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
𝐾𝐾2
1
𝑅𝑅
8.31
4.69𝑒𝑒3
∆𝐻𝐻 𝑜𝑜 1
� − � → ∆𝐻𝐻 𝑜𝑜 = �
� ln �
�
� ln � � = �
1
1
1
1
𝑅𝑅 𝑇𝑇1 𝑇𝑇2
𝐾𝐾1
2.48𝑒𝑒2
−
−
𝑇𝑇1 𝑇𝑇2
369 491
𝑘𝑘𝑘𝑘
= 36
𝑚𝑚𝑚𝑚𝑚𝑚
ln(𝐾𝐾2 ) = ln(𝐾𝐾1 ) +
Based on the higher temperature equilibrium constant, determine the value of
So=
1.461e+02
So for this reaction:
J/mol-K
∆S o ∆𝐻𝐻 𝑜𝑜
−
𝑅𝑅
𝑅𝑅𝑅𝑅
∆𝐻𝐻 𝑜𝑜
36.99𝑒𝑒3
𝐽𝐽
∆𝑆𝑆 𝑜𝑜 = 𝑅𝑅 ln 𝐾𝐾 +
= 8.31 ln(4.96𝑒𝑒3) +
= 146
𝑇𝑇
8.31 ∗ 491
𝑚𝑚𝑚𝑚𝑚𝑚 𝐾𝐾
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
𝑘𝑘𝑘𝑘
∆𝐺𝐺 𝑜𝑜 = −𝑅𝑅𝑅𝑅 ln 𝐾𝐾 = −8.31(300) ln �
� = 10.9
760
𝑚𝑚𝑚𝑚𝑚𝑚
So =
85.0
J/mol-K
∆𝐺𝐺𝑣𝑣𝑣𝑣𝑣𝑣 = ∆𝐻𝐻𝑣𝑣𝑣𝑣𝑣𝑣 − 𝑇𝑇∆𝑆𝑆𝑣𝑣𝑣𝑣𝑣𝑣 → ∆𝑆𝑆𝑣𝑣𝑣𝑣𝑣𝑣 =
∆𝐺𝐺𝑣𝑣𝑣𝑣𝑣𝑣 − ∆𝐻𝐻𝑣𝑣𝑣𝑣𝑣𝑣 10.9 − 36.4
𝐽𝐽
=
= 85.0
𝑇𝑇
300
𝑚𝑚𝑚𝑚𝑚𝑚 𝐾𝐾
Assume that Ho and So are temperature independent and estimate the normal boiling point of the
compound to the nearest degree Celsius.
t=
155
𝑇𝑇𝑏𝑏𝑏𝑏 =
o
C
∆𝐻𝐻𝑣𝑣𝑣𝑣𝑣𝑣 36400
=
= 428 𝐾𝐾 = 155𝑜𝑜 𝐶𝐶
∆𝑆𝑆𝑣𝑣𝑣𝑣𝑣𝑣
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?