CHEM 341
PHYSICAL CHEMISTRY
EXAM 2
Do not open this exam until
told to do so. The exam consists
of 6 pages, including this one. Count
them to insure that they are all there.
Constants and
conversions:
R = 8.31 J K-1
mole-1 NA
= 6.02 x 1023
1 atm = 101325 Pa =
101325 J/m3
Page |
Score |
|
2 |
/16 |
|
3 |
/16 |
|
4 |
/28 |
|
5 |
/28 |
|
6 |
/12 |
|
Total |
/100 |
Conceptual Problems
(each problem is worth 4 points).
1) As you increase the temperature (at constant pressure) of ice above 0 C, the molar gibbs energy of the ice becomes higher than that of liquid water and the ice melts. The reason why the molar gibbs energy of ice increases faster with temperature than does the molar gibbs energy of liquid water is:
a) Ice has a higher molar volume than liquid water.
b) Ice has a lower molar volume than liquid water.
c) Ice has a higher molar entropy than liquid water.
d)
Ice has a lower molar entropy than liquid water.
2) When you dissolve salt in an glass containing ice and water in equilibrium (assume that no heat enters or leaves the glass):
a) Some of the ice should melt and the temperature of the water/ice mixture should increase.
b)
Some of the ice should melt and the temperature of the
water/ice mixture should decrease.
c) Some of the water should freeze and the temperature of the water/ice mixture should increase.
d) Some of the water should freeze and the temperature of the water/ice mixture should decrease.
3) Picture two sealed flasks connected by a tube with a value in it. Which of the following situations will give rise to the greatest drop in gibbs energy upon opening the valve?
a)
One flask initially contains 1 mole of N2 and the
other 1 mole of O2.
b) One flask initially contains 1 mole of N2 and the other contains nothing (vacuum).
c) Both flasks initially each contain 0.5 moles of N2 and O2.
d) Both flasks initially contain 1 mole of N2.
4) When a reaction comes to equilibrium it is always the case that:
a) DG0 = 0
b) Keq = 1
c) DG = DG0
d)
DG = 0
5) Water has a number of unique properties. One of these is that the density of ice near atmospheric pressure is less than the density of liquid water (i.e., ice floats). If I were to take an equilibrium mixture of ice and liquid water and increase the pressure, holding the temperature constant, what would happen?
a) All of the liquid water would freeze
b) Some of the water would freeze
c)
All of the ice would melt
d) Some of the ice would melt
6) Water is also unique in its acid/base properties. In particular:
a) Water is a weak acid
b) Water is a weak base
c) Water is both a weak base and a weak acid
d) Water is a diprotic acid
7) When two ideal gases are mixed:
a) DSmix > 0; DHmix < 0
b) DSmix < 0; DHmix = 0
c) DSmix < 0; DHmix < 0
d)
DSmix
> 0; DHmix
= 0
8) Consider an open container of water which is split into two parts separated by a membrane (the pressure is constant at 1 atm). This membrane allows the water to pass freely between the two sides, but does not allow any solute to pass. If I dissolve salt in side A and nothing in side B.
a) Water will always pass from the A side to the B side.
b)
Water will always pass from the B side to the A side.
c) Water will pass from the A side to the B side only if the osmotic pressure is greater than 1 atm.
d) Water will pass from the B side to the A side only if the osmotic pressure is greater than 1 atm
Numerical Problems
(each problem is worth 14 points). All work must be shown for credit.
9) The
molar entropy of liquid ethanol is 160.7 J K-1mole-1? What would be the change in the molar Gibbs
energy of liquid ethanol when it is cooled from 25 C to 0 C?
DG = -SmDT = -(160.7 J K-1 mole-1)(-25K)
= 4018 J/mole
10)
When 40 grams of a compound was added to 1 kg of benzene, the
boiling point increased by 3.7 C. What
is the molar mass of the compound? The
ebullioscopic constant for benzene (Kb) is 2.53 K kg/mole.
DT = Kb mB = (2.53 K kg/mole) (40
grams/1 kg)(1/MW)
3.7 K = (101.2 K g/mole)/MW
MW = (101.2 K
g/mole)/(3.7 K) = 27.4 g/mole
11)
The standard reaction Gibbs energy for the conversion of
Glucose-6-phosphate to Fructose-6-phosphate is 1.7 kJ/mole at 37 C.
A) what is the equilibrium constant for this reaction?
KEQ = e-DG/RT =
exp(-(1700 J/mole)/(8.314 J K-1 mole-1 x 310K)) = 0.517
B) If a 10 mM solution of glucose-6-phosphate is allowed to convert to
fructose-6-phosphate until equilibrium is achieved, what is the final
concentration of glucose-6-phosphate?
0.517 = [F-6-P]/[G-6-P] = (0.01 – x)/x
12) Acetic acid is a weak acid with a pKA of 4.75. A solution was prepared by adding 0.1 moles of acetic acid and 0.03 moles of NaOH to water giving a final volume of 1 liter. What is the pH?
When you add 0.03 moles of NaOH to
0.1 moles of acetic acid, you end up with a solution that has 0.07 moles of
acetic acid and 0.03 moles of acetate ion.
Since this is in 1 liter, the respective molarities are 0.03 and
0.07. Plugging into the Henderson Hasselbach
equation:
pH = pKA + log(base/acid)
= 4.75 + log([0.03]/[0.07]) = 4.38
Quasi Real World
Problem (12 points)
13) In a very important metabolic pathway called the Krebs cycle, there is a reaction in which succinyl CoA is converted to succinate and acetyl CoA. This reaction is coupled to the formation of GTP (similar to ATP) from GDP and phosphate by the enzyme succinyl CoA synthetase. The two coupled reactions are (DRG0 values given at 37 C):
Succinyl CoA à Succinate + Acetyl CoA DRG0 = -33.8 kJ/mole
GDP + Phosphate à GTP + H2O DRG0 = 30.5kJ/mole
If the concentration of Succinyl CoA is 2 mM, Succinate is 5 mM, Phosphate is 10 mM, Acetyl CoA is 5 mM, GDP is 3 mM and GTP is 10 mM in the cell, what is DRG (NOT DRG0) for the coupled reaction at 37 C?
DRG = DRG0 + RTln(Q)
We need to determine DRG0 and Q
DRG0 is just
the sum of the two gibbs energy changes in the coupled equations or -3.3
kJ/mole.
Q = [Succinate][Acetyl
CoA][GTP]/[Succinyl CoA][GDP][Phosphate]
= ([0.005][0.005][0.010])/([0.002][0.003][0.01])
= 4.17
DRG = -3300 kJ/mole +
(8.314 J K-1 mole-1)(310K) ln(4.17) = 378 J/mole
Actually, given the significant
figures above, it should be 400 J/mole.