SATP
  • 20. ENTROPY AND GIBBS FREE ENERGY