Download 1) In the reaction H2O + CH3COOH H3O+ +

CHEM 341
PHYSICAL CHEMISTRY
EXAM 4
Name_____________________________________
ID#_________________KEY__________________
Do not open this exam until told to do so.
The exam consists of 9 pages,
including this one. Count them to insure that they are all there. The last four pages of the exam are
the lists of constants and equations you were given previously. No additional notes are allowed. You
should only have your exam, writing implements and a calculator on your desk.
Do not write in the area below
Page
2
3
4
5
Total
Score
/30
/30
/20
/20
/100
1
Conceptual Problems. For these five problems, you do not need to justify or explain
your answer. Each problem is worth 5 points
1) What is the OH concentration (in units of moles per liter) of an aqueous solution of
pH 6?
OH   HKO   10

W

8
M
3
2) The oxidation/reduction reaction
Cu2+(aq)+ SO42-+ Zn(s)  Cu(s) + Zn2+(aq) + SO42-(aq)
is spontaneous. Is the standard potential for this reaction greater than zero, less than
zero, equal to zero or impossible to determine?
Greater than zero.
3) In a galvanic cell such as a chemical battery, does an oxidation reaction occur at the
positive electrode, the negative electrode, both electrodes or neither electrode?
The negative electrode: the oxidation reaction gives up electrons to the electrode making
it negative.
4) What are the units of the rate constant for the irreversible reaction A  B+C?
s-1 : It is first order in this direction
5) In the irreversible reaction A + B  C, if one increases the concentration of C, will
the rate of change in the concentration of A increase, decrease, remain the same or is
it impossible to tell?
Remain the same: the reaction is irreversible and therefore the amount of product has no
effect on the rate.
2
Numerical Problems. You must show all your work for complete credit.
6) (20 points) Consider an aqueous solution which contains 0.010 M acetic acid
(CH3COOH), 0.010 M sodium acetate (Na+ CH3COO-) and 0.010 M sodium chloride
(Na+ Cl-). The pKA of acetic acid is 4.75.
a) Assuming that all activity coefficients are one, what is the pH of the solution ? Justify
your answer with an equation.
b) Determine the ionic strength of the solution.
c) Determine the average activity coefficient for sodium acetate in the solution.
d) Using the activity coefficient that you calculated, estimate the pH that you would
actually measure of this solution using a pH meter.
a) pH  pK A  log
CH COO   pK

3
CH 3COOH 
A
 4.75
b)
2
2
2
I  0.5 mi z i2  0.5 m Na  1  m Acetate  1  mCl   1  0.50.020  0.010  0.010  0.020


i
c)
log    10.509 0.020  0.0720
   0.847
d) pH  pK A  log
  CH 3COO  
CH 3COOH 
 4.68
3
7) (20 points) Let's say that a galvanic cell (spontaneous oxidation/reduction reaction
like a battery) is constructed using the half reactions Cu2+(aq) + 2e-  Cu(s) (E0 =
+0.34V) and Zn2+ + 2e-  Zn(s) (E0 = -0.76V). What ratio of concentrations,
Zn2+/Cu2+, in the electrolyte will be required if one wants a 1.10 V zero current
potential? Assume that the temperature is 298 K and that all activity coefficients are
one. Hint: What is Q in the Nernst Equation for this reaction?
The reaction is spontaneous, so the voltage must be positive. Therefore the reaction is:
Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)
E0 = 1.10
For this reaction: Q 
aCu ( s ) a Zn 2
aCu 2 a Zn ( s )

a Zn 2 
aCu 2 
Zn 2
If all of the activity coefficients are assumed to be 1.00 then: Q 
Cu 2
RT
Zn 2
Thus the Nernst equation becomes: E  E 0 
ln
F
Cu 2
Zn   e
Therefore:
Cu 
2
2

F
RT
E  E 
0


 1.00
4


8) (20 points) For the elementary reaction series A+B  C  D, for components C
and D only, give
a) the rate equations (differential equations describing the rate of change in
concentration of each component with time),
b) the values of the initial rates of change of C and D and
c) the final concentrations of C and D when the reaction comes to equilibrium.
Note that the initial concentrations are A0 = B0= 0.30 M, C0 = 0.20 M, D0 = 0.00 M. The
rate constant for the first reaction, k1, is 30 M-1 s-1, the forward rate constant for the
second reaction, kf2, is 15 s-1 and the reverse rate constant, kb2, is 10 s-1.
dC
 k1  AB   k f 2 C   k b 2 D 
dt
a)
dD
 k f 2 C   k b 2 D 
dt
initially :
b)
dC 0 
M
 30 M 1 s 1 0.30 M 0.30 M   15s 1 0.20 M   10 s 1 0.00 M   0.30
dt
s
dD0 
M
 15s 1 0.20 M   10 s 1 0.00 M   3.00
dt
s






D  k f 2  1.5
C  k b 2
D  C   0.3M  0.2M  0.0M
c) 1.5C   C   0.5
C   0.2M
D  0.3M


K eq 
 0.5M
5


Applied Problem. You must show all your work for complete credit.
9) (15 points) The purpose of a catalyst is to lower the activation energy of a reaction
(in fact, that is all that a catalyst does). The enzymes in your body which mediate
chemical reactions are catalysts. One of these enzymes is called catalase and it
catalyzes the breakdown of hydrogen peroxide: 2H2O2 (hydrogen peroxide)  2H2O
+ O2. At 20 C, the uncatalyzed rate of this reaction is quite slow, taking days to
weeks for an open bottle of hydrogen peroxide to decay. When the enzyme is added,
the rate increases by a factor of 108 at the same temperature (in other words,
k enz
 108 ). Assuming that the rate is related to the activation energy in the usual
k noenz
way (k is proportional to e  E A / RT ), determine how much the enzyme lowers the
activation energy for this reaction.
The ratio of the rates with and without enzyme is 108. Therefore:
k
e  E A enz  / RT
10 8  enz   E A noenz / RT  e E A  noenz E A enz  / RT
k noenz e
ln 10 8  E A noenz   E A enz  / RT
E A noenz   E A enz   RT ln 10 8  44.9
kJ
mole
6