Post on 14-Apr-2018
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Lecture 10: The Meaning of ΔG
• Reading: Zumdahl 10.10-10.13
• Outline
– A descriptive example of ΔG – Temperature Dependence of K
– van’t Hoff Relation for K and T.
• Problems:
– Z10.6, Z10.71 (Demo), Z10.72, Z10.74
– Z10.77-79, Z10.81, Z10.90 (qualitative)
7/27/2019 KF lecture
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A Descriptive Example
• Consider the following gas-phase
“reaction”:
OClO (g) ClOO (g)
• Now, ΔGrxn = ΔG°rxn +RTln(Q)= -RTln(K) + RTln(Q)
= RTln(Q/K)
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Review
OClO (g) ClOO (g)
Using ΔGrxn = RTln(Q/K)
Qualitatively:
If Q/K < 1 ΔGrxn < 0
If Q/K > 1 ΔGrxn > 0
OClO (g) ClOO (g)
OClO (g) ClOO (g)
If Q/K = 1 ΔGrxn = 0 equilibrium
Both Forms Exist in equimolar amounts
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Sample Calculations
• K = 0.005
• Reaction (as written) is Not Spontaneous in
standard state.
ΔGrxn = RTln(Q/K)
= RTln(.001/.005) = RTln(.2) = -1.6RT < 0
Q = PClOO/POClO = .001
ΔG°rxn = -RTln(K) = +13.1 kJ/mol
• Let’s start the rxn off with 5 atm of OClO and
0.005 atm of ClOO. (Lots of Reactants.)
ΔGrxn < 0, reaction is proceeding towards
products, at fixed T and fixed Total Pressureof 5.005 atm.
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Finding Equilibrium
• Where is equilibrium? Back to Chem 142 and
ICE table for OClO (g) ClOO (g)
x = 0.02
POClO = 5 - .02 = 4.98 atm
( )
PClOO = 0.025 atm
( )
( )
0.005
(5 )
@ 0.005
ClOO
P
OClO
P
xP
Q x P x
EQ Q x K
+
= = −
= =
Initial Pressure StoicCoeffs
Change New Conc
P(OClO)=5 1(-) -x 5-x
P(ClOO)=0.005 1(+) +x 0.005+x
For this reaction the total pressure will not change as moles of reactant go to product.
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Change initial conditions
ΔGrxn = RTln(Q/K) Q = PClOO/POClO = 10
• Let’s start the rxn off with 5 atm. of ClOO and
0.5 atm of OClO, total pressure now 5.5 atm.
= RTln(10/0.005)
= RTln(2000) >> 0
ΔGrxn > 0, reaction is proceeding towards
reactants (i.e. backwards) with about 8kJ/mole
of driving energy.
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Temperature Dependence of K
• K cannot depend on pressure (it is define for 1 Atm of eachspecies)
• However it can depend on temperature (here’s how)
• We have two definitions for ΔG°
• Rearranging:
Plot of ln(K) vs 1/T is a straight line, the slope gives the enthalpy and
the intercept (at infinite T) gives the entropy for the reaction. Weassume Enthalpy and Entropy are independent of Temperature.
lno o o
rxn rxn rxnG RT K H T S Δ = − = Δ − Δ
1ln
o o
rxn rxn H S
K R T R
y m x b
Δ Δ= − ⋅ +
= +
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Plot lnK(T)
• If we measure K as a function of T, we candetermine ΔH° by determining the slope of theline
ln(K
)=
−Δ H °
R
1
T
⎝ ⎜
⎠⎟+
ΔS °
R
slope
intercept
Work Problems 10.72 and 10.74
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T Dependence of K
• Once we know the Temperature dependence of K,we can predict K at another temperature:
2 1
2 1
2 12 1
2
1 2 1
1 1ln( ) ln( )
1 1
ln( ) ln( )
1 1ln
H S H S K Subtct K R T R R T R
H
K K R T T
K H
K R T T
⎛ ⎞ ⎛ ⎞−Δ ° Δ ° −Δ ° Δ °= + = +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎧ ⎫⎛ ⎞ ⎛ ⎞−Δ ° ⎪ ⎪
− = −⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭
⎛ ⎞ ⎧ ⎫−Δ °⎛ ⎞= −⎨ ⎬⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠ ⎩ ⎭
The van’t Hoff equation.
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K(T) For a Specific Reaction
• For the following reaction:
CO(g) + 2H2(g) CH3OH(l)
What is K at T=T2=340 K?
• First, what is K eq when T =T1= 298 K?
( ) 5
1 1
30 ln 2.48ln
29ln 11.7
2.48
1.2 10
o
rxnG RT K K
K
K K T
Δ = − = − = −
= =
= = ⋅
30o
rxnkJ G
molΔ = −
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Calculate K(T)
To use the van’t Hoff Eq., we need Δ
H°
CO(g) + 2H2(g) CH3OH(l)
ΔHf °(CO(g)) = -110. kJ/mol
ΔHf °(H2(g)) = 0
ΔHf °(CH3OH(l)) = -240 kJ/mol
ΔH°rxn = ΣΔH°f (products) - ΣΔH° f (reactants)
= ΔH°f (CH3OH(l)) - ΔH°f (CO(g))= -239 kJ - (-110.5 kJ)
= -130 kJ
130o
rxnkJ H
molΔ = −
K(T)
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K(T)
• With ΔH°, we’re ready for the van’t Hoff Eq.
( )
{ }
2 1
1 2 1 1 2
32
1
3 5 3
2 1
1 1ln 1
130 2981 52 1 .88 6.32.48 340
exp 6.3 1.8 10
1.8 10 1.2 10 1.8 10 200
o
rxn H K T H
K R T T RT T
K
K
K K
−
− −
⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ−Δ °= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
− ⎛ ⎞= − = − − = −⎜ ⎟⎝ ⎠
⎛ ⎞= − = ⋅⎜ ⎟
⎝ ⎠= ⋅ ⋅ = ⋅ ⋅ ⋅ =
Why is K reduced (a lot!)?Reaction is Exothermic!
LeChatelier: Increase T, Shift Eq. To React.
K eq (T) will then decrease with increasing T;
Entropy is not part of the reason for the Change in K with T
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The Entropy for this reaction
• Compute the Entropy change for the reaction
• To save problems with dimensions (becauseenergies are in kJ and entropy is in J/K)
3 3130 30 10010 10 330298 300
o oo o o o rxn rxnrxn rxn rxn rxn
orxn
H GG H T S S
T
J S mol K
Δ − ΔΔ = Δ − Δ ⇒ Δ =
− +Δ = ⋅ = − ⋅ = − −
@ 25 2.48
130 30 402.48
o o
o rxn rxnrxn
kJ T C RT mol
H GS R R R RT
= ° =
Δ − Δ − +⎛ ⎞Δ = ⋅ = =⎜ ⎟⎝ ⎠
This is a better way to report entropy; all units are in R; uses the
ratio of kJ in energies; less prone to mistakes.
G l C l i
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General Conclusions
• K and ΔGo
are independent functions of T.• How K changes with T depends only on the sign
of the Enthalpy
• How ΔGo changes with T depends only on the
sign of the Entropy.
• Principles of LeChatelier are born out by van’tHoff equation.
• Now know where the equilibrium constants came
from and why and how they depend on thereaction energy (viz. Enthalpy)
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Gibbs Energy and Work (at fixed P and T)
• Preparation to talk about batteries (next chapter)
• From Calorimetry: to actually get heat (or an
enthalpy change) you actually have to change the number of
moles of something (X).
• Now to get work you need to consume some amount of
reactant (X) :
• This shows that to do work you need a (chemical) potential
times a (chemical) distance. And the work you get is
proportional to the amount of stuff you use up.
• The “useful” work (text) is non-PdV work, and is electrical
work; like a battery
• Calling the work useful and not useful (or PdV) work, whichthe book does, is not useful. (Sec 10.12-13).
rxn H H X Δ = Δ ⋅
max
electrical
rxnw G G X = Δ = Δ ⋅
ΔG d W k (Fi d T d P)
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ΔG and Work (Fixed T and P)
• Consider a reversible reaction, and work is the sum of thePdV part and extra work. Then
) ( )
( ) x x
rev rev rev rev
x nonPdV
rev rev
G H TS E PV T S
q P V w P V q w
G w w
Δ = Δ − Δ = Δ + Δ − Δ
= − Δ + + Δ − =
Δ = =
We previously showed that the reversible work is the
maximum work (as a negative number) you can getout of a system. Now see Gibbs energy is the same as
the maximum non-PdV work (i.e. electrical) we can
get from a chemical reaction when run at constant Tand P.