Download Quiz Keys - Section 10

Physical Chemistry 418-010
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Name __________KEY_______________
Problem 1 (7 points).
a) Compare the pressure exerted by 30 mole of HCl enclosed in a 10 L cylinder at room temperature
predicted by the ideal gas law with that predicted by the van der Waals description.
Answer: Ideal gas law:
J
30 mol × 8.3144
× 298.15 K
nRT
mol
×
K
P(HCl )ideal =
=
= 7436815.1 Pa = 73.396 atm
V
1 m3
10 L ×
1000 L
P(HCl )vdW
Van der Waals law:
Pa × m 6
J
0.3700
× 298.15 K
8.3144
nRT
n2a
RT
a
mol × K
mol 2
=
− 2 =
− 2 =
−
=
3
3
3 2
V − nb V
Vm − b Vm
10 L
−3 m
−6 m
⎛
⎞
10
L
−3 m
× 10
− 40.6092 × 10
⎜
⎟
30 mol
L
mol ⎜ 30 mol × 10 L ⎟
⎝
⎠
= 8468513.8Pa − 3330000 Pa = 5138513.8 Pa = 50.713 atm
b) Based on the result of this comparison, what type of interaction dominates at these conditions? Be
concise in this answer; do not write more than a sentence or two.
Because the answer predicted by the vdW law is over 30% lower than that predicted by the ideal gas
law, attractive interactions between HCl molecules play the dominant role at these conditions reducing
the pressure predicted by the ideal gas description that does not take into account intermolecular
interactions.
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
Column B
1. zeroth law of thermodynamics states that ___e___
a) molecules are hard spheres
b) two systems in thermal equilibrium with one
another always represent an open system
c) gas can not be stable
d) liquid can not be produced at any pressure
e) two systems that are separately in thermal
equilibrium with a third system are also in
thermal equilibrium with one another
f) collisions with a solid wall are inelastic
g) molecules are point masses
h) U = q + w
i) is based on the activated complex theory
j) solid can not be in one piece at any pressure
2. One of the ideal gas law approximations
3. Above critical temperature ___d___
___g__
Physical Chemistry 418-010
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Name __________KEY_______________
Problem 1 (5 points). In class we calculated the heat required to raise the temperature of copper to 50°C
at atmospheric pressure. This is indeed a practical exercise since copper wiring is used in cables and it is
important to know the heat lost for cost estimates. The alternative material for cable manufacturing is
aluminum. Assuming that the heat capacity of aluminum does not depend too much on the temperature
within our range of interest, calculate the heat required to raise the temperature of a 53.96 g piece of
aluminum from room temperature to 50°C at atmospheric pressure. Heat capacity of aluminum at room
J
temperature is Cp = 24.55
mol × K
At constant pressure, the enthalpy change gives the heat required to change the state of the system:
⎛ ∂H ⎞
⎛ ∂H ⎞
⎛ ∂H ⎞
dH = ⎜
⎟ dT = C P dT = dq
⎟ dP = ⎜
⎟ dT + ⎜
⎝ ∂T ⎠ P
⎝ ∂P ⎠ T
⎝ ∂T ⎠ P
T2
q = ∫ C p dT = C P (T2 − T1 ) = 24.55
T1
J
J
× 25 K = 613.75
mol × K
mol
Since the molar mass of aluminum is 26.98 g/mol, the weight of the piece that we investigate corresponds
to exactly 2 moles and the heat required to raise its temperature to 50°C at atmospheric pressure is twice
what we have calculated for 1 mol. Thus q = 2 mol x 613.75 J/mol = 1227.5 J or 1.2275 kJ.
Problem 2 (5 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 process involving ideal gas dU = CvdT. Show
all your work clearly.
⎞
⎛ ⎛ ∂P ⎞
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜
⎟ dT + ⎜
⎟ dV = C v dT + ⎜⎜ T ⎜
⎟ − P ⎟⎟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 2010
TuTh 5:00-6:15 pm, 206 BrL
Quiz#3
Problem 1 (5 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 (5 points). Joule-Thompson experiment can be described as a study of an isenthalpic process.
Based on the expression for Joule-Thompson process given on page 5-6 of the Blue Book, find the exact
Joule-Thompson coefficient (μJT) for Argon at room temperature, assuming that it is an ideal gas.
μ JT
⎛ ∂V ⎞
T⎜
⎟ − V T nR − nRT
∂
T
∂
T
⎛
⎞
⎝
⎠P
P = 0 , which is correct for any ideal gas (argon or not…)
=⎜
= P
⎟ =
Cp
Cp
⎝ ∂P ⎠ H
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Quiz#4
Problem 1 (5 points). The standard enthalpy of formation of gaseous H2O
1
⎛
⎞
⎜ H 2 ( g ) + O2 ( g ) → H 2 O( g ) ⎟ at 25ºC is -241.82 kJ/mol. Estimate its value at 100ºC given the
2
⎝
⎠
following values of the molar heat capacities at constant pressure: Cp(H2O, gas) = 33.58 J/mol/K, Cp(H2,
gas) = 28.84 J/mol/K, Cp(O2, gas) = 29.37 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 ( H 2 O, g ) − 0.5C p (O2 , g ) − C 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 = −241.82 kJ / mol + (33.58 − 0.5 × 29.37 − 28.84 )J / mol / K × 75 K =
= −241.82 kJ / mol − 9.945 J / mol / K × 75 K = −241.82 kJ / mol − 745.845 J / mol = −242.57 kJ / mol
Although ~1 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.
Problem 2 (5 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. In a spontaneous process, entropy ___c___
2. Perpetual motion machine of the second kind
violates ___d__
3. Carnot cycle describes a(n) ___g___ process
4. When considering an irreversible process __m__
5. The expression of the second law stated as
dS ≥
dq
is referred to as ___j___
T
Column B
a) first law of thermodynamics
b) entropy must be calculated along an irreversible
path
c) increases
d) second law of thermodynamics
e) Joule-Thompson coefficient
f) entropic loss
g) reversible
h) decreases
i) third law of thermodynamics
j) Clausius inequality
k) entropy can not be calculated
l) does not change
m) entropy must be calculated along a reversible path
n) irreversible
o) perpetual
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Quiz#5
Problem 1 (6 points). Use the definition of Helmholtz free energy, first law of thermodynamics, and
your knowledge of the Clausius inequality to prove that dA≤0 at constant V and T, and that the equality
dA=0 is a characteristic of equilibrium at these conditions. Show all your work clearly.
dA = dU − d (TS ) = dU − TdS − SdT = dq − Pext dV − TdS − SdT
dA V ,T =Const = dq − TdS ≤ 0, since TdS ≥ dq, according to the Clausius inequality
Thus dA V ,T =Const = dq − TdS < 0, for any irreversible (spontaneous) process and
dA V ,T =Const = dq − TdS = 0, for a reversible process or for a condition of equilibrium
Problem 2 (4 points total). Briefly answer the following questions:
a) (2 points) State in a couple of sentences why Maxwell relations are useful
The four Maxwell relationships described during the lecture are very useful in transforming
seemingly obscure partial derivatives into the partial derivatives that can be directly measured.
b) (2 points) Describe in a couple of sentences why dS≥0 may not be a sufficient criterion for the
spontaneity of a process.
The special case considered previously for understanding the change in entropy for an
irreversible (spontaneous) transformation involved an isolated system, for which this is exactly
the condition of spontaneity. However, most of the time we are interested in systems that
actually interact with surroundings rather than in isolated systems, so a separate set of criteria
had to be developed.
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Quiz#6
Problem 1 (8 points). a) Calculate the Kp (as it is determined in the text) for the following reaction at
room temperature: CH 4 ( g ) + Cl 2 ( g ) → CH 3 Cl ( g ) + HCl ( g )
0
Answer: First we should calculate ΔGrxn
:
0
ΔG rxn
= 1 mol × ΔG 0f ,m (CH 3 Cl ( g )) + 1 mol × ΔG 0f , m ( HCl ( g )) − 1 mol × ΔG 0f ,m (CH 4 ( g )) − 1 mol × ΔG 0f ,m (Cl 2 ( g )) =
= 1 mol × ( −57.4kJ / mol ) + 1 mol × (−95.3) − 1 mol × (−50.5) − 1 mol × 0kJ / mol = −102.2 kJ
0
0
⎞
⎛ ΔG rxn
ΔGrxn
102200 J / mol
⎞
17
⎟ = exp⎛⎜
, K p = exp⎜⎜ −
⎟ = 8 × 10
⎟
RT
⎝ 8.3144 J / mol / K × 298.15K ⎠
⎝ RT ⎠
Thus, the reaction equilibrium is shifted towards the products.
0
0
b) For the reaction in part (a), calculate the ΔGrxn
at 400 K assuming that ΔH rxn
is independent of
temperature. How does this temperature change affect Kp?
Since ln K p = −
0
Answer: Since ΔH rxn
is independent of temperature, we can use the Gibbs-Helmholtz equation but first
0
we need to calculate ΔH rxn
:
0
ΔH rxn
= 1 mol × ΔH 0f ,m (CH 3 Cl ( g )) + 1 mol × ΔH 0f ,m ( HCl ( g )) − 1 mol × ΔH 0f ,m (CH 4 ( g )) − 1 mol × ΔH 0f ,m (Cl 2 ( g )) =
= 1 mol × (−81.9kJ / mol ) + 1 mol × (−92.3) − 1 mol × (−74.6) − 1 mol × 0kJ / mol = −99.6 kJ
Now, using Gibbs-Helmholtz equation:
0
0
(at 400 K ) ΔG rxn
(at 298.15 K )
ΔGrxn
1
⎛ 1
⎞
0
(at 298.15 K )⎜
=
+ ΔH rxn
−
⎟
400 K
298.15 K
⎝ 400 K 298.15 K ⎠
(−102200 J / mol )
1
⎛ 1
⎞
+ 400 K × (−99600 J / mol )⎜
−
⎟=
298.15K
⎝ 400 K 298.15 K ⎠
= -137112.2 J/mol + 34024 J/mol = -103088.2 J/mol
0
ΔGrxn
(at 400 K ) = 400 K
0
⎞
⎛ ΔG rxn
103088.2 J / mol
⎞
13
⎟ = exp⎛⎜
Then at 400 K: K p = exp⎜⎜ −
⎟ = 2.9 × 10
⎟
⎝ 8.3144 J / mol / K × 400 K ⎠
⎝ RT ⎠
The equilibrium is still shifted towards the products formation but not as much as at room temperature.
⎛ ∂G ⎞
Problem 2 (2 points). If x is the extent of reaction parameter and ⎜
⎟ > 0 , then (circle one):
⎝ dx ⎠ T , P
a) the reaction proceeds spontaneously as written;
b) the reaction proceeds spontaneously in the opposite direction;
c) the reaction has reached equilibrium.
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Quiz#7
Problem 1 (7 points). In class we discussed the problems associated with high altitude caused by lower
boiling points of liquids at lower pressures. A different kind of problem may be caused by changing a
boiling point of a liquid at altitudes below sea level. Northern Europe (especially Sweden and Norway)
are known for their fjords (or lochs as in Loch Ness), valleys that are located below the sea level and
may or may not be filled with water. Although the altitude change in this case is not really comparable
with that for the height of the highest mountain ranges, the barometric pressure at the bottom a 300
meters deep fjord is about 4% higher than at the sea level. Estimate boiling point of water at this altitude
assuming that the heat of vaporization for water is not affected substantially by this pressure change.
Answer: we can use Clausius-Clapeyron equation
Δ H
P
d (ln P) Δ v H
=
or ln 2 = − v
2
P1
R
dT
RT
⎛1 1⎞
⎜⎜ − ⎟⎟
⎝ T2 T1 ⎠
At 1 atm water boils at 100ºC or 373.15 K and ΔH vΘ = 40.657 kJ / mol (Table 6.1)
P
R
1
1
ln 2 + = ; −
Thus: −
Δ v H P1 T1 T2
J
1
1
mol × K ln 1.04 +
=
J
1
373.15 K T2
40657
mol
8.3144
T2 = 374.27 K. It only boils at a temperature that is about one degree higher than at a sea level. Probably
not much of an effect on your cooking habits.
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. The point on a P-T phase diagram, where the liquidgas line terminates because both phases have the same
density is called ___e___
2. For a first order phase transition, the Cp at the phase
transition temperature on a Cp vs. T diagram is ___i__
3. ___c___ can be used as appropriate units for surface
tension
Column B
a)
b)
c)
d)
e)
f)
g)
h)
i)
zero
moot point
J/m2
undefined
critical point
N/m2
standard boiling point
N2/m
infinity by definition of Cp
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Quiz#8
Problem 1 (7 points). According to Wikipedia (http://en.wikipedia.org/wiki/Mercury_(element)),
pressure of mercury gas above liquid mercury is 1 Pa around room temperature and its density is 13.534
g·cm−3. What would be the pressure of mercury dispersed into uniform droplets with diameter of 10 nm?
⎛ P ⎞ 2γM
and
ln⎜ • ⎟ =
⎝ P ⎠ ρrRT
⎛
⎞
⎜
⎟
⎜
⎟
⎜
⎟
N
kg
2 × 485.48 × 10 −3 × 200.59 × 10 −3
⎜
⎟
⎛ 2γM ⎞
m
mol
⎟
⎟⎟ = 1 Pa × exp⎜
P = P • × exp⎜⎜
kg
⎜
⎟
⎝ ρrRT ⎠
0.001
⎜
⎟
g
J
g
−9
×
×
×
5
10
m
8
.
3144
298
.
15
K
⎜ 13.534 3 ×
⎟
3
mol × K
cm
−6 m
⎜
⎟
10
⎜
⎟
3
cm
⎝
⎠
P = 1 Pa × exp(1.161) = 3.2 Pa , more than three times higher than over a puddle
Problem 2 (3 points). What would be the pressure inside the droplet of mercury described above in
Problem 1?
Pinside = Poutside
2γ
+
= 101325 Pa +
r
2 × 485.48 × 10 −3
5 × 10 −9 m
N
m = 101325 Pa + 194192000 Pa = 19429333000 Pa = ~ 1917atm
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Quiz#9
Problem 1 (5 points). Determine the molality and mole fraction of solvent and solute for a solution
prepared by dissolving 119 g of potassium bromide (KBr) in a liter of pure water.
M ( KBr ) = M ( K ) + M ( Br ) = (39.983 + 79.904) g / mol ≈ 119 g / mol
1000 g
= 55.56 mol of water
Thus we dissolve one mole of KBr in a
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 1 mole of KBr in 1 kg of water and the molality is 1 mol/kg.
b)
To determine the molar fraction we need to determine how many moles of either
substantce is present in the solution. The total is 56.56 mol. To fine 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
=
= 0.982;
X water + X KBr 56.56 mol
X KBr
1 mol
=
=
= 1 − X water = 0.018
X water + X KBr 56.56 mol
X water =
X KBr
Problem 2 (5 points). Apply your knowledge of Raoult’s law to determine the total pressure above a
mixture of benzene and toluene at room temperature with the molar fraction of 0.3 for benzene in a
liquid phase. Assume that the behavior of this mixture can be precisely described by Raoult’s law and
that the pressures of toluene above a pure toluene is 28.9 torr and the pressure of benzene above pure
benzene is 96.4 Torr.
overs olution
over solution
vapor
pure
vapor
pure
liquid
pure
liquid
pure
Ptotal = Pbenzene
+ Ptoluene
= y benzene
Pbenzene
+ y toluene
Ptoluene
= xbenzene
Pbenzene
+ xtoluene
Ptoluene
Ptotal = 0.3 × 96.4Torr + 0.7 × 28.9Torr = 49.15Torr = 6552.8 Pa
Physical Chemistry 418-010
Name __________KEY_______________
Fall 2010
TuTh 5:00-6:15 pm, 206 BrL
Quiz#10
Problem 1 (3 points). Estimate the molecular weight of a biopolymer if a solution of 20 mg of this
compound in 10 ml of pure deionized water gives 0.1 torr of osmotic pressure at room temperature.
40 mg
kg
J
× 10 −6
× 8.3144
× 298.15 K
Pa
nRT
MW
mg
mol × K
Π = CRT =
=
= 0.1 torr × 133.322
3
torr
V
m
10 ml × 10 −6
ml
MW = 743.7
kg
g
= 743700
mol
mol
Problem 2 (7 points). Using the data on solubility of N2 (g) in water at 25°C from Table 7.2, find the
ΔGfΘ(N2 (ao, m)) at 1 atm of N2 pressure and 25°C
N 2 ( g , P) → N 2 (ao, m)
mN2
ΔG
Θ
rxn
Θ
ΔGrzn
γN
1mol / kg
, but γ 2 m ≈ 1 and Φ ≈ 1
PN 2
Φ
PΘ
1 mol
g
g
⎛
⎞
⎛
⎞
×
⎜ 0.001751
⎟
⎜ 0.001751
⎟
100 g H 2 O ⎟
0.1 kg H 2 O 2 × 14.0067 g ⎟
⎜
⎜
⎜
⎟
⎜
⎟
1mol / kg
1mol / kg
= − RT ln⎜
⎟
⎟ = − RT ln⎜
PN 2
101325 Pa
⎜
⎟
⎜
⎟
Θ
⎜
⎟
⎜
⎟
100000
Pa
P
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
J
J
Θ
ΔGrzn
= −8.3144
× 298.15 K × (− 7.39 ) = 18319
mol × K
mol
J
J
J
Θ
ΔG Θf ( N 2 , ao) = ΔGrzn
+ ΔG Θf ( N 2 , g ) = 18319
−0
= 18319
mol
mol
mol
Θ
Solubility is taken from Table 7.2 and the ΔG f ( N 2 , g ) = 0
Θ
f
Θ
f
= − RT ln K a = ΔG ( N 2 , ao) − ΔG ( N 2 , g ) = − RT ln
2