Thermodynamics Chapter 19 Liquid benzene Production of quicklime Solid benzene ⇅ CaCO 3 (s) ⇌...
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Transcript of Thermodynamics Chapter 19 Liquid benzene Production of quicklime Solid benzene ⇅ CaCO 3 (s) ⇌...
![Page 1: Thermodynamics Chapter 19 Liquid benzene Production of quicklime Solid benzene ⇅ CaCO 3 (s) ⇌ CaO + CO 2.](https://reader035.fdocuments.us/reader035/viewer/2022062421/56649d445503460f94a215c6/html5/thumbnails/1.jpg)
ThermodynamicsThermodynamicsChapter 19Chapter 19
Liquid benzene
Production of quicklimeProduction of quicklime
Solid benzene
⇅
CaCO3 (s) CaO + CO⇌ 2
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Gibbs Energy
For a constant-pressure & constant temperature process:
G = Hsys - TSsysGibbs energy
(G)
G < 0 The reaction is spontaneous in the forward direction
G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction
G = 0 The reaction is at equilibrium
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Fig 19.17 Analogy between Potential Energy and Free Energy
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Fig 19.18 Free Energy and Equilibrium
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aA + bB cC + dD
G°rxn nG° (products)f= mG° (reactants)f-
Standard free-energy of reaction (Gorxn) ≡ free-energy
change for a reaction when it occurs under standard-state conditions.
Standard free energy of formation (G°)
• Free-energy change that occurs
when 1 mole of the compound
is formed from its elements
in their standard states.
f
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Fig 19.19 Energy Conversion
What’s “Free” About Gibbs Energy?
• ΔG ≡ the theoretical maximum amount of work that can bedone by the system on the surroundings at constant P and T
• ΔG = − wmax
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What’s “free” about the Gibbs energy?What’s “free” about the Gibbs energy?
• “ “Free” does not imply that the energy has no costFree” does not imply that the energy has no cost
• For a constant-temperature process, “free energy”For a constant-temperature process, “free energy”
is the amount available to do workis the amount available to do work
e.g., Human metabolism converts glucose toe.g., Human metabolism converts glucose to
COCO22 and H and H22O with a O with a ΔΔG° = -2880 kJ/molG° = -2880 kJ/mol
This energy represents approx. 688 CalThis energy represents approx. 688 Cal
or about two Snickers bars worth... or about two Snickers bars worth...
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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.
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(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.
G = Hsys - TSsys
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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.
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Go = Hsys - TSsys
• The reaction is nonspontaneous at 500 oC
• The reaction is spontaneous at 25 oC
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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°?
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Gibbs Free Energy and Chemical Equilibrium
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
When not all products and reactants are in their standard states:
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G° = - RT lnK
Table 19.5
RToGΔ
eK
or
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Calculate ΔG° for the following process at 25 °C:
BaF2 (s) ⇌ Ba2+(aq) + 2 F− (aq); Ksp = 1.7 x 10-6
Example
ΔG = 0 for any equilibrium, so:
ΔG° = − RT ln Ksp
Equilibrium lies to the left
ΔG° = − (8.314 J/mol∙K) (298 K) ln (1.7 x 10-6)
ΔG° = + 32.9 kJ/mol
ΔG° ≈ + 33 kJ/mol
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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
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CC66HH1212OO66 ((ss)) + 6O + 6O22 ((gg) ) 6CO 6CO22 ((gg)) + 6H + 6H22O O ((ll)
Metabolism of glucose in humansMetabolism of glucose in humans
ΔΔG° = -2880 kJ/molG° = -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:
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Fig 19.20 Free Energy and Cell Metabolism
ADP + H3PO4 → ATP + H2O ΔG° = +31 kJ/mol
(Free energy stored)