Tro Chapter 17 - Free Energy and Thermodynamics
•  Spontaneous and non-spontaneous processes
Thermodynamics vs. Kinetics…Equilibrium?
•  Entropy and the second law of thermodynamics
•  Gibbs free energy (ΔG˚), entropy (ΔS˚) and enthalpy (ΔH˚)
•  Gibbs free energy and K
On-line HW due
Thursday Nov. 11th
at 11:59 PM
Perpetual motion, energy transactions and nature’s “heat tax”
Tro 17.2
Spontaneous vs. nonspontaneous change
Tro 17.2
Tro 17.3
Entropy and the second law of thermodynamics
Predicting spontaneous change
S = K ln W
CH4 + 2 O2
CO2 + 2H2O
ΔH = -802 kJ
2Fe (s) + 3/2O2
Fe2O3
ΔH = -826 kJ
H2O (l)
H2O (s)
ΔH = -6.02 kJ (spontaneous at T < 0˚C)
H2O (s)
H2O (l)
ΔH = +6.02 kJ (spontaneous at T > 0˚C)
NaCl (s)
Na+
(aq) + Cl- (aq) ΔH = +3.9 kJ
Entropy and Temperature/Phase Change
S = entropy
K =Boltzmann’s constant (1.38e-23 J/K)
W is the number of energetically
equivalent ways to arrange the
components of a system
Tro 17.3 and 17.4
Entropy and the second law of thermodynamics
Predicting relative S0 values of a system
1.  Temperature changes and S for copper metal
The entropy of the universe is always increasing
ΔSuniv = ΔSsys + ΔSsurr > 0
Entropy and the third law of thermodynamics
A perfect crystal has zero entropy at absolute zero
Ssys = 0 at 0 K
Tro 17.4
Temperature and its effect on spontaneity
T (K)
273
295
298
S (J/molK):
31.0
32.9
33.2
2. Physical states and phase changes and entropy
Na
H2O
C (graphite)
S˚ (s or l)
51.4 (s)
69.9 (l)
5.7 (s)
S˚ (g)
153.6
188.7
158.0
3. Dissolution of a solid or liquid
NaCl
AlCl3
CH3OH
S˚ (s or l)
72.1 (s)
167 (s)
127 (l)
S˚ (aq)
115.1
-148
132
Tro 17.4
Temperature and its effect on spontaneity cont’d
• Product favored reactions for exothermic processes with net increase in
entropy (+ ΔS)
• Reactant favored reactions for endothermic processes with a net decrease in
entropy (- ΔS)
e.g. consider the freezing of water - spontaneous at low
temperatures, nonspontaneous at high temperatures
But, what happens with endothermic processes with a net increase
in entropy
Or
What happens with exothermic processes and a decrease in entropy
Tro 17.5
Entropy, free energy and work
Gibb’s free energy, G, combines a system’s enthalpy
and entropy
Remembering the second law…
ΔSuniv>0 spontaneous
ΔSuniv = equilibrium
G = H - TS
ΔSuniv < 0 non-spontaneous
ΔSuniv = ΔSsys + ΔSsurr
Since absolute temperature is always positive,
ΔSsurr = -ΔHsys/T
-TΔSuniv < 0 for spontaneous processes
ΔSuniv = ΔSsys -ΔHsys/T
and, since ΔG = -TΔSuniv
-TΔSuniv = -TΔSsys +ΔHsys
ΔG < 0 for a spontaneous process
ΔGsys = ΔHsys -TΔSsys = -TΔSuniv
ΔG = 0 for a process at equilibrium
ΔG > 0 for a non-spontaneous process
Gibbs Free energy - “A decrease in G corresponds
to a spontaneous process”
Tro 17.6
Calculating the change in entropy of a reaction
Tro 17.7
Entropy, free energy, and work: Calculating Standard
Free Energy changes (see Appendix II B)
(using table 17.2 or Appendix IIB)
4KClO3 (s)
ΔS˚rxn = ΣmS˚products - ΣnS˚reactants
N2 (g) + 3H2 (g)
2NH3 (g)
ΔS˚rxn = [(2 mol NH3)(S˚ of NH3)] - [(1 mol N2)(S˚ of N2) + (3 mol H2)(S˚ of H2)]
3KClO4
Δ
+
KCl
ΔH˚f = -397.7 kJ/mol
ΔH˚f =-432.8 kJ/mol ΔH˚f = -436.5 kJ/mol
S˚ = 143.1 J/molK
S˚ = 151.0 J/molK
S˚ = 82.6 J/molK
Calculate ΔG˚ at 25˚C
1) Calculate ΔH˚ (must know ΔH˚f for each species)
= [2(192.8)]-[1(191.6)+3(130.7)]
= -198.76 J/K
2) Calculate ΔS˚ (must know S˚ for each species)
3) Calculate ΔG using ΔG˚ = ΔH˚ - TΔS˚
Calculating ΔG˚rxn from ΔG˚f values
4KClO3 (s)
ΔG˚f = -296.3 kJ/mol
Δ
3KClO4
+
ΔG˚f = -303.1 kJ/mol
KCl
ΔG˚f = -408.5 kJ/mol
ΔG˚rxn = ΣmG˚products - ΣnG˚reactants
ΔG˚rxn = [3(ΔG˚f KClO4) + ΔG˚f KCl] - [4 ΔG˚f KClO3]
Calculating ΔG˚rxn at elevated temperatures
2SO2+ O2
2SO3
At 298 K, ΔG˚ = -141.6 kJ, ΔH˚ = -198.4 kJ and ΔS˚ = -187.9 J/K
Is the reaction spontaneous at 25˚C?
Is the reaction spontaneous at 900˚C (assume ΔH˚ and ΔS˚ don’t
change much with temperature)?
Tro 17.8 and 17.9
Free energy for Non-standard states: ΔG˚rxn and ΔGrxn
Sign of ΔG˚ allows us to predict spontaneity and
thus reaction direction…
If Q<K, ln Q/K < 0; ΔG < 0
If Q>K, ln Q/K > 0; ΔG > 0
If Q=K, ln Q/K = 0; ΔG = 0
ΔG = RT ln Q/K = RT ln Q - RT ln K
ΔG˚ = -RT ln K (by choosing standard conditions for Q)
ΔG = ΔG˚ + RT ln Q
Tro 17.9
Free Energy and Equilibrium
ΔG˚ = -RT ln K
Free Energy and Equilibrium
Calculating ΔGrxn at non-standard conditions
2SO2+ O2
2 SO3
At 298 K, ΔG˚ = -141.6 kJ
a) Calculate K at 298 K and at 973 K (ΔG˚298 = -141.6 kJ/mol and
ΔG˚973 = -12.12 kJ/mol)
b) Which direction would the above reaction proceed if we had 0.500
atm of SO2, 0.0100 atm of O2 and 0.100 atm of SO3 at 25˚C? At
700˚C?
ΔG = -138.16 kJ mol-1 @ 298 K
ΔG = -.906 kJ mol-1 @ 973 K
Useless expressions:
ΔG˚ = -RTlnK
ΔG = ΔG˚ + RTlnQ
1) For each of the following pairs, choose (circle) the substance
with the higher entropy per mole at a given temperature:
S˚ C2H4 = 219.4 Jmol-1K-1
S˚ C2H6 = 229.5 Jmol-1K-1
S˚ H2 = 130.58 Jmol-1K-1
S˚ N2O4 = 304.3 Jmol-1K-1
S˚ NO2 = 240.45 Jmol-1K-1
a) Ar (l) or Ar (g)
2) Using S˚ values provided above, calculate ΔS˚ values for the
following reactions. In each case, account for the sign of ΔS˚.
b) He (g) at 3 atm pressure or He (g) at 1.5 atm pressure
a)
C2H4 (g) + H2 (g)
b)
N2O4 (g)
C2H6 (g)
c) 1 mol of Ne (g) in 15.0 L or 1 mol of Ne (g) in 1.50 L
d) CO2 (g) or CO2 (s)
2NO2 (g)
3). For a certain chemical reaction, ΔH˚ = -35.4 kJ and ΔS˚ = -85.5 J/K.
Is the reaction exothermic or endothermic
Does the reaction lead to an increase or decrease in the disorder of the
system?
Calculate ΔG˚ for the reaction at 298 K
The value of Ka for nitrous acid (HNO2) at 25˚C is 4.5x10-4
By using the value of Ka, calculate ΔG˚ for the dissociation of nitrous acid in
aqueous solution.
What is the value of ΔG at equilibrium
Is the reaction spontaneous at 298 K
What is the value of ΔG when [H+] = 5.0 x 10-2 M, [NO2-] = 6.0 x 10-4 M, and
[HNO2] = 0.20 M?