Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz #1
Problem 1 (4 points).
As we have learnt, heat capacity is a function of temperature and Shomate equation is normally used to
describe this dependence. However, if the temperature range that we are interested in is relatively small,
we sometimes can use a value of the heat capacity that is given somewhere within this range. The molar
heat capacity of silver at a pressure of 1 atmosphere and room temperature is 23.35 J/mol/K. Calculate the
heat required to increase the temperature of a block of silver that weighs 215.74 g from room temperature
to 35.15ºC at 1 atm, assuming that within this temperature range the heat capacity of this materials does
not change significantly.
T2
q P = ∫ C P dT = C P (T2 − T1 ) = C P ,m × n (moles) ×(T2 − T1 ) = 23.35
T1
⎛ 215.74 g ⎞
J
⎟ × 10.15 K =
× ⎜⎜
mol × K ⎝ 107.87 g / mol ⎟⎠
= 474 J
Problem 2 (4 points). Use van der Waals law to predict the pressure exerted by 10 mole of CO2 enclosed
in a 1 L cylinder at room temperature.
P(HCl )vdW
Pa × m 6
J
0
.
3658
×
K
298
.
15
nRT
n2a
RT
a
mol × K
mol 2
=
− 2 =
− 2 =
−
V − nb V
Vm − b Vm
m3
m3 ⎛ 1 L
1L
m3
× 10 −3
− 42.8588 × 10 − 6
⎜⎜
× 10 −3
L
mol
10 mol
L
⎝ 10 mol
8.3144
= 43382679.4 Pa − 36580000 Pa = 6802679.4 Pa = 67 atm
⎞
⎟⎟
⎠
2
=
Problem 3 (2 points)
From the possible statements in column B, select the best match for each phrase in column A and put its
letter in the adjacent blank. There is only one best match for each phrase.
Column A
Column B
1. The system that allows no exchange of matter but
allows exchange of energy across the boundary is
called ___i___
2. One of the ideal gas law approximations
___f__
a)
b)
c)
d)
molecules are hard spheres
open system
isolated system
two systems that are separately in thermal
equilibrium with a third system are also in
thermal equilibrium with one another
e) collisions with a solid wall are inelastic
f) molecules are point masses
g) U = q + w
h) is based on the activated complex theory
i) closed system
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz#2
Problem 1 (8 points). Calculate the work for the following processes:
a) (2 points) Isothermal expansion of 1 mol of a gas that can be described by the van der Waals law
against vacuum
dw = − Pext dV = 0 × dV = 0 The work of expansion against vacuum is zero regardless of which gas does
the work
b) (3 points) Isothermal expansion of 1 mol of a gas that can be described by the van der Waals law
from 1 L to 10 L against 1 atm pressure
dw = − Pext dV
V2
w = − ∫ Pext dV = − Pext ΔV = −1 atm × 9 L = −101325 Pa × 9 L × 0.001
V1
m3
= −911.9 J
L
Negative sign is significant because the gas does the work.
c) (3 points) Isothermal reversible expansion of one mole of an ideal gas from 1 L to 10 L at room
temperature
dw = − Pext dV = − PdV for a reversible process and for ideal gas P =
V2
V2
nRT
V
J
nRT
⎛
⎞ dV
Thus, w = − ∫ PdV = − ∫
× 298.15 K ⎟ ∫
=
dV = −⎜1 mol × 8.3144
mol × K
V
⎝
⎠ 1L V
V1
V1
10 L
J
⎛
⎞ 10 L
− ⎜1 mol × 8.3144
× 298.15 K ⎟ ln
= −5707.97 J
mol × K
⎝
⎠ 1L
Negative sign is significant because the gas does the work.
Problem 2 (2 points). In class we showed that for a reversible process dU = cV dT − PdV (where heat is
described as cV dT and work is calculated as − PdV ) then immediately recognized this as a universal
formula that does not depend on the reversibility of the process. Explain in 2-3 short sentences why we
could state this.
Internal energy is a state function, which means that it does not depend on whether we choose reversible or
irreversible pathway to get from one point to another. Thus, the formula that is correct for a reversible path
holds true for an irreversible process. However, in that case cV dT is not heat and − PdV is not work.
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz#3
Problem 1 (6 points). It is possible that hydrogenation of coal was one of the important chemical
processes that ultimately led to biochemical reactions on Earth. However, this is the type of reactions,
for which determination of enthalpy (or heat at constant pressure) is difficult. Consider the reaction of
ethane formation:
2C ( s ) + 3H 2 ( g ) → C 2 H 6 ( g ) (1)
It is impossible to use simple calorimetry to do the measurement but the strategy similar to that
discussed during the lecture can be employed. Use Hess’s law and the following set of reactions with
Θ
experimentally determined enthalpies to calculate ΔH rxn
for reaction (1):
7
C 2 H 6 ( g ) + O2 ( g ) → 2CO2 ( g ) + 3H 2 O(l ) + ΔH I (2)
2
C ( s ) + O2 ( g ) → CO2 ( g ) + ΔH II (3)
1
H 2 ( g ) + O2 ( g ) → H 2 O(l ) + ΔH III (4), where
2
I
ΔH = −1560.4 kJ ; ΔH II = −393.5 kJ ; ΔH III = −285.8 kJ ;
Reaction (3) times 2 minus reaction (2) plus reaction (4) times 3 yields reaction (1) or:
ΔH rxn (1) = 2ΔH II − ΔH I + 3ΔH III = −2 × (393.5 kJ ) + 1560.4 kJ − 3 × (285.8 kJ ) = −84.0 kJ , which, of
course, also happens to be the standard enthalpy of ethane formation so you can easily check your answer
based on Table 5.8.
Problem 2 (4 points). Starting with the slope formula for dU and using thermodynamic derivatives of the
energy on page 5-2 of the Blue Book, prove that for any expansion/compressions process involving ideal
gas dU = CvdT. Show all your work clearly.
⎞
⎛ ⎛ ∂P ⎞
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜
⎟ dT + ⎜
⎟ dV = C v dT + ⎜⎜ T ⎜
⎟ − P ⎟⎟dT = C v dT = dq
⎝ ∂T ⎠V
⎝ ∂V ⎠ T
⎠
⎝ ⎝ ∂T ⎠V
Because for ideal gas:
⎛ ⎛ ⎛ nRT ⎞ ⎞
⎞
⎜ ⎜ ∂⎜
⎟
⎟⎟
⎛ ⎛ ∂P ⎞
⎞
⎛ ⎛ nR ⎞ nRT ⎞
nRT ⎟
⎜ ⎜ ⎝ V ⎠⎟
⎜⎜ T ⎜
dT = ⎜⎜ T ⎜
−
⎟dT = 0
⎟ − P ⎟⎟dT = ⎜ T ⎜
⎟−
⎟
∂T ⎟
V
V ⎠
V ⎟⎠
⎝
⎝
⎝ ⎝ ∂T ⎠V
⎠
⎜ ⎜
⎟
⎟
⎜
⎟
⎝
⎠
V
⎝
⎠
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz #4
Problem 1 (5 points). The standard enthalpy of formation of gaseous HCl
1
⎛1
⎞
⎜ H 2 ( g ) + Cl 2 ( g ) → HCl ( g ) ⎟ at 25ºC is -92.3 kJ/mol. Estimate its value at 100ºC given the
2
⎝2
⎠
following values of the molar heat capacities at constant pressure: Cp(HCl, gas) = 29.1 J/mol/K, Cp(H2,
gas) = 28.8 J/mol/K, Cp(Cl2, gas) = 33.9 J/mol/K. Assume that these values do not vary substantially
with temperature in the temperature interval studied.
T2
ΔH (T2 ) = ΔH (T1 ) + ∫ ΔC p ,reaction dT , where ΔC p , reaction = C p ( HCl , g ) − 0.5C p (Cl 2 , g ) − 0.5C p ( H 2 , g )
T1
Since we assume that heat capacities are temperature-independent, it is simplified to:
ΔH (T2 ) = ΔH (T1 ) + ΔC p ,reaction × ΔT = −92.3 kJ / mol + (29.1 − 0.5 × 28.8 − 0.5 × 33.9 )J / mol / K × 75 K =
= −92.3 kJ / mol − 2.25 J / mol / K × 75 K = −92.3 kJ / mol − 168.75 J / mol = −92.47 kJ / mol
Although ~0.2 kJ/mol difference does not seem substantial, it may translate into a real adjustment of the
reaction temperature to save energy for a given reaction. Also, this example indicates that for gases the
more pronounced change would occur is a total number of moles of the compounds involved would
change (which is not the case in this specific reaction).
Problem 2 (5 points). Calculate absolute molar entropy of NaCl (s) at 300°C and 1 bar. Use any
information available in the Blue Book and assume that the heat capacity of NaCl (s) can be
approximated as a constant within the temperature interval studied.
At constant pressure:
S mΘ,573.15 K = S mΘ, 298.15 K +
573.15 K
∫
298.15 K
C p ,m ( NaCl ( s ))
T
dT = 72.1
573.15 K
J
J
J
+ 50.5
= 105.1
ln
mol × K
mol × K 298.15 K
mol × K
This is a substantial change. Importantly, entropy increases as the temperature increases, as expected.
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz #5
Problem 1 (8 points). NO2 is a toxic brownish-yellowish gas but at room temperature it is also in
equilibrium with its colorless dimer, N2O4. The equilibrium constant of the dimerization reaction
2 NO2 ↔ N 2 O4
is Ka=57.6
If a sample of this equilibrium mixture placed in a cylinder exerted the pressure of 0.5 bar at room
temperature, what would be the partial pressure of the N2O4 in this mixture?
Answer:
If we have a total of 0.5 bar of gases, let’s set the partial pressure of N2O4 to be X bar. Then the pressure
of NO2 is (0.5-X) bar.
⎛ PN 2O4
⎞
⎜
Θ⎟
P ⎠
(a( N 2 O4 ) ⎝
X
K a = 57.6 =
=
=
2
2
(a( NO2 ) ⎛ PNO24 ⎞ (0.5 − X )2
⎜
⎟
PΘ ⎠
⎝
X
57.6 × 0.25 − X + X 2 = X or X 2 − X −
+ .25 = 0 This is a quadratic equation, so:
57.6
1
(
)
2
1 ⎞
1 ⎞
⎛
⎛
⎜1 +
⎟ ± ⎜1 +
⎟ − 4 × 0.25
⎝ 57.6 ⎠
⎝ 57.6 ⎠
It has two solutions: X1=0.6 and X2=0.4151. Since the total
X =
2
pressure is 0.5 bar, the first solution does not fit the physical description of the problem and X=0.4151. In
other words, the ratio of N2O4 to NO2 in the mixture is 0.4151/(0.5-0.4151) about 5 to 1.
Problem 2 (2 points). State and prove the condition of equilibrium for two phases of the same pure
compound.
dG = − SdT − VdP − μ α dn + μ β dn , where α and β denote two different phases and μ is chemical potential.
dG = 0 at equilibrium
However, T is a constant, otherwise we would have spontaneous heat transfer; P is also a constant,
otherwise we would have spontaneous movement possible. Thus, the condition of equilibrium is:
0 = − μ α dn + μ β dn or μ α dn = μ β dn or μ α = μ β
One can also use the expression for dA ( dA = − SdT − PdV − μ α dn + μ β dn ) to arrive at the same
conclusion.
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz #6
Problem 1 (8 points). Lead dioxide (PbO2) is an odorless dark-brown crystalline solid and it is a strong
oxidizing agent due to the reaction that can be expressed as:
2 PbO2 ( s ) → 2 PbO( s, red ) + O2 ( gas )
If the reaction is performed by placing 0.1 mole of PbO2 into a 100 L evacuated cylinder, calculate the
mass of gaseous O2 present at equilibrium at room temperature.
Θ
= − RT ln K a = ∆G Θf (O2 , gas ) + 2∆G Θf ( PbO, s, red ) − 2∆G Θf ( PbO2 , s )
1) ∆Grxn
Θ
∆Grxn
=0
kJ
kJ 
kJ 


+ 2 mol ×  − 188.9
 = 56.8 kJ per mole of O2
 − 2 mol ×  − 217.3
mol
mol 
mol 


Θ
 ∆Grxn

K a = exp −
 RT





56800 J / mol
 = exp(−22.91) = 1.119 × 10 −10
 = exp −
J



298.15 K 
 8.3144
mol × K


2)
2 PbO2 ( s ) → 2 PbO( s, red ) + O2 ( gas )
Initial moles 0.1
0
0
Final moles 0.1-2x
2x
x
Final Pressure 0
RT
x
V
0
3) Plug everything in
J
× 298.15 K
mol × K
x mol
3
−3 m
100 L × 10
L
= 1.119 × 10 −10
100000 Pa
8.3144
RT
x
V
×1
Θ
 a(O2 )a(PbO )2 
P


=
Ka = 
2
=
1

 a(PbO2 )
x = 4.516x10-10 mol
Molar mass of O2 is 32 g/mol so the total mass of O2 present in the cylinder at equilibrium is 1.445x10-8
g.
Problem 2 (2 points). Fill in the blanks. In class we discussed the concepts of fugacity and fugacity
coefficient for real gases. If a pressure of a real gas is approaching 0,
fugacity is approaching ___pressure____ and fugacity coefficient is approaching ____one_______
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz #7
Problem 1 (5 points). Calculate the height of a column of water in a capillary of 1 mm in diameter at
room temperature and atmospheric pressure
γ =
ρghr
2γ
or h =
=
ρgr
2 × 71.99 × 10 −3
N
m
= 0.029 m = 2.9 cm
kg
0.001
m2
g
g
×
× 0.5 × 10 −3 m
1 3×
9
.
8
3
s
m
cm
10 −6
3
cm
If you had a capillary of 1 µm in diameter, how long would it have to be for you to experimentally
measure the height of a column of water in a set up similar to that described above for 1 mm capillary?
2
The height of the column is inversely proportional to the radius of the capillary so the expected height of
a column in a 1 mm capillary would be about 29 meters, not a very realistic experimental measurement.
Problem 2 (5 points). Back to the example that we had in class about boiling liquids at high altitudes. So
what is the boiling temperature of water at 3500 meters above sea level where the barometric pressure
we calculated to be 68780 Pa?
P 
∆H v
∆H v
d ln P
or ln 2  = −
=−
R
d (1 / T )
R
 P1 
1 1
 − 
 T2 T1 
For water at 1 atm (101325 Pa) T1 = 100°C=373.15 K and ∆H v = 40.657
kJ
mol
Thus:
J
8.3144
 P2 
1
1
1
1
7.923 × 10 −5 2.759 × 10 −3
R
mol × K × ln 68780 Pa =
−
+
=
ln  =
= −
J
101325 Pa 373.15 K
T2 T1 ∆H v  P1  373.15 K
K
K
40657
mol
or
T2 = 362.43 K, more than 10 degrees lower than at the sea level
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz #8
Problem 1 (5 points). Use the diagram below for chloroform/acetone solution and Raoult’s law to
determine the activity coefficient for acetone at 0.4 acetone mole fraction in the solution.
Answer: Using the diagram, 0.4 mole fraction of
acetone corresponds to 0.6 mole fraction of
chloroform, thus the vertical line can be drawn
through x(CHCl3) = 0.6 to evaluate the activity
coefficient. At this molar fraction, the observed
pressure of acetone is approximately 95 Torr. The
pressure predicted by Raoult’s law is
approximately 135 Torr. That last point can be
determined even better by multiplying the
pressure of pure acetone (331 Torr) by 0.4, which
makes 132 Torr. The activity coefficient:
γ =
P(acetone, x a = 0,4) 95
=
= 0.72
132
Pid , RL
www.nuigalway.ie/chem/Donal/Chapter13.pp
Problem 2 (5 points). Determine the molality and mole fraction of solvent and solute for a solution
prepared by dissolving 16.6 g of potassium iodide (KI) in a liter of pure water.
M ( KBr ) = M ( K ) + M ( Br ) = (39.0983 + 126.90447) g / mol ≈ 166 g / mol
1000 g
Thus we dissolve 0.1 mole of KI in a
= 55.56 mol of water
18 g / mol
a)
Molality is defined as a number of moles of a solute per 1 kg of solvent. In this case we
dissolve 0.1 mole of KI in 1 kg of water and the molality is 0.1 mol/kg.
b)
To determine the molar fraction we need to determine how many moles of either
substance is present in the solution. The total is 55.66 mol. To find the molar fraction, we
simply need to divide the number of moles of each component by the total number of
moles in a solution:
X water
55.56 mol
=
= 0.9982;
X water + X KI 55.66 mol
X KI
1 mol
=
=
= 1 − X water = 0.0018
X water + X KI 55.66 mol
X water =
X KI
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz #9
Problem 1 (7 points). Solid magnesium hydroxide (Mg(OH)2) forms a suspension in water that is called
milk of magnesia because of its milk-like appearance. Suppose that an experimental measurement
produced a value of +64.5 kJ/mol for the standard Gibbs free energy change upon dissolution of
Mg(OH)2 in water according to the following reaction:
Mg (OH ) 2 ( solid ) → Mg 2+ (aq ) + 2OH − (aq )
The positive value for this process indicates that Mg(OH)2 has very low solubility, which is indeed the
case. Use your knowledge of thermodynamics of aqueous solutions and experimentally measured Gibbs
free energy change to find ∆G Θf ( Mg (OH ) 2 , solid ) .
Answer:
Θ
∆Grxn
= ∆G Θf , prod − ∆G Θf ,react = ∆G Θf ( Mg 2+ aq ) + 2∆G Θf (OH − aq ) − ∆G Θf ( Mg (OH ) 2 , solid ) or
Θ
∆G Θf ( Mg (OH ) 2 , solid ) = −∆Grxn
+ ∆G Θf ( Mg 2+ aq ) + 2∆G Θf (OH − aq )
∆G Θf ( Mg (OH ) 2 , solid ) = −64.5kJ / mol + (−454.8kJ / mol ) + 2(−157.2 Kj / mol ) = −833.7 kJ / mol
Problem 2 (3 points). From the possible statements in column B, select the best match for each phrase in
column A and put its letter in the adjacent blank. There is only one best match for each phrase.
Column A
1. Substances that dissociate into positively and
negatively charged mobile solvated ions when
dissolved in an appropriate solvent are called ___g__
2. By convention, the formation Gibbs free energy for
a H+ (aq) at unit activity is set equal to ___i__
3. The expression
(γ
ν+
+
γ ν−
−
)
1 /ν
gives ___f__
Column B
a)
b)
c)
d)
e)
f)
g)
h)
i)
electrophiles
mean ionic activity
average molal ionic coefficient
one kJ/mol
nucleophiles
mean ionic activity coefficient
electrolytes
one eV
zero
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2011
TuTh 5:00-6:15 pm, 206 BrL
Quiz #10
Problem 1 (6 points). Determine the half-cell reactions, the overall reaction, and the standard emf for the
following electrochemical cell:
Ag ( s ) AgCl ( s ) Cl − (aq, aCl − = 0.005) Cd 2+ (aq, aCd 2 + = 0.100) Cd ( s )
Answer:
Right: Cd 2+ + 2e → Cd ( s ) ;
Left: Cl − + Ag ( s ) → AgCl ( s ) + e
Overall Reaction: Cd 2+ + 2Cl − + 2 Ag ( s ) → Cd ( s ) + 2 AgCl ( s )
Θ
Θ
− ε Left
= X − 0.4030V − (−0.22233V ) = −0.18067V
Standard emf: ε Θ = ε Right
Problem 2 (4 points). Calculate the diffusion coefficient for Xe at room temperature and 1 atm.
D=
RT
1
1 8 RT
vλ =
×
=
2
2 πM
2 PN Aπd 2
J
J
298.15K
8.3144
298.15K
mol × K
mol × K
×
1
kg
2 × 101325Pa × 6.022 × 10 23
π 0.131293
π (0.4099 × 10 −9 m) 2
mol
mol
Here the d parameter for Xenon is taken from Table 8.2
1
2
8 × 8.3144
D = 5.967 × 10
−6
m2
s