Download CHM 3410

CHM 3410 - Physical Chemistry 1
Third Hour Exam
November 18, 2011
There are five problems on the exam. Do all of the problems. Show your work.
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R = 0.08206 L.atm/mole.K
NA = 6.022 x 1023
1 atm = 1.013 bar
.
.
R = 0.08314 L bar/mole K
1 L.atm = 101.3 J
1 atm = 760. torr
R = 8.314 J/mole.K
k = 1.381 x 10-23 J/K
1 atm = 1.013 x 105 Pa
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1. (24 points) Consider a sample of chlorine gas (Cl 2, M = 70.91 g/mol,  = 0.93 nm2) at a pressure p =
500.0 torr and a temperature T = 300.0 K. Find the value for the following:
a) crms, the rms average speed of a chlorine molecule, in units of m/s.
b) , the mean free path of a chlorine molecule, in units of nm.
c) f(100. m/s < v < 150. m/s), the fraction of chlorine molecules with speeds between 100. m/s and
150. m/s.
2. (24 points) Measurements on galvanic cells can be used to determine solubility products for slightly
soluble ionic compounds. As an example, consider the solubility of lead II bromide (PbBr 2) in water.
PbBr2(s)  Pb2+(aq) + 2 Br-(aq)
(2.1)
A galvanic cell whose net cell reaction is the solubility reaction (equn 2.1) is given below
Pb(s)|Pb(NO3)2(aq)||KBr(aq)|PbBr2(s)|Pb(s)
(2.2)
a) What are the half-cell oxidation reaction, the half-cell reduction reaction, and the net cell
reaction corresponding to the above galvanic cell?
b) Data for the cell potential for standard conditions for the above galvanic cell are given below.
T(C)
Ecell(V)
15.0
25.0
35.0
- 0.1539
- 0.1532
- 0.1525
Based on these data and additional information in Table 6.2 in Atkins (Table of standard half-cell potentials
at T = 298. K), find the following (at T = 25.0 C):
(1) Grxn, Hrxn, and Srxn for the solubility reaction (equn 2.1).
(2) Ksp, the numerical value for the solubility product for the solubility reaction (equn 2.1)
(3) Ered, the half-cell reduction potential, for the process
PbBr2(s) + 2 e-  Pb(s) + 2 Br-(aq)
(2.3)
3. (16 points) Acetic acid (CH3COOH) is a weak electrolyte. In water, acetic acid ionizes by the process
CH3COOH(aq)  H+(aq) + CH3COO-(aq)
K = (aH+) (aCH3COO-)
(3.1)
(aCH3COOH)
In this problem T = 25.0 C. The numerical value for K for the above reaction in pure water in the limit
[CH3COOH]  0 (where ideal behavior of all species may be assumed) is K = 1.754 x 10 -5.
In a 0.110 mol/L aqueous solution of sodium chloride (NaCl), the numerical value for K C, defined
as
KC = [H+] [CH3COO-]
(3.2)
[CH3COOH]
in the limit [CH3COOH]  0, is KC = 2.850 x 10-5.
a) Assuming that CH3COOH, the activity coefficient for acetic acid, is equal to 1.00, use the data
given above to find the experimental value for , the mean activity coefficient for H + and CH3COO- ions in
the 0.110 mol/L NaCl solution. Recall that for acetic acid, a 1:1 electrolyte,  is given by the expression
 = [ (H+)(CH3COOH-) ]1/2
(3.3)
b) Using Debye-Huckel theory find the theoretical value for  for a dilute solution of acetic acid
in a 0.11 mol/L aqueous solution of NaCl.
4. (18 points) Between 1400 K and 1700 K the rate constant for the gas phase decomposition of ethyl
alcohol (C2H5OH)
C2H5OH  “products”
d[C2H5OH]/dt = - k [C2H5OH]
(4.1)
A = 98.32
B = - 2.55 x 105 K
C = 1.81 x 108 K2
(4.2)
is given by the expression
ln k = A + B/T + C/T2
where k, the rate constant, has units of s-1.
a) What is the numerical value for k at T = 1500. K?
b) What is the numerical value for Ea, the activation energy, for the above reaction at T = 1500. K?
c) For an experiment at a particular constant temperature T the experimental value for the rate
constant for reaction 4.1 is k = 34000. s-1. A system at this temperature has an initial concentration of ethyl
alcohol [C2H5OH]0 = 3.84 x 10-4 mol/L. How long will it take the concentration of ethyl alcohol to
decrease to 1.00 x 10-5 mol/L? Give your answer in units of sec.
5. (18 points) Consider the following reaction mechanism for the conversion of monomers of A into a
tetramer.
stoichiometric
4 A  A4
step 1
A + A  A2
reversible (k1, k-1)
step 2
A2 + A2  A4
irreversible (k2)
Find the rate law for the following cases.
a) Step 1 is fast and reversible; step 2 is slow.
b) Step 1 is slow and reversible; step 2 is fast.
c) General case, where no assumptions are made about the relative rates of step 1 and step 2 (this
is difficult).
Solutions.
1)
a)
crms = (3RT/M)1/2 = [ 3(8.3145 J/mol.K)(300.0 K)/(70.91 x 10-3 kg/mol) ]1/2
= 325. m/s
b)
=
kT
(2)1/2p
=
(2)
1/2
(1.381 x 10-23 J/K) (300.0 K)
(0.93 x 10-18 m2) (500/760)atm (1.013 x 105 Pa/atm)
= 4.7 x 10-8 m = 47. nm
c) f(100 m/s < v < 150 m/s) = 100150 f(v) dv  (150. – 100.)m/s f(125. m/s)
where f(v) = (4) (M/2RT)3/2 v2 exp(- Mv2/2RT)
So f(100 m/s < v < 150 m/s) = (50. m/s) (4) [ (70.91 x 10-3 kg/mol)/2(8.3145 J/mol.K)(300.0 K) ]3/2
exp[ - (70.91 x 10-3 kg/mol)(125. m/s)2/2(3.3145 J/mol.K)(300.0 K) ]
= (50. m/s) (1.209 x 10-7 s3/m3)(1.5625 x 104 m2/s2) (8.008 x 10-1)
= 0.0757
2)
a)
ox
Pb(s)  Pb2+(aq) + 2 e-
E = 0.13 v
red
PbBr2(s) + 2 e-  Pb(s) + 2 Br-(aq)
E = ?
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net
PbBr2(s)  Pb2+(aq) + 2 Br-(aq)
Ecell = - 0.1532 v
b) Grxn = - FEcell = - 2 (96485 C/mol) (- 0.1532 v) = + 29.56 kJ/mol
Srxn = F (dEcell/dT)
(dEcell/dT)  Ecell/T = [ (- 0.1525 v) – (- 0.1539 v) ] = 7. x 10-5 V/K
[ 35.0 C – 15.0 C ]
So
Srxn = = (2) (96485. C/mol) (7. x 10 -5 V/K) = 13.5 J/mol.K
Since
Grxn = Hrxn - TSrxn
then
Hrxn = Grxn + TSrxn = (29.56 kJ/mol) + (298.2 K) (13.5 x 10 -3 kJ/mol.K)
= 33.58 kJ/mol
ln K = FEcell = (2) (96485. C/mol) (- 0.1532 V) = - 11.92
RT
(8.3145 J/mol.K)(298.2 K)
and so K = e-11.92 = 6.6 x 10-6
Finally, from the work in part a, we may say
Ecell = Eox + Ered
Ered = Ecell - Eox = ( - 0.1532 v) – (0.13 v) = - 0.2832 v
3)
a) K = (aH+) (aCH3COO-) =
(aCH3COOH)
But
KC = [H+][CH3COO-]
[CH3COOH]
and so
K = KC ()2
(H+[H+]) (CH3COO-)[CH3COO-])
(CH3COOH)[CH3COOH])
CH3COOH = 1
(H+)(CH3COO-) = ()2
() = (K/KC)1/2 = [(1.754 x 10-5)/(2.850 x 10-5)]1/2 = 0.784
b) log10() = - A | z+ z- | I1/2
I = (1/2) i zi2 bi
Since the only significant source of ions is from the NaCl (other ions have concentrations of at least 1000
times smaller than Na+ or Cl-), and since bNaCl  [NaCl]
I = (1/2) [ (1)2 (0.11) + (-1)2 (0.11) ] = 0.11
At T = 25.0 C, the value of A for water is A = 0.509
So
log10() = - (0.509) | (1) (-1) | (0.11)1/2 = - 0.1688
4)
a) ln k = A + B/T + C/T2 = (98.32) + (- 2.55 x 105 K)/(1500.0 K) + (1.81 x 108 K2)/(1500.0 K)2
= + 8.764
() = 10-0.1688 = 0.678
K = e8.764 = 6400. s-1
b) Ea = RT2 (d lnk/dT)
But
(d lnk/dT) = d/dT {A + B/T + C/T 2 } = [ - B/T2 – 2C/T3 ]
Ea = RT2 (d lnk/dT) = RT2 (-1) [ B/T2 + 2C/T3 ]
= - R [ B + 2C/T ] = - (8.314 J/mol.K) [ (- 2.55 x 105 K) + 2 (1.81 x 108 K2)/(1500.0 K) ]
= + 113.6 kJ/mol
c) The reaction is 1st order, and so
[C2H5OH]t = [C2H5OH]0 exp(- kt)
exp( - kt) = [C2H5OH]t / [C2H5OH]0
- kt = ln{ [C2H5OH]t / [C2H5OH]0 }
t = (1/k) ln{ [C2H5OH]0 / [C2H5OH]t } = (1/34000. s-1) ln [ (3.84 x 10-4) / (1/00 x 10-5) ]
= 1.07 x 10-4 s = 107. sec
5)
a) Step 2 is slow, and so the overall rate of reaction is
d[A4]/dt = k2 [A2]2
A2 is a reaction intermediate, and step 1 is fast and reversible, and so establishes a quasi-equilibrium, and
so
k1 [A]2 = k-1 [A2]
[A2] = (k1/k-1) [A]2
Substituting, we get
d[A4]/dt = k2 [ (k1/k-1) [A]2 ]2 = (k12k2/k-12) [A]4
b) Step 1 is slow and reversible. Notice that once A2 is formed by step 1 acting in the forward
direction, there are two things that can happen to it
disappearance by the back reaction, (k-1), which is slow
disappearance by step 2, (k2), which is fast
Since we have a fast and a slow process, we can assume that the fast process will dominate, and so the rate
of reaction is the rate of the first step in the forward direction
d[A]/dt = - 2 k1 [A]2
(the 2 is because two A disappear for each forward reaction)
From the stoichiometry of the reaction, 1 A4 appears for every 4 A that disappear, and so we could also
write
d[A4]/dt = (1/4) 2 k1 [A]2 = (k1/2) [A]2
c) If we cannot classify the steps in this mechanism as fast or slow, we can still say that the overall
rate of formation of A4 is
d[A4]/dt = k2 [A2]2
and we can still make the steady state approximation for [A2]
d[A2]/dt  0 = k1 [A]2 – k-1[A2] – 2 k2[A2]2
or, multiplying through by (-1)
- k1 [A]2 + k-1[A2] + 2 k2[A2]2 = 0
This is quadratic in [A2], and so has the form
ax2 + bx + c = 0
a = 2k2
b = k-1
c = - k1 [A]2
x = [A2]
Using the quadratic formula, we get
[A2] = - k-1  [ (k-1)2 + 8 k1 k2 [A]2 ]1/2
4 k2
Since the term in the square brackets must be larger than k-1, we know that it is the positive root that gives
a positive value for [A2]. Further, if we multiply and divide by k-1, we can make the expression for [A2] a
bit prettier. Doing this gives
[A2] = (k-1/4k2) { [ 1 + (8k1k2/k-12) [A]2 ]1/2 – 1 }
and so
d[A4]/dt = k2 [A2]2 = (k-12/16k2) { [ 1 + (8k1k2/k-12) [A]2 ]1/2 – 1 }2
We may check this result by examining the limiting cases.
Case 1 – k1, k-1 >> k2
In this case (8k1k2/k-12) << 1, and we may use the Taylor series expansion for (1 + x) 1/2
(1 + x)1/2 = 1 + x/2 + .....
d[A4]/dt = (k-12/16k2) [ (4k1k2/k-12) [A]2 ]2 = (k12k2/k-12) [A]4
Case 2 – k2 >> k1, k-1
In this case (8k1k2/k-12) >> 1, and so
d[A4]/dt = (k-12/16k2) { [ (8k1k2/k-12) [A]2 ]1/2 }2
= (k-12/16k2) [ (8k1k2/k-12) [A]2 ] = (k1/2) [A]2
Notice that these are the same results we obtained in parts a and b of this problem, suggesting that our
solution for the general case is correct.