21 Lecture Some Thermodynamics
clarifications
Thursday November 17, 2011
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Method 1 Determine the free energy change in
the following reaction at 298 K:
2 H2O(g) + O2(g) 2 H2O2(g).
G, kJ/mol
105.6
0
228.6
Substance
H2O2(g)
O2(g)
H2O(g)
2
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PracticeDetermine the free energy change in the
following reaction at 298 K:
2 H2O(g) + O2(g) 2 H2O2(g).
Given: standard free energies of formation from Appendix IIB
Find: G (kJ)
Conceptual
Plan:
Relationships:
Gf of prod and react
Solution:
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G
Method 2 Determine the free energy change in
the following reaction at 298 K:
2 H2O(g) + O2(g) 2 H2O2(g).
Substance
H2O2(g)
O2(g)
H2O(g)
4
H, kJ/mol
136.3
0
241.8
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S, J/mol
232.7
205.2
188.8
PracticeDetermine the free energy change
in the following reaction at 298 K:
2 H2O(g) + O2(g) 2 H2O2(g).
Given: standard energies from Appendix IIB, T = 298 K
H = 211.0 kJ, S = 117.4 J/K, T = 298 K
Find: G (kJ)
G (kJ)
Conceptual H S of prod & react
f
Plan:
Relationships:
Solution:
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H, S
G
Methods comparison
Method 1: Use the tabulated
values for standard G.
G
0
rxn
= mG
0
(f)products
nG (f)reactants
0
Method 2: Use the Gibbs free
energy equation:
Gsys = Hsys TSsys
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PracticeDetermine the free energy change in
the following reaction at 298 K:
2 H2O(g) + O2(g) 2 H2O2(g).
Substance
H2(g) + O2(g) H2O2(g)
2 H2(g) + O2(g) 2 H2O(g)
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G, kJ/mol
105.6
457.2
PracticeDetermine the free energy change in
the following reaction at 298 K:
2 H2O(g) + O2(g) 2 H2O2(g).
Given: H2(g) + O2(g) H2O2(g)
2 H2(g) + O2(g) 2 H2O(g)
Find: G of 2 H2O(g) + O2(g) 2 H2O2(g)
G = 105.6 kJ
G = 457.2 kJ
Rel: Manipulate the given reactions so they add up to the
reaction you wish to find. The sum of the Gs is the
G of the reaction you want to find.
Solution: 2 H2(s) + 2 O2(g) 2 H2O2(g)
G = 211.2 kJ
2 H2O(g) 2 H2(g) + O2(g)
G = +457.2 kJ
2 H2O(g) + O2(g) 2 H2O2(g)
G = +246.0 kJ
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G under Nonstandard Conditions
G = G only when the reactants and
products are in their standard states.
their normal state at that temperature
partial pressure of gas = 1 atm
concentration = 1 M
Under nonstandard conditions, G = G +
RTlnQ.
Q is the reaction equilibrium, quotient.
At G = 0.
G = RTlnK
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Example 22.2 Calculate G at 298 K for
the reaction under the given conditions:
2 NO(g) + O2(g) 2 NO2(g) G = 71.2 kJ.
Substance
P, atm
NO(g)
O2(g)
0.100
0.100
NO2(g)
2.00
Given: nonstandard conditions, G, T
Find: G (kJ)
Conceptual
Plan:
Relationships:
G, PNO, PO2, PNO2
Solution:
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G
PracticeCalculate Grxn for the given Substance
reaction at 700 K under the given
N2(g)
conditions: N2(g) + 3 H2(g) 2 NH3(g)
H2(g)
G =+46.4 kJ.
NH (g)
P, atm
33
99
2.0
3
Given: nonstandard conditions, G, T
Find: G (kJ)
Conceptual
Plan:
Relationships:
G, PNO, PO2, PNO2
Solution:
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G
PracticeCalculate Grxn for the given
reaction at 700 K under the given
conditions:
N2(g) + 3 H2(g) 2 NH3(g) G =+46.4 kJ.
Substance
N2(g)
H2(g)
NH3(g)
12
P, atm
33
99
2.0
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Example 17.10 Calculate K at
298 K for the reaction
N2O4(g) 2 NO2(g).
Substance Gf, kJ/mol
N2O4(g)
+99.8
+51.3
NO2(g)
Given: standard free energies of formation from Appendix IIB
Find: K
Conceptual
Plan:
Relationships:
Gf of prod & react
K
G = RTlnK
Solution:
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G
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Relationship of Go and Keq
G = 0 is a criterion for equilibrium at a
single temperature, the one temperature at
which the equilibrium state has all reactants
and products in their standard states
G and Go are related through the reaction
quotient, Q
G = Go + RT ln Q
Under the conditions of G = 0
and Q = Keq, the equation above
Go = RT ln Keq
becomes
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The Dependence of G and Keq on Temperature
To obtain equilibrium constants at different
temperatures, it will be assumed that H does
not change much with temperature.
To obtain Keq at the desired temperature, the
vant Hoff equation is used:
H
ln Keq = + constant
RT
K2
H
ln =
K1
R
or
[]
1
1
T1
T2
The form used depends on whether we have a single value of Keq
available, or multiple values.
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Example 22.10
Consider this reaction at 298 K:
CO(g) + H2O(g)
CO2(g) + H2(g)
H298 = 41.2 kJ
Determine Keq for the reaction at 725 K.
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Real Reactions
In a real reaction, some of the free
energy is lost as heat.
if not most
Therefore, real reactions are
irreversible.
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