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Chapter 16.4-16.6 in our book
Spontaneity, Entropy, and Free Energy
2
Section 16.4
Free Energy
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Spontaneous Reactions
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3
Section 16.4
Free Energy
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Concept Check ?
A liquid is vaporized at its boiling point. Predict the signs of:
wqΔHΔSΔSsurr
ΔG
Explain your answers.
–
+
+
+
– ?
0
w = –PΔV
4
Section 16.5
Entropy Changes in Chemical Reactions
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Concept Check ?
Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant.
Predict the signs of:
ΔH ΔSsurr ΔSΔSuniv
Explain.
– + 0 +
Temperature Dependence of KTemperature Dependence of K
So, ln(K) ∝ 1/T
Free Energy and WorkFree Energy and Work The maximum possible useful work obtainable
from a process at constant temperature and pressure is equal to the change in free energy
The amount of work obtained is always less than the maximum
Henry Bent's First Two Laws of Thermodynamics
First law: You can't win, you can only break even
Second law: You can't break even
7
Chapter 16
Table of Contents
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16.4 Free Energy
16.5Entropy Changes in Chemical Reactions
16.6Free Energy and Chemical Reactions
16.7The Dependence of Free Energy on Pressure
16.8 Free Energy and Equilibrium
16.9 Free Energy and Work
8
Chapter 16
Table of Contents
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• CW: Finish Notes 16.3-16.6• CW: 2 Online Quizzes Ch. 16 and ch. 6 (Links on
podcast page)• Turn in Calorimetry Lab report IN BOX before leaving
today.• HW: Ch. 16 question packet due Thursday• HW: Make sure you completed the 9 question
QUIA.com quiz. You may link from podcast page for Tests/Quizzes.
9
Section 16.3
The Mole The Effect of Temperature on Spontaneity
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ΔSsurr
Heat flow (constant P) = change in enthalpy = ΔH
In terms of temperature, how would you describe an object that has an entropy value of 0?
0 KPerfect solid crystal with no
motion
Only Theoretical
It is not possible to reach absolute 0!
Entropy of universe is always increasing!
3rd Law of Thermodynamics
the entropy of a perfect crystalline substance is zero at absolute zero
See appendix in book for values
*Based on 0 entropy as a reference point, and calculations involving calculus beyond the scope of this
course, data has been tabulated for
Standard Molar Entropies
ΔSº
Pure substances, 1 atm pressure, 298 K
Standard Molar Entropies
ΔSº
1) Standard molar entropies of elements are not 0 (unlike ΔHºf).
(0 entropy is only theoretical; not really possible)
2) S.M.E of gases > S.M.E of liquids and solids.
(gases move faster than liquids)
3) S.M.E. increase with increasing molar mass.
(more potential vibrational freedom with more mass)
4) S.M.E. increase as the number of atoms in a formula increase.
(same as above)
Calculating Entropy Change in a Calculating Entropy Change in a ReactionReaction
Entropy is an extensive property (a function of the number of moles)
Generally, the more complex the molecule, the higher the standard entropy value
Calculating Entropy Change in a Calculating Entropy Change in a ReactionReaction
Units for ΔS
ΔS=J/mol•K
Since we are considering ΔS°
J/K are often used because moles are assumed and cancel in the calculations when considering
standard states.
Calculate the standard entropy change (ΔSº) for the following reaction at 298K
Al2O3(s) + 3H2(g)→2Al(s) + 3H2O(g)
Substance Sº(J/mol-K) at 298K
Al 28.32
Al2O3 51.00
H2O(g) 188.8
H2(g) 130.58
ΔSo = [2Sº(Al) + 3Sº(H2O)] - [Sº(Al2O3) + 3Sº(H2)]
ΔSorxn = ∑n So
(products) - ∑m So(reactants)
Al2O3(s) + 3H2(g)→2Al(s) + 3H2O(g)
= 180.3 J/K
Predict the sign of ΔSº of the following reaction.
2SO2(g) + O2(g)→ 2SO3(g)
Entropy decreases, -
Lets’ Calculate
Calculate the standard entropy change (ΔSº) for the following reaction at 298K
2SO2(g) + O2(g)→ 2SO3(g)
Substance Sº(J/mol-K) at 298K
SO2(g) 248.1
SO3(g) 256.7
O2(g) 205.0
ΔSº = -187.8 J K-1
• Spontaneous reactions result in an increase in entropy in the universe.
• Reactions that have a large and negative ΔΗ tend to occur spontaneously.
• Spontaneity depends on enthalpy, entropy, and temperature.
Predicting spontaneous reactions
Provides a way to predict the spontaneity of a reaction using a combination of enthalpy and entropy of a reaction.
Gibbs Free Energy(G)
21
Section 16.4
Free Energy
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Free Energy (G)
• A process (at constant T and P) is spontaneous in the direction in which the free energy decreases. Negative ΔG means positive ΔSuniv.
22
Section 16.4
Free Energy
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Free Energy (G)
• ΔG = ΔH – TΔS (at constant T and P)• all of these are in reference to the system
ΔΔH, H, ΔΔS, S, ΔΔG and SpontaneityG and Spontaneity
Value of ΔH Value of TΔS
Value of ΔG
Spontaneity
Negative Positive Negative Spontaneous
Positive Negative Positive Nonspontaneous
Negative Negative ??? Spontaneous if the absolute value of ΔH is greater than the absolute value of TΔS (low temperature)
Positive Positive ??? Spontaneous if the absolute value of TΔS is greater than the absolute value of ΔH (high temperature)
ΔΔG = G = ΔΔH - TH - TΔΔSSH is enthalpy, T is Kelvin temperatureH is enthalpy, T is Kelvin temperature
24
Section 16.4
Free Energy
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Effect of DH and DS on Spontaneity
ΔΔG = G = ΔΔH - TH - TΔΔSSH is enthalpy, T is Kelvin temperatureH is enthalpy, T is Kelvin temperature
Standard Free Energy ChangeStandard Free Energy Change ΔG0 is the change in free energy that will
occur if the reactants in their standard states are converted to the products in their standard states
ΔG0 cannot be measured directly The more negative the value for ΔG0, the
farther to the right the reaction will proceed in order to achieve equilibrium
Equilibrium is the lowest possible free energy position for a reaction
26
Section 16.4
Free Energy
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Exercise
The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.
Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm.
27
Section 16.4
Free Energy
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ExerciseThe value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.
Calculate ΔSsurr,
ΔSsurr = (-45,700 J/mol)/(72.5 +273 K) = -132.3
=-132 J/K·mol
28
Section 16.4
Free Energy
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Exercise
The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.
Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm.
ΔS = 132 J/K·mol
ΔSsurr = -132 J/K·mol
ΔG = 0 kJ/mol
29
Section 16.4
Free Energy
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FOR ME• At the boiling point of a substance, the liquid and gaseous forms of the substance are in equilibrium. That means that for the reaction:
X_liquid = X_gas
the free energy change is zero:
0 = ΔG = ΔH_vap - T_boiling*ΔS_vap
where ΔH_vap and ΔS_vap are the enthalpy and entropy of vaporization, respectively, and T_boiling is the boiling temperature (all this assume that a specific pressure has been picked, and is held constant).
Rearranging this, we have:
ΔH_vap = T_boiling*ΔS_vap
ΔH_vap/ΔS_vap = T_boiling
The boiling temperature is therefore given by the ratio of the enthalpy to the entropy of vaporization. If the enthalpy change required to vaporize a material increases, so does the boiling temperature. If the entropy change accompanying vaporization increases, the boiling temperature decreases.
Note that ΔH_vap is the energy needed to convert a specified amount of the substance from a liquid to a gas at constant pressure. Because it always takes energy to do this (boiling is always an endothermic reaction), ΔH_vap is always positive.
T is measured on the thermodynamic scale (in kelvins), and is always positive. Because ΔH_vap is always positive, this means that ΔS_vap is also always positive; the molar entropy of a gas is always larger than the molar entropy of the liquid.
For reactions at constant temperature:ΔG0 = ΔH0 - TΔS0
Calculating Free Energy Calculating Free Energy Method #1Method #1
Calculating Free Energy: Method #2Calculating Free Energy: Method #2
An adaptation of Hess's Law:Cdiamond(s) + O2(g) CO2(g) ΔG0 = -397 kJ
Cgraphite(s) + O2(g) CO2(g) ΔG0 = -394 kJ
CO2(g) Cgraphite(s) + O2(g) ΔG0 = +394 kJ
Cdiamond(s) Cgraphite(s) ΔG0 =
Cdiamond(s) + O2(g) CO2(g) ΔG0 = -397 kJ
-3 kJ
Calculating Free Energy Calculating Free Energy Method #3Method #3
Using standard free energy of formation (ΔGf0):
ΔGf0 of an element in its standard state is zero
The Dependence of Free Energy on The Dependence of Free Energy on PressurePressure
Enthalpy, H, is not pressure dependent Entropy, S
entropy depends on volume, so it also depends on pressure
Slarge volume > Ssmall volume
Slow pressure > Shigh pressure
Free Energy and EquilibriumFree Energy and Equilibrium
Equilibrium point occurs at the lowest value of free energy available to the reaction system
At equilibrium, ΔG = 0 and Q = K
ΔG0 KΔG0 = 0 K = 1ΔG0 < 0 K > 1ΔG0 > 0 K < 1
35
Section 16.5
Entropy Changes in Chemical Reactions
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Third Law of Thermodynamics
• The entropy of a perfect crystal at 0 K is zero.• The entropy of a substance increases with
temperature.
36
Section 16.5
Entropy Changes in Chemical Reactions
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Standard Entropy Values (S°)
• Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure.
ΔS°reaction = ΣnpS°products – ΣnrS°reactants
37
Section 16.5
Entropy Changes in Chemical Reactions
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Exercise
Calculate DS° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information:
S° (J/K·mol) Na(s) 51
H2O(l) 70
NaOH(aq) 50
H2(g) 131
DS°= –11 J/K
38
Section 16.5
Entropy Changes in Chemical Reactions
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Basically equilibrium and pressure effects will be covered at a later time
• We will mainly cover all of chapter 6 and 16.1-16.6 for test.
• ASSIGNMENTS: Hess’s Quiz graded; see score.• QUIA.com HW looked over (9 questions)• CW: PROBLEMS CH. 6 AND 16 PRACTICE ONLINE• SHOW ME SCORES• TURN IN ALL PREVIOUS ASSIGNMENTS• HW: SG ch 16 handout” pg. 389-391 #1-32 due Thursday
39
Section 16.5
Entropy Changes in Chemical Reactions
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WEDNESDAY - FEB. 13, 2013
• We will mainly cover all of chapter 6 and 16.1-16.6 for test - TEST is TUESDAY. Be well prepared. See ch. reviews also.
• Calorimetry Problems #4-#5 from LAB discussion.• ASSIGNMENTS: 2 online quizzes - show me answers• QUIA.com HW looked over (9 questions) - not for grade but
must complete - only 1/2 of you have done this.• CW/HW: Problems handout sheet from ch. 16 #1-32 (also be
sure you completed the problem packet from ch. 6• TURN IN ALL PREVIOUS ASSIGNMENTS
40
Section 16.6
Free Energy and Chemical Reactions
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Standard Free Energy Change (ΔG°)
• The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.
ΔG° = ΔH° – TΔS°
ΔG°reaction = ΣnpG°products – ΣnrG°reactants
41
Section 16.6
Free Energy and Chemical Reactions
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Concept Check
A stable diatomic molecule spontaneously forms from its atoms.
Predict the signs of:
ΔH° ΔS° ΔG°
Explain.
– – –
42
Section 16.6
Free Energy and Chemical Reactions
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Concept Check
Consider the following system at equilibrium at 25°C.
PCl3(g) + Cl2(g) PCl5(g)
ΔG° = −92.50 kJ
What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain. The ratio will decrease.
43
Section 16.7
The Dependence of Free Energy on Pressure
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Free Energy and Pressure
G = G° + RT ln(P)
or
ΔG = ΔG° + RT ln(Q)
I am not covering section 16.7-16.9 on this test.
44
Section 16.7
The Dependence of Free Energy on Pressure
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Concept Check
Sketch graphs of:
1. G vs. P2. H vs. P3. ln(K) vs. 1/T (for both endothermic and
exothermic cases)
45
Section 16.7
The Dependence of Free Energy on Pressure
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The Meaning of ΔG for a Chemical Reaction
• A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion.
46
Section 16.8
Free Energy and Equilibrium
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• The equilibrium point occurs at the lowest value of free energy available to the reaction system.
ΔG = 0 = ΔG° + RT ln(K)
ΔG° = –RT ln(K)
47
Section 16.8
Free Energy and Equilibrium
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Change in Free Energy to Reach Equilibrium
48
Section 16.8
Free Energy and Equilibrium
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Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction
49
Section 16.9
Free Energy and Work
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• Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy.
wmax = ΔG
50
Section 16.9
Free Energy and Work
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• Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.
• All real processes are irreversible.• First law: You can’t win, you can only break
even.• Second law: You can’t break even.