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Entropy, Free Energy, and Equilibrium

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... energy of formation (DG0) is the free-energy change that occurs when 1 mole of ... [ 2( 16.66 kJ/mole) 0 0 = - 33.32 kJ ... – PowerPoint PPT presentation

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Title: Entropy, Free Energy, and Equilibrium


1
System strives for minimum Free Energy
2
(No Transcript)
3
What is the standard free-energy change for the
following reaction at 25 0C?
DG0 -6405 kJ
lt 0
spontaneous
18.4
4
DG DH - TDS
DG gives direction of reaction
If A ? B DG lt 0 then B ? A
DG gt 0 A ? B DG 0
DG does not indicate the rate of the process.
5
Temperature and Spontaneity of Chemical Reactions
DH0 177.8 kJ
DS0 160.5 J/K
DG0 DH0 TDS0
At 25 0C, DG0 130.0 kJ
When does reaction switch from being spontaneous
to non-spontaneous?
DG0 0 DHsys -TDSsys DHsys / DSsys T 177.8
kJ / 0.1605 kJ/K T T 835 0C
18.4
Watch units !!!
6
Gibbs Free Energy and Phase Transitions
DG0 0
DH0 TDS0
109 J/K
18.4
7
Gibbs Free Energy and Chemical Equilibrium
DG DG0 RT lnQ
R is the gas constant (8.314 J/Kmol)
T is the absolute temperature (K)
Q is the reaction quotient Products0 or
PProducts
Reactants0 PReactants
Can correct Gibbs Free Energy change for
non-standard state
18.4
8
What is the standard free-energy change for the
following reaction at 25 0C?
N2 (g) 3H2 (g) 2 NH3 (g)
DG0 (N2)
-
3DG0 (H2)
-

f
f
What is the standard free-energy change for the
following mixture of gases at 25 0C 0.10 atm
N2, 0.10 atm H2 and 0.10 atm NH3 ?
DG DG0 RT lnQ DG0 RT ln PNH32
PN2PH23
-33.32 kJ (0.008314 kJ /molK)(298K) ln
0.102
(0.10)(0.10)3
-33.32 kJ 11.4 kJ -21.92 kJ
9
Gibbs Free Energy and Chemical Equilibrium
DG DG0 RT lnQ
R is the gas constant (8.314 J/Kmol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
DG 0
Q K
0 DG0 RT lnK
DG0 - RT lnK
18.4
10
What is Kp for the following reaction at 25 0C?
N2 (g) 3H2 (g) 2 NH3 (g)
- (-33.32 kJ)/(0.008314 kJ /molK)(298K)
Kp e
13.4
Kp e
Kp 6.6 x 105
11
DG0 lt 0
DG0 gt 0
18.4
12
DG0 - RT lnK
18.4
13
DG0 - RT lnK
Consider the reaction
NH3 (aq) H2O (l) NH4 (aq) OH-
(aq)
Kb 1.8 x 10-5
What is DG0 for the reaction at 25 0C?
DG0 - RT lnK -(0.008314 kJ /molK)(298K)
ln(1.8 x 10-5)
DG0 27.08 kJ /mol
K lt 1, reinforces that this is a weak base and
left side is favored.
14
How does K change with Temperature?
At equilibrium
- RT lnK DG0 DH0 TDS0
RT lnK DH0 TDS0
constant
R lnK DH0/T DS0
Compare 2 Temperatures
R lnK1 DH0/T1 R lnK2 DH0/T2
R(lnK2 - lnK1) DH0/T1 - DH0/T2
ln(K2 / K1) DH0/R (1/T1 - 1/T2)
Similar to Clausius-Clapeyron Equation for vapor
pressure and rate constant change equation
15
What is Kp for the following reaction at 25 0C?
H2 (g) I2 (g) 2 HI (g)
DG0 2.6 kJ
- (2.6 kJ)/(0.008314 kJ /molK)(298K)
-1.05
e
e
Kp 0.350
What is Kp at 50 0C?

2(25.9 kJ/mole) 0 0 51.8 kJ
ln(K2 / K1) DH0/R (1/T1 1/T2)
ln(K2 / 0.350 (51.8 kJ )/(0.008314 kJ
/molK)(1/298K 1/323K)
lnK2 - ln0.350 (6.23 x 103)(2.60 x 10-4)
lnK2 1.62 - 1.05 0.60
K2 1.82
16
Calculate DG0 for the following reaction at 25
0C?
2 NO (g) O2 (g) 2 NO2 (g)
Given
f
DG0 DH0 TDS0
f
f
f
2(33.84 kJ) - 2(90.37 kJ) - 0 -113.06 kJ
2(240.45J/molK) - 2(210.62 J/molK) - 205.0
j/molK -145.34 J/K
17
DG0 DH0 TDS0
-113.06 kJ - (298 K)(- 0.14534 kJ/K -113.06
kJ 43.31 kJ -69.75 kJ
Can also calculate with DG0 s from tabulated data
f
f
f
f
2(51.84 kJ/mol) - 2(86.71 kJ/mol) - 0 103.68
kJ - 173.42 kJ -69.74 kJ
Get same answer either way DG0, DH0 and DS0 are
all state functions
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