NAME:
Section Number:
KEY
CHEMISTRY 418, Fall, 2011(11F)
Midterm Examination #2, November 10, 2011
10
Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY
understand your work; only work on the exam sheet will be considered. Write answers, where appropriate, with reasonable numbers of
significant figures. You may use only the handbook of “Essential Data and Equations for a Course in Physical Chemistry”, a calculator,
and a straight edge.
1. (20 points) The Gibbs-Helmholtz equation is useful for calculating changes in the Gibbs free energy
DO
NOT
of a system as a function of temperature. Use this equation (If you do not remember it, this equation can WRITE IN THIS
be derived based on the equations presented in the sections on Thermodynamic Basis for Equilibrium
SPACE
Θ
and Temperature Dependence of Equilibrium Constant) to determine the ∆G f (ethanol, liq) at 340 K.
p. 1________/20
Θ
You can use any thermochemical information from the handbook and assume that ∆H f (ethanol, liq)
p. 2________/20
does not depend significantly on temperature within the temperature range studied.
p. 3________/20
Answer: First, let’s derive the Gibbs-Helmholtz equation.
p. 4________/20
Θ
Page 5-11 of the handbook has: ∆G rxn (T ) = − RT ln K a (T ) and
Θ

 ∆Grxn


d
Θ
RT
−
 = − ∆H rxn (T ) and

Thus,
d (1 / T )
R
Θ
rxn
∆H (T )
d ln K a (T )
=−
d (1 / T )
R
=============
Θ
 ∆G rxn
d 
 T

1
Θ
 = ∆H rxn
(T )d for the enthalpy of the process

T

independent of temperature.
Then:
p. 5________/20
Θ
Θ
1 1
(T2 ) ∆G rxn
(T1 )
∆G rxn
Θ
(T ) ×  −  , which is the Gibbs-Helmholtz equation.
=
+ ∆H rxn
T2
T1
 T2 T1 
Plugging in the actual numbers from Table 5.8:
kJ


 − 174.8

kJ
kJ
1
1


Θ
mol − 277.6
(T2 = 340 K ) = 340 K × 
∆Grxn
×
−
  = −160.4
mol  340 K 298.15 K  
mol
 298.15 K




This is noticeably different from the room temperature value.
p. 6_______/5
(Extra credit)
=============
TOTAL PTS
/100
NAME:
CHEM 418, Midterm Exam #2, Fall, 2011, page 2
2. (20 points) Fill in the blanks
a)
(4 points). The change in the Gibbs free energy per mole of substance i added at a constant concentration of all other
substances to the single phase reaction mixture at constant pressure and temperature is defined as__CHEMICAL
POTENTIAL______ of i.
b)
(4 points). The compression (or compressibility) factor is defined as____(PVm)/(RT)___________
c)
(4 points). On a P vs. T phase diagram of water, if one moves upwards along the line dividing solid and liquid, the
condition applied to the chemical potentials of these two phases is (circle one) ___µ(liq)
d)
=
<
>
µ(solid)___
(4 points). In living organisms molecules that are composed of a polar segment (hydrophilic part) and a non-polar
segment (lypophilic part) often play major roles in the formation of membranes and surfaces. This class of molecules
is called __amphiphiles___
e)
(4 points). The statement that the activity coefficient for one of two completely miscible solvents in a solution can be
several times higher than 1 is (circle one) __true
false___
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NAME:
CHEM 418, Midterm Exam #2, Fall, 2011, page 3
3. (20 points) Derive Clausius-Clapeyron equation starting with Clapeyron equation. State all the assumptions and show
all your work clearly.
Clapeyron equation is given on page 6-1 of the Handbook:
dP ∆ Φ H
=
dT T∆ ΦV
The assumptions for deriving Clausius-Clapeyron equation are:
1) One of the two phases at equilibrium is a gas phase (sublimation or evaporation can be considered)
2) The gas phase has low enough pressure so that the Ideal Gas Law can be used
3) ∆ vap H (or ∆ subl H ) does not depend on temperature changes within the temperature intervals of interest (this is
actually not a necessary condition for the derivation, just for practical use, as described below)
Thus:
∆ ΦV = ∆ VapV = V gas − Vsolid or liquid ≈ V gas =
nRT
RT
or ∆ Φ , mV ≈ V gas , m =
P
P
Substituting this into Clapeyron equation, one obtains:
∆ vap H m
P × ∆ vap H m
dP ∆ Φ H
=
=
=
dT T∆ ΦV T∆ gas ,mV
RT 2
and
dP ∆ vap H m
=
dT , which is the same as the Clausius-Clapeyron equation given in the Handbook:
P
RT 2
d ln P ∆ vap H m
=
dT
RT 2
The “molar” subscripts are often omitted in these formulas but it is assumed that you operate with molar heats of vaporization
or submimation. If one assumes that
∆ vap H is independent of temperature (which allows you to do a very simple
integration), this is converted to a practical equation:
ln
∆H m ,Φ
P2
=−
P1
R
1 1
 − 
 T2 T1 
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NAME:
CHEM 418, Midterm Exam #2, Fall, 2011, page 4
4. (20 points). Calculate the pressure drop across the border of the bubble of air with a diameter of 1 µm inside a beaker
of liquid mercury at room temperature.
Single surface : ∆P =
2γ
485.48 × 10 −3 N / m
N
= 2×
= 194192 2 = 194192 Pa = 1.917 atm
−7
r
5 × 10 m
m
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NAME:
CHEM 418, Midterm Exam #2, Fall, 2011, page 5
5. (20 points). Using the data on solubility of N2(g) in water 25°C from Table 7.2, find the ∆GfΘ(N2 (ao, m)) at 1 atm of N2
pressure and 25°C
N 2 ( g , P ) → N 2 (ao, m)
m o2
1mol / kg
, but γ 2 m ≈ 1 and Φ ≈ 1
PO2
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
PO2
101325 Pa




Θ




100000
Pa
P








J
J
Θ
× 298.15 K × (− 7.39 ) = 18319
∆Grzn
= −8.3144
mol × K
mol
J
J
J
Θ
∆G Θf ( N 2 , ao) = ∆Grzn
− ∆G Θf ( N 2 , g ) = 18319
−0
= 18319
mol
mol
mol
Θ
∆Grxn
= − RT ln K a = ∆G Θf ( N 2 , ao) − ∆G Θf ( N 2 , g ) = − RT ln
Θ
∆Grzn
γO
Solubility is taken from Table 7.2 and the
2
∆G Θf ( N 2 , g ) = 0
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NAME:
CHEM 418, Midterm Exam #2, Fall, 2011, page 6
6. (5 points, extra credit). Absolute zero is the point where no more heat can be removed from a system, according to the
absolute or thermodynamic temperature scale. This corresponds to 0 K or -273.15°C. Several approaches can be used to
calculate the work required to reach this temperature and this number is finite for every system in question. However,
actually reaching this temperature proved to be impossible despite much effort. The NIST achieved a record cold
temperature of 700 nK (billionths of a Kelvin) in 1994. MIT researchers set a new record of 0.45 nK in 2003. Use your
knowledge of the second law of thermodynamics to explain why it is so difficult despite the fact that we can easily get the
amount of work required to reach it from many different sources.
Even though the work required to reach absolute zero is finite, the time to do it is actually infinite. In any refrigeration
system, the heat will flow only if there exists some temperature difference between the cooling reservoir and the sample.
Thus, it becomes progressively more and more difficult to remove the heat from the sample as the temperature
approaches zero K. The second law does not forbid approaching the temperature of 0 K, it just forbids us actually reaching
it. This is similar to reaching the limit of a geometrical progression. We may know what it is but actually reaching it within
any finite number of operations is impossible.
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