Exam 4
General Chemistry
Spring 2009
Circle the best answer:
1. Entropy is:
a)
b)
c)
d)
a measure of spontaneity in a system
a measure of the possible arrangements of a system
the sum of heat and work
all of the above
2. Enthalpy is:
a)
b)
c)
d)
a measure of spontaneity in a system
a measure of the possible arrangements of a system
the sum of heat and work
none of the above
3. A reaction must be spontaneous when:
a)
b)
c)
d)
Ssys is positive
Ssurr is positive
both (a) and (b)
none of the above
4. In: Ti(s) + O2 (g)  TiO2 (s), the reaction entropy is expected to:
a)
b)
c)
d)
increase
decrease
remain unchanged
there is too little info to tell
5. If G = 0 for a system then
a)
b)
c)
d)
Stotal > 0
Stotal < 0
the system is at equilibrium
the temperature of the system is 0 K.
6. Which of the following mixtures would be a useful buffer?
a) NaNO3 + HNO3
b) KH2PO4 + Na2HPO4
c) NaOH + NaCl
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d) Fe(OH)3 + HCl
7. The standard free energy of formation for N2O4 is +97 kJ/mol and indicates that
N2O4 is thermodynamically unstable towards decomposition to the elements.
a) Explain what the term “standard free energy of formation “ means.
Standard Free Energy refers to the energy required to form a molecule from the elements in
their standard states in a reaction that proceeds under standard conditions of 25 deg C, 1 atm pressure.
b) Write the reaction referred to in the phrase “decomposition to the elements”.
N2O4  N2 + 2O2
c) Why is decomposition of N2O4 likely to happen?
The reaction N2O4  N2 + 2O2
is the reverse of the standard formation which has a positive free energy (+97 kJ/mol).
Therefore the decomposition has a free energy cahneg of -97 J/mol and should happen spontaneousl.
8. The reaction PCl3(g) + Cl2(g)  PCl5(s) is spontaneous at room temperature.
Predict whether the reaction is exothermic or endothermic and explain your prediction.
The change in enthalpy for this reaction will be negative since two gases form a solid. That will not
contribute to a negative free energy. So the enthalpy must be negative so that G = H - TS is a
negative number and the reaction spontaneous.
9. Using the thermodynamic values provided for ethanol, calculate its boiling point, that is, the
Temperature (in deg C) where:
C2H5OH (l)  C2H5OH (s)
Starting with G = H - TS,
Then recognizing that at the boiling point, G = 0,
So now you can write 0 = H - TS,
Or H = TS
And H/S =T
Finding H = -235.1 –(-277.7) and S = (287.7-160.7) x 1/1000 kJ/J
Gives t= 76 deg C
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10. The pale green precipitate you may find around the faucets in your bathroom is Cr(OH)3.
Can 1.0 mg (0.001 g) of Cr(OH)3 be completely dissolved in 100 mL water?
Ksp for Cr(OH)3 is 6 x 10-31
Reaction: Cr(OH)3 <==> Cr(3+) + 3 OHKsp = [Cr(3+)][OH-]3
Solving for [Cr(3+)]: set [Cr(3+)] = x and [OH-] = 3x
Then Ksp = (x)(3x) 3 = 27x 4
x = 1.22 x 10-8 = [Cr(3+)], M
This is the maximum solubility of Cr(OH)3
How much mass in g is this?
1.22 x 10-8 M x 0.100L x 103 g/mol = 1.26 x 10-6 g Cr(OH)3 can be dissolved in 100 mL.
This is much almost 1000x smaller than the 1.0 mg asked in the question, so NO, it cannot be dissolved.
Alternatively, you can calculate Q, after converting 1.0 mg in 100mL to concentration :
Q = [Cr(3+)][OH-]3
Q = 27x4 = 27 (.001g/ 103 g/mol / 0.100L)4
Q = 2.4 x 10-19 >>>Ksp = 6 x 10-31
So again, it cannot all dissolve; here reaction will go backwards and a ppt will form.
11. Mother Nature is so clever! A buffer was needed to the maintain a constant pH for her mammalian
life-forms and she made use of one of the most abundant gases that blankets the earth: CO2.
a. Write all of the equilibrium equations that are involved in converting the gas to a useful buffer
for the blood plasma of mammals.
CO2(aq) + H2O(l)

H2CO3(aq)
H2CO3(aq) + H2O(l)
 HCO3- (aq) + H3O+
HCO3 (aq) + H2O  CO32- (aq) + H3O+
b. Identify which species make up the buffer system in mammalian blood.
H2CO3(aq) and HCO3- (aq)
c. If normal blood pH is 7.43, what is the ratio of buffer species, A- / HA, in blood?
For HA: Ka1 = 4.2 x 10-7. Note that Ka2 is not needed in this problem.
pH = pKa + log[A]/[AH]
so 7.43 = -log(4.2 x 10-7) + log[A]/[AH]
7.43 = 6.38 + log[A]/[AH]
[A]/[AH] =11.3
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d. If the blood pH falls to the dangerous (and possibly fatal) level of pH 7.0, what is the ratio of
buffer species, A- / HA?
pH = pKa + log[A]/[AH]
so 7.00 = -log(4.2 x 10-7) + log[A]/[AH]
7.00 = 6.38 + log[A]/[AH]
[A]/[AH] =4.16
12. Keq = 2.6 x 108 for this reaction:
[Ni(H2O)6]2+ + 6 H3N  [Ni(NH3)6]2+ + 6 H2O
a) What is the free energy change for this reaction at equilibrium? G = 0 at equilibrium
(b) When the concentrations of all species are at the standard concentration of 1.00 M in water, what is
the free energy for the reaction? Go = -RTlnK, so Go = -2.48 kJ/mol ln (2.6 x 108); Go = -48 kJ/mol
(c) When the concentrations of all species are 0.00100 M, what is the free energy for the reaction?
G = Go + RTlnQ, G = -48 kJ/mol + 2.48 kJ/mol ln ( .001/[.001x(.001)6]) = +54 kJ/mol
(d) using your results from parts (b) and (c), under which conditions—(b) vs (c)— is the product complex
[Ni(NH3)6]2+ more stable?
G is more negative in (b) under standard conditions so the reaction is more favored to go forward and
the products are more stable.
13. Many cultures worship the sun as a god.
Propose a thermodynamic justification for these spiritual practices.
These cultures are humans who live by respiration, a process that produces energy (free energy of
respiration is negative and spontaneous) to keep them alive through the oxidation of carbohydrates, like
glucose. The carbohydrates originate from plants, which make the carbohydrates through the process of
photosynthesis. Photosynthesis is not a spontaneous process (free energy is positive) but is made possible
through the energy in the form of light from the sun. The sun is responsible for all life on earth.
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14. Minerals exposed to the atmosphere can change over time. For example, the mineral magnetite,
Fe3O4 in contact with oxygen in the air will form the mineral hematite, Fe2O3.
(a)
Calculate the free energy change for the conversion of the mineral magnetite into the mineral
hematite. Gfo (Fe3O4 ) = -1015.4 kJ/mol; Gfo (Fe2O3 ) = -742.2 kJ/mol
You may assume that the only species involved in the reaction are those shown below:
Fe3O4 and O2 form Fe2O3
First balance the equation!
2Fe3O4 + 1/2 O2  3Fe2O3
G (rxn) = G (products) - G (rgts)
G (rxn) = 3 G (Fe2O3) – 2 G (Fe3O4) – 1/2 G (O2) =
G (rxn) = 3 (-742.2) – 2(-1015.4) – 1/2 (O) = -195.8 kJ (for formation of 3 mole Fe2O3 )
If the reaction is balanced as: 4Fe3O4 + O2  6Fe2O3
G (rxn) = G (products) - G (rgts)
G (rxn) = 6 G (Fe2O3) – 4 G (Fe3O4) – G (O2) =
G (rxn) = 6(-742.2) – 4(-1015.4) – 1/2 (O) = -391.6 kJ (for formation of 6 mole Fe2O3 )
(b) Is this reaction predicted by thermodynamics to spontaneously occur? Yes, G (rxn) <0
(c) Does entropy contribute to making this reaction spontaneous? (explain why or why not)
NO- because S (rxn) < 0 since a gas is a reagent but the product is a solid.
(d) Predict whether the conversion of magnetite to hematite is exothermic or endothermic.
Therefore once again, the reaction must be exothermic to be spontaneous.
15. You're helping a classmate finish her lab report and all she has to do is label the titration plot.
But she cannot remember if the titration was sodium phosphate to which was added hydrochloric acid or
phosphoric acid to which was added sodium hydroxide. Good thing she has you for a friend! Help her out
then you can both go sit in the sun.
a) Label each aspect of the titration plot (found at end of exam) that helps to identify which titration
reaction was done.
b)
The titration was NaOH added to phosphoric acid, H3PO4, a triprotic acid.
You can determine this because the pH begins acidic and because there are two equivalence points,
meaning the acid is at least dibasic.
The species in the first buffer region are H3PO4/ H2PO4-.
The species in the second buffer region are H2PO4 - / HPO4 2-.
The species in the third buffer region are HPO4 2-/ PO4 3-.
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b) My graduate student Shannon has heard of your extensive knowledge of the phosphate system!
You can help her make a buffer suitable for her experiments on DNA. DNA needs a buffer that holds pH
at 7.0 in order for it to be stable. Which phosphate species can be combined to make a buffer at pH 7?
Give the molecular formulas of the sodium salts that Lindsay should use to make her buffer.
The second buffer region falls at pH 7 so the appropriate salts to be used to make the buffer are NaH2PO4
and Na2HPO4 .
Extra Credit —Only ONE (1) will be graded and added to your exam grade.
You may do both problems if you please.
16. You have just purchased (for $20,000) a marble stature of an angel to grace your garden.
Unfortunately, you live in an area of very acid rain. Rain falling on your garden ( and your angel statue)
typically has a pH of 4.5.
a) Remembering that marble is calcium carbonate, write the reaction(s) involved as acid rain dissolves
marble.
CaCO3(s) <==> Ca(2+) + CO32CO32- + H+ <==> HCO3HCO3- + H+ <==> H2CO3
H2CO3 <=> H2O + CO2
b) If your angel weighs half a ton, how many liters of rain at pH 4.5 will dissolve all of it? (one ton = 907
kg)
calc the # moles of calcim carbonate:
[0.5 ton x 907 kg/ton x 1000g/1 kg] / 100 g/mol CaCO3 = 4.53 x 103 moles
CaCO3.
calc the concentration of acid rain:
pH 4.5 => 3.16 x 10-5 M H+
calc the number of volume of rain that ha s 2 x eq of H+ to neutralize carbonate:
[ 4.53 x 103 moles CaCO3 x 2H+/CO3 ] / 3.16 x 10-5 M H+ in acid rain = 287,000,000 L rain!
17. If there are 5 L of blood in a mammal and, at pH 7.43, the concentration of the acidic buffer
component HA = 3.72 x 10-6 M, how much strong acid (such as HCl) or strong base (such as NaOH)
must be added to the 5 L of blood to make the pH fall from 7.43 to 7.0?
For HA: Ka1 = 4.2 x 10-7. Note that Ka2 is not needed in this problem.
At a pH 7.43 = 6.38 + log( [A]/ 3.72 x 10-6)
so [A] = 4.17 x 10-5 M initially/
When the pH decreases, the ratio A/HA will decrease; the adjusted concentrations are;
[A] = 4.17 x 10-5 -x and [HA] = 3.72 x 10-6 +x.
So at a pH 7.00 = 6.38 + log(4.17 x 10-5 - x / 3.72 x 10-6 + x)
0.62 =log (4.17 x 10-5- x / 3.72 x 10-6 + x)
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Can now convert concentrations to moles, since both are in 5L:
0.62 =log (2.04 x 10-4 moles A - x moles H+ / 1.85 x 10-5 moles HA + x moles H+)
x = 2.5 x 10-5 moles H+ in a strong acid.
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