ΔG
o
= -RT lnK
Slightly soluble
K is a small number
If K <1 (but bigger than 0), ΔG
o
= -RT ln(<1)
ln(<1) = negative numbr
ΔG
o
= positive number

Clicker Q10
Using the data provided estimate ΔH°
vap
for CCl
4
?
A.
2.86 kJ/mol
B.
-344 J/mol
C.
35 kJ/mol
D.
3.44 J/mol
Temp (°C)
Temp (°C)
P
P
vap
vap
(atm)
(atm)
4.3
4.3
0.0526
0.0526
23.0
23.0
0.131
0.131

Clicker Q10: Solution
Using the data provided estimate ΔH°
vap
for CCl
4
?
A.
2.86 kJ/mol
B.
-344 J/mol
C.
35 kJ/mol
D.
3.44 J/mol
Temp (°C)
Temp (°C)
P
P
vap
vap
(atm)
(atm)
4.3
4.3
0.0526
0.0526
23.0
23.0
0.131
0.131
)
1
1
(
ln
1
2
1
2
T
T
R
H
K
K
o
)
3
.
277
1
295
1
(
0526
.
0
131
.
0
ln
R
H
o
ΔH° = ln(2.49)(0.0083)/(2.16 x 10
-4
)
= 35 kJ/mol

Clicker Q11
Using your answer in Clicker Q10 and the data provided
estimate ΔS°
vap
for CCl
4
?
Temp (°C)
Temp (°C)
P
P
vap
vap
(atm)
(atm)
4.3
4.3
0.0526
0.0526
23.0
23.0
0.131
0.131
A.
0.0102 J/K
B.
101.7 J/K
C.
150.6 J/K
D.
I am lost

Clicker Q11: Solution
Using your answer in Clicker Q10 and the data provided
estimate ΔS°
vap
for CCl
4
?
A.
0.0102 J/K
B.
101.7 J/K
C.
150.6 J/K
D.
I am lost
Temp (°C)
Temp (°C)
P
P
vap
vap
(atm)
(atm)
4.3
4.3
0.0526
0.0526
23.0
23.0
0.131
0.131
R
S
T
R
H
K
o
o
)
1
(
)
ln(
ln(0.0526)
35
0.0083
(
1
277.3
)
S
o
0.0083
ΔS° = (ln(0.052) + 15.2)*(0.0083)
ΔS° = 0.101 kJ/K

For the fusion reaction of water:
H
2
0(s)
H
⇄
2
O(l)
At what temperature does ΔG = 0?
(ΔH°
fus
= 6.02 kJ/mol,
ΔS°
fus
= 22 J/mol-K)
A. 0.27°C
B.
0.27 K
C.
274 K
D.
Need more information
Clicker Q12

A. 0.27°C
B.
0.27 K
C.
274 K
D.
Need more information
Clicker Q12: Solution
T = ΔH°
fus
/ΔS°
fus
= 6.02/0.022
= 273.6 K
ΔG
o
= -RT lnK
= ΔH° – TΔS°
=0
ΔG
= 0 at equilibrium, so
T = ΔH° /ΔS°
For the fusion reaction of water:
H
2
0(s)
H
⇄
2
O(l)
At what temperature does ΔG = 0?
(ΔH°
fus
= 6.02 kJ/mol,
ΔS°
fus
= 22 J/mol-K)

A. ΔG
> 0
B.
ΔG = 0
C.
ΔG < 0
Clicker Q13
For
the fusion reaction of water:
H
2
0(s)
H
⇄
2
O(l)
What would you expect ΔG to be at T = 10°C?
(ΔH°
fus
= 6.02 kJ/mol,
ΔS°
fus
= 22 J/mol-K)

A. ΔG
> 0
B.
ΔG = 0
C.
ΔG < 0
Clicker Q13: Solution
For
the fusion reaction of water:
H
2
0(s)
H
⇄
2
O(l)
What would you expect ΔG to be at T = 10°C?
(ΔH°
fus
= 6.02 kJ/mol,
ΔS°
fus
= 22 J/mol-K)
At 10
o
C you are not at equilibrium:
ΔG = ΔG
o
= ΔH° – TΔS°
ΔG = ΔG
o
+ RT ln(1)
ΔG = ΔG
o
ΔG = ΔH° – TΔS°
ΔG = 6.02 – (283)(0.022)
= -0.206 kJ

A.
100 °C
B.
93 °C
C.
85 °C
D.
78 °C
Clicker Q14: Difficult one!
What is the boiling point of water in Denver, where
P
atm
is 0.823 atm?

Clicker Q14: Solution
What is the boiling point of water in Denver, where
P
atm
is 0.823 atm?
A.
100 °C
B.
93 °C
C.
85 °C
D.
78 °C
H
2
O(l)
H
⇄
2
O(g)
NOTE: Pvap is not = 1, ΔG
o
is NOT = 0
We cannot use the normal bp equation:
T = ΔH° /ΔS°
R
S
T
R
H
K
o
o
1
ln
If we can find ΔH° and
ΔS° we can find T

Clicker Q14: Solution
What is the boiling point of water in Denver, where
P
atm
is 0.823 atm?
ln(0.823)
44
0.0083
1
T
0.1189
0.0083
H
2
O(l)
H
⇄
2
O(g)
R
S
T
R
H
K
o
o
1
ln
ΔH
o
rxn
= 44 kJ
ΔS
o
rxn
= 0.119 kJ/mol K
(ln0.823) – 14.32 = -5301(1/T)
T = -(5315)/(-14.5)
T = 366 K – 273 = 93 °C

A.
The entropy of the system will increase under standard
conditions at 25
C.

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- Fall '13
- Thermodynamics, Reaction, Energy, Entropy, ΔG