# S = J/mol-K

```HW 12
1.
The equilibrium constant of a reaction is 2.48e+02 at 369 K and 4.96e+03 at 491 K. Determine the value
of Ho for this reaction:
Ho=
3.699e+01
kJ/mol
πΎπΎ2
1
ππ
8.31
4.69ππ3
βπ»π» ππ 1
οΏ½ − οΏ½ → βπ»π» ππ = οΏ½
οΏ½ ln οΏ½
οΏ½
οΏ½ ln οΏ½ οΏ½ = οΏ½
1
1
1
1
ππ ππ1 ππ2
πΎπΎ1
2.48ππ2
−
−
ππ1 ππ2
369 491
ππππ
= 36
ππππππ
ln(πΎπΎ2 ) = ln(πΎπΎ1 ) +
Based on the higher temperature equilibrium constant, determine the value of
So=
1.461e+02
So for this reaction:
J/mol-K
βS o βπ»π» ππ
−
ππ
ππππ
βπ»π» ππ
36.99ππ3
π½π½
βππ ππ = ππ ln πΎπΎ +
= 8.31 ln(4.96ππ3) +
= 146
ππ
8.31 ∗ 491
ππππππ πΎπΎ
lnK =
2. The heat of vaporization of a compound at 300. K is 36.4 kJ/mol and its vapor pressure is 9.6 torr.
Determine the following thermodynamic properties of the vaporization at 300K:
10.90
Go =
kJ/mol
Hint : The vaporization reaction of material A is A(l) A(g). Think about the expression for the
equilibrium constant for this reaction, and what its value would be based on its vapor pressure given
above.
9.6
ππππ
βπΊπΊ ππ = −ππππ ln πΎπΎ = −8.31(300) ln οΏ½
οΏ½ = 10.9
760
ππππππ
So =
85.0
J/mol-K
βπΊπΊπ£π£π£π£π£π£ = βπ»π»π£π£π£π£π£π£ − ππβπππ£π£π£π£π£π£ → βπππ£π£π£π£π£π£ =
βπΊπΊπ£π£π£π£π£π£ − βπ»π»π£π£π£π£π£π£ 10.9 − 36.4
π½π½
=
= 85.0
ππ
300
ππππππ πΎπΎ
Assume that Ho and So are temperature independent and estimate the normal boiling point of the
compound to the nearest degree Celsius.
t=
155
ππππππ =
o
C
βπ»π»π£π£π£π£π£π£ 36400
=
= 428 πΎπΎ = 155ππ πΆπΆ
βπππ£π£π£π£π£π£
85
Hint: At the normal boiling point, the vapor pressure equals 1 atm. Given the value of K at 1 atm
pressure, what do you know about Go at the normal boiling point?
```