Transcript System strives for minimum

System
strives
for
minimum
Free
Energy
The standard free-energy of reaction (DG0rxn) is the freeenergy change for a reaction when it occurs under standardstate conditions.
aA + bB
cC + dD
0
DGrxn
= [cDG0f (C) + dDG0f (D) ] - [aDG0f (A) + bDG0f (B) ]
0
DGrxn
= S nDG0f (products) - S mDG0f (reactants)
Standard free energy of
formation (DG0f ) is the free-energy
change that occurs when 1 mole
of the compound is formed from its
elements in their standard states.
DG0f of any element in its stable
form is zero.
What is the standard free-energy change for the following
reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
0
DGrxn
= S nDG0f (products) - S mDG0f (reactants)
0
DGrxn
= [12DG0f (CO2) + 6DG0f (H2O)] - [ 2DG0f (C6H6)]
0
DGrxn
= [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is the reaction spontaneous at 25 0C?
DG0 = -6405 kJ < 0
spontaneous
18.4
DG = DH - TDS
DG gives direction of reaction
If A  B
then B  A
A  B
DG < 0
DG > 0
DG = 0
DG does not indicate the rate of the process.
Temperature and Spontaneity of Chemical Reactions
CaCO3 (s)
CaO (s) + CO2 (g)
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 nonspontaneous?
DG0 = 0 = DHsys -TDSsys
DHsys / DSsys = T
177.8 kJ / 0.1605 kJ/K = T
T = 835 0C
Watch units !!!
18.4
Gibbs Free Energy and Phase Transitions
DG0 = 0 = DH0 – TDS0
H2O (l)
DS =
H2O (g)
DH
40.79 kJ
=
T
373 K
= 109 J/K
18.4
Gibbs Free Energy and Chemical Equilibrium
DG = DG0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient = [Products]0 or P°Products
[Reactants]0
P°Reactants
Can correct Gibbs Free Energy change for non-standard state
18.4
What is the standard free-energy change for the following
reaction at 25 0C?
N2 (g) + 3H2 (g)
2 NH3 (g)
0
DGrxn
= S nDG0f (products) - S mDG0f (reactants)
0
DGrxn
=
2DGf0 (NH3) -
DG0f (N2) -
3DGf0 (H2)
0
DGrxn
= [ 2(–16.66 kJ/mole) – 0 – 0 = - 33.32 kJ
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
= -33.32 kJ + 11.4 kJ = -21.92 kJ
0.102
(0.10)(0.10)3
Gibbs Free Energy and Chemical Equilibrium
DG = DG0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
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
What is Kp for the following reaction at 25 0C?
N2 (g) + 3H2 (g)
2 NH3 (g)
-DG0 / RT
Kp = e
- (-33.32 kJ)/(0.008314 kJ /molK)(298K)
Kp = e
13.4
Kp = e
Kp = 6.6 x 105
DG0 < 0
DG0 > 0
18.4
DG0 = - RT lnK
18.4
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 < 1, reinforces that this is a
weak base and left side is favored.
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
What is Kp for the following reaction at 25 0C?
H2 (g) + I2 (g)
-DG0 / RT
Kp = e
2 HI (g)
DG0 = 2.6 kJ
- (2.6 kJ)/(0.008314 kJ /molK)(298K)
=e
=e
-1.05
Kp = 0.350
What is Kp at 50 0C?
DH0rxn = 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
Calculate DG0 for the following reaction at 25 0C?
2 NO (g) + O2 (g)
Given:
DH°
f
2 NO2 (g)
S°
NO 90.37 kJ/mole 210.62 j/molK
0
205.0
O2
NO2
33.84
240.45
DG0 = DH0 – TDS0
DH0rxn = 2 x DH0(NO2) – [2 x DH0(NO) + DH0(O2)]
f
f
f
= 2(33.84 kJ) - 2(90.37 kJ) - 0 = -113.06 kJ
DS0rxn = 2 x S0(NO2) – [2 x S0(NO) – S0(O2)]
= 2(240.45J/molK) - 2(210.62 J/molK) - 205.0 j/molK
= -145.34 J/K
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
0
DGrxn
= 2 x DG0f (NO2) – [2 x DG0f (NO) + DGf 0(O2)]
= 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