CHM 3400 – Fundamentals of Physical Chemistry
Third Hour Exam
April 10, 2013
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
R = 0.08314 Lbar/moleK
1 L.atm = 101.3 J
R = 8.314 J/moleK
1 atm = 1.013 bar = 1.013 x 105 N/m2
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1. (30 points) The reduction of iron III oxide (Fe2O3) by carbon monoxide (CO) is one way in which iron
metal can be produced. The reaction that takes place is
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
(1.1)
a) Write the expression for the equilibrium constant for the above reaction in terms of the activities
of the reactants and products.
b) Write the expression for the equilibrium constant for the above reaction assuming ideal
behavior.
c) Find the numerical value for K for the above reaction at T = 298. K.
d) Find the numerical value for K for the above reaction at T = 500. K
Thermochemical data for the reactants and products needed to do parts c and d of the above problem are
given below for T = 298. K.
Substance
Hf (kJ/mol)
Gf (kJ/mol)
S(J/molK)
CO(g)
CO2(g)
- 110.53
- 393.51
- 137.17
- 394.36
197.67
213.74
Fe(s)
Fe2O3(s)
0.0
- 824.2
0.0
- 742.2
27.28
87.40
2. (15 points) Consider the following galvanic cell
Bi(s) | Bi3+(aq) || Zn2+(aq) | Zn(s)
(2.1)
Find the half-cell oxidation reaction, the half-cell reduction reaction, the net cell reaction, and the cell
voltage (for standard conditions and T = 25.0 C) for the above cell. Half-cell reduction potentials are
given in the Appendix of Atkins.
3. (15 points) In a disproportionation reaction the same metal ion is both oxidized and reduced. As an
example of such a reaction, consider the disproportionation reaction of Cr 2+ ion
3 Cr2+(aq)  Cr(s) + 2 Cr3+(aq)
(3.1)
Using the standard table of half-cell reduction potentials in the Appendix of Atkins, find the numerical
value for the equilibrium constant for the above reaction at T = 25.0 C.
4. (20 points) Ozone (O3) in the gas phase will slowly decompose into molecular oxygen (O 2). The
reaction taking place is irreversible, and may be written as
O3(g)  3/2 O2(g)
(4.1)
In a particular experiment, carried out in a system at T = 320. K, the initial concentration of ozone was [O 3]
= 2.65 x 10-5 mol/L. At t = 3.00 hours the concentration of ozone was found to be 1.88 x 10 -5 mol/L. What
will the concentration of ozone be in the system at t = 10.00 hours for the following two cases
a) The reaction is first order homogeneous in ozone, that is
d[O3]/dt = - k [O3]
(4.2)
b) The reaction is second order homogeneous in ozone, that is
d[O3]/dt = - k [O3]2
(4.3)
5. (20 points) On rare occasions chemical reactions are observed that follow the following rate law
d[A]/dt = - k/[A]
(5.1)
The expression involving the concentration of A as a function of time obtained from the above rate law (for
times t < [A]02/2k ) is
[A]t2 = [A]02 – 2kt
(5.2)
where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, and k is the rate
constant for the reaction.
a) Starting with eq 5.2 find an expression for t 1/2, the half-life for a reaction following the above
rate law.
b) For a particular system the rate constant for the reaction is found to be k = 5.4 x 10 -6 mol2/L2s
at T = 300.0 K and k = 2.1 x 10 -2 mol2/L2s at T = 400.0 K. Assuming the Arrhenius equation applies, find
the value for Ea, the activation energy, for the above reaction.
Solutions.
1)
a) K = (aFe)2 (aCO2)3
(aFe2O3)(aCO)3
b) K = (pCO2)3
(pCO)3
Activities for solids are equal to 1; activities for gases are equal to the
partial pressure (divided by a standard pressure of 1 bar).
c) For the next two parts of the problem it will be useful to find Grxn and Hrxn.
Grxn = [ 2 (Gf(Fe(s)) + 3 (Gf(CO2(g)) ] – [ (Gf(Fe2O3(s)) + 3 (Gf(CO(g)) ]
= [ 2 (0.0) + 3 ( - 394.36) ] – [ ( - 742.2) + 3 ( - 137.17) ] = - 29.37 kJ/mol
Hrxn = [ 2 (Hf(Fe(s)) + 3 (Hf(CO2(g)) ] – [ (Hf(Fe2O3(s)) + 3 (Hf(CO(g)) ]
= [ 2 (0.0) + 3 ( - 393.51) ] – [ ( - 824.2) + 3 ( - 110.53) ] = - 24.74 kJ/mol
Then
ln K = - Grxn
RT
= -
( - 29370. J/mol)
= 11.85
(8.3145 J/molK)(298.2 K)
And so K = e11.85 = 1.4x 105
d) ln(K2/K1) = - (Hrxn/R) [ (1/T2) – (1/T1) ]
= - [ ( - 24740. J/mol)/ (8.3145 J/molK) ] [ (1/500. K) – (1/298. K) ] = - 4.03
K2 = K1 e-4.03 =(1.4 x 105)(e-4.03) = 2.5 x 103
2)
ox
2 x ( Bi(s)  Bi3+(aq) + 3 e- )
red
3x ( Zn2+(aq) + 2 e-  Zn(s) )
____________________________________________
net
2 Bi(s) + 3 Zn2+(aq)  2 Bi3+(aq) + 3 Zn(s)
3)
ox
2 Cr2+(aq)  2 Cr3+(aq) + 2 e-
red
Cr2+(aq) + 2 e-  Cr(s)
___________________________________
net
3 Cr2+(aq)  Cr(s) + 2 Cr3+(aq)
E = - 0.20 v
E = - 0.76 v
_____________
Ecell = - 0.96 v
E = + 0.41 v
E = - 0.91 v
____________
Ecell = - 0.50 v
ln K = FEcell = (3) (96485. C/mol) ( - 0.50 v) = - 38.91
RT
(8.3145 J/molK) (298.2 K)
K = e-38.91 = 1.3 x 10-17
4)
a) If the reaction is first order homogeneous
ln( [O3]t/[O3]0 ) = kt, and so
k = (1/t) ln( [O3]t/[O3]0 ) = (1/3.0 hr) ln[(2.65 x 10 -5)/(1.88 x 10-5)] = 0.114 hr-1
Since
[O3]t = [O3]0 e-kt
Then after 10.0 hours
[O3]t = (2.65 x 10-5 mol/L) exp[ - (0.114 hr-1)(10.0 hr) ] = 0.84 x 10-5 mol/L
b) If the reaction is second order homogeneous
( 1/[O3]t ) – ( 1/[O3]0 ) = kt, and so
k = (1/t) [ ( 1 [O3]t ) – ( 1/[O3]0 ) ] =
= (1/3.0 hr) [ ( 1/1.88 x 10-5 mol/L ) – (1/2.65 x 10-5 mol/L) ] = 5152. L/molhr
Then after 10.0 hours
( 1/[O3]t ) = ( 1/[O3]0 ) + kt = ( 1/2.65 x 10-5 mol/L) + (5152 L/molhr)(10.0 hr)
= 89260 L/mol
And so [O3]t = 1/(89260. L/mol) = 1.12 x 10 -5 mol/L
5)
a) At t = t1/2 [A]t = [A]0/2. If we substitute into our expression for concentration vs time, we get
( [A]0/2 )2 = [A]02 – 2kt1/2
t1/2 = 3 [A]02
8k
b) If the reaction obeys the Arrhenius equation, then
ln(k2/k1) = - (Ea/R) [ (1/T2) – (1/T1) ]
or
Ea = -
R ln(k2/k1)
= - (8.3145 J/molK) ln[ (2.1 x 10-2)/(5.4 x 10-6) ] = 82.5 kJ/mol
[ (1/T2) – (1/T1) ]
[ (1/400.0 K) – (1/300.0 K) ]