CHM 3410 – Problem Set 5
Due date: Wednesday, October 3rd
Do all of the following problems. Show your work.
“When I thought I was stupid, I acted like a stupid person. And when I thought I was smart, I acted like a smart person and achieved like a smart person.” - Benjamin S. Carson
1) Starting with the relationship
dH = V dp + T dS (1.1)
derive the following.
a) (¶H/¶p)S = V (1.2)
b) (¶H/¶S)p = T (1.3)
c) (¶V/¶S)p = (¶T/¶p)S (1.4)
2) In class we derived the following "T dS" equation
T dS = Cp dT - aVT dp a = (1/V) (¶V/¶T)p
There is a second "T dS" equation that is often useful in thermochemistry
T dS = T (¶p/¶T)V dV + CV dT (2.1)
a) Starting with S º S(V,T) (that is, writing entropy as a function of V and T) derive equn 2.1.
b) Use equn 2.1 to find q when the volume of one mole of a gas obeying the van der Waals equation of state is changed isothermally and reversibly from an initial value Vi to a final value Vf.
3) Based on the Clapeyron equation (equn 4.6 of Atkins) there is a general relationship between the slope of the boundary between two phases (in a p vs T phase diagram) and the relative density of the phases. There are two cases to consider. If dp/dT > 0 then the lower temperature phase has a higher density (smaller molar volume) than the higher temperature phase. If dp/dT < 0 then the lower temperature phase has a lower density (larger molar volume) than the higher temperature phase.
Using the phase diagram below, determine which phase for water has the higher value for density.
a) ice II or ice III
b) ice II or ice V
c) ice VI or liquid
4) The normal freezing point for benzene (C6H6, M = 78.12 g/mol) is T°fus = 5.50 °C. The density of liquid and solid benzene at the normal freezing point are r(s) = 0.891 g/cm3, r() = 0.879 g/cm3, and the enthalpy of fusion for benzene is DH°fus = 10.59 kJ/mol.
a) Based on the above data, estimate the freezing point of benzene when p = 1000. atm (be careful about using consistent units).
b) The vapor pressure of liquid benzene above the triple point temperature (in units of torr) is given by the expression
log10p = 7.960 – 1780. K (4.1)
T
Using equn 4.1, find the pressure of benzene at the triple point, and the normal boiling point for benzene.
5) As shown in class, we can, by making a few reasonable assumptions, transform the Clapeyron equation (which applies to all first order phase transitions) into the Clausius-Clapeyron equation (which applies to phase transitions where the initial state is a solid or liquid and the final state is a gas).
a) Starting from the Clausius-Clapeyron equation
d ln(p) = - DH°vap (5.1)
d(1/T) R
derive the relationship
ln(p2/p1) = - DH°vap [ (1/T2) – (1/T1) ] (5.2)
R
Note that this is a more useful than the equivalent expression given by Equation 4.12 of Atkins.
b) The vapor pressure of pure nitric acid (HNO3) is pvap = 47.9 torr at T = 20.0 °C, and pvap = 208. torr at T = 50.0 °C. Using equn 5.2, estimate the value for DH°vap, and the normal boiling point for pure nitric acid.
EXTRA CREDIT The following data are for mercury II chloride (HgCl2 M = 271.50 g/mol), and are taken from Chase, M. W. The NIST-JANAF Thermochemical Tables, 4th Edition, 1998.
DH°fus = 19.41 kJ/mol DH°vap = 58.9 kJ/mol
T°fus = 550. K T°vap = 577. K
Using the above data, give the phase diagram for HgCl2. Include temperatures in the range 500 K - 600 K, and pressures in the range 0.0 atm - 2.0 atm. Include in your phase diagram the correct location (pressure and temperature) of the triple point. (Hint – For the solid-gas and liquid-gas phase boundaries use the Clausius-Clapeyron equation to find values of p and T at several different temperatures).
Solutions.
1) a) dH = V dp + T dS divide by dp
(dH/dp) = V + T (dS/dp) assume S is constant
(¶H/¶p)S = V
b) dH = V dp + T dS divide by dS
(dH/dS) = V (dp/dS) + T assume p is constant
(¶H/¶S)p = T
c) (¶H/¶p)S = V take ¶/¶S)p of both sides
(¶2H/¶S¶p)p,S = (¶V/¶S)p
(¶H/¶S)p = T take ¶/¶p)S of both sides
(¶2H/¶p¶S)S,p = (¶T/¶p)S
The order of taking the partial derivatives does not matter, and so the left hand sides of the above two relationships are equal. Therefore the right sides are also equal, and so
(¶V/¶S)p = (¶T/¶p)S
2) If we write entropy as a function of volume and temperature, that is S º S(V,T), then
dS = (¶S/¶V)T dV + (¶S/¶T)V dT Multiply both sides by T
T dS = T (¶S/¶V)T dV + T (¶S/¶T)V dT
But (from the Chapter 3 handout) (¶S/¶V)T = (¶p/¶T)V
T = (¶U/¶S)V
Substitution of the above gives
TdS = T (¶p/¶T)V dV + (¶U/¶S)V (¶S/¶T)V dT
= T (¶p/¶T)V dV + (¶U/¶T)V dT
= T (¶p/¶T)V dV + CV dT
b) For an isothermal reversible expansion dT = 0, and so
dq = T dS = T (¶p/¶T)V dV This is because dqrev = T dS
For a van der Waals gas
p = nRT - a n2
(V - nb) V2
So
(¶p/¶T)V = nR
(V - nb)
Integrating both sides gives us
òif dq = òif nRT/(V - nb)
q = nRT ln[ (Vf - nb)/(Vi - nb) ]
As a check, note that for b = 0 we get the result for an isothermal reversible expansion of an ideal gas.
3) a) For the ice II/ice III boundary dp/dT > 0, and so r(ice II) > r(ice III)
b) For the ice II/ice V boundary dp/dT < 0, and so r(ice II) < r(ice V)
c) For the ice VI/liquid boundary dp/dT > 0, and so r(ice VI) > r(liquid)
4) a) From the Clapeyron equation
dp = DHsub
dT T DVsub
For a small change in temperature dp/dT @ Dp/DT, and so
Dp = DHsub
DT T DVsub
or DT = (Dp) T (DVsub)
DHsub
To make sure the final answer will come out in appropriate units, all quantities in the above equation will be entered in MKS fundamental or derived units.
Dp = 1000. atm – 1. atm = 999. atm (1.0135 x 105 Pa/atm) = 1.012 x 108 Pa
T = 5.50 °C = 278.65 K
Vs = 78.12 g/mol = 87.68 cm3/mol (1 m3/106 cm3) = 87.68 x 10-6 m3/mol
0.891 g/cm3
V = 78.12 g/mol = 88.87 cm3/mol (1 m3/106 cm3) = 88.87 x 10-6 m3/mol
0.879 g/cm3
DV = V - Vs = (88.87 x 10-6 m3/mol) – (87.68 x 10-6 m3/mol) = 1.19 x 10-6 m3/mol
DHsub = 10.59 kJ/mol = 10590. J/mol
So DT = (1.012 x 108 Pa) (278.65 K) (1.19 x 10-6 m3/mol) = 3.17 K
(10590. J/mol)
Since the size of a °C and K are the same, the estimated freezing point of benzene at p = 1000. atm is
Tfus = (5.50 + 3.17) °C = 8.67 °C
b) The normal fusion point for benzene is T°fus = 5.50 °C = 278.65 K. Since the solid-liquid phase boundary is approximately vertical (see Figure 4.4, page 137 of Atkins for a “generic” phase diagram), and since the triple point occurs at a pressure less than 1.0 atm (because benzene is a liquid at room temperature) we can assume Ttpt @ T°fus = 278.65 K. (As a test, note that if we made these approximations for th triple point for water, we would be off by only 0.01 °C in the triple point temperature).
For the vapor pressure at this temperature we can use equn 4.1
log10p = 7.960 - 1780 K = 1.572
278.65 K
So p = 101.572 = 37.3 torr.
At the normal boiling point p = 760. torr (log10p = 2.881). If we solve equn 4.1 for T, we get
log10p = 7.960 - 1780 K
T
log10p - 7.960 = - 1780 K
T
T°vap = - 1780 K = 1780 K = 1780 K = 350.5 K ( = 77.3 °C).
(log10p - 7.960) 7.960 - log10p (7.960 – 2.881)
Note that the actual value for the normal boiling point is T°vap = 353.2 K. The small differencein our value is due to the failure to account to the temperature dependence of DH°vap and nonideal behavior of the vapor phase.
5) a) d ln(p) = - DH°vap
d(1/T) R
So d ln(p) = - (DH°vap/R) d(1/T)
If we integrate both sides of this equation from some initial state “1” to some final state “2”, then
òp1p2 d ln(p) = - òT1T2 (DH°vap/R) d(1/T)
ln(p2/p1) = - DH°vap [ (1/T2) – (1/T1) ]
R
b) If we solve equn 5.2 for DH°vap we get
DH°vap = - R ln(p2/p1)
[ (1/T2) – (1/T1) ]
If we substitute the information in the problem, we get
DH°vap = - (8.3145 J/mol.K) ln(208/47.9) = 38.550 kJ/mol
[ (1/323.15 K) – (1/293.15 K)]
We can now use the Clausius-Clapeyron equation to find the normal boiling point temperature. Since
ln(p2/p1) = - DH°vap [ (1/T2) – (1/T1) ]
R
Then
[ (1/T2) – (1/T1) ] = - R ln(p2/p1)
DH°vap
or, finally
(1/T2) = (1/T1) - R ln(p2/p1)
DH°vap
We can let p1 and T1 be one of the original data points, and p2 be the pressure at the normal boiling point. Substituting, we get
(1/T2) = (1/323.15 K) - (8.3145 J/mol.K) ln(760/208) = 0.002815 K-1
38550 J/mol
and so T2 = T°vap = (1/.002815 K-1) = 355.2 K ( = 82.1 K), the normal boiling point.
EXTRA CREDIT
To do the plot of the phase diagram we need to generate points on the boundaries between phases. Since the solid-liquid boundary is, for the range of pressures used in this problem, almost a straight vertical line, we can assume Ttpt = T°fus = 550. K.
For the solid-gas and liquid-gas boundaries we may use the Clausius-Clapeyron equation (equn 5.2 of the problem set). For T < 550. K we will have solid in equilibrium with vapor, while at T > 550. K we will have liquid in equilibrium with vapor. Since we know the enthalpy change for both of these phase transitions, the only other piece of information we need is the pressure and temperature for one point on the phase boundary.
Let’s start with the liquid-gas boundary first. We know the normal boiling point of the compound, and so p = 1.000 atm when T = 577. K. Substituting this information into the Clausius-Clapeyron equation gives
ln(p2) = ln(p1) - DH°vap [ (1/T2) – (1/T1) ] = - 7084.4 K [ (1/T2) – 0.0017331 ]
R
where the equation furthest to the right above is obtained by substituting values in for p1, T1, DH°vap, and R.
A table of temperatures and pressures for the liquid-gas boundary is given below
T(K) p(atm) T(K) p(atm) T(K) p(atm)
550. 0.547 570. 0.860 590. 1.311
555. 0.615 575. 0.958 595. 1.450
560. 0.689 580. 1.066 600. 1.601
565. 0.770 585. 1.183
Since we know Ttpt = 550. K, and that the triple point is on the liquid-gas phase boundary, we can say that ptpt = 0.547 atm, the value of pressure on the liquid-gas boundary at that temperature. If we use that point as p1 and T1 in the Clausius-Clapeyron equation, we can now construct the solid-gas phase boundary in the same way we previously constructed the liquid-gas phase boundary. We do need a value for DH°sub, which we can get by using the approximate relationship
DH°sub @ DH°fus + DH°vap = (19.41 kJ/mol + 58.9 kJ/mol) = 78.3 kJ/mol
ln(p2) = ln(p1) - DH°sub [ (1/T2) – (1/T1) ] = ( - 0.603) - 9419. K [ (1/T2) – 0.0018182 ]
R
where the equation furthest to the right above is obtained by substituting values in for p1, T1, DH°sub, and R.
A table of temperatures and pressures for the solid-gas boundary is given below
T(K) p(atm) T(K) p(atm) T(K) p(atm)
500. 0.099 520. 0.204 540. 0.399
505. 0.119 525. 0.242 545. 0.468
510. 0.143 530. 0.287
515. 0.171 535. 0.339
The data above are used to construct a phase diagran for HgCl2, which is given on the following page.