CHM 3400 – Problem Set 5
Due date: Wednesday, February 22nd
Do all of the following problems. Show your work.
“Stupid entropy ruins everything.” – Jennifer Ouellette
1) The pressure of 2.000 moles of nitrogen gas (N2) is changed from an initial value pi = 10.00 atm to a final value pf
= 3.000 atm by some unspecified irreversible process. The process is isothermal, with T = T surr = 320.0 K. During
the process, 4955. J of heat is transferred from the surroundings to the system. Find Ssyst, Ssurr, and Suniv for the
process.
2) The temperature and pressure of 5.000 moles of Ar(g) are changed from initial values p i = 2.000 atm, Ti = 300.0 K
to final values pf = 1.000 atm, Tf = 400.0 K by some unspecified process. Over this range of pressures and
temperatures argon behaves like an ideal gas, and has a temperature independent constant pressure molar heat
capacity Cp,m = 20.78 J/molK. Find U, H, and S for the process.
3) Consider the process where heat is slowly added to a 1.000 mole sample of bromine (Br2) at a constant pressure p
= 1.000 atm. The initial and final temperature of the bromine are T i = 300.0 K and Tf = 400.0 K. The process can be
represented as
Br2(l, T = 300.0 K)  Br2(g, T= 400.0 K)
(3.1)
What is S for this process? Note that the normal boiling point of liquid bromine is Tvap = 332.4 K, and that the
enthalpy of vaporization for bromine is Hvap(Br2) = 29.45 kJ/mol. Also note that the constant pressure molar heat
capacities for liquid and gaseous bromine are Cp,m(Br2(l)) = 75.69 J/molK and Cp,m(Br2(g)) = 36.02 J/molK, and
that these values can be assumed to be constant over the range of temperatures in the problem.
HINT – Break the above process into three parts, one for the heating of the liquid to its normal boiling point, the
second for the   g phase transition, and the third for the heating of the vapor.
4) Using the data in the appendix of Atkins, find the value for Grxn, Hrxn, and Srxn the following reaction (at T
= 25.0 C).
CO(NH2)2(s) + H2O()  CO2(g) + 2 NH3(g)
(4.1)
Check to see if the relationship Grxn = Hrxn - TSrxn is obeyed, and also indicate whether or not the reactions is
spontaneous for standard conditions at T = 25.0 C.
5) For the reaction
MgCO3(s)  MgO(s) + CO2(g)
(5.1)
use the data in the appendix of Atkins to find the value for Grxn at T = 25.0 C and T = 100.0 C. To do the
calculation for T = 100.0 C you will need to first find the value for Hrxn and Srxn at that temperature, and then
use the equation relating to Grxn to Hrxn and Srxn to find Grxn. Also indicate whether or not the reaction is
spontaneous for standard conditions at T = 25.0 C and at T = 100.0 C. You may assume in this problem that the
heat capacities of the reactants and products are independent of temperature.
6) There are several methods that have been developed to estimate the normal boiling point for a pure liquid.
Using the thermochemical data given below, estimate the normal boiling point temperature (in C) for
benzene (C6H6) and ethyl alcohol (CH3CH2OH) by the following two methods
a) By finding the values for Hvap and Svap at T = 25.0 C, and assuming that these values are
approximately constant.
b) By using Trouton’s rule
Hvap/Tvap  85. J/molK
(6.1)
Compare your answers to the experimental values Tvap(benzene) = 80.1 C and Tvap(ethyl alcohol) = 78.5 C.
Data needed to do this problem are given below, at T = 25.0 C
Substance
C6H6()
C6H6(g)
CH3CH2OH()
CH3CH2OH(g)
Hf (kJ/mol)
Gf (kJ/mol)
S (J/molK)
49.0
82.93
124.3
129.72
173.3
269.31
- 277.69
- 235.10
- 174.78
- 168.49
160.7
282.70
7) The phase diagram for sulfur is given below. Note that there are two solid phases for sulfur, labeled rhombic and
monoclinic. Based on the phase diagram, put the four phases of sulfur (rhombic, monoclinic, liquid, gas) in order
from most dense to least dense.
Solutions.
1)
For the entropy change for the system we need a pathway that is reversible and which has the same initial
and final state as the irreversible process taking place. Since
n = 2.000 mol
T = 320.0 K
pi = 10.00 atm
pf = 3.000 atm
and since n and T are constant, an isothermal reversible expansion from p i = 10.00 atm to pf = 3.000 atm has the same
initial and final state as the actual process taking place, but is reversible. If we assume that N 2 behaves like an ideal
gas for the conditions of the problem, then we can use our previous result for the entropy change for the system for
an isothermal change in pressure for an ideal gas (derived in class, and also in “step 1” of problem 2, below)
Ssyst = nR ln(pi/pf)
= (2.000 mol) (8.3145 J/molK) ln(10.00/3.000) = 20.02 J/K
Since all that occurs during the process from the point of view of the surroundings is a transfer of heat from the
surroundings to the system, the process is reversible from the point of view of the surroundings. Therefore
Ssurr = if (dqrev,surr)/T = qsurr = - 4955. J = - 15.48 J/K
Tsurr
320.0 K
Finally, Suniv = Ssyst + Ssurr = (20.02 J/K) = ( - 15.48 J/K) = 4.54 J/K
2)
Since the gas is ideal, and Cp,m = 20.78 J/molK, then
CV,m = Cp,m - R = 20.78 J/molK - 8.3145 J/molK = 12.47 J/molK
For an ideal gas,
U = if CV dTU = n if CV,m dT
H = if Cp dTU = n if Cp,m dT
Since both heat capacities are constant over the temperature range of the problem
U = n CV,m (Tf - Ti) = (5.000 mol)(12.47 J/molK)(400.0 K - 300.0 K) = 6235. J
H = n Cp,m (Tf - Ti) = (5.000 mol)(20.78 J/molK)(400.0 K - 300.0 K) = 10390. J
To find the entropy change for the system we need a reversible pathway connecting the initial and final states of the
system. One such pathway is the following:
Step 1
Isothermal reversible expansion of the gas, from pi = 2.000 atm to pf = 1.000 atm, at T = 300. K
Step 2
Constant pressure heating of the gas from T i = 300.0 K to Tf = 400.0 K, at p = 1.000 atm
Step 1
S1 = if (dqrev/T)
The process is isothermal and so T can be taken out of the integral. This gives
S1 = if (dqrev/T) = (1/T) if (dqrev) = qrev/T
The gas is ideal, and so U = 0. From the first law, U = q + w, and so qrev = - wrev.
The work done by the hypothetical reversible expanion is
w = - if pex dV
The gas is ideal and the process is reversible and isothermal, and so p ex = p = nRT/V. Substituting, we get
wrev = - if (nRT/V) dV = - nRT ln(Vf/Vi)
From Boyle's law, for an isothermal process involving an ideal gas p iVi = pfVf, or (Vf/Vi) = (pi/pf)
wrev = - nRT ln(pi/pf)
qrev = - wrev = nRT ln(pi/pf)
S1 = qrev/T = (1/T) nRT ln(pi/pf) = nR ln(pi/pf)
= (5.000 mol)(8.3145 J/molK) ln(2.000/1.000) = 28.82 J/K
Step 2
S2 = if (dqrev/T)
The process is constant pressure, and so dqrev = n Cp,m dT
S2 = if (dqrev/T) = if (nCp,m/T)dT
Since Cp,m is constant over the temperature range of the process it can be taken outside of the integral, and so
S2 = if (nCp,m/T) dT = nCp,m if (dT/T) = nCp,m ln(Tf/Ti)
And so S2 = (5.000 mol)(20.78 J/molK) ln(400.0/300.0) = 29.89 J/K
Since our hypothetical two step process has the same initial and final state as the actual process that takes
place, and since entropy is a state function, it follows that
S = S1 + S2 = 28.82 J/K + 29.89 J/K = 58.71 J/K
3)
For heating a pure substance under conditions of constant pressure, we showed in class that
S = if (dqrev/T) = if (nCp,m/T)dT +  n Hpt/Tpt
where the first term on the right accounts for heating the substance in the absence of phase transitions, and the second
term (the summation) accounts for the entropy change for the phase transitions that take place.
We can divide the process of adding heat to Br 2 into three steps, with S = S1 + S2 + S3
Step 1 - Heat Br2() from 300.0 K to 332.4 K under conditions of constant pressure
Step 2 - Convert Br2() into Br2(g) at 332.4 K and constant pressure
Step 3 - Heat Br2(g) from 332.4 K to 400.0 K under conditions of constant pressure
Since Cp,m is constant for both Br2() and Br2(g), we can use the result from step 2 of problem 2 for step 1 and step 3
above
S1 = n[Cp,m(Br2(l))]ln(Tf/Ti) = (1.000 mol)(75.69 J/molK) ln(332.4/300.0) = 7.763 J/K
S2 = n Hvap/Tvap = (1.000 mol) [(29450. J/mol)/(332.4 K)] =88.60 J/K
S3 = n[Cp,m(Br2(g))]ln(Tf/Ti) = (1.000 mol)(36.02 J/molK) ln(400.0/332.4) = 6.668 J/K
And so S = S1 + S2 + S3 = 7.763 J/K + 88.60 J/K + 6.668 J/K = 103.03 J/K
4) The expressions for Grxn, Hrxn, and Srxn are as follows:
Grxn = [ Gf(CO2(g)) + 2 Gf(NH3(g)) ] – [ Gf(CO(NH2)2(s)) + Gf(H2O()) ]
= [ (- 394.36) + 2 ( - 16.45) ] – [ ( - 197.33) + ( - 237.13) ] = + 7.2 kJ/mol
Hrxn = [ Hf(CO2(g)) + 2 Hf(NH3(g)) ] – [ Hf(CO(NH2)2(s)) + Hf(H2O()) ]
= [ (-393.51) + 2 ( - 46.11) ] – [ ( - 333.51) + ( - 285.83 ) ] = + 133.6 kJ/mol
Srxn = [ S(CO2(g)) + 2 S(NH3(g)) ] – [ S(CO(NH2)2(s)) + S(H2O()) ]
= [ (213.74) + 2 (192.45) ] – [ (104.60) + (69.91) ] = + 424.13 J/molK
As a check, we can say
Grxn = Hrxn – T Srxn = (133.6 kJ/mol) – (298. 2 K) (424.13 x 10-3 kJ/molK) = + 7.1 kJ/mol
which is the same value (to within roundoff error) found directly using the free energy of formation data.
5) At T = 25.0 C
Grxn = [ Gf(MgO(s)) + Gf(CO2(g)) ] – [ Gf(MgCO3(s)) ]
= [ ( - 569.43) + ( - 394.36) ] – ( - 1012.1) ] = + 48.3 kJ/mol
Hrxn = [ Hf(MgO(s)) + Hf(CO2(g)) ] – [ Hf(MgCO3(s)) ]
= [ ( - 601.70) + ( - 393.51) ] – ( - 1095.8) ] = + 100.6 kJ/mol
Srxn = [ S(MgO(s)) + S(CO2(g)) ] – [ S(MgCO3(s)) ]
= [ (26.94) + (213.74) ] – [ (65.7) ] = + 175.0 J/molK
To find Grxn at T = 100.0 C, we first need to find Hrxn and Srxn at this temperature. To do this, we need to
know the value for Cp,m.
Cp,m = [ Cp,m(MgO(s)) + Cp,m(CO2(g)) ] – [ (Cp,m(MgCO3(s)) ]
= [ (37.15) + (37.11) ] – [ (75.52) ] = - 1.26 J/molK
where we have assumed that the heat capacity values are constant. Then
Hrxn(T2) = Hrxn(T1) + T1T2 Cp,m dT = Hrxn(T1) + (Cp,m) (T2 – T1)
and so
Hrxn(100.0 C) = (100.6 kJ/mol) + (- 1.26 x 10-3 kJ/molK) (75.0 K) = + 100.5 kJ/mol
Srxn(T2) = Srxn(T1) + T1T2 (Cp,m/T) dT = Srxn(T1) + (Cp,m) ln(T2/T1)
and so
Srxn(100.0 C) = (175.0 J/molK) + ( - 1.26 J/molK) ln(373.2/298.2) = + 174.7 J/molK
Notice that neither Hrxn or Srxn change much in this particular problem. This is why we often assume these two
values are approximately independent of temperature.
Finally, since Grxn = Hrxn – T Srxn , the value for Grxn at T = 100.0 C is
Grxn(100.0 ) = 100.5 kJ/mol – (373.2 K) (174.7 x 10-3 kJ/molK) = + 35.3 kJ/mol
If you compare this value to the value for Grxn at T = 25.0 C, you will see that they are significantly different.
This is why, even when Hrxn and Srxn are approximately independent of temperature (as in the above reaction) it
is never a good idea to assume Grxn is temperature independent.
6)
a) The values for Hvap and Svap are as follows (at T = 25.0 C)
Hvap(benzene) = Hf(C6H6(g)) - Hf(C6H6())
= 82.93 kJ/mol - 49.0 kJ/mol = 33.9 kJ/mol
Svap(benzene) = S(C6H6(g)) - S(C6H6())
= 269.31 J/molK - 173.3 J/molK = 96.0 J/molK
Hvap(ethanol) = Hf(C2H5OH(g)) - Hf(C2H5OH())
= (- 235.10) kJ/mol - (- 277.69) kJ/mol = 42.59 kJ/mol
Svap(ethanol) = S(C2H5OH(g)) - S(C2H5OH())
= 282.70 J/molK - 160.7 J/molK = 122.0 J/molK
So the estimated normal boiling points are
benzene
Tvap  (33900. J/mol)/(96.0 J/molK) = 353.1 K = 80.0 C (exp = 80.1 C)
ethanol
Tvap  (42590. J/mol)/(122.0 J/molK) = 349.1 K = 75.9 C (exp = 78.5 C)
The above values are within a few degrees of the experimental values for the normal boiling point. Note the method
works better for benzene than for ethanol, probably because benzene is nonpolar and
b) Using Trouton’s rule (eq 6.1)
Hvap/Tvap  85. J/molK , and soTvap  Hvap/(85. J/molK)
So the estimated normal boiling points are
benzene
T  (33900. J/mol)/(85. J/molK) = 398.8 K = 125.7 C (exp = 80.1 C)
ethanol
T  (42590. J/mol)/(85. J/molK) = 501.1 K = 227.9 C (exp = 78.5 C)
This is terrible agreement, particularly for ethanol. Trouton’s rule tends to not work well for strongly interacting
molecules, like ethanol, which can hydrogen bond. In fairness, Trouton’s rule is usually used to estimate the value for
Hvap, since the normal boiling point for a liquid is much easier to determine experimentally than the enthalpy of
vaporization.
7) When dp/dT > 0 then the initial phase is more dense than the final phase, and when dp/dT < 0 the final phase is
more dense than the initial phase.
If we look at the phase boundaries, we see
rhom > gas
rhom > mono
rhom > liq
mono > gas
mono > liq
liq > gas
So the phases, in order of density, are (rhom) > (mono) > (liq) > (gas)