Chemistry 360
Dr. Jean M. Standard
Spring 2015
Name ______________KEY_________________
Exam 3 Solutions
1.) (14 points) Consider the reaction
NO ( g)
NO 2 ( g)
+
1O
2 2
( g) .
At 500 K and a total pressure of 2 bar, pNO 2 / pNO = 2.20. Determine the equilibrium constant K eq for
the reaction at 500 K, assuming that there
€ is 1 mole of NO2 present in the system initially.
€
We first need to determine the mole fractions at equilibrium and the equilibrium constant expression. Starting
€
from 1 mole of NO2 initially, we€have
NO ( g) + 0
NO 2 ( g)
1
moles init.
moles reacted − ξ
moles€equil.
xi
€
1− ξ
1+ ξ / 2
O 2 ( g) .
0
ξ
1ξ
2
1ξ
2
ξ
1+ ξ / 2
ξ /2
1+ ξ / 2
ξ
1− ξ
1
2
The equilibrium constant is
€
1/ 2
K eq
" PNO %" PO 2 %
$ ! '$ ! '
# P &# P &
=
" PNO %
$ !2 '
# P &
1/ 2
" x NO P %" xO 2 P %
$
'
$ ! '
x NO xO 2
# P &# P ! &
=
=
" xNO P %
xNO 2
2
$
'
!
# P
&
( )
1/ 2
" P %1/ 2
$ !' .
#P &
Substituting the mole fractions expressed in terms of the extent of reaction yields
€
K eq =
( )
x NO xO 2
xNO 2
1/ 2
" P %1/ 2
$ !'
#P &
%1/ 2
" ξ %"
ξ
''
$
'$$
# 1+ ξ / 2 &# 2(1+ ξ / 2) & " P %1/ 2
=
$ !' .
" 1− ξ %
#P &
$
'
# 1+ ξ / 2 &
Simplifying the expression above (not a required step but helpful later) leads to the result
€
%1/2 " P %1/2
" ξ %"
ξ
$
' $ ' .
K eq = $
'
# 1− ξ &$# 2 (1+ ξ / 2 ) '& # P! &
The information in the problem does not include the value of the equilibrium constant; however, the ratio
pNO 2 / pNO is provided.
€
2
1.
Continued
Expressing the ratio pNO 2 / pNO in terms of the extent of reaction will allow us to determine ξ,
pNO2
€
pNO
x NO2 P
= x NO P
= x NO2
x NO
" 1− ξ %
$
'
# 1+ ξ / 2 &
= " ξ %
$
'
# 1+ ξ / 2 &
pNO2
pNO
= 1− ξ
.
ξ
Solving for ξ,
pNO2
pNO
= 2.20 = 1− ξ
ξ
1− ξ
ξ
2.20ξ = 1− ξ
3.20ξ = 1
ξ = 0.3125.
Finally, the equilibrium constant may be calculated by substitution,
%1/2 " P %1/2
" ξ %"
ξ
' $ '
K eq = $
'$
# 1− ξ &$# 2 (1+ ξ / 2 ) '& # P! &
%1/2 " 2 bar %1/2
" 0.3125 %"
0.3125
' $
= $
'$$
'
# 1− 0.3125 &# 2 (1+ 0.3125 / 2 ) '& # 1 bar &
K eq = 0.236.
3
2.) (14 points) The temperature dependence of the chemical potential for a single component system is
shown below for a pressure of 1 atm.
S
L
µ
V
Tmelt
Tboil
T
Indicate on the diagram above the temperature at which solid-liquid equilibrium occurs (the melting
point) and the temperature at which liquid-vapor equilibrium occurs (the boiling point).
The S-L and L-V equilibria occur at the temperatures where the chemical potential curves intersect, as noted
above on the diagram.
Now, consider what happens to the diagram above if the pressure is increased. Does the boiling point of
the system increase or decrease relative to its 1 atm value? Explain your reasoning on the basis of
changes in the chemical potentials. Sketch the behavior of the chemical potentials as a function of
temperature for a higher pressure. You may want to use the diagram below and draw in the changes in
the chemical potentials for the higher pressure relative to those shown for P=1 atm.
We expect that as the pressure is increased, the boiling point will increase (since the boiling point is where the
vapor pressure of the liquid equals the applied pressure). To explain this on the basis of the chemical potential,
consider the fundamental equation for G,
dG = − SdT + VdP.
Dividing both sides of the equation by the number of moles yields the differential of the chemical potential,
€
dµ = − Sm dT + Vm dP.
Relating this to the exact differential for the chemical potential,
€
# ∂µ &
# ∂µ &
dµ = % ( dT + % ( dP,
$ ∂T ' P
$ ∂P 'T
we see that the pressure dependence of the chemical potential is related to the molar volume,
€
# ∂µ &
% ( = Vm .
$ ∂P 'T
4
2.) Continued
The molar volume of any phase is positive, so if the pressure increases the chemical potential must also increase
in order for the derivative to be positive. This suggests that for the graph of chemical potential as a function of
temperature, if the graph is drawn at a higher pressure, then the lines for each phase will be shifted up to higher
values of the chemical potential. This is sketched on the figure below.
S
µ
L
V
T
Tboil1 Tboil2
The solid lines show the original chemical potentials for the solid, liquid, and vapor phases at a pressure of 1
atm. The dashed lines show the chemical potentials of the three phases for a higher pressure, so the chemical
potentials values of each phase are larger. The effect of pressure on the solid and liquid phases is not very great
since the molar volumes of those phases are not affected much by increasing the pressure. The vapor phase
molar volume varies significantly with pressure, though, so the chemical potential of the vapor phase would
experience a larger increase with pressure, as shown in the diagram.
From the sketch, we see that the boiling point for the system at higher pressure, Tboil2, is greater than the
boiling point for the system at 1 atm pressure, Tboil1, as expected.
5
3.) (14 points) The vapor pressure of liquid sulfur dioxide obeys the relation
ln P = 17.9962 − 2999.8
,
T
where the pressure is in torr and temperature is in Kelvin.
a.) Determine the normal boiling point of sulfur dioxide in degrees Kelvin.
The liquid-vapor equilibrium coexistence curve is given by the Clausius-Clapeyron equation as
ln P = −
ΔH vap,m
RT
+ C .
The normal boiling point can be determined by solving the equation given for T and substitution of P = 760
torr,
€
ln P = 17.9962 − ln P − 17.9962 = −
T = 2999.8
T
2999.8
T
−2999.8
.
ln P − 17.9962
Then, for P = 760 torr,
T = −2999.8
ln 760 − 17.9962
T = 264.0 K (or − 9.2 ! C).
b.) Calculate the molar enthalpy of vaporization of sulfur dioxide in kJ/mol.
Using the Clausius-Clapeyron equation given above, we have that
−
ΔH vap,m
R
= − 2999.8K,
or ΔH vap,m = R ( 2999.8K )
(
)
= 8.314 J mol−1K −1 ( 2999.8K )
ΔH vap,m = 24940 J/mol or 24.9 kJ/mol.
6
4.) (15 points) True/false, short answer, multiple choice.
a.) True or False : The Clapeyron equation is valid only for the liquid-vapor phase coexistence curve in
one-component systems.
b.)
True or False: The reaction CO (g) + 3 H2 (g)
to the left if carbon monoxide is removed from the system.
CH4 (g) + H2O (g) is expected to shift
c.) Short answer
A maximum or minimum boiling solution for a binary liquid-vapor system is referred to as an
________azeotrope________ .
d.) Short answer
The _______chemical potential_________ is defined to be the molar Gibbs free energy of a substance.
e.) Multiple Choice
For the three graphs shown below, circle the one that exhibits the correct behavior for a typical reaction
which reaches chemical equilibrium.
G
G
ξ
(a)
G
ξ
(b)
ξ
(c)
7
5.) (14 points) Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created
containing 1.5 moles of liquid benzene and 2.5 moles of liquid cyclohexane, and at 50ºC the total vapor
pressure of the solution was measured to be 340 torr. Another solution was created containing 1.5 moles
of liquid benzene and 3.5 moles of liquid cyclohexane, and at 50ºC the measured total vapor pressure was
370 torr. Calculate the vapor pressures of pure benzene and pure cyclohexane at 50ºC.
The ideal solution can be described by Raoult's Law,
Pi = x i Pi* .
The total pressure of the mixture is given by the sum of partial pressures,
€
P = Pb + Pc
or P = x b Pb* + x c Pc* .
Here b refers to benzene and c refers to cyclohexane. We are given the total pressure for two different liquid
phase compositions, and are asked€to determine the pure vapor pressures. In the first case, the mole fractions
are
1.5mol
1.5mol + 2.5mol
= 0.375
xb =
xb
and
€
xc = 1 − xb
= 1 − 0.375
x c = 0.625.
In the second case, the mole fractions are
€
1.5mol
1.5mol + 3.5mol
= 0.300
xb =
xb
and
€
xc = 1 − xc
= 1 − 0.300
x c = 0.700.
*
*
The total pressure, given by P = x b Pb + x c Pc , in the first case is
€
340 torr = 0.375Pb* + 0.625Pc* .
€
The total pressure in the second case is
€
€
370 torr = 0.300 Pb* + 0.700 Pc* .
8
5.) Continued
*
There are two equations and two unknowns. One way to solve this is to solve the first equation for Pb and
*
substitute into the second equation. So, solving the first equation for Pb yields
0.375 Pb* = 340 − 0.625Pc*
or
Pb*
€
= 906.67€− 1.667 Pc* .
Substituting this result into the second equation leads to the vapor pressure of pure cyclohexane,
€
370 = 0.300 Pb* + 0.700 Pc*
(
)
370 = ( 0.300) 906.67 − 1.667 Pc* + 0.700 Pc*
370 = 272 − 0.500 Pc* + 0.700 Pc*
98 = 0.200 Pc*
Pc* = 490 torr .
*
*
From the first equation, we had Pb = 906.67 − 1.667 Pc . Substituting the vapor pressure of pure
cyclohexane allows us
€ to determine the vapor pressure of pure benzene,
Pb* = 906.67 − 1.667 Pc*
€
Pb* = 906.67 − 1.667 ( 490 torr )
Pb* = 90 torr .
€
9
6.) (14 points) Chlorine oxide, ClO, a key component in the destruction of the ozone layer in the polar
stratospheric regions, dimerizes to form Cl2O 2 according to the reaction
2ClO ( g)
Cl2O 2 ( g) .
At 310 K and a total pressure of 1 bar, ΔH R! = –72.43 kJ/mol and ΔSR! = –144 J mol–1K–1.
€
€
(a.) Determine the standard molar Gibbs free energy ΔGR! and equilibrium constant K eq at 310 K and 1
bar.
We can use the equation ΔGR! = ΔH R! − TΔSR! . to calculate the standard €
molar Gibbs free energy.
Substituting,
ΔGR! = ΔH R! − TΔSR!
(
= −72.43×10 3 Jmol−1 − (310 K ) −144 Jmol−1K −1
ΔGR! = −27790 Jmol−1, or −27.79 kJmol−1.
Next, the following expression may be used to determine the equilibrium constant,
ΔGR! = − RT ln K eq ,
# ΔG ! &
or K eq = exp $− R '.
% RT (
Substituting,
# ΔG ! &
R
K eq = exp $−
'
% RT (
#
&
)
)
(−27790 J/mol)
= exp $−
'
−1 −1
)% 8.314 J mol K (310 K ) )(
(
= e+10.740
K eq = 4.82 ×10 4 .
)
)
10
6.) Continued
(b.) Discuss whether increasing the total pressure (at constant temperature) will increase or decrease the
extent of reaction. Explain your answer.
The effect of pressure on the extent of reaction depends on whether there are more moles of gas on the
product side or on the reactant side. In this case, there is one mole of gas on the product side and there are
two moles of gas on the reactant side. Therefore, increasing the total pressure will put stress on the reactant
side, and the system will shift to the right to alleviate this stress. As a result the extent of reaction will
increase if the pressure is increased.
An alternate explanation is based upon the expression for the equilibrium constant. For the reaction
2ClO ( g)
Cl2O 2 ( g) .
the equilibrium constant expressed in terms of mole fractions is
€
€
! PCl2O2 $
! xCl2O2 P $
# ! &
#
&
!
" P %
" P %
K eq = = .
! PClO $2
! xClO P $2
# ! &
# ! &
" P %
" P %
Simplifying, we have
K eq = xCl2O2 P!
⋅ .
2
P
xClO
Notice that the equilibrium constant expression for this reaction includes the total pressure in the
P!
denominator. Therefore, if the total pressure increases, the
term will decrease. In order for the
P
numerical value of the equilibrium constant K eq to remain the same, that means that the other term,
xCl2O2
2
xClO
must increase. For this to occur, xCl2O2 (the mole fraction of the product) must increase and xClO (the mole
fraction of the reactant) must decrease; this has the effect of increasing the extent of reaction.
,
11
6.) Continued
(c.) Discuss whether increasing the temperature (at constant pressure) will increase or decrease the
equilibrium constant. Explain your answer.
The temperature dependence of the equilibrium constant depends on the sign of the molar enthalpy of
reaction. We are given above that ΔH R! = –72.43 kJ/mol, so it is an exothermic reaction. One way to
write the reaction including the heat released is
Cl2O 2 (g) + heat.
2ClO ( g)
Increasing the temperature puts stress on the product side (since that is where the heat resides). To
alleviate this stress,
€ the reaction shifts to the left, and the equilibrium constant decreases.
An alternative explanation can be given using the van't Hoff plot, shown below.
Exothermic case
lnKeq
slope = -ΔHo/R > 0
Thigh
Tlow
1/T
Since the enthalpy of reaction is negative, the slope of the van't Hoff plot is positive. When the
temperature increases, 1/T decreases. We can see that this leads to a smaller value of ln K eq and hence a
smaller value of K eq ; thus, the equilibrium constant decreases as the temperature increases.
€
€
12
7.) (15 points) True/false, short answer, multiple choice.
a.) True or False : For the liquid-to-vapor phase transition in a one-component system, the Gibbs free
energy change is negative ( ΔG < 0 ).
b.) True or False: A chemical equilibrium is an example of a dynamic equilibrium.
c.) Short answer
The ________van't Hoff________ Equation gives the temperature dependence of the equilibrium
constant.
d.) Short answer
The _______extent________ of reaction provides a measure of how far a chemical reaction has
progressed from reactants to products.
e.) Multiple Choice
Henry's Law is generally expected to be valid for which component or components of a solution?
1) the solute
2) the solvent
3) both the solute and the solvent