Transcript Thermodynamics Chapter 19 Liquid benzene ⇅

Thermodynamics
Chapter 19
Production of quicklime
Liquid benzene
⇅
Solid benzene
CaCO3 (s) ⇌ CaO + CO2
Gibbs Energy
For a constant-pressure & constant temperature
process:
Gibbs energy
(G)
DG = DHsys - TDSsys
DG < 0
The reaction is spontaneous in the forward direction
DG > 0
The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction
DG = 0
The reaction is at equilibrium
Fig 19.17 Analogy between Potential Energy and Free Energy
Fig 19.18 Free Energy and Equilibrium
Standard free-energy of reaction (DGorxn) ≡ free-energy
change for a reaction when it occurs under standard-state
conditions.
aA + bB
cC + dD
DG°rxn = S nDG°f (products) - S mDG°f (reactants)
Standard free energy of formation (DG°)
f
• Free-energy change that occurs
when 1 mole of the compound
is formed from its elements
in their standard states.
What’s “Free” About Gibbs Energy?
• ΔG ≡ the theoretical maximum amount of work that can be
done by the system on the surroundings at constant P and T
• ΔG = − wmax
Fig 19.19 Energy Conversion
What’s “free” about the Gibbs energy?
• “Free” does not imply that the energy has no cost
• For a constant-temperature process, “free energy”
is the amount available to do work
e.g., Human metabolism converts glucose to
CO2 and H2O with a ΔG° = -2880 kJ/mol
This energy represents approx. 688 Cal
or about two Snickers bars worth...
Sample Exercise 19.9 Determining the Effect of
Temperature on Spontaneity
The Haber process for the production of ammonia
involves the equilibrium
Assume that ΔH° and ΔS° for this reaction do not change
with temperature.
(a) Predict the direction in which ΔG° for this reaction
changes with increasing temperature.
(b) Calculate the values ΔG° of for the reaction at 25 °C
and 500 °C.
DG = DHsys - TDSsys
(a) The temperature dependence of ΔG° comes from the
entropy term.
•
We expect ΔS° for this reaction to be negative because
the number of molecules of gas is smaller in the products.
•
Because ΔS° is negative, the term –T ΔS° is positive and
grows larger with increasing temperature.
•
As a result, ΔG° becomes less negative (or more positive)
with increasing temperature.
•
Thus, the driving force for the production of NH3 becomes
smaller with increasing temperature.
Sample Exercise 19.9 Determining the Effect of
Temperature on Spontaneity
The Haber process for the production of ammonia
involves the equilibrium
Assume that ΔH° and ΔS° for this reaction do not change
with temperature.
(a) Predict the direction in which ΔG° for this reaction
changes with increasing temperature.
(b) Calculate the values ΔG° of for the reaction at 25 °C
and 500 °C.
DGo = DHsys - TDSsys
• The reaction is spontaneous at 25 oC
• The reaction is nonspontaneous at 500 oC
Gibbs Free Energy and Chemical Equilibrium
• We need to distinguish between ΔG and ΔG°
• During the course of a chemical reaction, not all
products and reactants will be in their standard states
• In this case, we use ΔG
• When the system reaches equilibrium, the sign of ΔG°
tells us whether products or reactants are favored
• What is the relationship between ΔG and ΔG°?
Gibbs Free Energy and Chemical Equilibrium
When not all products and reactants are in their
standard states:
ΔG = ΔG° + RT lnQ
R ≡ gas constant (8.314 J/K•mol)
T ≡ absolute temperature (K)
Q ≡ reaction quotient = [products]o / [reactants]o
At Equilibrium:
ΔG = 0
Q=K
0 = ΔG° + RT lnK
ΔG° = − RT lnK
DG° = - RT lnK
or
Table 19.5
Ke
 ΔGo
RT
Example
Calculate ΔG° for the following process at 25 °C:
BaF2 (s) ⇌ Ba2+(aq) + 2 F− (aq);
Ksp = 1.7 x 10-6
ΔG = 0 for any equilibrium, so:
ΔG° = − RT ln Ksp
ΔG° = − (8.314 J/mol∙K) (298 K) ln (1.7 x 10-6)
ΔG° = + 32.9 kJ/mol
ΔG° ≈ + 33 kJ/mol
Equilibrium lies to the left
Thermodynamics in living systems
• Many biochemical reactions have a positive ΔGo
• In living systems, these reactions are coupled to a
process with a negative ΔGo (coupled reactions)
• The favorable rxn drives the unfavorable rxn
Metabolism of glucose in humans
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l)
ΔG° = -2880 kJ/mol
• Does not occur in a single step as it would in simple
combustion
• Enzymes break glucose down step-wise
• Free energy released used to synthesize ATP from ADP:
Fig 19.20 Free Energy and Cell Metabolism
ADP + H3PO4 → ATP + H2O
ΔG° = +31 kJ/mol
(Free energy stored)