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Lecture 9: Gibbs Free Energy

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пЃ„ пЃ€ В° sys < 0
пЃ„ SВ° sys
Lecture 8 Summary
пЃ„ SВ° univ
пЃ„ SВ° s urr
Spontaneous, exothermic
пЃ„ пЃ€ В° sys < 0
пЃ„ SВ° s urr
пЃ„ SВ° univ
Spontaneous, exothermic
пЃ„ SВ° sys
пЃ„ пЃ€ В° sys
>
пЃ„ SВ° sys
0
пЃ„ SВ° univ
пЃ„ SВ° s urr
Spontaneous, endothermic
Lecture 9: Gibbs Free Energy G
Reading: Zumdahl 10.7, 10.9
Outline
Defining the Gibbs Free Energy (пЃ„G)
Calculating пЃ„G (several ways to)
Pictorial Representation of пЃ„G
Defining пЃ„G
Recall, the second law of thermodynamics:
пЃ„Suniv = пЃ„Stotal = пЃ„Ssys + пЃ„Ssurr
Also recall:
пЃ„Ssurr = -пЃ„Hsys/T
Substituting,
пЃ„Stotal = пЃ„Ssys +-пЃ„H
пЃ„S
/T
syssurr
Multipling all by (-T) gives
-TпЃ„Stotal = -TпЃ„Ssys+ пЃ„Hsys
We then define:
пЃ„G = -TпЃ„Stotal
Substituting
пЃ„G = -TпЃ„Ssys+ пЃ„Hsys
Giving finally
пЃ„G = пЃ„H - TпЃ„S
пЃ„G = The Gibbs Free Energy @const P
пЃ„G and Spontaneous Processes
Recall from the second law the conditions of spontaneity:
Three possibilities:
If Suniv > 0…..process is spontaneous
If Suniv < 0…..process is spontaneous in opposite
direction.
If Suniv = 0….equilibrium
In our derivation of пЃ„G, we divided by -T;
therefore, the direction of the inequality
relative to entropy is now reversed.
Three possibilities in terms of пЃ„S:
If Suniv > 0…..process is spontaneous.
If Suniv < 0…..process is spontaneous in opposite
direction.
If Suniv = 0…. system is in equilibrium.
Three possibilities in terms of пЃ„G:
If G < 0…..process is spontaneous.
If G > 0…..process is spontaneous in opposite
direction.
If G = 0….system is in equilibrium
Spontaneous Processes: temperature dependence
Note that пЃ„G is composed of both пЃ„H and пЃ„S terms
пЃ„G = пЃ„H - TпЃ„S
A reaction is spontaneous if пЃ„G < 0. So,
If H < 0 and S > 0….spontaneous at all T
If H > 0 and S < 0….not spontaneous at any T
If H < 0 and S < 0…. becomes spontaneous at low T
If H > 0 and S > 0….becomes spontaneous at high T
Example: At what T is the following reaction
spontaneous?
Br2(l)
Br2(g)
where пЃ„HВ° = 30.91 kJ/mol, пЃ„SВ° = 93.2 J/mol.K
пЃ„GВ° = пЃ„HВ° - TпЃ„SВ°
Try 298 K just to see result at standard conditions
пЃ„GВ° = пЃ„HВ° - TпЃ„SВ°
пЃ„GВ° = 30.91 kJ/mol
- (298K)(93.2 J/mol.K)
пЃ„GВ° = (30.91 - 27.78) kJ/mol
= 3.13 kJ/mol > 0
Not spontaneous at 298 K
At what T then does the process become spontaneous?
пЃ„GВ° = пЃ„HВ° - TпЃ„SВ° = 0
T = пЃ„пЃ€п‚°/пЃ„Sп‚°
T = (30.91 kJ/mol) /(93.2 J/mol.K)
T = 331.65 K
Just like our previous calculation
Calculating пЃ„GВ°
In our previous example, we needed to determine
пЃ„HВ°rxn and пЃ„SВ°rxn separately to determine пЃ„GВ°rxn
But ∆ G is a state function; therefore, we can
use known пЃ„GВ° to determine пЃ„GВ°rxn using:
пЃ„ G п‚° rxn =
 G 
prod .
-   G  react
.
Standard пЃ„G of Formation: пЃ„GfВ°
Like пЃ„HfВ° and SВ°, the standard Gibbs free
energy of formation пЃ„GfВ° is defined as the
“change in free energy that accompanies the
formation of 1 mole of that substance for its
constituent elements with all reactants and
products in their standard state.”
As for пЃ„HfВ° , пЃ„GfВ° = 0 for an element in its standard
state:
Example: пЃ„GfВ° (O2(g)) = 0
Example
• Determine the G°rxn for the following:
C2H4(g) + H2O(l)
C2H5OH(l)
• Tabulated G°f from Appendix 4:
пЃ„GВ°f(C2H5OH(l)) = -175 kJ/mol
пЃ„GВ°f(C2H4(g)) = 68 kJ/mol
пЃ„GВ°f(H2O (l)) = -237 kJ/mol
• Using these values:
C2H4(g) + H2O(l)
пЃ„ G п‚° rxn =
 G 
prod .
C2H5OH(l)
-   G  react
.
пЃ„GВ°rxn = пЃ„GВ°f(C2H5OH(l)) - пЃ„GВ°f(C2H4(g))
-пЃ„GВ°f(H2O (l))
пЃ„GВ°rxn = -175 kJ - 68 kJ -(-237 kJ)
пЃ„GВ°rxn = -6 kJ < 0 ; therefore, spontaneous
More пЃ„GВ° Calculations
• Similar to H°, one can use the G° for
various reactions to determine пЃ„GВ° for the
reaction of interest (a “Hess’ Law” for G°)
• Example:
C(s, diamond) + O2(g)
CO2(g) пЃ„GВ° = -397 kJ
C(s, graphite) + O2(g)
CO2(g) пЃ„GВ° = -394 kJ
C(s, diamond) + O2(g)
CO2(g) пЃ„GВ° = -397 kJ
C(s, graphite) + O2(g)
CO2(g) пЃ„GВ° = -394 kJ
CO2(g)
C(s, graphite) + O2(g) пЃ„GВ° = +394 kJ
C(s, diamond)
C(s, graphite)
пЃ„GВ° = -3 kJ
G°rxn < 0…..rxn is spontaneous
пЃ„GВ°rxn в‰ Reaction Rate
• Although G°rxn can be used to predict if a
reaction will be spontaneous as written, it
does not tell us how fast a reaction will
proceed.
• Example:
C(s, diamond) + O2(g)
CO2(g)
пЃ„GВ°rxn = -397 kJ
<<0
But diamonds are forever…. G°rxn ≠rate
Example Problem
• Is the following reaction spontaneous under standard conditions?
4 KClO 3 ( s) п‚ѕ
п‚ѕ п‚® 3KClO 4 (s) пЂ« KCl ( s)
п‚ пЃ„HВ°f (kJ/mol)
SВ° (J/mol.K)
KClO3(s)
-397.7
143.1
KClO4(s)
-432.8
151.0
KCl (s)
-436.7
82.6
Example Problem Solution
• Calulating H°rxn
4 KClO 3 ( s) п‚ѕ
п‚ѕ п‚® 3KClO 4 (s) пЂ« KCl ( s)
пЃ„ H п‚° rxn = 3пЃ„ H п‚° f (KClO
4
) пЂ« пЃ„ H п‚° f (KCl ) - 4 пЃ„ H п‚° f (KClO 3 )
= 3(-432 .8 kJ ) пЂ« (- 436 .7 kJ ) - 4 ( 397 .7 kJ )
п‚ = -144 kJ
• Calulating S°rxn
п‚ пЃ„ S п‚° rxn = 3S п‚° (KClO
= 3(151 .0 J
= -36 .8 J
K
4
K
)пЂ«
S п‚° (KCl ) - 4 S п‚° (KClO
) пЂ« (82 .6 J
K
3
) - 4 (143 .1 J
)
K
)
Example Problem Solution
• Calulating G°rxn
пЃ„ G п‚° rxn = пЃ„ H п‚° rxn - T пЃ„ S п‚° rxn
(
= -144 kJ - (298 K )
-38 .6 J
пѓ¦ 1kJ пѓ¶
пѓ§
пѓ·
K пѓЁ 1000 J пѓё
)
= -133 kJ
п‚ пЃ„GВ°rxn < 0 ;therefore, reaction is spontaneous.
Example Problem Continued
For what temperatures will this reaction be spontaneous?
Answer: For T in which пЃ„Grxn < 0.
пЃ„ G rxn = пЃ„ H rxn - T пЃ„ S rxn
0 = пЃ„ H rxn - T пЃ„ S rxn
пЃ„ H rxn
пЃ„ S rxn
=
- 133 kJ
= 3446 K = T
пѓ¦ 1kJ пѓ¶
- 38 .6 J
пѓ§
пѓ·
K пѓЁ 1000 J пѓё
(
)
Spontaneous as long as T < 3446 K.
п‚ 
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