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Study Guide Key for Final Exam Fall 2010.
1. Write equilibrium constant expressions for the following reactions:
a) 2 H2 (g) + 2 NO (g) 2 H2O (g) + N2 (g)
b) Fe (s) + CO2 (g) FeO (s) + CO (g)
a) Keq =
[H 2 O] 2 [N 2 ]
[H 2 ] 2 [NO]2
b Keq =
[CO]
[CO 2 ]
2 Shown below is a balanced equation for the decomposition of H2S to form H2 and S2.
2 H2S (g) 2 H2 (g) + S2 (g)
Keq =
a) Write an equilibrium constant expression for the reaction.
[H 2 ] 2 [S2 ]
[H 2 S]2
b) Given the equilibrium concentrations: [H2S] = 0.1007 M, [H2] = 0.0219 M, and [S2] = 3.30 × 10-3 M, calculate the
numerical value of Keq.
[0.0219]2 [3.30 x 10 -3 ]
= 1.56 x 10-4
2
[0.1007]
Keq =
c) Assume the equilibrium is perturbed. When equilibrium is reestablished, the following concentrations are observed:
[H2] = 0.00287 M and [S2] = 0.171 M. Calculate [H2S] under these new conditions.
[H2S]2 =
Rearrange the equation in (a)
[0.00287]2 [0.171]
[H 2 ] 2 [S2 ]
=
= 9.0289 x 10-3 M2
-4
K eq
1.56 x 10
[H2S] =
9.0289 x10 −3 = 0.0950 M
d) What can you say about the forward and reverse reaction rates when the system is at equilibrium?
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
e) What can you say about the forward and reverse rate constants, whether the system is at equilibrium or not?
kf
kr
= Keq
3. Qualitatively, for the reaction shown #2, would the ∆G° value be positive or negative? Explain your logic
∆G would be positive, because Keq << 1.
4. Assuming I told you the reaction in #2, was relatively fast. Draw a reaction coordinate diagram that describes that
system. Then draw another showing how the system would be changed in the presence of a catalyst. Be sure to clearly
label all important quantities.
transition state (top of hill)
transition state (top of hill)
A relatively fast reaction
means a relatively low
hill.
Eact uncatalyzed
2H2 + S2
2H2 + S2
Eact catalyzed
Go
Change in
G is positive
2 H2S
uncatalyzed reaction (small hill)
Reaction Progress
Go
Change in
G is positive
2 H2 S
catalyzed reaction (smaller hill)
Reaction Progress
5. What is the definition of a Brønsted-Lowry acid?
A Brønsted-Lowry acid is a substance that can donate an H+.
6. Identify the acid and base on the reactant side of the equation shown below. Predict the products
of the reaction and indicate the conjugate of each reactant.
NH3
base
+
HBr
acid
NH4+
Br -
+
conjugate
acid
conjugate
base
7.(a) Write a balanced chemical reaction for the dissociation of acetic acid (CH3COOH) in water, and then (b) write the
Ka expression for that rxn. Acetic acid is a relatively weak acid. (c) What could you say about the strength of its
conjugate base? Formic acid is a slightly stronger acid than acetic acid. (d) Would formic acid dissociate to a greater
or lesser extent in water compared to acetic acid? (e) Which acid would have the larger pKa?
a) CH3COOH + H2O CH3OO-
+ H3O+
b) Ka =
[CH 3 COO - ] [H 3 O + ]
CH 3 COOH
c) A relatively strong conjugate base
d) Formic acid would dissociate to a greater extent
e) Acetic acid would have the larger pKa
8.(a) Calculate the [H3O+], [OH-], and pH of a 7.00 x 10-3 M solution of HNO3.
HNO3 is a strong acid and completely dissociates, so [H3O+] = 7.00 x 10-3 M.
pH = -log (7.00 x 10-3) = 2.155
+
-
Kw = [H3O ] [OH ]
Kw
1x10 −14
Rearrange to get [OH ] =
=
= 1.43 x 10-12 M
−3
+
[H 3 O ]
7.00 x10
-
(b) Is this solution acidic, basic or neutral? Acidic
(c) Would the pH be the same for a 7.00 x 10-3 M solution of acetic acid? Explain.
No. The pH would be higher (less acidic), because acetic acid is a weak acid and would not
dissociate as much.
9. How does a buffer function to keep the pH of a solution relatively constant when strong acid or base is added? Make
sure your answer includes equations for chemical reactions.
A buffer is composed of a weak acid (HA) and its conjugate base (A-).
The weak acid neutralizes added strong base (for example OH-).
HA + OH- A- + H2O
The conjugate base, A-, neutralizes added strong acid (for example H3O+).
A- + H3O+ HA + H2O
10. What weak acid could be used to make a buffer that was effective around pH 4.25? You must explain your answer
thoroughly to receive full credit.
An effective buffer must have reasonable quantities of both weak acid and conjugate base components.
Ideally [HA] = [A ].
From the Henderson-Hasselbalch expression, when [HA] = [A-], then pH = pKa.
Because the desired pH of the buffer is 4.25, we need a weak acid with a pKa near 4.25. Looking at the
accompanying list of pKa for various acids, we can see that benzoic acid, with a pKa of 4.19 would be
suitable.
11. Three coal miners are trapped in a closed space within the mine after a cave-in. Although they have oxygen tanks
with enough O2 to provide for their respiratory needs, they are still likely to incur health risks associated with gas
exchange. Describe one risk that is likely to occur. Be sure to include chemical rxns. as necessary to support your
answer.
There is a risk of respiratory acidosis. The reason for this is that CO2, released by the miner’s normal
respiratory processes will build up in the air of the closed space (i.e. the concentration of CO2 in the air will
increase.) As external CO2 concentrations rise, the rate of efflux (outflow) of CO2 from the miner’s lungs will
decrease. This would cause the miner’s internal CO2 concentration to go up, causing more CO2 to dissolve in
their blood. The equilibrium point of the following equations would then shift to the right (towards products).
H2O + CO2
W
H2CO3 + H2O
H2CO3
W
HCO3- + H3O+
The acid, H3O+, produced would lower the blood pH, causing acidosis.
12. How does a buffer function to keep the pH of a solution relatively constant when strong acid or base is added? Make
sure your answer includes equations for chemical reactions.
This is the same as question 9! Silly me!
13. Write a balanced chemical reaction for the dissociation of water, and then write the Kw expression for that rxn.
H2O + H2O
W
H3O+ + OH-
Kw = [H3O+][OH-]
The answers to questions 14-21 and 23-27 follow. (Note: The numbers are not the same as the questions on
the study guide.)
22. a) Human venous blood is usually slightly more acidic than arterial blood. Explain why this is so (part of your
answer should include one or more balanced equations for chemical reactions).
b) How is this difference in pH used by humans to improve the efficiency of our O2 transport by hemoglobin (Hb)?
a) Humans produce carbon dioxide (CO2) as one of the products in the breakdown of molecules.
Carbon dioxide reacts with water to produce carbonic acid. H2O(l) + CO2 (aq) H2CO3(aq)
The acid then dissociates in water to yield hydronium ions (H3O+), which causes the venous blood to
become more acidic:
H2O + H2CO3 H3O+ + HCO3
As the blood passes through the lungs, it picks up O2, but more importantly, for the problem at hand,
it releases CO2. This causes both of the above reactions to flow to the left. The decrease in H3O+
causes arterial blood to be more basic than venous blood.
b) Hemoglobin (Hb) has two different conformations (shapes). One of these conformations has
higher affinity for O2 and the other has lower affinity for O2. Hemoglobin also has acid-base
properties such that under more acidic conditions, it shifts to the lower affinity form.
Hb·(O2)4
+ H3O+
H2O +
H·Hb+
+
4O2
high affinity
low affinity
(lower acid conditions)
(higher acid conditions)
The acidic form of hemoglobin (H·Hb+) is the form of hemoglobin that has lower affinity for O2.
This shift is known as the Bohr effect, and it is among a number of factors that act to shift Hb to the
lower affinity form to increase its release of O2 in the tissues.
28. Given a chiral molecule, be able to circle the chiral carbon. A chiral carbon can be recognized because it has 4
atoms or groups of atoms attached to it. For example the α-carbon in all 20 common amino acids (except gly)
is a chiral carbon. (See the first part of the “Amino Acids and Proteins” notes and p. 6-7 of the “Representing
Chemical Structures Lab.”