Chemistry 181
Professor S. Alex Kandel
Fall 2015
Due 12/9/2015
Problem Set 12
Discussion problems:
We will discuss these in class (3:30) on Wednesday the 9th. Be prepared to present
answers to and to discuss any of these up at the blackboard.
1. The reaction enthalpy change at any temperature (T2 ) can be calculated from a
reaction enthalpy change at any other temperature (T1 ), along with the heat capacities of the reactants and products. (This calculation is not simple if the heat
capacities change with temperature, but we will assume for this problem that
they are more or less constant.) A diagram using Hess’s Law looks something
like this:
A @ T2
|
Cp (A)∆T
↓
A @ T1
∆H
@ T
rxn
−−−−
−−−−2→
B @ T2
↑
Cp (B)∆T
|
∆Hrxn @ T1
−−−−−−−−→
B @ T1
Note that the ∆T values above are opposite in sign (depending on whether you
are going from T1 to T2 or from T2 to T1 ).
Use the following data to calculate the heat of vaporization of water at 20 ◦ C.
Enthalpy of vaporization of water at 100 ◦ C:
Heat capacity of H2 O(`):
Heat capacity of H2 O(g):
40.68 kJ mol−1
75.33 J mol−1 K−1
37.47 J mol−1 K−1
Using the diagram, the reaction is
water @ 20
◦
C → water @ 100
◦
C → steam @ 100
1
◦
C → steam @ 20
◦
C
∆Hvap @ 20 ◦ C = q(heat water to 100 ◦ C) + ∆Hvap @ 100 ◦ C
+q(cool steam to 20 ◦ C)
= 40.68 kJ mol−1
+75.33 J mol−1 K−1 · 80 K + 37.47 J mol−1 K−1 · (−80) K
= 40680 J mol−1 + (75.33 − 37.47)(80) J mol−1
= 40680 J mol−1 + 3029 J mol−1
= 43.71 kJ mol−1
2. Heats of formation:
∆Hf◦ (kJ/mol)
CH3 Li
C4 H9 Li
LiC2 H
LiC6 H5
112.5
−21.5
271.5
228.9
CH3 Cl
−83.7
C4 H9 Cl −147.3
C2 HCl
53.8
C6 H5 Cl
52.0
Use these, values in Appendix D, and lithium’s heat of sublimation (about
150 kJ/mol) to estimate a C–Li molar bond energy.
One thing to keep track here is the definition of the bond enthalpy, which is
the energy put into the system to break a bond and form gas-phase fragment
products. So, for CH3 Li, the reaction is:
CH3 Li(g) −→ CH3 (g) + Li(g)
Starting at this point, there are many ways to do this problem. This one is
arithmetically neat. The general reaction to consider (for R = CH3 , C4 H9 ,
C2 H, or C6 H5 ) is
RCl + Li(g)
∆Hbond (C−Cl)
−→
R + Cl(g) + Li(g)
2
−∆Hbond (C−Li)
−→
RLi + Cl(g)
Written this way, we have
∆Hrxn = ∆Hbond (C − Cl) − ∆Hbond (C − Li)
But also
∆Hrxn =
=
=
=
∆Hbond (C − Cl) =
−∆Hbond (C − Li)
∆Hbond (C − Li) =
∆Hf (prod) − ∆Hf (react)
∆Hf (RLi) + ∆Hf (Cl(g)) − ∆Hf (RCl) − ∆Hf (Li(g))
∆Hf (RLi) − ∆Hf (RCl) + 121.7 − 150
∆Hf (RLi) − ∆Hf (RCl) − 29.3 kJ mol−1
∆Hf (RLi) − ∆Hf (RCl) − 29.3 kJ mol−1
∆Hbond (C − Cl) − ∆Hf (RLi) + ∆Hf (RCl) + 29.3 kJ mol−1
Plugging in values from the table, using a C–Cl bond enthalpy of 328 kJ/mol:
C–Li bond in
∆Hf◦ (kJ/mol)
CH3 Li
C4 H9 Li
LiC2 H
LiC6 H5
161.1
231.5
139.6
180.4
How much does the bond energy depend on the hybridization of the carbon?
The first two compounds are sp3 , while acetylene and benzene are sp and sp2 ,
respectively. The bond enthalpies are all over the map with respect to hybridization. A much better correlation is to the pKa values of the corresponding alkane:
pKa (C4 H10 ) > pKa (CH4 ) > pKa (C6 H6 ) > pKa (C2 H2 )
The anomalous value for C4 H10 is because the molecule given is tert-butyl (as
opposed to n-butyl) lithium.
3. Use your table of molar bond energies (page 494 in your text) to calculate the
enthalpy change required to convert 1-hexene to cyclohexane:
3
H2
C
H2
C
H2C
C
H
H2
C
C
H2
CH3
H2C
CH2
H2C
CH2
C
H2
This requires breaking one C=C bond, but forming two C–C bonds (it also
involves moving an H atom from one C to another, but this does nothing so far
as bond enthalpies are concerned), so
∆Hrxn = ∆Hbond (C = C) − 2∆Hbond (C − C) = 614 − 2 · 348 = −82 kJ mol−1
(a) Does the answer you got in part (a) depend on the length of the carbon
chain (i.e., propene to cylcopropane, octene to cylcooctane, etc.)?
No—if we’re only treating the bond broken and bonds formed, these are
always the same.
(b) Use the following enthalpies of combustion:
cyclopropane
cylcobutane
cyclohexane
cyclooctane
H2
C
H2C
H2
C
H2C
∆Hc◦ @ 298 K (kJ/mol):
H2C
CH2
H2C
CH2
CH2
H2C
C
H2
CH2
CH2
H2C
CH2
H2C
CH2
−2091
−2720
−3920
−5265
1-propene
1-butene
1-hexene
1-octene
H
C
H
C
H2C
∆Hc◦ @ 298 K (kJ/mol):
CH2
H2C
CH2
H2
C
CH3
−2059
H2C
H
C
CH3
C
H2
−2717
H2
C
H2C
C
H2
H2C
−4003
to calculate the conversion enthalpy for each α-olefin to each cycloalkane.
∆Hrxn = ∆Hc (olefin) − ∆Hc (cycloalkane)
4
H2
C
H
C
CH3
C
H2
C
H2
H2
C
C
H2
−5310
CH3
C
H2
n ∆Hrxn
3
4
6
8
32
3
−83
−45
(c) Comment. (Some time with plastic models might give relevant information.) The phenomenon at work here is ring strain—very small cycloalkanes force the C–C–C bond angle to be very small (as low as 60◦ for cyclopropane) instead of the 109.5◦ optimal for sp3 carbons. The larger rings
are not planar, however, and the three-dimensional staggering of carbons
allows for nearly perfect 109.5◦ angles for cyclohexane.
It’s interesting that the geometry of 8-membered rings isn’t as good for
tetrahedral carbons—this is just the way things fit together in three dimensions—
so the reaction enthalpy goes back up. We’d expect that as the ring gets
very large, it will eventually be possible to get perfect 109.5◦ angles again,
so the reaction enthalpies should go back down to the −82 kJ mol−1 predicted by bond enthalpies. (Very large ring alkanes are strongly disfavored
because of entropy, but are just fine as far as enthalpy is concerned.)
4. Elemental sulfur in its most stable form exists as cyclic S8 molecules, and solid
S8 is found in two different crystalline phases, depending on conditions. Use
the following data for solid and gaseous sulfur:
∆H◦f (kJ mol−1 )
S8 (s, rhombic)
0.
S8 (s, monoclinic)
2.40
S8 (g)
102.3
S(g)
278.80
S◦ (J mol−1 K−1 )
254.4
260.8
430.87
?.
(a) Which phase is stable at lower temperature, which at higher, and at what
temperature are the two phases in equilibrium?
Always write out a chemical equation to get started on a problem like this
one.
S8 (s, monoclinic) −→ S8 (s, rhombic)
If ∆G◦rxn < 0, monoclinic sulfur will spontaneously convert to rhombic
sulfur, but rhombic will not convert back to monoclinic. This would mean
5
that rhombic was the stable phase under standard conditions.
Calculate ∆H and ∆S for this reaction:
◦
∆Hrxn
=
=
=
◦
∆Srxn =
=
=
∆Hf◦ (products) − ∆Hf◦ (reactants)
0 − 2.40
−2.40 kJ mol−1
S ◦ (products) − S ◦ (reactants)
254.4 − 260.8
−6.4 J mol−1 K−1
With
◦
◦
∆G◦rxn = ∆Hrxn
− T ∆Srxn
then
◦
, so forming
• At low temperature (i.e., as T → 0), ∆G◦rxn ≈ ∆Hrxn
rhombic from monoclinic is spontaneous: i.e., rhombic is stable. (You
could guess this because it’s the phase with the zero enthalpy of formation.)
◦
, so rhombic
• At high temperature (i.e., as T → ∞), ∆G◦rxn ≈ −T ∆Srxn
reacts spontaneously to form monoclinic (the reverse of the reaction
written), and monoclinic is the more stable phase.
• The two phases are in equilibrium when ∆G◦rxn = 0.
∆G◦rxn = 0
◦
◦
∆Hrxn
− T ∆Srxn
= 0
◦
◦
∆Hrxn = T ∆Srxn
◦
∆Hrxn
T =
◦
∆Srxn
−2.40 kJ mol−1 1000 J
=
−6.4 kJ mol−1 K−1 1 kJ
= 375 K
(b) Estimate the sublimation temperature of monoclinic sulfur. Would the
rhombic form sublime at higher or lower temperature?
6
Again, always write out the chemical equation. Here, we want to convert
solid monoclinic sulfur to gaseous S8 , so
S8 (s, monoclinic) −→ S8 (g)
At the sublimation temperature, ∆G◦ for this reaction will again be zero.
The calculation follows the one in the previous part, with new ∆H and
∆S values.
∆Hf◦ (products) − ∆Hf◦ (reactants)
102.3 − 2.40
99.9 kJ mol−1
S ◦ (products) − S ◦ (reactants)
430.87 − 260.8
170.1 J mol−1 K−1
◦
∆Hrxn
T =
◦
∆Srxn
99.9 kJ mol−1
1000 J
=
−1
−1
1 kJ
170.1 kJ mol K
= 587 K
◦
=
∆Hrxn
=
=
◦
∆Srxn =
=
=
At this temperature, converting rhombic to monoclinic is spontaneous.
This means:
∆G(rhombic → monoclinic) < 0
∆G(monoclinic → gas) = 0
∆G(rhombic → gas) = ∆G(rhombic → monoclinic) + ∆G(monoclinic → gas)
= (< 0) + 0
< 0
So ∆G will be 0 at a lower temperature, and rhombic sulfur will sublime
at a lower temperature.
(c) Calculate the average bond dissociation enthalpy for an S–S bond in S8 (g).
The reaction is
S8 (g) −→ 8S(g)
7
and
◦
∆Hrxn
= 8 · 278.80 − 102.3 = 2128 kJ mol−1
This is the enthalpy for breaking 8 S–S bonds, so the average bond enthalpy
is an eighth of this: 266 kJ mol−1 .
(d) Other forms of sulfur with different ring sizes from n = 5–12 are known,
but all of these are thermodynamically unstable with respect to S8 . What
conclusions (if any) can you draw about the bond dissociation enthalpy
of the S–S bond in these other cyclo-sulfurs compared to the S–S bonds
in S8 ? Does your conclusion vary with the ring size of the cyclo-sulfur in
question?
Thermodynamically unstable means that conversion to S8 is spontaneous,
e.g.:
8S7 (g) −→ 7S8 (g)
∆G◦rxn < 0
For n < 8 (smaller rings), this reaction decreases entropy (less gas), so
the enthalpy change must be negative: that is, the S–S bonds in S8 are
stronger than those in S5 , S6 , or S7 .
We can’t say much about S9 or bigger, because entropy favors S8 , so enthalpy could either favor it as well (stronger bonds in S8 ), or disfavor it
but be outweighed by entropy (weaker bonds in S8 ).
Graded problem:
This should be written up and handed in before class begins on Wednesday the 9rd.
This problems may also be discussed in class.
1. Use the values in Appendix D:
∆Hf◦ (kJ mol−1 ) S ◦ (J mol−1 K−1 )
Ca2+ (aq)
CaCO3 (s)
Ba2+ (aq)
BaCO3 (s)
CO2−
3 (aq)
−542.8
−1207.6
−537.6
−1213.0
−677.1
−53.1
91.7
9.6
112.1
−56.9
8
For CaCO3 ,
◦
∆Hrxn
=
=
=
◦
∆Srxn =
=
=
◦
∆Grxn =
=
=
∆Hf◦ (products) − ∆Hf◦ (reactants)
−542.8 − 677.1 − (−1207.6)
−12.3 kJ mol−1
S ◦ (products) − S ◦ (reactants)
−53.1 − 56.9 − 91.7
−201.7 J mol−1 K−1
◦
◦
∆Hrxn
− T ∆Srxn
−12.3 kJ mol−1 − (298 K)(−201.7 J mol−1 K−1 )
47.5 kJ mol−1
Similarly, for BaCO3 ,
◦
∆Hrxn
= −1.7 kJ mol−1
◦
∆Srxn
= −159.4 J mol−1 K−1
∆G◦rxn = 45.8 kJ mol−1
(a) What are the solubilities (in mol/L) of BaCO3 and CaCO3 in water? The
relevant Ksp is for the reaction
XCO3 (s) −→ X2+ (aq) + CO2−
3 (aq)
For calcium,
∆G◦rxn = −RT ln K
∆G◦rxn
ln K = −
RT
47.5 kJ mol−1
= −
8.314 J mol−1 K−1 · 298 K
= −19.17
K = e−19.17
= 4.7 × 10−9
2+
2−
Ca (aq) CO3 (aq) = 4.7 × 10−9
9
Since Ca2+ and CO2−
3 concentrations are equal,
−5
M
[CaCO3 (aq)] = [Ca2+ (aq)] = [CO2−
3 (aq)] = 6.86 × 10
Similarly for barium,
K = 9.4 × 10−9
[BaCO3 (aq)] = 9.7 × 10−5 M
(b) Is there a temperature where the two compounds are equally soluble?
We’re looking for a temperature where Ksp (CaCO3 ) = Ksp (BaCO3 ). In
order for this to be true, then also
∆G◦rxn (CaCO3 ) = ∆G◦rxn (BaCO3 )
◦
◦
We assume that the values for ∆Hrxn
and ∆Srxn
don’t change (much) with
temperature, so the only temperature dependence is
∆G = ∆H − T ∆S
Solve for T :
−12.3 kJ mol−1 − T · (−201.7 J mol−1 K−1 ) = −1.7 kJ mol−1 − T · (−159.4 J mol−1 K−1 )
(−1.7 + 12.3) kJ mol−1
T =
(201.7 − 159.4) J mol−1 K−1
= 258 K
(c) What are the Ba2+ , Ca2+ , and CO2−
3 concentrations in a solution with an
excess of both BaCO3 (s) and CaCO3 (s) at 25 ◦ C?
This is an equilibrium (as opposed to a free energy) problem. We have
equilibrium constants for both solvation reactions:
−9
[Ca2+ (aq)][CO2−
3 (aq)] = 4.7 × 10
−9
[Ba2+ (aq)][CO2−
3 (aq)] = 9.4 × 10
10
2−
and we also know that any CO2−
or
3 was created along with either a Ca
2+
a Ba , and so
[Ca2+ (aq)] + [Ba2+ (aq)] = [CO2−
3 (aq)]
Divide the first two equations to get:
[Ca2+ (aq)]
4.7 × 10−9
=
= 0.5
[Ba2+ (aq)]
9.4 × 10−9
So if we set
x = [Ca2+ (aq)]
then
[Ba2+ (aq)] = 2x
and
[CO2−
3 (aq)] = 3x
so
[Ba2+ (aq)][CO2−
3 (aq)]
2x · 3x
6x2
x
2−
[CO3 (aq)]
11
=
=
=
=
=
9.4 × 10−9
9.4 × 10−9
9.4 × 10−9
4.0 × 10−5
1.2 × 10−4