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Unit V - Chemical Thermodynamics Chapters 6 & 18
18.1. Determine H° for a chemical
reaction from Hf° values of
reactants and products.
[Readings 6.8 Problems 76, 78, 80,
81, 82, & 102]
Review
Hºrxn =
Hºf(prod) -
Hºf(react)
CO(g) + NO(g) --> CO 2(g) + 1/2 N 2(g)
Hºf = -110.5 kJ
+90 kJ
mol
mol
x 1 mol
x 1 mol
-110.5 kJ
+90 kJ
|__________|
-20.5 kJ
-393.5 kJ
0 kJ
mol
mol
x 1 mol
x 1/2 mol
-393.5 kJ
0 kJ
|__________|
-393.5 kJ
Hºrxn = (-393.5 kJ) - (-20.5 kJ) = -373 kJ
Since the natural tendency is for spontaneous
processes to be exothermic -- then the reaction is
spontaneous???
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18.2. Recognize favorable
conditions for spontaneous
reactions. [Readings 18.4
Problems 15, 16, 17, 18, 56, & 60]
Spontaneity in Chemical Reactions
Chemical and physical processes that occur
without continued external influence -Spontaneous
Other processes occur only with external
influence -- Nonspontaneous
Spontaneous might be slow or fast.
Nonspontaneous reactions do occur.
The reverse of a spontaneous process is a
Nonspontaneous process (v.v.).
Examples of Spontaneous Processes
Ice Melting:
H2O(s) --> H 2O(l) @25ºC
Iron Rusting:
4Fe(s) + 3O2(g) + 6H2O(l) --> 4Fe(OH) 3(s)
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What drives a chemical reaction?
Enthalpy, H
Entropy, S
Enthalpy, H (see chapter 8)
Flow of energy (heat)
CH4(g) + 2 O2(g) --> CO2(g) +H 2O (l) + heat
CH4(g) + 2 O2(g) --> CO2(g) +H 2O (l))
H = - heat kJ
Many exothermic reactions are spontaneous
The natural tendency is to lower energy states
Enthalpy, H
Water freezing is an exothermic process.
Is it a spontaneous process?
At – 10°C?
At + 10°C?
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Enthalpy, H
NH 4NO3 (s) ––> NH4+ (aq) + NO 3– (aq) H° = + 26.2 kJ/mol
The reaction is endothermic
Is it spontaneous?
Entropy, S
A measure of “randomness” or molecular disorder
(degrees of freedom).
A gas has more disorder than a liquid, thus more
entropy.
evaporation:
H2O(l) --> H2O(g)
at 25ºC
Natural tendency is to increased disorder
S = Sproducts - Sreactants
Increased dissorder give positive S
Factors Affecting Entropy
Volume
increasing volume increases entropy
Temperature
increases the kinetic energy (motion)
Physical State
compare the entropy of a gas to a solid.
Molecular complexity
Cl 2 (g) ––> 2 Cl (g)
KHT (s) ––> K+ (aq) + HT– (aq)
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Predicting whether a reaction is
spontaneous
Example:
Ba(OH) 2•8 H 2O (s) + 2 NH 4Cl (s) ––> BaCl 2 (aq) + 2 NH 3 (aq)
+ 10 H 2O (l)
H° = + 80.3 kJ unfavorable
S° = + 428 J favorable
Predicting whether a reaction is
spontaneous
Spontaneous process:
Favored by decrease in H
( H< 0)
Favored by an increase in S ( S > 0)
Nonspontaneous process:
Favored by increase in H
( H > 0)
Favored by an decrease in S ( S< 0)
18.3. Determine S° for a
chemical reaction from S° values
of reactants and products.
[Readings 18.5 Problems 62, 64,
& 66]
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Standard Molar Entropies
S°
• The entropy of 1 mol of a pure substance at:
• 1 atm
• 25°C
• Units are J K-1 mol-1
• See values in table 17.1 and Appendix B
Sºrxn = Sº(prod) - Sº(react)
CO(g) + NO(g) --> CO 2(g) + 1/2 N 2(g)
Sº =
+198 J
+211 J
+214 J
+192 J
mol-K
mol-K
mol-K
mol-K
x 1 mol
x 1 mol
x 1 mol
x 1/2 mol
+198 J
+211 J
+214 J
+96 J
K
K
K
K
|__________|
|__________|
+409 J/K
+310 J/K
Sºrxn = (+310 J/K) - (+409 J/K) = -99 J/K
Since the natural tendency is for spontaneous
processes to increase entropy -- then is the reaction
nonspontaneous???
CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g)
Hºrxn = -373 kJ
Sºrxn = -99 J/K
Driving Forces Oppose
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18.4. Determine G° for a
chemical reaction from H° and
S° values. Also calculate the
temperature at which a process
becomes spontaneous.
[Readings 18.4 – 18.6 Problems
62, 64, 66, 68, 70, 72, & 74]
Gibbs Free–energy
• Gibbs free-energy predicts the spontaneity of
a chemical or physical process
• G= H–T S
• G<0
The process is spontaneous
• G=0
The process is at equilibrium
• G>0
The process is nonspontaneous
Example problem
• Determine whether a process will be
spontaneous at 500K if H = – 128 kJ and S
= + 35 J/K
• G = – 128 kJ – 500K * .035 kJ/K
• G = – 144.5 kJ
• Will the reaction ever be nonspontaneous?
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•Example problem
• Determine temperature at which a process
will become spontaneous if H = – 50 kJ and
S = – 35 J/K
• At equilibrium, G = 0
• H=T S
•Example problem
• Determine temperature at which a process
will become spontaneous if H = – 50 kJ and
S = – 35 J/K
• At equilibrium, G = 0
• H=T S
T=
∆H
−50kJ
=
−1 = 1430K
∆S −.035kJK
Free Energy:
energy available to do work
Standard State Conditions
1 atm pressure
1 M concentration
Temperature of 25°C
G° = Hº - T Sº
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CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g)
Hºrxn = -373 kJ
Sºrxn = -99 J/K
Gº = Hº - T Sº
Gº = (-373 kJ) - {298 K(-99 J/K)}
Gº = (-373 kJ) - (- 29,500 J)
Gº = (-373 kJ) - (- 29.5 kJ)
Gº = -343.5 kJ
18.5. Predict whether a chemical
reaction, as written, is spontaneous,
non-spontaneous, or at equilibrium.
[Readings 18.4 Problems]
Consider both driving forces
Hº
Sº
–
+
+
–
+
+
–
–
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Consider both driving forces
Hº
Sº
–
--->
+
--->
+
<---
–
<---
+
<---
+
--->
–
--->
–
<---
Consider both driving forces
Hº
Sº
–
--->
+
--->
+
<---
–
<---
+
<---
+
--->
–
--->
–
Spontaneous
<---
Consider both driving forces
Hº
Sº
–
--->
+
--->
Spontaneous
+
<---
–
<---
Nonspontaneous
+
<---
+
--->
–
--->
–
<---
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Consider both driving forces
Hº
Sº
–
--->
+
--->
Spontaneous
+
<---
–
<---
Nonspontaneous
+
<---
+
--->
Spontaneous
@ high Temp
–
--->
–
<---
Consider both driving forces
Hº
Sº
–
--->
+
--->
Spontaneous
+
<---
–
<---
Nonspontaneous
+
<---
+
--->
–
--->
–
Spontaneous
@ high Temp
Spontaneous
@ low Temp
<---
18.6. Determine G° for a
chemical reaction from Gf°
values of reactants and
products.
[Readings 18.69 Problems 70, 71,
72, 73, & 74]
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Gºrxn =
Gºf(prod) -
Gºf(react)
CO(g) + NO(g) --> CO 2(g) + 1/2 N 2(g)
Gºf =
-137 kJ
+87 kJ
mol
mol
x 1 mol
x 1 mol
-137 kJ
+87 kJ
|__________|
-50 kJ
-394 kJ
0 kJ
mol
mol
x 1 mol
x 1/2 mol
-394 kJ
0 kJ
|__________|
-394 kJ
Gºrxn = (-394 kJ) - (-50 kJ) = -344 kJ
Three Possible Conditions
Gº < 0 reaction is spontaneous
R --> P
Gº > 0 reaction is nonspontaneous
Gº = 0 reaction is at equilibrium
R <-- P
R <==> P
18.7. Calculate the standard free
energy for a chemical reaction from
the equilibrium constant (vv).
[Readings 18.9 Problems 86, 87,
88, 90, 92, & 94]
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Nonstandard conditions of
concentration
G = Gº + RT ln Q
Nonstandard conditions of
concentration
G = Gº + RT ln Q
@ equilibrium, G = 0, Q = K
G = Gº + RT ln Q
0 = Gº + RT ln K
Gº = -RT ln K
G° and K
Consider the equation at standard conditions (unit
concentration)
A + B <==> C + D
1M 1M
1M 1M
Kc = [C] [D] / [A] [B]
if Gº < 0;
reaction shifts to right; K c > 1
if Gº > 0;
reaction shifts to left;
if Gº = 0;
reaction does not shift; Kc = 1
Kc < 1
The free energy drives a chemical reaction toward
equilibrium
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G° and K
G° = -RT ln K
(R = 8.31 J / mol K)
CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g)
Gºrxn = -344 kJ / mol
-344 kJ / mol = -(8.31 J / mol K) (298 K) ln K
344 kJ / mol / (2480 J / mol) (1kJ / 10 3 J)= ln K
139 = ln K
K = 2 x 10 60
CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g)
Hºrxn = -373 kJ
Sºrxn = -99 J/K
Gºrxn = -343.5 kJ
K = 2 x 10 60
18.8. Determine the equilibrium
temperature, Te, for a chemical
reaction from H° and S° (vv).
[Problems web page]
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Consider the hypothetical reaction:
A <==> B
Hºrxn = -150 kJ
drives reaction R --> P
Sºrxn = -600 J/K drives reaction R <-- P
Therefore, the reaction is spontaneous at low
temperatures.
@ 25°C
Gºrxn = 28.8 kJ
@
0°C
Gºrxn = 13.8 kJ
@ -25°C
Gº rxn = -1.2 kJ
At Equilibrium, G = 0 kJ and T = T e
Gº = Hº - T Sº
0 = Hº - Te Sº
Hº = Te Sº
Hº / Sº = Te
-150 kJ / -600 J/K = T e
Te = -150 kJ / -0.600 kJ/K
Te = 250 K or -23°C
CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g)
Hºrxn = -373 kJ
Sºrxn = -99 J/K
Hº / Sº = T e
-373 kJ / -99 J/K = T e
-373 kJ / -0.099 kJ/K = T e
Te = 3770 K or 3500°C
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What is the melting point of NaCl(s)?
NaCl (s) <==> NaCl (l)
Hºrxn = 30.3 kJ
Sºrxn = .0282 kJ/K
Hº / Sº = T e
Te = 1074 K or 801°C
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