Post on 19-Dec-2015
Useful Information
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ΔE =q+ w
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ΔE =nCvΔT
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ΔH =nCpΔT
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w=−PextΔV
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wrev=−nRTlnVfinalVinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
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ΔS=dqrevT∫
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S=kln Ω( )
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Ω= A!ai!
i∏
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ΔS =nRlnVfinalVinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
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ΔS=nCP lnTfinalTinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
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ΔS=nCv lnTfinalTinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
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ΔG=ΔG°+RT (lnQ) and
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K=Q ( )at equilibrium
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aA+bB ⏐ → ⏐ ← ⏐ ⏐ cC+ dD
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K =C[ ]
c D[ ]d
A[ ]a B[ ]
b
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ΔH°rxn= cΔH°fproducts∑ − cΔH°f
reactants∑
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ΔS°rxn= cS°products∑ − cS°
reactants∑
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ΔG°rxn= cΔG°fproducts∑ − cΔG°f
reactants∑
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(lnK)=−ΔH°R
1T ⎛ ⎝ ⎜
⎞ ⎠ ⎟+ΔS°R
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lnK2
K1
⎛
⎝ ⎜
⎞
⎠ ⎟=−ΔH°R
1T2
−1T1
⎛
⎝ ⎜
⎞
⎠ ⎟
Lecture 10: ΔG, Q, and K
• Reading: Zumdahl 10.10, 10.11
• Outline– Relating ΔG to Q– Relating ΔG to K– The temperature dependence of K
Relating ΔG to Q
• Recall from Lecture 6: ΔS = R ln (Ωfinal/Ωinitial)
• For the expansion of a gas⏐⏐⏐ Ωfinal Volume
Relating ΔG to Q (cont.)
• Given this relationship ΔS = R ln (Vfinal/Vinitial)
€
ΔS = R ln
nRTPfinal
nRTPinitial
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟= R ln
PinitialPfinal
⎛
⎝ ⎜
⎞
⎠ ⎟= −R ln
PfinalPinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
Relating ΔG to Q (cont.)
• This equation tells us what the change in entropy will be for a change in concentration away from standard state.
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ΔSfi = ΔSfi°− R lnPfinalPinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
Entropy change for processoccurring under standardconditions
Additional term for changein concentration.
(1 atm, 298 K) (P ≠ 1 atm)
Relating ΔG to Q (cont.)
• How does this relate to ΔG?
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ΔGrxn = ΔH rxn − TΔSrxn
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ΔGrxn = ΔH orxn − TΔSorxn + RT ln
PfinalPinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
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ΔGrxn = ΔGrxno + RT ln
PfinalPinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
Relating ΔG to Q (cont.)
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ΔGrxn = ΔGrxno + RT ln
PfinalPinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
€
ΔGrxn = ΔGrxno + RT ln Q( )
• Generalizing to a multicomponent reaction:
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Q =
C[ ]
C[ ]reference
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
coeff
prod .
∏
C[ ]
C[ ]reference
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
coeff
react .
∏
• Where
An Example
• Determine ΔGrxn at 298 K for:
C2H4(g) + H2O(l) C2H5OH(l)
where PC2H4 = 0.5 atm (others at standard state)
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ΔGrxn = ΔGrxno + RT ln Q( )
ΔG°rxn = -6 kJ/mol (from Lecture 9)
An Example (cont.)C2H4(g) + H2O(l) C2H5OH(l)
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ΔGrxn = ΔGrxno + RT ln Q( )
ΔGrxn = -6 kJ/mol + (8.314 J/mol.K)(298K)ln(2)
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Q =1PC2H4
1atm
⎛
⎝ ⎜
⎞
⎠ ⎟
=1
0.5atm
1atm
⎛
⎝ ⎜
⎞
⎠ ⎟= 2
= -4.3 kJ/mol
ΔG and K
• The Reaction Quotient (Q) corresponds to a situation where the concentrations of reactants and products are not those at equilibrium.
• At equilibrium, we have K.
• What is the relationship between ΔG and K?
ΔG and K (cont.)
• At equilibrium, ΔGrxn = 0
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ΔGrxn = ΔGrxno + RT ln Q( )
0 K
0 = ΔG°rxn +RTln(K)
ΔG°rxn = -RTln(K)
ΔG and K (cont.)• Let’s look at the interaction between ΔG° and K
ΔG°rxn = -RTln(K)
If ΔG° < 0
then > 1
Products are favored over reactants
reactants
products
ΔGrxn
ΔG and K (cont.)• Let’s look at the interaction between ΔG° and K
ΔG°rxn = -RTln(K)
If ΔG° = 0
then = 1
Products and reactants are equally favored
reactants
products
ΔGrxn = 0
ΔG and K (cont.)• Let’s look at the interaction between ΔG° and K
ΔG°rxn = -RTln(K)
If ΔG° > 0
then < 1
Reactants are favored over products
reactantsΔGrxn
An Example
• For the following reaction at 298 K: HBrO(aq) + H2O(l) BrO-(aq) + H3O+(aq)
Ka = 2.3 x 10-9 What is ΔG°rxn?
ΔG°rxn = -RTln(K)= -RTln(2.3 x 10-9)
= 49.3 kJ/mol
An Example (cont.)
• What is ΔGrxn when pH = 5, [BrO-] = 0.1 M, and [HBrO] = 0.2 M ?
HBrO(aq) + H2O(l) BrO-
(aq) + H3O+(aq)
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Q =H 3O
+[ ] BrO
−[ ]
HBrO[ ]=
10−5( ) 0.1( )
(0.2)= 5x10−6
An Example (cont.)• Then:
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ΔGrxn = ΔGrxno + RT ln Q( )
= 49.3 kJ/mol + (8.314 J/mol.K)(298 K)ln(5 x 10-6)
= 19.1 kJ/mol
ΔGrxn < ΔG°rxn “shifting” reaction towards products
Temperature Dependence of K
• We now have two definitions for ΔG°
ΔG°rxn = -RTln(K)= ΔH° - TΔS°
• Rearranging (dividing by -RT)
y = m x + b
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ln(K ) =−ΔH°
R
1
T
⎛
⎝ ⎜
⎞
⎠ ⎟+
ΔS°
R
• Plot of ln(K) vs 1/T is a straight line
T Dependence of K (cont.)
• If we measure K as a function of T, we can determine ΔH° by determining the slope of the line
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ln(K ) =−ΔH°
R
1
T
⎛
⎝ ⎜
⎞
⎠ ⎟+
ΔS°
R
slope
intercept
T Dependence of K (cont.)
• Once we know the T dependence of K, we can predict K at another temperature:
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ln(K2 ) =−ΔH°
R
1
T2
⎛
⎝ ⎜
⎞
⎠ ⎟+
ΔS°
R-
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ln(K1) =−ΔH°
R
1
T1
⎛
⎝ ⎜
⎞
⎠ ⎟+
ΔS°
R
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lnK2
K1
⎛
⎝ ⎜
⎞
⎠ ⎟=
−ΔH°
R
1
T2
−1
T1
⎛
⎝ ⎜
⎞
⎠ ⎟
the van’t Hoff equation.
An Example• For the following reaction:
CO(g) + 2H2(g) CH3OH(l) ΔG° = -29 kJ/mol
What is K at 340 K?
• First, what is Keq when T = 298 K?
ΔG°rxn = -RTln(K)= -29 kJ/mol
ln(K298) = (-29 kJ/mol)
-(8.314 J/mol.K)(298K) = 11.7
K298 = 1.2 x 105
An Example (cont.)• Next, to use the van’t Hoff Eq., we need ΔH°
CO(g) + 2H2(g) CH3OH(l)
ΔHf°(CO(g)) = -110.5 kJ/mol
ΔHf°(H2(g)) = 0ΔHf°(CH3OH(l)) = -239 kJ/mol
ΔH°rxn = ΔH°f (products) - ΔH° f (reactants)
= ΔH°f(CH3OH(l)) - ΔH°f(CO(g)) = -239 kJ - (-110.5 kJ) = -128.5 kJ
An Example (cont.)• With ΔH°, we’re ready for the van’t Hoff Eq.
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lnK2
K1
⎛
⎝ ⎜
⎞
⎠ ⎟=
−ΔH°
R
1
T2
−1
T1
⎛
⎝ ⎜
⎞
⎠ ⎟
K340 = 2.0 x 102
Why is K reduced?
Reaction is Exothermic.
Increase T, Shift Eq. To React.
Keq will then decrease
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lnK340
1.2x105
⎛
⎝ ⎜
⎞
⎠ ⎟=
−(−128500 J)
(8.314 J /K)
1
340K−
1
298K
⎛
⎝ ⎜
⎞
⎠ ⎟
€
lnK340
1.2x105
⎛
⎝ ⎜
⎞
⎠ ⎟= −6.4