Chemistry 471
Midterm
10/26/06
Page 1 of 6
Name:____________________________
(Please print clearly on all pages)
Please leave the exam pages stapled together. The formulas are on a separate sheet.
This exam has 5 questions. You must answer at least 4 of the questions. You may answer more
questions if you wish.
Answering 5 questions can be an advantage if you are unsure of some of your answers (this will
distribute the “risk”). Answering 4 questions is advantageous if you are very sure of your
answers.
Each page is worth 20 points. The total exam grade will be normalized so that the maximum
number of course points for this exam will be 25. For example, getting 80 points on 4 questions
equals 100 points on 5 questions equals 25 points toward the final grade. Getting 80 points on 5
questions would be worth 80% of the maximum grade.
If you leave a page blank, it will not be included in the grading. If you work on a page and then
decide that you do not want it to be graded, be sure to mark the “DO NOT GRADE THIS PAGE
…” box at the bottom of the page. If you work on the page and fail to mark the box, the page
will be graded.
Work at least 4 problems (of your choosing) or more, as you prefer.
For grading purposes:
question
1
score
2
3
4
5
Tot
Chemistry 471
Midterm
10/26/06
Page 2 of 6
KEY
Name:____________________________
(Please print clearly on all pages)
1. Consider the gas phase reaction shown below in which two formic acid molecules dimerize.
The bonding interaction is via hydrogen bonds (dotted lines).
O
2
H
O
H
C
O
ΔHo = -60.08 kJ/mol
3 pts
ΔSo = -154.3 J K-1 mol-1
5 pts
ΔGo = -14.10 kJ/mol
5 pts
O
C
C
O
H
a) Some thermodynamic values for these
components are given in the table to the
right. Calculate ΔHo, ΔSo, and ΔGo for
the gas-phase dimerization of formic acid
at T = 25 °C. Be sure to include units.
3 pts
…H
H
…O
ΔHfo (kJ mol-1)
-362.63
-785.34
HCOOH (g)
(HCOOH)2 (g)
H
S (J K-1 mol-1)
251.0
347.7
ΔHo = ΔHf,dimero – 2 ΔHf,monomero = (-785.34 kJ/mol) –(2)(-362.63 kJ/mol)
ΔSo = Sdimero – 2 Smonomero = 347.7 J/K⋅mol) –(2)(251.0 J/K⋅mol)
ΔGo = ΔHo – T ΔSo = (-60.08 kJ/mol) –(298 K)(-154.3 J K-1 mol-1)
b) Calculate the enthalpy change per hydrogen bond formed in the gas phase for this reaction.
The enthalpy change per hydrogen bond formed is just half of the enthalpy
change in the reaction. The only bonds formed are H bonds, and they must be
the source of the net ΔHo. Thus, the ΔHH-bondo must be half of the net ΔHo, so
ΔHH-bondo = (-60.08/2) kJ/mol = -30.04 kJ/mol.
4 pts
c) Can you calculate the entropy of hydrogen bond formation in this reaction? Explain why or
why not.
No. The net entropy change in the reaction includes contributions from Hbond formation as well as changes in translation and rotation of the molecules.
One cannot separate these contributions.
The same answer also holds for the ΔGo for H-bond formation.
DO NOT GRADE THIS PAGE …
Chemistry 471
Midterm
10/26/06
Page 3 of 6
KEY
Name:____________________________
(Please print clearly on all pages)
2. Consider the gas phase dimerization of formic acid in problem 1.
5 pts
a) What is the value of the equilibrium constant for this dimerization reaction at T = 25 °C?
Because K = e
K =e
5 pts
− ΔG °
− ( −14 ,100 J / mol )
RT
we can compute K from the value of ΔGo.
( 8.3145 J / K ⋅mol )( 298.15 K )
, so K = e5.69 = 295
b) If you find that the partial pressure of formic acid (monomer) is 0.5 atm when the system is at
equilibrium at T = 25 °C, what is the partial pressure of the formic acid dimer?
K = adim/amon2 = 295, so adim = 295(0.5)2 = 73.75, so the pressure is the
activity times 1 atm, or Pdim = 74 atm. The reaction favors dimer formation at
25 °C, so we would expect the dimer concentration to be higher than the
monomer concentration.
c) If, in part b, the gasses are in a 2 L container, what will be the effect of reducing the volume
of the container reversibly to 1 L if T remains at 25 °C? (Note: you may assume the gasses to be
ideal and that no gas escapes from the container. You do not need to calculate actual numbers
here, but you do need to explain your answer.)
2 pts
3 pts
5 pts
{ The reaction will not change.
{ More monomer will be produced at the expense of dimer.
X More dimer will be produced at the expense of monomer.
{
Explanation:
There are two moles of monomer to one mole of dimer in the reaction. Increasing pressure (by reducing V
with T held constant) will cause the chemical potential of the monomer to increase twice as much as for the
dimer. Thus, the chemical driving force will increase from monomer toward dimer. This is just Le
Chatelier’s principle. An alternate explanation: using ΔΔG = ΔnRT ln(P2/P1) , we see that Δn = -1, and P2 is
larger than P1. Thus ΔΔG will be negative, and the reaction will be more favored in the direction as written.
d) What will be the value of ΔG for the dimerization reaction in part c? Be sure to include units.
Zero. This is an equilibrium situation because the compression was done reversibly.
DO NOT GRADE THIS PAGE …
Chemistry 471
Midterm
10/26/06
Page 4 of 6
KEY
Name:____________________________
(Please print clearly on all pages)
3. The half cell redox reactions of cytochrome c and FAD are shown below, with pH 7 standard
state electrochemical potentials. Answer the following based on these half-reactions.
Cyt[Fe(III)] + e- → Cyt[Fe(II)]
FAD + 2 H+ + 2 e- → FADH2
5 pts
Εo’ = +0.254 V
Εo’ = -0.219 V
a) Write the balanced chemical equation for the reduction of FAD by Cyt[Fe(II)].
There are two electrons in the bottom reaction, so we need to multiply the top equation by 2 to
balance electrons. If FAD is being reduced, the top equation must be reversed. So the net equation is
FAD + 2 H+ + 2 Cyt[Fe(II)] → 2 Cyt[Fe(III)] + FADH2
b) What is the value of Εo’ for the chemical reaction in part a? Be sure to include units.
6 pts
The top reaction is reversed, so its electrochemical potential is negated and added to the bottom
reaction’s electrochemical potential: Εo’ = -(+0.254 V) + (-0.219 V) = -0.473 V
c) Calculate the value of the equilibrium constant at T = 25 °C for the reaction in part a.
4 pts
K =e
ne FE 0 '
RT
)( −0.473V )
mol ⋅V
)( 298.15 K )
K ⋅mol
( 2 )( 96 , 500 J
=e
( 8.3145 J
= e −36.8 = 1.02 × 10 −16 . You can also compute from ΔGo’ = -neFEo’
= -(2)(96,500 J/mol⋅V)(-0.473 V) = +91.3 kJ/mol and K = e
reactants over products.
5 pts
−
ΔG 0 '
RT
. This reaction strongly favors
d) If Cyt[Fe(III)] and Cyt[Fe(II)] at standard state concentrations and pH 7 were in one
electrochemical cell while FAD and FADH2 at standard state concentrations and pH 7 were in
another electrochemical cell, would electrons flow from the cytochrome cell to the FAD cell or
vice versa. You may assume that T = 25 °C and that the two cells are connected by an ionconducting frit between solutions and by a current meter outside the solutions. Explain your
answer.
If the cytochrome cell is on the left and the FAD cell is on the right, a voltmeter will show that the
cytochrome cell is positive (by 0.473 volts) compared to the FAD cell. That is, the FAD cell will be
the source of electrons, and electrons will run from FAD to cytochrome through the current meter.
That is, FADH2 is producing electrons and Cyt[Fe(III)] is accepting electrons. This is consistent with
our value of ΔGo’ in part c, which says that the reaction in part a tends to run in the reverse direction.
DO NOT GRADE THIS PAGE …
Chemistry 471
Midterm
KEY
Name:____________________________
(Please print clearly on all pages)
4 pts
10/26/06
Page 5 of 6
4. a) In a series of experiments, you find that addition of 5 kJ of heat to pure liquid ethanol kept
at T = 25 °C causes the vaporization of 0.13 moles of the ethanol. What is the enthalpy of
vaporization of ethanol at T = 25 °C?
The enthalpy of vaporization is the same thing as the latent heat of vaporization, so q = nΔHvap, or
ΔHvap = q/n = (5 kJ)/(0.13 mol) = 38.5 kJ/mol.
6 pts
b) When you apply heat to a flask of ethanol and allow the temperature to rise, you find that
pure ethanol will boil at T = 78.3 °C when the pressure is 1 atm. When you take the ethanol to
the top of Mt. Denali, ethanol boils at T = 54.3 °C. What is the pressure at the top of Mt. Denali?
Be sure to include units.
Use ln P2 = − ΔH vap ⎛⎜ 1 − 1 ⎞⎟ . We know P1 = 1 atm when T1 = 78.3 °C = 351.45 K. We also
⎜
⎟
P1
R
⎝ T2
T1 ⎠
know T2 = 327.45 K and ΔHvap= 38,500 J/mol. Plug into the formula all the values known and
solve for P2. You get ln (P2) = -0.966 so P2 = 0.38 atm.
c) Suppose you place a 0.1 L sample of pure ethanol in a sealed flask. What will the vapor
pressure be in the flask when the system comes to equilibrium at T = 25 °C?
5 pts
Again, use ln P2 = − ΔH vap ⎛⎜ 1 − 1 ⎞⎟ . We know P1 = 1 atm when T1 = 351.45 K. We also know
⎜
⎟
P1
R
⎝ T2
T1 ⎠
T2 = 25 °C = 298.1 K and ΔHvap= 38,500 J/mol. Plug into the formula all the values known
and solve for P2. You get ln (P2) = -2.355 so P2 = 0.095 atm.
5 pts
d) If you add 100 g of an unknown, ethanol-soluble compound to the ethanol in part c, you find
that the vapor pressure of the solution drops to 87% of what it was in part c. The molar mass of
ethanol is 46.07 g/mol and the density of pure ethanol is 0.789 g/mL. What is the molar mass of
the unknown compound you added to the ethanol? (You may assume ideal solution behavior.)
This is Raoult’s law. So Psolvent = asolventPsolvent,pure = XsolventPsolvent,pure. The second step comes
from the ideal solution statement. We are told that the vapor pressure drops by 87%, so that
means that Xethanol = 0.87. The 0.1 L of ethanol has a mass of 78.9 g (from the given density).
That means there are (78.9 g)/(46.07 g/mol) = 1.71 mol of ethanol in the solution. Thus, the
mole fraction of ethanol is X = (1.71)/(1.71 + n) = 0.87 where n is the number of moles of
solute. Solving for n gives n = 0.25, so MW = (100 g)/0.25 = 400 g/mol.
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Chemistry 471
Midterm
10/26/06
Page 6 of 6
KEY
Name:____________________________
(Please print clearly on all pages)
2 pts
each
5. For the following processes, determine whether each of the thermodynamic quantities listed is
either greater than, less than, or equal to zero for the system, which is shown in italics. Consider
all gases to be ideal. For each answer briefly state the reason for your answer.
a) An ideal gas expands adiabatically and reversibly.
<0, =0, >0?
Reason
ΔT
<0
We see that ΔE is negative, and for an ideal gas ΔE = CVΔT.
w
<0
Volume increases and dw = -PdV.
q
=0
This is an adiabatic process.
ΔE
<0
ΔE = q + w; q is zero and w is negative.
ΔH
<0
For an ideal gas ΔH = CPΔT and we know that ΔT is negative.
b) Samples of hot water and cold water are mixed together in a thermally insulated, closed
container of fixed volume.
<0, =0, >0?
Reason
q
=0
This is an adiabatic process (the container is insulated).
w
=0
Since liquids are essentially incompressible, volume does not change and dw = -PdV.
ΔE
=0
ΔE = q + w; q is zero as is w.
ΔH =0
ΔH = ΔE + Δ(PV); ΔE equals zero and there is essentially no volume change, so P won’t change.
ΔS
This is an irreversible (spontaneous) process.
>0
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