Download 1) In the reaction H2O + CH3COOH H3O+ + CH3COO

CHEM 341
PHYSICAL CHEMISTRY
EXAM 4
Name_____________________________________
ID#_______________________________________
Do not open this exam until told to do so.
The exam consists of 10 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?
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?
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?
4) What are the units of the rate constant for the irreversible reaction A  B+C?
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?
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 and c) 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.
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 = -0.76). 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 the Q in the Nernst equation for this reaction?
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. 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.
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  e  E A / RT ), determine how much the enzyme lowers the activation energy for
this reaction.
6
Summary of Equations and Constants for Exam 1
Constants
J
L  atm
 0.08206
K  mole
K  mole
J
k Boltz  1.381x10  23
K
1
N A  6.022 x10 23
mole
N
J
J
 1 3  10 3
2
L
m
m
1atm  101325Pa
1Pa  1
R  8.314
1bar  100000 Pa
Equations
PV  nRT
q const.V  CV T
3RT
M
8 RT
c
M
c P
z  rel
kT
kT

2P
 U 

  CV
 T V
dU const.V  CV dT
c
nRT
n
 a 
V  nb
V 
U  q  w
dU   T dV  CV dT
1  V 


V  T  P
 H 
 H 
dH  
 dP  
 dT
 P  T
 T  P

dU ideal  CV dT
dH   T dP  C P dT
 H 
T  
  C P 
 P  T
H  U  PV
H ideal  U ideal  nRT
q const. P  C P T
2
P
dw   Fdx
VF
w    Pex dV
VI
wconst. Pex   Pex V
V
wconst.T ,revers.,ideal   nRT ln  F
 VI
wad ,ideal  CV T
 T 

 P  H
1  V 
T   

V  P  T
H const. P  q const. P
 
 H 

  CP
 T  P
C P ,ideal  CV ,ideal  nR
1
CV , M
 V c
TF ,ad  TI ,ad  I  , c 
R
 VF 


 r H   f H   f H 
Pr oducts
Re ac tan ts
  H  T    H  T   T  T  C 
 r
2
r
1
2
1
r P

 U 
 U 
dU  
 dV  
 dT
 V  T
 T V
7
dV
 dT   T dP
V
C P , general  CV , general 
 2TV
T
Summary of equations and constants for Exam 2
For a spontaneous reaction:
S tot  0
S
r S 
dq
T
0
products
dS sys 
dAT ,V
For chemical reactions:

S

reac tan ts
In general:
A=U-TS
G=H-TS
dGT , P  0
Const. Pressure:
dG=dH-TdS
For a reversible reaction:
S tot  0
f  P
Vi
 U 
 U 
dU  
 dS  
 dV
 S V
 V  S
 U 

 T
 S V
Pi
Pf
 U 

  P
 V  S
For an isothermal ideal gas:
Vf
S sys  nR ln
In general:
dS sys
q sys
dG = VdP - SdT
Tsur
 1
1
 dq 

 TC Th
 G 
 G 
dG  
 dP  
 dT
 P  T
 T  P
 G 

 V
 P  T



At const. Pressure
S sys  C P ln
Tf
 G 

  S
 T  P
Ti
At const. Volume
S sys  CV ln
Tf
  
  V
 P  T
  

  S
 T  P
Ti
Carnot Engine:
 rev  1 
TC
Th
wrev   nR (Th  Tc ) ln
VA
VB
G ( P )  G   nRT ln
     RT ln
P
P
P
P
Useful Constants For Water --
CS (specific heat capacity of ice) =
1.95 J/( K g)
J
C P ,liquid  75.5
K  mole
J
C P , gas  33.58
K  mole
kJ
H fus  6.008
mole
kJ
H vap  40.656
mole
g
 liquid  density  1.000
ml
g
 ice  density  0.917
ml
Boiling point = 100 C
Freezing point = 0 C
Molecular weight = 18
For a monoatomic ideal gas:
3
R
2
5
CP  R
2
For a transition:
S trs 
f
P
Isothermal, ideal gas
In General:
dU=TdS - PdV
dGT , P  0
S sur  
  G / T  

  H
  1 / T   P
 G 
 

 n  T , P
     RT ln
Const. Volume
dA=dU-TdS
dq
dS sys 
T
dAT ,V  0
S sys  nR ln

CV 
H trs
Ttrs
8
Equations and Constants for Exam 3
Molar volume:
 V 

Vi  
 ni  T , P ,n '
colligative properties
2
RTvap
T  KX B 
X B  Kbb
H vap
2
RT fus
Gibbs free energy (const T,P)
G  1n1   2 n2  ...
T  K ' X B 
dG  1dn1   2 dn2  ...
Gibbs free energy (general)
dG  Vdp  SdT  1dn1   2 dn2  ...
for ideal gases
P
P 

Gmix  RT  n A ln A  n B ln B 
P
P

Gmix  nRT  X A ln X A  X B ln X B 
S mix  nR X A ln X A  X B ln X B 
Gmix  TS mix

H fus  1
  1 
R  T T fus 
  BRT
Gibbs phase rule
F=C-P+2
For reactions
 G 

   B   A
   P ,T
XB  Kfb
ln X B  
rG  B   A
for ideal gas reactions
P
 r G   r G  RT ln B
PA
P
K eq  B
PA
general reactions
 r G   r G   RT ln Q
H mix  0
for ideal solutions
PA  PA* X A
 A l    *A l   RT ln  X A 
Gmix  nRT  X A ln X A  X B ln X B 
S mix  nR X A ln X A  X B ln X B 
ideal dilution solutions (solute)
PB  K B X B
X K
 B l    B* l   RT ln  B * B
 PB
m 
 B l    B l   RT ln  B 
m 
more generally
 B l    B l   RT ln a B 
for water:
the cryoscopic constant,
K  kg
K f  1.86
mole
the ebullioscopic constant,
K  kg
K b  0.51
mole
H fus
 r G    RT ln K eq



K eq  e

 r G
RT
d ln K eq 

r H 
R
1
d 
T 
 r G    r H   T r S 
RT ln K eq   r H   T r S 
ln K eq 
r H  r S

RT
R
r H 
ln K eq (T2 )  ln K eq T1  
R
9
1 1
  
 T2 T1 
Summary of Equations for Exam 4
Acid/Base
F = 9.65 x 104 C/mole
pK W  pK A  pK B
solubility :
pK W  pH  pOH
MX  M+(aq) + X-(aq)
KS = aM+ aX-
pK W  14
Titrations (weak acid titrated with strong
base). Given in terms of activities:
Start point: pH  12 pK A  12 log acid 
Near the half stochiometry point:
base
pH  pK A  log
acid
At the stochiometry point:
pH  12 pK A  12 pKW  12 log base 
Past the stochiometry point:
pH  pKW  log excess _ base
S  KS
Chemical dynamics:
v
1 d J 
 j dt
First order:
A B
dAt 
 kAt 
dt
dAt 
 k  dt
At 
At   A0 e  kt
Activity coef.
for monovalent salt:
     
second order:
For Mn+ Xm- :
dAt 
 kAt Bt 
dt
A B C
    n m n  m
1
I
Equilibrium approximation
A  B  C
B  K
A eq
1
 mi zi2
2 i
log     z  z  A I
dC
 k 2 K eq  A
dt
for an aqueous solution:
Steady state approximation
activity    concentration
B   k f 1
 A k b1  k 2 
Electrochemistry
rG = -FE
k f1
dC
 k2
A
k b1  k 2 
dt
E  E 
RT
ln Q
F
Reaction rate vs. temperature and EA.
k  e  E A / RT
10