Download CHEM-UA 652: Thermodynamics and Kinetics

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CHEM-UA 652: Thermodynamics and Kinetics
Problem set #8: due 4/17/2014
Practice problems: 11.13, 11.14, 11.15, 11.22, 12.18, 12.16
Additional Practice Problem:
i. Derive the expression given in class for the boiling point elevation starting from equalization of
chemical potentials between the solution and vapor phases, and show that the elevation of the
boiling point is predicted as part of your derivation.
Graded problems
1. A solution is prepared by mixing liquid benzene and liquid carbon tetrachloride in equal amounts
at 25◦ C. When equilibrium is established, the vapor above the solution is collected and analyzed
for its content. Determine to at least three significant figures the mole fraction of each gas in
the vapor using only the following data: The standard enthalpies of formation for benzene in
the gas and liquid phases are 82.93 kJ/mol, 49.03 kJ/mol, respectively. The standard entropies
for benzene in the gas and liquid phases are 269.2 J/mol·K, 172.8 J/mol·K, respectively. The
standard enthalpies of formation for carbon tetrachloride in the gas and liquid phases are -102.9
kJ/mol and -135.44 kJ/mol, respectively. The standard entropies for carbon tetrachloride in
the gas and liquid phases are 309.74 J/mol·K and 216.4 J/mol·K, respectively.
2. When BaCl2 and K2 SO4 solutions are mixed together, a solid precipitate, BaSO4 , is formed
according to the equation:
BaCl2 (aq) + K2 SO4 (aq) −→ BaSO4 (s) + 2KCl(aq)
In a certain experiment, 0.2000 L of a 1.6000 M BaCl2 solution are combined with 0.3000 L
of a K2 SO4 solution of unknown molarity. The final solution is analyzed and found to contain
unreacted Barium ions. In addition, the boiling point of the solution is measured and found to
be 101.4245 o C, and the mass of the final solution is determined to be 599.2300 g. What is the
molarity of the K2 SO4 solution?
3. The activity coefficient γ of a nonideal gas is defined to be a correction factor, which, when
multiplied by the observed pressure of the gas gives the pressure it would have if it were an ideal
gas.
Consider the gas phase reaction:
H2 (g) + F2 (g) ⇀
↽ 2HF(g)
whose equilibrium constant at a high temperature is K = 115.00. Suppose that H2 , F2 and
HF gases are to be treated as nonideal gases with activity coefficients 1.111, 1.180 and 1.250,
respectively. A 50 L vessel contains HF gas at an observed partial pressure of 2.4 atm and H2
gas at an observed partial pressure of 4.5 atm.
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a. It is desired to add enough fluorine gas to the vessel to obtain a total ideal gas pressure of
10 atm. To what observed partial pressure must fluorine gas be added?
b. What will be the observed partial pressures of each gas in the vessel after equilibrium is
established?
c. What will be the total ideal gas pressure in the container at equilibrium?
d. Suppose the volume of the vessel is doubled. What effect does this have on the equilibrium?
(Think carefully about this one!) Determine the sign of the Gibbs free energy change for
this volume doubling process.
4. In 1918, the German chemist Fritz Haber won the Nobel prize for the nitrogen fixation process, through which ammonia for nitrogen-based fertilizers is synthesized from its elemental
components, N2 and H2 gases, via the equilibrium
N2 (g) + 3H2 (g) ⇀
↽ 2NH3 (g)
In this problem, assume all gases can be treated as ideal gases.
a. Suppose a reaction vessel is filled with N2 and H2 gases to partial pressures of 3.0 atm
and 2.0 atm, respectively, but suppose the the temperature at which the reaction occurs
is unknown. When equilibrium is established in the vessel, it is observed that the total
pressure in the vessel is 4.50 atm. What is the temperature in the vessel? The following
data might be helpful:
∆Hf◦ (NH3 (g)) = −46.11 kJ/mol
∆S ◦ (N2 (g)) = 191.50 J/mol
∆S ◦ (NH3 (g)) = 192.34 J/mol
∆S ◦ (H2 (g)) = 130.57 J/mol
b. What would the total pressure in the vessel be if the same reaction were carried out at
25◦ C? You may assume that the standard enthalpy and entropy are roughly temperature
independent.
c. Despite the high yield at 25◦ C, the reaction proceeds very slowly at this temperature.
For this reason, high temperatures are used as a compromise between yield and speed.
However, in order that yields not be unreasonably low, very high pressures must also be
used. Suppose the reaction is carried out at 800 K, and we wish to use the same ratio, 3/2,
of initial partial pressures of N2 and H2 gases. What would the initial pressures of these
gases need to be in order that the yield of NH3 at equilibrium be 15% of what it would
be if the reaction went to completion? You may assume that the standard enthalpy and
entropy are roughly temperature independent.
5. Recall that the inhibition constant Ki for an enzyme E complexed with an inhibitor I is the
equilibrium constant for the dissociation of the enzyme-inhibitor complex EI into free enzyme
and free inhibitor in solution at body temperature:
EI(aq) ⇀
↽ E(aq) + I(aq)
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a. Suppose a medical patient is diagnosed with a viral infection, the treatment for which is
the administration of a drug I that inhibits and enzyme E produced by the virus that is
essential in the viral replication cycle. The viral load is such that the concentration of the
enzyme is 1.0×10−5 M in the patient’s body. It is known that the drug has an inhibition
constant of 2.0×10−8 . You need to calculate a drug dose such that the at least 99.9% of
the enzyme is inhibited at equilibrium for the dose to be considered effective. If the drug
has a molar mass of 698.14 g/mol, how many milligrams of the drug should be given to
the patient (in pure crystalline form) such that the dose is effective? Assume the patient’s
blood volume is 3.79 L. Warning: There is no credit for killing your patient with an
overdose of the drug.
b. Over a period of time, the drug is metabolized, and the concentration of the enzyme returns
to the original 1.0×10−5 M, so that a new dose of the drug is needed. However, it is found
that the virus has created a mutant enzyme E′ that is more resistant to the drug, such
that 60% of the total enzyme concentration (1.0×10−5M) is the mutant, while just 40%
is the original enzyme E. The inhibition constant of the drug I for the mutant enzyme
is 50 times larger than that for the origianl enzyme E. Remembering that 99.9% of the
total enzyme content must be in the form of enzyme-inhibitor complexes for the dose to be
effective, how many milligrams of the drug must be administered to the patient? Assume
the patient’s blood volume is 3.79 L. Warning: There is no credit for killing your patient
with an overdose of the drug.