SCHM 109
Study Guide Key Exam II
Spring 2013
Remember you will need to show your work for full credit. On the real exam always work the problems you know best
first. If you get hung up on a problem, you should move on and come back to it at the end. If you have time, check over
your work. To test your speed, work this study guide as if it was the exam. How long does it take to finish?
1. Determine the type of reaction (like single replacement), predict the products of the following reactions & balance the equations.
Type
Single replacement
a) FeCl3 + 3 K →
Double replacement
b) 2 K3PO4(aq) + 3CoCl2(aq) → 6 KCl (aq) + Co3(PO4)2 (s)
Combination or synthesis
c) 2 Ca + O2 → 2 CaO
Double replacement
d) HCl(aq) + NaOH(s)
Combustion
e) C5H12 +
Single replacement
f) 2 K + 2 H2O → 2 KOH + H2
3 KCl
→
+ Fe
NaCl (aq) +
H2O (l)
8 O2 → 5 CO2 + 6 H2O
2. Write a net-ionic equation for the following full balanced equations:
a) MgCl2(aq) + K2CO3(aq) → MgCO3(s) + 2KCl(aq)
Ionic eq: Mg2+ + 2 Cl- + 2 K+ + CO32- MgCO3 + 2 K+ + 2 ClNet ionic eq:
Mg2+ (aq)
+ CO32- (aq) MgCO3 (s)
b) 2Cs(s) + CaBr2(aq) Ca(s)
Ionic eq: 2Cs + Ca2+ + 2Br1Net ionic eq:
(2 Cl- & 2 K+ are spectator ions & cancel)
Ca
+ 2CsBr(aq)
+ 2Cs+ + 2Br1-
2Cs(s) + Ca2+(aq) Ca(s)
(2 Br- is a spectator ion)
+ 2Cs+ (aq)
3. a) How many moles of propane (C3H8) are present in 500.8 g propane? b) How many propane molecules are present
in this 500.8 g? c) How many C atoms are present in this sample?
a) First find the molecular weight of C3H8.
500.8 g x
.
3 C 3 x 12.011 = 36.033
8 H 8 x 1.00794 = 8.06352
44.09652 g/mole
= 11.3569 mol C3H8. or 11.36 mol C3 H8.
b) (11.3569 mol C3H8 )(6.022 x 1023) = 6.839125 x 1024 or 6.839 x 1024 molecules C3H8
c) (6.839125 x 1024 molecules C3H8 )(
3 C atoms
1 molecule
) = 2.051737 x 1025 or 2.052 x 1025 C atoms
4. How many anions are in 0.0135 moles of MgBr2?
First convert moles to number of formula units (what we use for an ionic compound instead of molecule) of
MgBr2 using Avogadro’s number. Then convert to number of anions.
(0.0135 moles MgBr2)(
6.022 x 1023 formula units
2 anions
1 mole
1 formula unit
)(
) = 1.62594 x 1022 anions 1.63 x 1022 anions.
5. Calculate the formula weight (molecular weight) of Ca(NO3)2.
1 Ca
2N
6O
40.07838 g
2 x 14.0067 = 28.0134
6 x 15.9994 = 95.9964
164.08818 g = 1 mole 164.0882 g/mol
6. Given the following reaction between nitrogen and oxygen to form nitric oxide: N2(g) + O2 (g) 2 NO(g)
How many grams of NO are obtained when 8.6 g of N2 are completely reacted?
Plan: g N2 mol N2 mol NO g NO
First need to determine molecular weight of N2:
8.6 g x
1 mole
28.0134 g
2 x 14.0067 = 28.0134 g in 1 mole of N2.
= 0.3069959 mol N2
0.3069959 mol N2 x
2 mol NO
1 mol N2
= 0.613399 mol NO
MW of NO 14.0067 + 15.9994 = 30.0061 g
0.613399 mol NO x
30.0061g
1 mol
30.0061 g = 1 mol NO
= 18.4057 g NO 18 g NO
7. Magnesium metal reacts with iron (II) chloride according to the equation
Mg(s) + FeCl2(aq) MgCl2(aq) + Fe(s).
Is the magnesium oxidized or reduced?
Is the iron (II) ion oxidized or reduced?
There are several ways one could proceed with this question. One way is to write the net ionic equation so as to see
charges on the ions.
I.E.
Mg(s) + Fe2+(aq) + 2 Cl-(aq) Mg2+(aq) + 2 Cl-(aq) + Fe(s)
(The 2 Cl- cancel.)
N.I.E.
Mg(s) + Fe2+(aq) Mg2+(aq) + Fe(s)
Now you can see that Mg is going from Mg0 to Mg2+. This means Mg is losing electrons and is oxidized.
And you can see that Fe is going from Fe2+ to Fe0. This means Fe is gaining electrons and is reduced.
8. Define reduction
The gain of electrons, the gain of H atoms, and/or the loss of O atoms
K2 Cr2 O7
9. In the following reaction CH2CH2OH + O2 ())))* CH2CH2O, is CH2CH2OH oxidized or reduced? How do you know?
CH2CH2OH is being oxidized. You can tell because it is losing H atoms.
10. Briefly describe a human genetic disease based on oxidation-reduction chemistry of hemoglobin.
A full credit answer should include one or more equations for chemical reactions.
There are inherited blood disorders, methemoglobinemia, that exist because a patient has problems with
maintaining the required charge on the Fe ion in hemoglobin. One type occurs due to a mutation in the beta
chain of Hb that makes the iron heme more easily oxidized: Hb(Fe2+) Hb(Fe3+) + e-. Hb(Fe3+) doesn’t
bind O2, so the hemoglobin can’t carry oxygen through the body.
Everyone’s Hb(Fe) is occasionally oxidized to the Fe3+ when it releases oxygen. So another type of
methemoglobinemia results from a lack of an enzyme (cytochrome b5 reductase). This enzyme is necessary to
3+
Hb(Fe ) + e
-
cytochrome b5
reductase
Hb(Fe 2+)
convert Fe3+ in methemoglobin back to Fe2+, so that it can bind oxygen.
11. A sample of 235.00 g of Mg reacts completely with HCl (aq) to produce an explosive gas.
a) Write a balanced chemical equation for the reaction.
b) How many grams of gaseous product are produced?
a) Mg + 2 HCl → MgCl2 + H2
b) Use Mg’s weight to convert g to moles: (235.00 g Mg)(
(9.6687924 mole Mg) (
(9.6687924 mole H2)(
1 mol Hydrogen gas
1 mole Mg
2.01588 g
1 mole
.+ ) = 9.6687924 mole Mg
) = 9.6687924 mole H2
) = 19.49112528 g H2 or 19.491 g H2
(You could also do these three steps as one continuous problem if you wanted.)
12. For the following oxidation-reduction reaction: Ca + Fe3+ Ca2+ + Fe
a) Write the half-reactions for this reaction.
b) Which reactant is being oxidized (How do you know?) and which is being reduced?
a) Ca → Ca2+ + 2e-
Fe3+ + 3e- Fe
b) Ca is being oxidized, because it is losing electrons. Fe3+ is being reduced; it gains electrons.
13. What is the concentration in molarity units of a solution that has 7.00 g of NaCl dissolved in water to give 520.6 mL of solution?
M = mol/liter, First convert 7.00 g of NaCl into moles of NaCl using the molecular weight of NaCl.
Na 22.98977 g
Cl 35.4527 g
58.44247 g in 1 mol of NaCl
520.6 mL x
0.1198 mol
0.5206 L
1L
1000 mL
7.00 g x (
mol NaCl
1.2 g
) = 0.1197759 mol
= 0.5206 L
= 0.2301 mol/L
or 0.230 M
14. If 83.00 mL of a 3.075 M solution of NaOH is diluted to a final volume of 12.88 L, what will its concentration be?
M1V1 = M2V2 .
Rearrange to get
M1V1
V2
Convert 83.00 mL to liters. 83.00 mL x
= M2. The volume amounts must be expressed in the same units.
1L
1000 mL
= 0.08300 L; M2 =
+.2 6 7 .1+ 8
.11 8
= 0.01982 M
15. Based on the following reaction, HCl + NaOH H2O + NaCl,
determine the concentration of hydrochloric acid, HCl, if 10.00 mL of acid required 15.51 mL of 0.2500 M NaOH to reach the
endpoint of the titration.
Follow the four steps for titration. Step 1. The equation is already balanced.
Step 2. Find the moles of NaOH:
(15.51 mL)(
1L
1000 mL
)(
Step 3. Find the moles of HCl: (0.0038775 mol NaOH)(
Step 4. Find the concentration (M) of HCl: (10.00 mL)(
.+122 :;< HC<
. L
0.2500 mol NaOH
) = 0.0038775 mol NaOH
L
:;< HC<
) = 0.0038775 mol HCl
:;< N@OH
1L
1000 mL
) = 0.01000 L
= 0.38775 mol/L 0.3878 M HCl
16. What volume of a 15% (w/v) solution of LiOH would be produced if you used 8750 g of LiOH?
% (w/v) =
x=
8750
15
g solute
x 100
mL solution
Rearrange to get mL of solution by itself. BmL solutionC =
x 100 = 58333 mL or 58,000 mL
g solute
% (w/v)
x 100
17. Which of the following compounds would you predict to have the highest, intermediate, and lowest solubility in water? Explain
your answer, using structures where appropriate.
H
H
H
C
H
C
C
H
C
O
b.
a.
H
H
H
H
H
H
O
C
H
H
c.
H
C
H
H
H
C
C
H
H
H
C
H
H
Solubility: a > b> c
C is a non-polar molecule. It would not be soluble in polar water.
Both a. and b. have a non-polar region and a polar region. The non-polar regions have similar surface areas for
both molecules. Next, compare the polar regions.
a. can form hydrogen bonds with water. Each of its propanol molecules has two H-bonding acceptor sites
and 1 donor site.
b. can also form hydrogen bonds w/ water. It has two acceptor sites and no donor sites. This has fewer hydrogen
bonding sites than molecule a.
a. is most soluble. b. has intermediate solubility. c. is least soluble.
Drawings follow.
Indicates hydrophobic surface area
H
H
H
H
H
C
C
C
C
H
H
H
H
H
c. Lowest solubility. All of
it's surface is non-polar.
H
O
Indicates hydrophobic surface
H
H
H
C
C
H
H
a. Highest solubility. It has about
the same size non-polar surface as
c, but it has more sites for
hydrogen-bonding with water.
H
H
O
H
O
H
H
O
O
H
H
H
H
H
H
C
H
b. Intermediate solubility. It has fewer
hydrogen bonding sites than a.
C
H
O
C
H
H
H
O
18. The following mechanism has been proposed:
k1
Step 1 O3 O2 + O
k2
Step 2 O + O3 2 O2
a) Write a balanced equation for the overall reaction. b) Write the predicted rate law for the overall reaction if k2 is much larger than k1.
c) Identify any reaction intermediates.
a) Add the equations: O3 + O + O3 O2 + O + 2 O2 Combine terms and cancel out what is the same on both sides.
the result is: 2 O3 3 O2
b) Rate = k1[O3]1
c) The only intermediate (produced in an early step and used in a later step) is the oxygen atom, O.
19. Given these data for the reaction of Fe3+ with I- :
3+
-
2+
2 Fe (aq) + 2 I (aq) 2 Fe (aq) + I2 (aq)
Exp#
1
2
3
[Fe3+] (M)
0.0400
0.0800
0.0400
-
[I ] (M)
0.0300
0.0300
0.0600
Initial rate (M/s)
8.10 x 10-4
-3
1.62 x 10
-3
3.24 x 10
Be sure to explain the logic for your answers.
a) Write a general rate law for the reaction.
b) Use the initial rate data to write the specific rate law consistent with the data. Make sure you solve to obtain a numerical value for k.
c) Write a chemical rxn. for the reactant side of the rate limiting step in the rxn. mechanism.
d) What is the initial rate of the rxn. with 0.090 M Fe3+ and 0.50 M I-?
e) If I told you that the value of k that you obtained is relatively large, what height hill would you draw for this reaction on a reaction progress
diagram (vertical axis = G°).
a) Rate = k [Fe3+]m[I-]n
b) Compare trials 1 and 2:
Rate2 = k [Fe3+]m[I-]n
Rate1 = k [Fe3+]m[I-]n
k/k cancels as does [0.0300]n/[0.0300]n, so
Compare trials 1 and 3:
Specific rate law is
1.62 x 10-3 = [0.0800]m
8.10 x 10-4 = [0.0400]m
Rate3 = k [Fe3+]m[I-]n
Rate1 = k [Fe3+]m[I-]n
k/k cancels as does [0.0400]m/[0.0400]m, so
k = Rate
[Fe3+]m[I-]n
1.62 x 10-3 = k [0.0800]m[0.0300]n
8.10 x 10-4 = k [0.0400]m[0.0300]n
or 2 = 2m , m = 1
3.24 x 10-3 = k [0.0400]m[0.0600]n
8.10 x 10-4 = k [0.0400]m[0.0300]n
3.24 x 10-3 = [0.0600]n
8.10 x 10-4 = [0.0300]n
filling in data from trial 1
k =
or 4 = 2n , n = 2
8.10 x 10-4
[0.0400]1[0.0300]2
= 22.5
Rate = 22.5[Fe3+]1[I-]2
c) Fe3+ + 2I- d) Rate = 22.5[0.090]1[0.50]2 = 0.50625 or 0.51 M/s
e) If k is relatively large, than the rate is relatively fast and the activation energy hill would be relatively small.
20. Explain how/why changes in temperature result in a change in rate constant, k. Include an appropriate graph.
Molecules move with a distribution of speeds. When temperaturechanges, the distribution changes. For higher
temperatures, the average speed and the top speed of a population of molecules would increase. The highest speeds are
particularly imprtant because only they have suffiennt energy to reach the transition state in a collision (in other words reach
the top of the energy barrier, the top of Ea hill). Because a larger fraction of molecules is moving at a high enough speed to
Boltzmann Distributions
colder temperature
warmer temperature
frequency
have their collisions reach the transition state, a larger fraction of
collisions will be productive, and the rate of the reaction will
increase. For lower temperatures, there is a decrease in the
fraction of molecules that can reach the transition state.
speeds that could
overcome Ea barrier.
21. Write equilibrium constant expressions for the following reactions:
speed
a) 2 H2 (g) + 2 NO (g) 2 H2O (g) + N2 (g)
b) Fe (s) + CO2 (g) FeO (s) + CO (g)
[H 2 O]2 [N 2 ]
a) Keq =
[H 2 ]2 [NO]2
b Keq =
[CO]
[CO 2 ]
22. Shown below is a balanced equation for the decomposition of H2S to form H2 and S2.
2 H2S (g) 2 H2 (g) + S2 (g)
a) Write an equilibrium constant expression for the reaction
.
Keq =
[H 2 ]2 [S2 ]
[H 2S] 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.
Keq =
[0.0219]2 [3.30 x 10-3 ]
= 1.56 x 10-4
2
[0.1007]
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.
[H 2 ]2 [S 2 ]
[0.00287] 2 [0.171]
= 9.0289 x 10-3 M2
Rearrange the equation in (a) [H2S] =
=
-4
K eq
1.56 x 10
2
[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) ) How is Keq defined?
kf
kr
= Keq
23. a) Qualitatively, for the reaction shown #6, would the ∆G° value be positive or negative? Explain your logic.
∆G would be positive, because Keq < 1.
b) Assuming I told you the reaction in #6, 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.
(The Keq determined in
#6 was less than 1, so the
reaction is reactant
favored.
transition state (top of hill)
transition state (top of hill)
Eact uncatalyzed
2H2 + S2
A relatively fast
reaction means a
relatively small hill.
(Actually write
“relatively small hill”
next to drawing.)
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