F2 g( ) F2 g( )

Chemistry 360
Fall 2014
Dr. Jean M. Standard
November 14, 2014
Name ____________KEY_______________
Exam 3 – 100 points
1.) (14 points) Diatomic and atomic fluorine gas exist in equilibrium according to the following reaction,
F2 ( g)
2 F ( g) .
(a.) The standard molar Gibbs free energy of reaction, ΔG r! at 298 K is 5.25 kJ/mol at 298 K. Use this
information to calculate the
for the reaction.
€ equilibrium constant
€
Using the equation ΔG r! = − RT ln K eq , the
€ equilibrium constant is
ΔG r!
,
RT
$& ΔG ! (&
r ).
= exp %−
'& RT *&
ln K eq = −
€
or K eq
Substituting, the equilibrium constant is
€
$ ΔG ! '
r
K eq = exp %−
(
& RT )
$
'
5.25 × 10 3 J/mol
+
+
= exp %−
(
−1 −1
+& 8.314 J mol K ( 298 K) +)
K eq = 0.120 .
(
)
(
)
€
(b.) Calculate the extent of reaction ξ for this reaction if 1.0 mole of F2 and 0.0 moles of F are present
initially. The equilibrium pressure is 1.3 bar.
The equilibrium constant can be expressed in terms of the extent of reaction,
€
moles init.
moles equil.
xi €
€
F2 ( g)
1
1−ξ
1−ξ
1+ξ
€
2 F ( g) .
0
2ξ
2ξ
1+ξ
2
1 b.)
Continued
The equilibrium constant expression is
K eq
" PF %2
$ !'
# &
= P
" PF %
$ !2 '
#P &
K eq
" x F P %2
$ ! '
# P &
=
.
" xF P %
2
$ ! '
# P &
The equilibrium constant can be written terms of the extent of reaction,
€
K eq
" x F P %2
$ ! '
# P &
=
" xF P %
$ 2! '
# P &
" 2ξ %2 " P %2
$
' $ '
# 1+ ξ & # P ! &
=
" 1− ξ %" P %
$
'$ '
# 1+ ξ &# P ! &
K eq =
"P%
4ξ 2
$ '.
(1+ ξ )(1− ξ ) # P ! &
!
Substituting the value for the equilibrium constant, equilibrium pressure (P=1.3 bar), and P = 1bar yields
€
K eq =
$P'
4ξ 2
& )
(1+ ξ )(1− ξ ) % P ! (
2
$ 1.3bar '
4ξ
0.120 =
&
).
(1+ ξ )(1− ξ ) % 1.0 bar (
Solving for the extent of reaction,
€
0.120 =
0.0924 =
$ 1.3 bar '
4ξ 2
&
)
(1 + ξ )(1 − ξ ) % 1.0 bar (
4ξ 2
(1 + ξ )(1 − ξ )
0.0924 (1 + ξ )(1 − ξ ) = 4ξ
(
0.0924 1 − ξ 2
)
= 4ξ
2
2
0.0924 − 0.0924ξ 2 = 4ξ 2
0.0924 = 4.0924ξ 2
0.02258 = ξ 2
ξ = 0.150 .
€
€
3
2.) (14 points) The vapor pressure of liquid sulfur dioxide is described by the equation
ln P = 8.3186 −
1425.7
.
T
where the pressure is expressed in units of torr and temperature is in degrees Kelvin. The vapor
pressure of solid sulfur dioxide is given by the relation
€
ln P = 10.592 −
1871.2
.
T
Determine the temperature and pressure of the triple point.
€
The triple point occurs at the intersection of the S-V and L-V coexistence curves. Equating these from above,
8.3186 −
1425.7
1871.2
= 10.592 −
T
T
445.5
2.2734 =
T
T = 196.0 K .
Thus, 196.0 K is the triple point temperature. Substituting this temperature back into either of the two
coexistence equations and solving for pressure give the triple point pressure.
€
Using the solid-vapor equilibrium equation,
1871.2
T
1871.2
ln P = 10.592 −
196.0 K
ln P = 1.045
P = 2.84 torr .
ln P = 10.592 −
€
4
3.) (14 points) Sketch a typical phase diagram for a one-component system. Make sure to carefully label all
the appropriate sections and points on the phase diagram. Explain the Gibbs phase rule and the role that
it plays in determining the degrees of freedom for each region or feature of the phase diagram.
A typical phase diagram for a one-component system is shown below.
Critical Point
L
S
Coexistence curves
P
V
Triple Point
T
Note that all phase diagrams of this type are graphs in which pressure is plotted on the y-axis with temperature
on the x-axis.
The Gibbs Phase Rule is F = C – Φ + 2, where F is the number of degrees of freedom, C is the number of
components, and Φ is the number of phases. For a one-component system, C = 1, so the phase rule reduces to F
= 1 – Φ + 2, or F = 3 – Φ.
The regions labeled S, L, and V correspond to regions in which only one phase is present, solid, liquid, or
vapor, respectively. In these regions, the Gibbs phase rule tells us that F = 3 – 1 = 2. This means that we are
free to select both P and T in those regions.
The regions on the phase diagram corresponding to the coexistence curves are lines along which two phases are
in equilibrium, either S-L, S-V, or L-V. The Gibbs phase rule tells us in this case that F = 3 – 2 = 1. This
means that if we select the temperature, then the pressure is fixed because it must lie on the coexistence curve.
(The same number of degrees of freedom holds for the critical point as well since it lies on the liquid-vapor
coexistence curve.)
Finally, at the triple point, all three phases are in equilibrium. The Gibbs phase rule in this case gives F = 3 – 3
= 0. This means that neither temperature nor pressure can be selected for a three-phase equilibrium. Both the
temperature and pressure are fixed at the triple point.
5
4.) (15 points) True/false, short answer, multiple choice.
a.) True or False : The chemical potential provides a measure of the progress of a chemical reaction from
reactants to products.
b.)
-4
True or False: The equilibrium constant of a particular reaction is determined to be 7.8×10 at 300 K.
The magnitude of this equilibrium constant implies that reactants are favored.
c.) Short answer
The notion that a chemical system in equilibrium will shift in order to relieve an applied stress is known as
_______LeChatelier's Principle___________ .
d.) Short answer
In a one-component system, the ________triple point__________ corresponds to solid, liquid, and vapor
all in equilibrium.
e.) Multiple Choice: For the two graphs shown below, circle the one that exhibits the correct behavior.
S
µ
V
L
µ
L
S
V
T
T
(a)
(b)
6
5.) (14 points) Nitrogen tetroxide decomposes according to the reaction
N 2O 4 ( g)
2 NO 2 ( g) .
At 55˚C and 1 atm, the average molecular weight of N2O4 and NO2 present at equilibrium is 61.2 g/mol.
Determine the equilibrium€constant. [Note: atomic
mass N = 14.007 g/mol; atomic mass O = 15.999
€
g/mol.]
We can use the average molecular weight to determine the equilibrium mole fractions. The equation for the
average molecular weight M is
M = x NO2 M NO2 + x N 2O4 M N 2O4 ,
€
where x is the gas phase mole fraction and M is the molecular weight. Since the sum of the mole fractions must
equal 1, we have
€
x NO2 + x N 2O4 = 1 ,
or
€
x NO2 = 1 − x N 2O4 .
Substituting into the expression for the average molecular weight yields
€
M = 1 − x N 2O4 M NO2 + x N 2O4 M N 2O4
61.2 g/mol =
(
)
(1 − x N O )(46.005 g/mol) +
2
4
x N 2O4 ( 92.010 g/mol)
61.2 = 46.005 x N 2O4 + 46.005
15.195 = 46.005 x N 2O4
x N 2O4 = 0.330 .
Then,
€
x NO2 = 1 − x N 2O4
x NO2
= 1 − 0.330
= 0.670 .
Knowing the mole fractions, we can now determine the equilibrium constant. Treating the gases as ideal gases,
the equilibrium constant can be determined as
€
K eq
€
" PNO %2
$ !2 '
# P &
=
.
" PN O %
2 4
$ ! '
# P &
7
5.) Continued
In terms of mole fractions, the equilibrium constant becomes
K eq
" x NO P %2
2
$
! '
# P &
=
" xN O P %
$ 2 !4 '
# P
&
=
2
x NO
P
2
x N 2O4 P !
2
=
( 0.670) (1.013bar )
( 0.330)(1bar )
K eq = 1.38.
Here, we have used the equilibrium pressure, P =1 atm = 1.013 bar.
€
8
6.) (14 points) The melting point of mercury is –38.9˚C at 1 bar and –19.9˚C at 3540 bar. The density of
liquid mercury is 13.69 g/mL and the density of solid mercury is 14.19 g/mL. Determine the molar
enthalpy of fusion. [The atmoic mass of Hg is 200.59 g/mol.]
The Clapeyron equation for solid-liquid phase equilibrium is
ΔH fus,m
dP
=
.
dT
T fusΔVm
We can approximate the left side of the equation as
€
dP
ΔP
≈
.
dT
ΔT
Substituting,
€
ΔH fus,m
ΔP
=
.
ΔT
T fus ΔVm
We can now solve for the desired molar enthalpy of fusion,
€
ΔH fus,m =
ΔP
⋅T fusΔVm .
ΔT
The molar volume of each phase can be calculated from the molecular weight M and the density D. For solid
mercury,
€
M # 1L &
⋅%
(
D $ 1000 mL '
# 200.59 g mol−1 & # 1 L &
= %
(⋅%
(
−1
$ 14.19 g mL ' $ 1000 mL '
Vs,m =
Vs,m = 0.01414 L/mol.
For liquid mercury, the molar volume is
€
M # 1L &
⋅%
(
D $ 1000 mL '
# 200.59 g mol−1 & # 1 L &
= %
(⋅%
(
−1
$ 13.69 g mL ' $ 1000 mL '
= 0.01465 L/mol.
Vl,m =
Vl,m
Substituting the information for the solid to liquid phase transition,
€
# ΔP &
ΔH fus,m = %
( T fusΔVm
$ ΔT '
# 3540 − 1 bar &
= %
(( 234.25 K)( 0.01465 − 0.01414 L/mol)
$ 253.25 − 234.25 K '
# 100 J &
= 22.53 Lbar/mol %
(
$ 1 Lbar '
ΔH fus,m = 2253 J/mol .
€
9
7.) (15 points) True/false, short answer, multiple choice.
a.)
True or False: For two phases to be in equilibrium, their chemical potentials must be equal.
b.) True or False : The reaction CO (g) + 3 H2 (g)
to the right if carbon monoxide is removed from the system.
CH4 (g) + H2O (g) is expected to shift
c.) Short answer
The _______Clausius-Clapeyron_____________ Equation gives an expression for the vapor pressure of
a solid or liquid in a one-component system.
d.) Short answer
The ________activity___________ of a solid or liquid takes a value of approximately unity in the
equilibrium constant expression for a heterogeneous reaction.
Cl2O 2 ( g) is 2.48 × 1011 at
e.) Multiple Choice: The equilibrium constant for the reaction 2 ClO ( g)
200 K. The standard molar enthalpy of reaction is –72.4 kJ/mol. If the temperature is increased to 500 K,
the equilibrium constant is expected to:
€
€
€
(a) increase.
(b) decrease.
(c) remain the same.

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