atkins & de paula: elements of physical chemistry: 5e
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Atkins & de Paula: Elements of Physical Chemistry: 5e. Chapter 7: Chemical Equilibrium: The Principles. End of chapter 7 assignments. Discussion questions: 2 Exercises: 1, 2, 3, 4, 5, 6, 7, 10, 11, 15, 18 Use Excel if data needs to be graphed. Homework Assignment. - PowerPoint PPT PresentationTRANSCRIPT
Atkins & de Paula: Elements of Physical Chemistry:
5e
Chapter 7: Chemical Equilibrium: The
Principles
End of chapter 7 assignmentsDiscussion questions:• 2
Exercises:• 1, 2, 3, 4, 5, 6, 7, 10, 11, 15,
18
Use Excel if data needs to be graphed
Homework Assignment
• How many of you have already read all of chapter 7 in the textbook?
• In the future, read the entire chapter in the textbook before we begin discussing it in class
Homework Assignment
• Connect to the publisher’s website and access all “Living Graphs”
• http://bcs.whfreeman.com/elements4e/
• Change the parameters and observe the effects on the graph
• Sarah: these “Living Graphs” are not really living; this is just a hokey name!
Homework Assignments
• Read Chapter 7.• Work through all of the
“Illustration” boxes and the “Example” boxes and the “Self-test” boxes in Chapter 7.
• Work the assigned end-of-chapter exercises by the due date
Principles of chemical equilibrium
Central Concepts:• Thermodynamics can predict
whether a rxn has a tendency to form products, but it says nothing about the rate
• At constant T and P, a rxn mixture tends to adjust its composition until its Gibbs energy is at a minimum
Gibbs Energy vs Progress of Rxn
• Fig 7.1 (158)• (a) does not go• (b) equilibrium with
amount of reactants ~amount of products
• (c) goes to completion
Example Rxns
• G6P(aq) F6P(aq)• N2(g) + 3 H2(g) 2 NH3(g)• Reactions are of this form: aA + bB cC +
dD• If n is small enough, then,
G = (F6P x n) – (G6P x n) --now divide by n
rG = G/n = F6P – G6P
The Rxn Gibbs Energy
rG = G/n = F6P – G6P rG is the difference of the chemical
potentials of the products and reactants at the composition of the rxn mixture
• We recognize that rG is the slope of the graph of the (changing) G vs composition of the system (Fig 7.1, p154)
Effect of composition on rG
• Fig 7.2 (154)• The relationship
of G to composition of the reactions
rG changes as n (the composition) changes
Reaction Gibbs energy
• Consider this reaction:aA + bB cC + dD
rG = (cC + dD) – (aA + bB) μJtμJ+ RT ln aJ (derived in sec 6.6)
• Chemical potential (μ) changes as [J] changes• The criterion for chemical equilibrium at
constant T,P is: rG = 0 (7.2)
Meaning of the value of rG
• Fig 7.3 (155)• When is rG<0? • When is rG=0?• When is rG>0?• What is the
signifi-cance of each?
Variation of rG with composition
For solutes in an ideal solution:• aJ = [J]/c, the molar concentration of J
relative to the standard value c = 1 mol/dm3
For perfect gases: • aJ = pJ/p, the partial pressure of J relative
to the standard pressure p = 1 barFor pure solids and liquids, aJ = 1
p155f
Variation of rG with composition
rG = (cC + dD) – (aA + bB) (7.1c)
rG = (cC + dD) – (aA + bB) (7.4a)
rG = {cGm(C)+dGm(D)} – {(aGm(A)+bGm(B)} (7.4b)
• 7.4a and 7.4b are the same
Is there an error in 7.1c in the textbook?
Variation of rG with composition
rG = rG + RT ln aA aB
a b
aC aDc d
Q = aA aB
a b
aC aDc d
( )
rG = rG + RT ln Q
Since
Then
Reactions at equilibrium
• Again, consider this reaction:aA + bB cC + dD
• Q, arbitrary position; K, equilibrium• 0 = rG + RT ln K and rG = –RT ln K
Q = aA aB
a b
aC aDc d
K = aA aB
a b
aC aDc d
equilibrium( )
Equilibrium constant
• With these equations….0 = rG + RT ln K rG = –RT ln K (7.8)
• We can use values of rG from a data table to predict the equilibrium constant
• We can measure K of a reaction and calculate rG
Relationship between rG and K• Fig 7.4 (157)• Remember, rG = –RT ln K• So, ln K = –(rG/RT)• If rG<0, then K>1; &
products predominate at equilibrium
• And the rxn is thermo-dynamically feasible
At K > 1, rG < 0At K = 1, rG = 0At K < 1, rG > 0
Relationship between rG and K• On the other hand…• If rG>0, then K<1 and the
reactants predominate at equilibrium…
• And the reaction is not thermo-dynamically feasible HOWEVER….
• Products will predominate over reactants significantly if K1 (>103)
• But even with a K<1 you may have products formed in some rxns
Relationship between rG and K
• For an endothermic rxn to have rG<0, its rS>0; furthermore,
• Its temperature must be high enough for its TrS to be greater than rH
• The switch from rG>0 to rG<0 corresponds to the switch from K<1 to K>1
• This switch takes place at a temperature at which rH - TrS = 0, OR….T = rH
rS
Table 7.1 Thermodynamic criteria of spontaneity
G = H – TS
Table 7.1 Thermodynamic criteria of spontaneity
G = H – TS
4. If H is positive and S is negative, G will always be positive—regardless of the temperature.
These two statements are an attempt to say the same thing.
GG = = HH –– T TSS1. If H is negative and S is positive, then G will
always be negative regardless of temperature.
2. If H is negative and S is negative, then G will be negative only when TS is smaller in magnitude than H. This condition is met when T is small.
3. If both H and S are positive, then G will be negative only when the TS term is larger than H. This occurs only when T is large.
4. If H is positive and S is negative, G will always be positive—regardless of the temperature.
GG = = HH –– TTSS
Factors Affecting the Sign of G
Gibbs Free Energy (Gibbs Free Energy (GG))
For a constant-temperature process:
G = Hsys – TSsys
The change in Gibbs free energy (G)
18.4
If G is negative (G<0), there is a release of usable energy,
and the reaction is spontaneous!
If G is positive (G>0), the reaction is not spontaneous!
G = H – TS
All quantities in the above equation refer to the system
For a constant-temperature process:
G = Hsys – TSsys
G < 0 The reaction is spontaneous in the forward direction.
G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.
G = 0 The reaction is at equilibrium.
18.4
Gibbs Free Energy (Gibbs Free Energy (GG))
aA + bB cC + dD
G0rxn dG0 (D)fcG0 (C)f= [ + ] – bG0 (B)faG0 (A)f[ + ]
G0rxn nG0 (products)f= mG0 (reactants)f–
The standard free-energy of reaction (G0 ) is the free-energy change for a reaction
when it occurs under standard-state conditions.
rxn
Gibbs Free Energy (Gibbs Free Energy (GG))
7.10p158
Who will explain this graph to the class?
Relationship between rG and K
• For an endothermic rxn to have rG<0, its rS>0; furthermore,
• Its temperature must be high enough for its TrS to be greater than rH
• The switch from rG>0 to rG<0 corresponds to the switch from K<1 to K>1
• This switch takes place at a temperature at which rH - TrS = 0, OR….T = rH
rS
Reactions at equilibrium• Fig 7.5 (162)• An endothermic
rxn with K>1 must have T high enough so that the result of subtract-ing TrS from rH is negative
• Or rH–TrS < 0• Set rH–TrS=0
and solve for TT = rH
rS
equilibrium
rG = rH – TrS
Reactions at equilibrium
equilibrium
rG = rH – TrS
Table 7.2 Table 7.2 Standard Gibbs energies of Standard Gibbs energies of formation at 298.15 K* (gases)formation at 298.15 K* (gases)
Table 7.2 Standard Gibbs energies of formation at 298.15 K* (liquids & solids)
Standard Gibbs Energy of Formation
• Fig 7.6 (159)• Analogous to
altitude above or below sea level
• Units of kJ/mol
The equilibrium composition• The magnitude of K is a qualitative indicator• If K 1 (>103) then rG < –17 kJ/mol @ 25ºC,
the rxn has a strong tendency to form products
• If K 1 (<10–3) then rG > +17 kJ/mol @ 25ºC, the rxn will remain mostly unchanged reactants
• If K 1 (10–3-103), then rG is between –17 to +17 kJ/mol @ 25ºC, and the rxn will have significant concentrations of both reactants and products p.160
Calculating an equilibrium concentration
•Example 7.1 (p165)
•Example 7.2 (p166)
Standard reaction Gibbs energy
rG = Gm(products) – Gm(reactants)
rG = rH – TrS
7.6 Kc and Kp
aA + bB cC + dD
Kc = [C]c[D]d
[A]a[B]b
In most cases
Kc Kp
aA (g) + bB (g) cC (g) + dD (g)
Kp = Kc(RT)n
Kp = pC
dpD
pA pBa b
c
Kp = Kc(RT)n
When does Kp = Kc ?
Derivation 7.1: Kc and Kp Atkins uses
Work through Derivation 7.1, p.162
K = Kc cRT
p ][vgas
K = Kc T
12.07K ][vgas
Substituting values for c, p, and R, we get
What is this K?
What is vgas?
Coupled reactions• Box 7.1 (164)• Weights as analogy
to rxns• A rxn with a large
rG can force another rxn with a smaller rG to run in its nonspontan-eous direction
• Enzymes couple biochemical rxns
Coupled reactions
• Biological standard state (pH = 7)• Typical symbols for standard state: ¤ ´ °
Read the last paragraph in Box 7.1 on p164 regarding ATP and the “high energy” bond
+¤
Equilibrium response to conditions
• What effect will a change in temperature, in pressure, or the presence of a catalyst have on the equilibrium position?
• Presence of a catalyst? None. Why? rG is unchanged, so K is not changed
• How about a change in temperature? • Or a change in pressure? Let’s see…
The effect of temperature
• Fig 7.7 (163) rG of a rxn that
results in fewer moles of gas increases with increasing T
rG of a rxn with no net change…
rG of a rxn that produces more moles of gas decreases with increasing T
Equilibrium response to conditions
• Le Chatelier’s principle suggests
When a system at equilibrium is compressed, the composition of a gas-phase equilibrium adjusts so as to reduce the number of molecules in the gas phase
p.172
The effect of pressure• Fig 7.9 (174)• A change in
pressure does not change the value of K, but it does have other consequences (composition)
• As p0, xHI1• What is [I2]?
H2(g) + I2(s) 2 HI(g)
Key Key IdeasIdeas
Key Key IdeasIdeas
The End…of this chapter…”
Spare parts to copy and paste
• μJtμJ+RT ln aJ
• Chemical potential (μ) changes as [J] changes
Box 7.1 pp.172f
• O2 binding in hemoglobin and myoglobin…
• …In resting tissue and in lung tissue
Chemical Potential
• Review pp.128-130, Partial molar properties (e.g. partial molar volume)
• Read p.129, last two paragraphs• Read handout, “Chemical Potential” by
Philip A. Candela• Chemical potential ( ) is usually
described as the “partial molar Gibbs function” or “partial molar Gibbs energy”
Chemical Potential
• The quantity G/n is so important that it is given a special symbol () and its own name (chemical potential)
• As the symbols G/n above indicate, chemical potential is the Gibbs free energy per mole of substance
• The chemical potential is an indication of the potential of a substance to be chemically active (p.130)
Excursus: Chemical Potential
• The standard chemical potential of a gas (μJ), is identical to its standard molar Gibbs energy (Gm) at 1 bar
• The greater the partial pressure of a gas, the greater its chemical potential
Excursus: Chemical Potential• Common expressions of chemical
potential:
μJtμJ+ RT ln aJ
μJtμJ+ RT ln
μJtμJ+ RT ln p
pp
p = 1 bar