Florida International University s18
CHM 3400 – Problem Set 4
Due date: Wednesday, February 22nd
Do all of the following problems. Show your work.
“...we should not be like children with their parents, simply accepting what we are told.” – Marcus Aurelius
1) 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) 2 O3(g) ® 3 O2(g) (thermal decomposition of ozone)
b) Fe2O3(s) + 3 CO(g) ® 2 Fe(s) + 3 CO2(g) (reduction of iron)
c) 2 H2O2() ® 2 H2O() + O2(g) (decomposition of hydrogen peroxide)
2) For nearly 100 years scientists have tried (with limited success) to synthesize diamonds in the laboratory. In this problem we consider the thermodynamics of the process.
a) Consider the formation of the diamond form of carbon from the graphite form, for standard conditions
C(s,graphite) ® C(s,diamond) (2.1)
What is the value for DG°rxn for the above process? Is the process spontaneous? Is the reverse process (diamond ® graphite) spontaneous? Should you buy your significant other a diamond ring?
b) The following general relationship for free energy was derived in class, and in the Chapter 5 handout
dG = V dp – S dT (2.2)
Starting with equn 2.2 show that if the pressure applied on a pure substance is changed isothermally, and assuming that there is no change in the volume of the substance (as is approximately true for solids) that
Gp = G° + DG = G° + Vm Dp (2.3)
where Gp is the free energy for the substance at pressure p, G° is the free energy of the substance for standard conditions, and Vm is the molar volume of the substance.
c) Using the relationship found in part b, derive the following result for the free energy change for reaction 2.1 when carried out at some pressure p and a standard temperature T = 298. K.
DGrxn(p) = DG°rxn + (Vm,dia – Vm,gra) Dp (2.4)
where Vm,dia and Vm,gra are the molar volume for diamond and graphite, respectively.
d) We know that DGrxn = 0 for a system at equilibrium at a particular pressure and temperature. Use this condition and equn 2.4 to find the value for pressure at which the diamond form of carbon becomes more stable than the graphite form. Note that rdia = 3.513 g/cm3 and rgra = 2.260 g/cm3. Be careful about dimensional analysis in your calculation.
e) Can we make diamonds out of graphite simply by increasing the pressure applied to a sample of graphite (at room temperature) until it is larger than the value found in part d of this problem? Why or why not?
3) When a phase transition occurs between two phases at equilibrium we may say
DGpt = 0 = DHpt - TptDSpt (3.1)
where “pt” indicates “phase transition. Since DHpt and DSpt are, to a first approximation, independent of temperature, we can use the values for these two quantities found for standard conditions, and, based on this approximation, find the normal transition temperature for the substance
T°pt = DHpt/DSpt (3.2)
a) Starting with equn 3.1 derive equn 3.2.
b) Use equn 3.2 and the information in the Appendix of Atkins to estimate the value for the normal boiling point for Br2 and C6H6. Compare your results to the experimental values, T°vap(Br2) = 58.8 °C and T°vap(C6H6) = 80.0 °C.
4) A phase diagram is given below for some ficticious pure chemical substance, and can be used to answer the questions given below. Note that this substance has two different solid phases, and that the pressure axis is logarithmic.
a) Which phase is more dense: s1 or s2? s2 or ? or g? Justify your answer.
b) How many triple points are there in the phase diagram? Where is each triple point located (T and p), and what three phases are at equilibrium at each triple point?
c) How many normal phase transition points are there in the phase diagram? Where is each of these points located (T and p), and what phase transition is occurring at each point?
d) The temperature of 1.00 mol of the above substance is changed isothermally and reversibly from 270. K to 370. K, at a constant pressure p = 10.0 atm. Describe what happens as the process is carried out.
Also do the following problems from Atkins:
5.6 Asample of water vapor at T = 200.0 °C is compressed isothermally from 350. cm3 to 120. cm3. What is the change in the molar Gibbs free energy? (You may assume 1.00 mol of water, and also that the vapor behaves ideally).
5.8 The standard molar entropy of liquid benzene is 173.3 J/mol.K. Calculate the change in the standard molar Gibbs free energy when 1.00 mol of benzene is heated at constant pressure from 25. °C to 45. °C.
5.15 The vapor pressure of pyridine (C5H5N) is 50.0 kPa at T = 365.7 K, and the normal boiling point is 388.4 K. What is the DH°vap, enthalpy of vaporization for pyridine?
Solutions.
1) a) DG°rxn = [ 3 DG°f(O2(g)) ] – [ 2 DG°f(O3(g)) ]
= [ 3 (0.0 kJ/mol) ] – [ 2 (163.2 kJ/mol) ] = - 326.4 kJ/mol
DH°rxn = [ 3 DH°f(O2(g)) ] – [ 2 DH°f(O3(g)) ]
= [ 3 (0.0 kJ/mol) ] – [ 2 (142.7 kJ/mol) ] = - 285.4 kJ/mol
DS°rxn = [ 3 S°(O2(g)) ] – [ 2 S°(O3(g)) ]
= [ 3 (205.138 J/mol.K) ] – [ 2 (238.93 J/mol.K) ] = 137.55 kJ/mol
DG°rxn = DH°rxn – T DS°rxn = ( - 285.4 kJ/mol) – (298. K) (0.13755 kJ/mol.K) = - 326.4 kJ/mol
The result agrees with the value found directly from the free energy data.
b) 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.
c) 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 all three reactions DG°rxn < 0, all three reactions are spontaneous for standard conditions.
2) a) DG°rxn = [ DG°f(C(s,dia)) ] – [ DG°f(C(s,gra)) ] = (2.900 kJ/mol) – (0.0 kJ/mol) = 2.900 kJ/mol
Since DG°rxn > 0, the reaction is not spontaneous as written, for standard conditions. For the reverse process, DG°rxn = - 2.900 kJ/mol (since Gibbs free energy is a state function) so the reverse process is spontaneous for standard conditions. Nonetheless, I would buy a diamond ring for your significant other, because even though the graphite form of carbon is thermodynamically more stable than the diamond form of carbon for standard conditions, the reaction diamond ® graphite is kinetically a very slow process. Bottom line, you don’t have to worry about the diamond spontaneously reverting to the graphite form of carbon.
b) The general relationship for dG is
dG = V dp – S dT
If the process is isothermal, then dT = 0, and so
dG = V dp
If we integrate from standard pressure (1 bar) to some final pressure p, then
ò1 barp dG = ò1 barp V dp
If we assume that Vm is independent of pressure we can take it outside the integral, to get
Gp - G° = Vm (p – 1 bar)
Gp = G° + Vm (p – 1 bar) = G° + Vm Dp
c) DGrxn(p) = G(C(s,dia,p)) – G(C(s,gra,p))
= [G°(C(s,dia) + Vm,dia Dp] - [G°(C(s,gra) + Vm,gra Dp]
= [G°(C(s,dia) - G°(C(s,gra)] + (Vm,dia – Vm,gra) Dp
= DG°rxn + (Vm,dia – Vm,gra) Dp
d) To find the pressure where the diamond and graphite forms of carbon are at equilibrium means finding the value for pressure where DGrxn(p) = 0. As a precaution, we will put all terms in MKS units as a way of making sure the dimensions work out in the calculation.
0 = DG°rxn + (Vm,dia – Vm,gra) Dp
Dp = - DG°rxn
(Vm,dia – Vm,gra)
Vm,dia = (12.01 g/mol) 1 cm3 = 3.419 cm3/mol = 3.419 x 10-6 m3/mol
3.513 g
Vm,gra = (12.01 g/mol) 1 cm3 = 5.314 cm3/mol = 5.314 x 10-6 m3/mol
2.260 g
So Dp = - DG°rxn = - (2900. J/mol)
(Vm,dia – Vm,gra) [(3.419 x 10-6 m3/mol) – (5.314 x 10-6 m3/mol)]
= 1.53 x 109 N/m2 1 bar = 15300. bar @ 15100. atm
105 Nt/m2
Since 1 bar is small compared to this number, we can say p @ Dp = 15100. atm. So for p > 15100. atm the diamond form of carbon will be thermodynamically more stable than the graphite form.
e) The above is approximately correct (the factors we have ignored, such as the change in the molar volume of diamond and graphite with pressure, are relatively small, and can be accounted for). It is not difficult to achieve a pressure of 15000 atm in the laboratory. So what is the problem? Well...although the diamond form of carbon is thermodynamically more stable at room temperature and high pressures, the rate for the graphite ® diamond reaction is extremely slow. Scientists have tried a variety of methods to overcome this difficulty, such as working at higher temperatures, using catalysts for the reaction, or exotic methods such as carbon deposition, with limited success.
3) a) DGpt = 0 = DHpt - TDSpt
So DHpt = TptDSpt
Tpt = DHpt/DSpt
If we assume p = 1.00 atm (standard pressure for phase transitions) then
T°pt = DH°pt/DS°pt
b) For Br2
DH°vap = DH°f(Br2(g)) - DH°f(Br2()) = (30.907 kJ/mol) – (0.0 kJ/mol) = 30.907 kJ/mol
DS°vap = S°(Br2(g)) - S°(Br2()) = (245.46 J/mol.K) – (152.23 J/mol.K) = 93.23 J/mol.K
T°vap = (30907 J/mol)/(93.23 J/mol.K) = 331.5 K = 58.4 °C (exp is 58.8 °C)
For C6H6
DH°vap = DH°f(C6H6(g)) - DH°f(C6H6()) = (82.93 kJ/mol) – (49.0 kJ/mol) = 33.93 kJ/mol
DS°vap = S°(C6H6(g)) - S°(C6H6()) = (269.31 J/mol.K) – (173.3 J/mol.K) = 96.0 J/mol.K
T°vap = (33930 J/mol)/(96.0 J/mol.K) = 353.4 K = 80.3 °C (exp is 80.0 °C)
In both cases the value for the normal boiling point calculated by the approximation method derived here is within 1 °C of the actual value. This is in part because both substances are nonpolar. The agreement would not be quite as good if this method were applied to a polar liquid such as water or ethyl alcohol.
4) a) If dp/dT > 0 then the phase to the right of the boundary between phases is lower in density, while if dp/dT < 0 then the phase to the left of the boundary between phases is lower in density. Using this
s1 or s2 dp/dT > 0, and so r(s1) > r(s2)
ss or dp/dT > 0, and so r(s2) > r()
or g dp/dT > 0, and so r() > r(g)
b) There are two triple points shown in the phase diagram. Their locations are
p = 0.1 atm, T = 265 K s1, s2, g
p = 0.3 atm, T = 290 K s2, , g
There is clearly a third triple point that will occur, with s1, s2, and existing simultaneously at equilbrium, but it falls outside of the range of values shown in the phase diagram.
c) A “normal” phase transition occurs when two phases exist simultaneously at equilibrium when p = 1.00 atm. For this phase diagram three such points occur
p = 1.00 atm, T = 275 K s1 ® s2 transition (solid to solid)
p = 1.00 atm, T = 292 K s2 ® transition (solid to liquid, or melting)
p = 1.00 atm, T = 332 K ® g transition (liquid to gas, or vaporization)
d) The substance is in the s1 solid phase at p = 10.0 and T = 270 K. The substance warms until T = 284 K, when the s1 ® s2 phase transition occurs. Between 284 K and 300 K the substance remains in the s2 solid phase. At T = 300 K the s2 ® phase transition occurs. Between 300 K and 360 K the substance remains in the liquid phase. At T = 360 K the ® g phase transition occurs. Between 360 K and 370 K, the final temperature, the substance remains a gas.
5.6 The general relationship for dG is
dG = V dp – S dT
The process is isothermal, and so dT = 0. Therefore
dG = V dp
If we integrate both sides of this equation from the initial to the final state
òif dG = DG = òif V dp
Since the gas is ideal, V = nRT/p. So
DG = òif V dp = òif (nRT/p) dp = nRT òif (dp/p) = nRT ln(pf/pi)
Since the gas is ideal and thr process is isothermal, piVi = pfVf, and so (pf/pi) = (Vi/Vf)
So DG = nRT ln(Vi/Vf) = (1.00 mol) (8.314 J/mol.K) (473. K) ln(350/120) = + 4210. J
5.8 We can again use the general relationship for dG
dG = V dp – S dT
In this process pressure is held constant, and so dp = 0. Therefore
dG = - S dT
If we integrate both sides of this expression, we get
òif dG = DG = - òif S dT
We know that the value for the molar entropy of a substance changes slowly with temperature (as long as no phase transition occurs). If we assume S° @ constant between 25. °C and 45. °C, then