Three Laws of Thermodynamics 1. ΔE univ = 0 For the system: ΔE = q + w 2. ΔS univ ≥ 0 For the system: ΔS ≥ q/T, where the “=“ applies to reversible processes and the “>” applies to irreversible processes. 3. The entropy of a pure substance in perfect crystalline form at absolute zero is 0 J/K.
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Three Laws of Thermodynamics 1. ΔE univ = 0 For the system: ΔE = q + w 2. ΔS univ ≥ 0 For the system: ΔS ≥ q/T, where the “=“ applies to reversible processes.
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Three Laws of Thermodynamics
1. ΔEuniv = 0 For the system: ΔE = q + w
2. ΔSuniv ≥ 0 For the system: ΔS ≥ q/T,
where the “=“ applies to reversible processes and the “>” applies to irreversible processes.
3. The entropy of a pure substance in perfect crystalline form at absolute zero is 0 J/K.
Gibbs Energy Change at Constant T,P
ΔG < 0
Process is spontaneous. ΔG is the maximum P-V work that can be obtained from the reaction at constant T or is all of the non-PV work that can be obtained at constant T and P.
ΔG = 0
Process is at equilibrium. (at constant T and P.)
ΔG > 0
Process is not spontaneous. A minimum energy ΔG (at constant T and P) must be supplied to the system for the reaction to take place (but the reverse reaction is spontaneous).
Gibbs EnergyThe change in Gibbs Energy ΔG is the maximum non-PV work that can be obtained from a chemical reaction at constant T and P (or the minimum energy that must be supplied to make the reaction happen).
There two ways to obtain ΔG°:
1. ΔrxnG° = ΣmΔfG°(products) – ΣnΔfG°( reactants)
(m and n represent the stoichiometric coefficients)
2. ΔG° = ΔH° – TΔS°
Conventions Used in Establishing Standard Gibbs Energy G°
State of Matter Standard State
Element ΔfG° is 0 if it is in its standard state at the specified T.
Solution 1 M
Gas 1 bar
Liquid pure liquid
Solid pure solid
Finding ΔG° at a T other than 25°C
ΔG°(T) = ΔH°(25°C) – TΔS°(25°C)
You may not use the ΔG° values in the Appendix. You must calculate ΔG° from ΔH°(25°C) and ΔS°(25°C).
(We make the assumption that ΔH°(25°C) and ΔS°(25°C) don’t change with T.)
Finding ΔG° at a T other than 25°C
How much energy must be added to 1 mole of water at 75°C to form 1 mole of water vapor at 1 bar and 75°C?
Answer: The exothermicity of the process is enough to overcome the decrease in entropy at 65°C.
How do ΔH and ΔS Affect ΔG?
ΔG = ΔH - TΔS
ΔH ΔS ΔG - - - at low T, + at higher T - + - at all T (always spontaneous) + + + at low T, - at higher T + - + at all T (never spontaneous)
Finding ΔG Under Nonstandard Conditions
ΔG(T) = ΔG°(T) + RT ln Q
Q is the reaction quotient
R is the gas constant
R = 8.3145
The Reaction Quotient Q
Q is found by putting values into the expression for the equilibrium
constant K.
N2(g) + 3H2(g) 2NH3(g)
The values here do NOT have to be equilibrium values.
The values here MUST BE equilibrium values.
3
2
)()(
)(
22
3
eqHeqN
eqNHP PP
PK
3
2
)()(
)(
22
3
HN
NH
PP
PQ
Finding ΔG Under Nonstandard Conditions
ΔG(T) = ΔG°(T) + RT ln QFind the change in Gibbs Energy for the formation of
ammonia at 298 K when the reaction mixture consists of 0.50 bar N2, 0.75 bar H2, and 2.00 bar NH3.
You must always write the equation for the reaction first.
ΔG is an extensive property, like ΔH and ΔS. Its value depends on the stoichiometry.
Finding ΔG Under Nonstandard Conditions
ΔG(T) = ΔG°(T) + RT ln QFind the Gibbs Energy for the formation of ammonia at 298 K when the reaction mixture consists of 0.50 bar N2,
0.75 bar H2, and 2.00 bar NH3.
N2(g) + 3H2(g) 2NH3 (g) 3
2
)()(
)(
22
3
HN
NH
PP
PQ
Finding ΔG Under Nonstandard Conditions
ΔG(T) = ΔG°(T) + RT ln QFind the Gibbs Energy for the formation of ammonia at 298 K when the reaction mixture consists of 0.50 bar N2, 0.75 bar H2, and 2.00 bar NH3.
N2(g, 1bar, 25°C) + 3H2(g, 1 bar, 25°C) 2NH3 (g, 1 bar, 25°C)
The reaction is not spontaneous at 450°C under standard conditions. It takes 25.47 kJ of energy to make the reaction happen at this temperature.
23
2
21
2
3
)()(
)()450(
eqHeqN
eqNHP
PP
PCK
)723(3145.8
)47.25(
KKmol
JkJ
e
0145.0237.4 e
Calculating K from ΔG° The CRC Handbook of Chemistry and Physics gives ΔfG° = -623.42 kJ for carbonic acid and ΔfG° = -587.06 kJ for the bicarbonate ion, both at 25°C. Find Ka1 using this data.
First, we write the equation for the reaction:
H2CO3(aq) H+(aq) + HCO3-(aq)
ΔG°(1M, 25°C) = -587.06 - (-623.42) = 36.36 kJ
767.14)298(3145.8
)36360(
102.41
xeeeKK
Kmol
JJ
RT
G
a
Appendix D, Table 1 has Ka1 = 4.3 x 10-7
Calculating ΔG° from K Appendix D Table 3 gives the solubility product Ksp for copper(II) carbonate. Use this Ksp to find ΔG° for copper(II) carbonate dissolving in water at 25°C.