CHM 3410 – Problem Set 4
Due date: Monday, September 17th (NOTE: The first hour exam is Friday, September 21st. It will cover Chapter 1 (all), Chapter 2 (all), and Chapter 3 (3.1 to 3.6).
Do all of the following problems. Show your work.
“The Second Law of Thermodynamics is, without a doubt, one of the most perfect laws in physics. Any reproducible violation of it, however, small, would bring the discoverer great riches as well as a trip to Stockholm.”
- Ivan Bazarov
1) Find DSsyst, DSsurr, and DSuniv for the following processes. For the pressures and temperatures of the problem you may assume that argon behaves ideally. Note that these processes were previously dealt with in earlier problem sets. You may use any results obtained in the earlier problem sets in this problem without rederiving them.
a) A reversible and isothermal expansion of 1.000 mol of Ar gas from an initial pressure pi = 1.000 atm to a final pressure pf = 0.250 atm, at a constant temperature T = 300.0 K. (Problem 4a, problem set 2).
b) An irreversible and isothermal expansion of 1.000 mol of Ar gas from an initial pressure pi = 1.000 atm to a final pressure pf = 0.250 atm, at a constant temperature T = 300.0 K. (Problem 4b, problem set 2).
c) The adiabatic irreversible expansion of 1.00 mol of Ar gas (CV,m = 3/2 R), from an initial temperature Ti = 300.0 K and an initial pressure pi = 2.00 atm, against a constant external pressure equal to the final pressure of the gas, pex = pf = 1.00 atm. (Problem 2, problem set 3).
2) An insulated container holds 100.00 g of liquid water (H2O(), M = 18.02 g/mol, Cp,m = 75.29 J/mol.K) at an initial temperature of 20.0 °C. A 10.00 g block of copper metal (Cu(s), M = 63.54 g/mol, Cp,m = 24.44 J/mol.K) at a temperature T = 80.0 °C is dropped into the container.
a) What is the temperature of the water and copper when equilibrium is achieved?
b) What are DSsyst, DSsurr, and DSuniv for the process?
3) The compound 1,3,5-trichloro-2,4,6-trifluorobenzene is an intermediate in the conversion of hexachlorobenzene into hexafluorobenzene. Its thermodynamic properties have been examined by measuring its heat capacity over a wide temperature range (R. L. Andon, J. F. Martin, J.Chem.Soc.Faraday Trans. I, 871 (1973)). Some of the data are as follows:
T (K) 14.14 16.33 20.03 31.15 44.08 64.81
Cp,m (J/mol.K) 9.492 12.70 18.18 32.54 46.86 66.36
T(K) 100.90 140.86 183.59 225.10 262.99 298.06
Cp,m(J/mol.K) 95.05 121.3 144.4 163.7 180.2 196.4
Calculate the molar entropy at T = 298.15 K. (HINT: See Fig. 3.14 of Atkins. Below the first data point, 14.14 K, you may use the Debye approximation for Cp,m.)
4) The constant pressure molar heat capacity of a substance is often written as
Cp,m = a + bT + c/T2 (4.1)
where a, b, and c are constants.
a) Use equn 4.1 to find a general expression for DS when the temperature of 1.00 mol of a substance is changed from an initial value Ti to a final temperature Tf at a constant pressure p = 1.00 atm.
b) Using your answer to part a, find DS when the temperature of 1.00 mol of graphite (C(s)) is changed at constant pressure from an initial value Ti = 300.0 K to a final value Tf = 400.0 K. Note that for graphite a = 16.86 J/mol.K, b = 4.77 x 10-3 J/mol.K2, c = - 8.54 x 105 J.K/mol.
5) Using the information in the appendix of Atkins, find DG°rxn, DH°rxn, and DS°rxn for each of the reactions given below, carried out for standard conditions. Check to confirm that DG°rxn = DH°rxn - TDS°rxn for each reaction. Which reactions are spontaneous for standard conditions?
a) Fe2O3(s) + 3 CO(g) ® 2 Fe(s) + 3 CO2(g) (reduction of iron)
b) 2 H2O2() ® 2 H2O() + O2(g) (decomposition of hydrogen peroxide)
Solutions.
1) a) The process is reversible and isothermal. So
DSsyst = 0.3846 J/g.K (1/T) òif (đq)rev = qrev/T
From Problem set 2, problem 4a, we found q = 3458. J. Therefore
DSsyst = (3458. J)/(300.0 K) = 11.53 J/K
Since the process is reversible, DSuniv = 0. Therefore DSsurr = - DSsyst = - 11.53 J/K.
b) The process is irreversible and isothermal. It has the same initial and final state for the system as in part a of this problem, and so DSsyst = 11.53 J/K.
For DSsurr, we use the relationship
DSsurr = òif (đq)rev,surr/T = (1/T) òif (đq)rev,surr = qrev,surr/T = - qsyst/T
From problem set 2, problem 4b, we found q = 1871. J
So DSsyst = - (1871. J)/(300.0 K) = - 6.24 J/K
Finally, DSuniv = DSsyst + DSsurr = 11.53 J/K + (- 6.24 J/K) = 5.29 J/K
c) The process is adiabatic and irreversible. For the entropy change of the surroundings, we may use the relationship
DSsurr = òif (đq)rev,surr/T
The process is adiabatic, and so đq= 0, and, therefore, DSsurr = 0 (Note that for any adiabatic process DSsurr = 0 )
For the system, we may say DSsyst = òif (đq)rev/T. The problem is that this particular process is irreversible from the point of view of the system. We therefore need a process or combination of processes that has the same initial and final state as the process that actually occurs. Since entropy is a state function, the entropy change for the hypothetical reversible pathway will be the same as for the actual pathway followed by the system.
The process is (according to the results from problem set 3, problem 2)
pi = 2.00 atm pf = 1.00 atm
Ti = 300.0 K Tf = 240.0 K
One possible reversible pathway connecting the above initial and final states is as follows
Step 1 – Isothermal reversible expansion of the gas from pi = 2.00 atm to pf = 1.00 atm, at T = 300.0 K.
Step 2 – Constant pressure cooling of the gas from Ti = 300.0 K to Tf = 240.0 K, at p = 1.00 atm.
For step 1, the isothermal reversible expansion of an ideal gas, we previously showed in class (and also problem set 2, problem 4a) that
q = - nRT ln(pf/pi).
And so DS1 = q/T = - nR ln(pf/pi) = - (1.000 mol)(8.3145 J/mol.K) ln(1.00/2.00) = 5.763 J/K
For step 2, the constant pressure cooling of an ideal gas, we may say (since Cp,m is constant)
DS2 = òif (đq)rev/T = òif (nCp,m dT)/T = nCp,m òif dT/T = nCp,m ln(Tf/Ti)
So DS2 = (1.000 mol)(5/2)(8.3145 J/mol.K) ln(240.0/300.0) = - 4.638 J/K
And so DSsyst = DS1 + DS2 = 5.763 J/K = (- 4.638 J/K) = 1.12 J/K
Finally, since DSsurr = 0, DSuniv = DSsyst = 1.12 J/K
2) It is convenient (though not required) to work this problem using the specific heat capacity (J/g)
Cp,s(H2O) = (75.29 J/mol.K)(1. mol/18.02 g) = 4.178 J/g.K = 4.178 J/g.°C
Cp,s(Cu) = (24.44 J/mol.K)(1. mol/63.54 g) = 0.3846 J/g.K = 0.3846 J/g.°C
a) The process is adiabatic (since the container is insulated)
So q = 0 = qH2O + qCu
For a constant pressure change in temperature, and assuming that the specific heat capacities are both constant over the temperature range of the problem, and calling the final temperature of the water and copper Tf
qH2O = mH2OCp,s(H2O)(Tf – Ti,H2O) mH2O = 100.00 g Ti,H2O = 20.0 °C
qCu = mCuCp,s(Cu)(Tf – Ti,Cu) mCu = 10.00 g Ti,Cu = 80.0 °C
So 0 = qH2O + qCu = mH2OCp,s(H2O)(Tf – Ti,H2O) + mCuCp,s(Cu)(Tf – Ti,Cu)
0 = mH2OCp,s(H2O)Tf + mCuCp,s(Cu)Tf – [mH2OCp,s(H2O)Ti,H2O + mCuCp,s(Cu)Ti,Cu ]
mH2OCp,s(H2O)Tf + mCuCp,s(Cu)Tf = mH2OCp,s(H2O)Ti,H2O + mCuCp,s(Cu)Ti,Cu
or, finally Tf = mH2OCp,s(H2O)Ti,H2O + mCuCp,s(Cu)Ti,Cu
mH2OCp,s(H2O) + mCuCp,s(Cu)
Substituting, we get
Tf = (100.00 g)(4.178 J/g.°C)(20.0 °C) + (10.00 g)(0.3846 J/g.°C)(80.0 °C) = 20.55 °C
(100.00 g)(4.178 J/g.°C) + (10.00 g)(0.3846 J/g.°C)
b) DSsyst = DSH2O + DSCu
For both the water and the copper we have a constant pressure temperature change with a constant specific heat capacity. So
DSH2O = òif (đq)rev,H2O/T = òif (mH2OCp,s(H2O)dT)/T = mH2OCp,s(H2O)òif dT/T = mH2OCp,s(H2O) ln(Tf/Ti)
= (100.00 g)(4.178 J/g.K) ln(293.70/293.15) = 0.783 J/K
DSCu = òif (đq)rev,Cu/T = òif (mCuCp,s(Cu)dT)/T = mCuCp,s(Cu)òif dT/T = mCuCp,s(Cu) ln(Tf/Ti)
= (10.00 g)(0.3846 J/g.K) ln(293.70/353.15) = - 0.709 J/K
DSsyst = DSH2O + DSCu = 0.783 J/K + (- 0.709 J/K) = 0.074 J/K
DSsurr = 0 (because the process is adiabatic), and so DSuniv = DSsyst = 0.074 J/K
3) S°(298.15 K) = ò0298.15 (Cp,m/T) dT
= ò014.14 (Cp,m/T) dT + ò14.14298.15 (Cp,m/T) dT
For temperatures below 14.14 K we assume that Cp,m = a T3. The value for a is found using the heat capacity at 14.14 K
(9.492 J/mol.K) = a (14.14 K)3
a = (9.492 J/mol.K) / (14.14 K)3 = 3.36 x 10-3 J/mol.K4
So ò014.14 (Cp,m/T) dT = ò014.14 (a T3/T) dT = ò014.14 (a T2) dT = (a T3)/3
= (3.36 x 10-3 J/mol.K4) (14.14 K)3 / 3 = 3.166 J/mol.K
For the second integral, ò14.14298.15 (Cp,m/T) dT, we will find the value for the integral numerically using the trapezoid rule. For adjacent temperatures we say
òT1T2 @ (T2 - T1) { [ Cp,m(T2)/T2 ] + [ Cp,m(T1)/T1 ] }/2
Since our data ends at T = 298.06 K we have added a short extrapolation to T = 298.15 K. The data and the results of the data analysis are given on the next page along with the plot used to find S°. Note that the area under the curve up to any temperature T represents the value for S° at that temperature.
The contribution to the entropy in going from 14.14 K to 298.15 K is
ò14.14298.15 (Cp,m/T) dT @ S DS = 239.322 J/mol.K
And so
S°(298.15 K) = ò014.14 (Cp,m/T) dT + ò14.14298.15 (Cp,m/T) dT = (3.166 + 239.322) J/mol.K = 242.49 J/mol.K
T (K) / Cp,m (J/mol.K) / Cp,m/T (J/mol.K2) / DT (K) / DS (J/mol.K)14.14 / 9.492 / 0.6713 / 2.19 / 1.587
16.33 / 12.70 / 0.7777 / 3.70 / 3.118
20.03 / 18.18 / 0.9076 / 11.12 / 10.854
31.15 / 32.54 / 1.0446 / 12.93 / 13.626
44.08 / 46.86 / 1.0631 / 20.73 / 21.632
64.81 / 66.36 / 1.0239 / 36.09 / 35.475
100.90 / 95.05 / 0.9420 / 39.96 / 36.026
140.86 / 121.3 / 0.8611 / 42.73 / 35.201
183.59 / 144.4 / 0.7865 / 41.51 / 31.417
225.10 / 163.7 / 0.7272 / 37.89 / 26.758
262.99 / 180.2 / 0.6852 / 35.07 / 23.569
298.06 / 196.4 / 0.6589 / 0.09 / 0.0593
298.15 / 0.658
4) a) From the general definition of the entropy change for the system
DSsyst = òif đqrev/T.
The process is reversible and carried out at constant pressure, so đqrev = n Cp,m dT = n (a + bT + c/T2) dT.
So DSsyst = òif đqrev/T = òif n [ (a/T) + b + (c/T3) ] dT
= n { a ln(Tf/Ti) + b (Tf – Ti) – (c/2) [ (1/Tf2) – (1/Ti2) ] }
b) Using the general result for this type of process found in part a, we may say
DSsyst = (1.00 mol) { (16.86 J/mol.K) ln(400./300.) + (4.77 x 10-3 J/mol.K2) (400. K – 300. K)
- [(- 8.54 x 105 J.K/mol)/2] [ 1/(400. K)2 - 1 (300. K)2 ] }
= (1.00 mol) { (4.850 J/mol.K) + (0.477 J/mol.K) – (2.076 J/mol.K) } = + 3.251 J/K
5) a) DG°rxn = [ 2 DG°f(Fe(s)) + 3 DG°f(CO2(g)) ] – [ DG°f(Fe2O3(s)) + 3 DG°f(CO(g)) ]
= [ 2 (0.0 kJ/mol) + 3 ( - 394.36 kJ/mol) ] – [ ( - 742.2 kJ/mol) + 3 ( - 137.17 kJ/mol) ]
= - 29.4 kJ/mol
DH°rxn = [ 2 DH°f(Fe(s)) + 3 DH°f(CO2(g)) ] – [ DH°f(Fe2O3(s)) + 3 DH°f(CO(g)) ]
= [ 2 (0.0 kJ/mol) + 3 ( - 393.51 kJ/mol) ] – [ ( - 824.2 kJ/mol) + 3 ( - 110.53 kJ/mol) ]
= - 24.7 kJ/mol
DS°rxn = [ 2 S°(Fe(s)) + 3 S°(CO2(g)) ] – [ S°(Fe2O3(s)) + 3 S°(CO(g)) ]
= [ 2 (27.28 J/mol.K) + 3 (213.74 J/mol.K) ] – [ (87.40 J/mol.K) + 3 (197.67 J/mol.K) ]
= 15.37 J/mol.K
DG°rxn = DH°rxn – T DS°rxn = ( - 24.7 kJ/mol) – (298. K) (0.01537 kJ/mol.K) = - 29.3 kJ/mol
The result agrees with the value found directly from the free energy data to within roundoff error.
b) DG°rxn = [ 2 DG°f(H2O()) + DG°f(O2(g)) ] – [ 2 DG°f(H2O2()) ]
= [ 2 (- 237.13 kJ/mol) + (0.0 kJ/mol) ] – [ 2 ( - 120.35 kJ/mol) ] = - 233.56 kJ/mol
DH°rxn = [ 2 DH°f(H2O()) + DH°f(O2(g)) ] – [ 2 DH°f(H2O2()) ]
= [ 2 (- 285.83 kJ/mol) + (0.0 kJ/mol) ] – [ 2 ( -187.78 kJ/mol) ] = - 196.10 kJ/mol
DS°rxn = [ 2 S°(H2O()) + S°(O2(g)) ] – [ 2 S°(H2O2()) ]
= [ 2 (69.91 J/mol.K) + (205.138 J/mol.K) ] – [ 2 (109.6 J/mol.K) ] = 125.76 J/mol.K
DG°rxn = DH°rxn – T DS°rxn = ( - 196.10 kJ/mol) – (298. K) (0.12576 kJ/mol.K) = - 233.58 kJ/mol
The result agrees with the value found directly from the free energy data to within roundoff error.
Since for both reactions DG°rxn < 0, both reactions are spontaneous for standard conditions.