Einführung in die Physikalische Chemie, Herbstsemester 2013
(EPC 2013)
Solutions to Exercise sheet 6
Prof. Dr. Anatole von Lilienfeld
16 December 2013
1. Entropies
For molecules with nearly equal mass, and nearly equal number of atoms, we can predict the
trend in entropies from the vibrational modes. Let us recall (problem 1b of exercise sheet 5)
that an increase in entropy (at a particular Temperature) implies an increase in the accessibility
(or Boltzmann population) of more energy levels (i.e. accessibility of states that have more
closely spaced energy levels). Hence entropy is large for a molecule that has more low frequency
vibrational modes (or the soft modes). To a good approximation, one can use the following
trend in the vibrational frequencies
a.
νtorsion > νbend > νstretch
1. CO vs. CO2 : CO has one stretch mode, but CO2 has two stretch modes and a bend mode,
hence S (CO2 ) > S (CO). In other words, CO2 has a low energy mode (bending) which is absent
in CO.
2. propane vs. cyclopropane: The linear hydrocarbon propane has two low energy torsional
modes (internal rotation of methyl groups across the two C-C bonds) which are absent in cyclopropane (molecule is constrained, no rotation of methyl groups across C-C bonds is not possible),
hence S (propane) > S (cyclopropane).
3. pentane CH3 CH2 CH2 CH2 CH3 vs. neopentane C (CH)4 : Both the molecules have four torsional modes (internal rotation of methyl groups across the C-C bonds). However, in neopentane
internal rotation across the C-C bonds is sterically hindered by the presence of the three bulky
methyl groups, hence the torsional modes of neopentane have higher frequencies than the torsional modes of pentane. Thus S (pentane) > S (neopentane).
The trend in the entropy of reactions can be predicted by counting the increase in the number
of gas, liquid, and solid molecules, gained when going from the reactant to the product. Using
the fact that
b.
Sgas Sliquid > Ssolid
we can predict that the reaction in which more gas molecules are gained (or less gas molecules
are lost) will relatively have the higher entropy.
1
Reaction
Reactant (R)
sol. liq. gas
1
0
1
0
0
2
0
0
4
1
0
1
R1
R2
R3
R4
Product (P)
sol. liq. gas
0
0
1
0
1
0
0
1
1
0
0
2
Reaction (P-R)
sol. liq. gas
-1
0
0
0
1
-2
0
1
-3
-1
0
1
Hence ∆S (R4) > ∆S (R1) > ∆S (R2) > ∆S (R3).
The standard molar entropy of substance S ◦ in various phases is usually given at room
temperature, 298.15K. However, at this temperature, the substance may not exist in some of
the phases. In this problem, we are asked to calculate the standard molar entropy of gas
phase methanol (Sg◦ ), although methanol is a liquid at room temperature (Tboil = 337.7 K). By
considering the cyclic process as shown in the following gure, Sg◦ can be estimated using Tboil ,
gas
liq
∆H boil , C P (T )and C P (T ).
C.
∆S 2 = ∆H boil /Tboil
liquid, Tboil
∆S 1 =
R Tboil
T298.15K
gas, Tboil
∆S 3 =
liq
dT C P (T )/T
liquid, 298.15 K
R 298.15K
Tboil
gas
dT C P (T )/T
gas, 298.15 K
∆S 4 = Sl◦ − Sg◦
Since entropy is a state function, the change in entropy of a cyclic process vanishes, i.e.,
∆S 1 + ∆S 2 + ∆S 3 + ∆S 4 = 0
Using the relation ∆S 4 = Sl◦ − Sg◦ in the above equation, we arrive at
Sg◦
=
Sl◦
ˆ
Tboil
+
dT
T1
liq
C P (T )/T
ˆ
+ ∆H boil /Tboil +
T1
Tboil
gas
dT C P (T )/T
In this problem, we use constant heat capacities that are independent of the temperature, hence
liq
gas
Sg◦ = Sl◦ + C P ln (Tboil /298.15) + ∆H boil /Tboil + C P ln (298.15/Tboil )
liq
gas
= Sl◦ + ∆H boil /Tboil + C P − C P
ln (Tboil /298.15) .
Using the values (given in the problem), Tboil = 337.7 K, S liq (T1 ) = 126.8 J K−1 mol−1 , ∆H boil =
liq
gas
36.5 kJ mol−1 , C P = 81.12 J K−1 mol−1 , and C P = 43.8 J K−1 mol−1 , we get
◦
Sg,estimated
= 239.51 J K−1 mol−1
◦
which is in good agreement with the experimentally determined value Sg,exp.
= 239.8 J K−1 mol−1 .
2. Free energies
In this problem, using ∆H boil , and Tboil let us estimate ∆Gboil at temperatures near Tboil . For
this we assume that both ∆H boil , and ∆S boil are independent of Tboil . For benzene, the problem
states that Tboil = 80.09◦ C = 353.24 K, ∆H boil (353.24 K) = 30.72 kJ mol−1 , and ∆Gboil (353.24
K) = 0. Using these facts, and the relation ∆G = ∆H − T ∆S , we get
a.
∆S boil = ∆H boil /Tboil = 30.72 kJ mol−1 / (353.24 K) J K−1 mol−1 = 86.97 J K−1 mol−1 .
Now we can get ∆Gboil for the temperatures 75◦ C (= 348.15 K) and 85◦ C (= 358.15 K).
2
• ∆Gboil (348.15 K) =∆H boil − 348.15 ∆S boil = 441.4 J K−1 mol−1 > 0. Hence vaporization
of benzene does not happen spontaneously below the boiling point.
• ∆Gboil (358.15 K) =∆H boil − 358.15 ∆S boil = −428.4 J K−1 mol−1 < 0. Hence vaporization
of benzene happens spontaneously above the boiling point.
• ∆Gboil (353.24 K) =0. Hence gas and liquid phase of benzene are in equilibrium at the
boiling point.
Let us nd an expression for ∂V U |T starting from the Helmholtz free energy A = U − T S ⇒
U = A + T S.
b.
∂V U |T
= ∂V A|T + T ∂V S|T + S∂V T |T
= ∂V A|T + T ∂V S|T
(1)
Now let us use dA = dU − T dS − SdT and dU = δq + δw = T dS − P dV to get
dA = T dS − SdT − T dS − P dV
= −SdT − P dV
(2)
Writing dA as an exact dierential, we also have
dA = ∂T A|V dT + ∂V A|T dV
(3)
Using (2) and (3), we get the rst term on the right side of (1), i.e., ∂V A|T = −P . The condition
for exact dierential also yields that ∂V (∂T A|V )|T = ∂T (∂V A|T )|V ⇒ −∂V S|T = −∂T P |V which
gives us the second term on the right side of (1). Thus we can rewrite (1) as
∂V U |T
= −P + T ∂T P |V
• If P = RT /f (V ), we have T ∂T P |V = T R/f (V ) = P . Hence ∂V U |T = −P + T ∂T P |V =
−P + P = 0.
• Example for an equation of state where ∂V U |T vanishes: P V = αT (if α = R, we get the
ideal gas equation). In this case T = P V /α and P = αT /V ⇒ ∂T P |V = α/V . Hence
∂V U |T = −P + T ∂T P |V = −P + (P V /α) (α/V ) = −P + P = 0.
• Example for an equation of state where ∂V U |T does not vanish: (P + β) V = αT (if
α = R, β = a/V 2 , we get van der Waals equation state with b = 0). In this case
T = (P + β) V /α and P = αT /V − β ⇒ ∂T P |V = α/V . Hence ∂V U |T = −P + T ∂T P |V =
−P + ((P + β) V /α) (α/V ) = −P + P + β = β 6= 0.
• If you try P (V − β) = αT (if α = R, β = b, we get van der Waals equation state with
a = 0), ∂V U |T will vanish as in the ideal gas case!
We have ∂V U |T = −P + T ∂T P |V from the previous exercise. Writing dU as an exact
dierential, we have dU = ∂T U |V dT + ∂V U |T dV = CV dT + ∂V U |T dV . The condition for exact
dierential also yields that ∂V (∂T U |V )|T = ∂T (∂V U |T )|V ⇒ ∂V (CV ) |T = ∂T (∂V U |T ) |V =
c.
∂T (−P + T ∂T P |V ) |V = −∂T P |V +T ∂T2 P |V +(∂T T |V ) (∂T P |V ) = −∂T P |V +T ∂T2 P |V +∂T P |V =
T ∂T2 P |V .
• For an ideal gas, we have P = RT /V , hence ∂V (CV ) |T = T ∂T2 P |V = T ∂T (R/V ) |V = 0.
• For a van der Waals gas, we have P = RT / (V − b)−a/V 2 , hence ∂V (CV ) |T = T ∂T2 P |V =
T ∂T (R/ (V − b)) |V = 0.
3
√
• For a Redlich-Kwong gas, we have P = RT / (V − b)−a/
T V (V + b) , hence ∂V (CV ) |T =
√
√
T ∂T2 P |V = T ∂T R/ (V − b) + a/ 2T T V (V + b) |V = T −3a/ 4T 2 T V (V + b) =
√
−3a/ 4T T V (V + b) .
To show that dH = [V − T ∂T V |P ] dP + CP dT , it is sucient to show that ∂P H|T =
V − T ∂T V |P , because dH can be written as the exact dierential dH = ∂P H|T dP + ∂T H|P dT =
∂P H|T dP + CP dT . Now to nd an expression for ∂P H|T let us proceed as in exercise 2b, but
by starting from the Gibbs free energy G = H − T S ⇒ H = G + T S .
d.
∂P H|T
= ∂P G|T + T ∂P S|T + S∂P T |T
= ∂P G|T + T ∂P S|T
(4)
Now let us use dG = dH − T dS − SdT and dH = d(U + P V ) = dU + P dV + V dP = T dS + V dP
to get
dG = T dS + V dP − T dS − SdT
= −SdT + V dP
(5)
Writing dG as an exact dierential, we also have
dG = ∂T G|P dT + ∂P G|T dP
(6)
Using (5) and (6), we get the rst term on the right side of (1), i.e., ∂P G|T = V . The condition
for exact dierential also yields that ∂P (∂T G|P )|T = ∂T (∂P G|T )|P ⇒ −∂P S|T = ∂T V |P which
gives us the second term on the right side of (4). Thus we can rewrite (4) as
∂P H|T
= V − T ∂T V |P
The equation dH = [V − T ∂T V |P ] dP +CP dT implies that H is dependent on P and T , however,
if we use the relation T dS = CP dT , we get dH = [V − T ∂T V |P ] dP +T dS , where H is dependent
on its natural variables P and S .
4