Thermodynamics
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Production of quicklime. Thermodynamics. Liquid benzene. ⇅. Solid benzene. Chapter 19. CaCO 3 (s) ⇌ CaO + CO 2. Gibbs Energy. For a constant-pressure & constant temperature process:. Gibbs energy (G). D G = D H sys - T D S sys.

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Thermodynamics

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Thermodynamics

Production of quicklime

Thermodynamics

Liquid benzene

Solid benzene

Chapter 19

CaCO3 (s) ⇌ CaO + CO2


Thermodynamics

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




Thermodynamics

Standard free-energy of reaction (DGorxn)≡ free-energy change for a reaction when it occurs under standard-state conditions.

aA + bB cC + dD

-

mDG° (reactants)

S

S

=

f

DG°

rxn

nDG° (products)

f

  • Standard free energy of formation (DG°)

  • Free-energy change that occurs

  • when 1 mole of the compound

  • is formed from its elements

  • in their standard states.

f


Thermodynamics

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


Thermodynamics

  • 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...


Thermodynamics

  • The Haber process for the production of ammonia involves the equilibrium

  • Assume that ΔH° and ΔS° for this reaction do not change with temperature.

  • 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.

Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity


Thermodynamics

D equilibriumG = DHsys - TDSsys

  • 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.


Thermodynamics

  • The Haber process for the production of ammonia involves the equilibrium

  • Assume that ΔH° and ΔS° for this reaction do not change with temperature.

  • 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.

Sample Exercise 19.9 Determining the Effect of Temperature on Spontaneity


Thermodynamics

D equilibriumGo = DHsys - TDSsys

  • The reaction is spontaneous at 25 oC

  • The reaction is nonspontaneous at 500 oC


Thermodynamics

Gibbs Free Energy and Chemical Equilibrium 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°?


  • Thermodynamics

    Gibbs Free Energy and Chemical Equilibrium 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:

    Q = K

    ΔG = 0

    0 = ΔG° + RT lnK

    ΔG° = −RT lnK


    Thermodynamics

    D equilibriumG° = -RT lnK

    or

    Table 19.5


    Thermodynamics

    Example equilibrium

    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

    Thermodynamics in living systems equilibrium

    • 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


    Thermodynamics

    C equilibrium6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O (l)

    Metabolism of glucose in humans

    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:


    Thermodynamics

    Fig 19.20 Free Energy and Cell Metabolism equilibrium

    ΔG° = +31 kJ/mol

    ADP + H3PO4→ ATP + H2O

    (Free energy stored)


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