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Table of Contents – Exam Review Packet (Teacher’s Version)
Equilibrium
Concept List
Free Response Questions
2
3
Acid / Base
Concept List
Free Response Questions
8
8
Kinetics
Concept List
Free Response Questions
20
21
Electrochemistry
Concept List
Free Response Questions
28
29
Thermodynamics
Concept List
Free Response Questions
42
43
Atomic Theory, Bonding, and Intermolecular Forces
Concept List
Free Response Questions
50
51
Concentration and Colligative Properties
Concept List
Free Response Questions
58
59
Laboratory
Free Response Questions
63
Nuclear
Free Response Questions
76
1
AP Chemistry Concept List – EQUILIBRIUM
All Problems are equilibrium problems because
All problems involve stoichiometry: soluble salts, strong acids, strong bases
Some problems involve equilibrium: “insoluble” salts, weak acids, weak bases
For chemical reactions – Keq, Kc, and Kp are the important quantities
For physical changes – Ka, Kb, Ksp, Kionize, and Kdissocation are the important quantities
Important points
1.
Law of mass action
2.
Kc for molarity for ions and gases
3.
Kp with atm, or mmHg for gases
Relationship / connection between these Kp = Kc (RT)Δn
4.
Shifting equilibrium – Le Chatlier’s Principle
a.
solid
b.
liquid
c.
catalyst
d.
inert gas added
e.
temperature changes (increasing T favors endothermic processes)
f.
only factors in equation constant will affect Keq eg. CaCO3(s)  CaO(s) +
CO2(g)
g.
pressure / volume changes
5.
Orientation of collisions
2
2003B #1
After a 1.0 mole sample of HI(g) is placed into an evacuated 1.0 L container at 700. K,
the reaction represented occurs. The concentration of HI(g) as a function of time is
shown below.
2 HI(g)  H2(g) + I2(g)
a.
Write the expression for the equilibrium constant, Kc, for the reaction.
b.
What is [HI] at equilibrium?
c.
Determine the equilibrium concentrations of H2(g) and I2(g).
d.
On the graph above, make a sketch that shows how the concentration of H2(g)
changes as a function of time.
e.
Calculate the value of the following equilibrium constants at 700. K.
f.
i.
Kc
ii.
Kp
At 1,000 K, the value of Kc for the reaction is 2.6 × 10-2. In an experiment, 0.75
mole of HI(g), 0.10 mole of H2(g), and 0.50 mole of I2(g) are placed in a 1.0 L
container and allowed to reach equilibrium at 1,000 K. Determine whether the
3
equilibrium concentration of HI(g) will be greater than, equal to, or less than the
initial concentration of HI(g). Justify your answer.
10 points total
a.
Kc 
[H 2 ][I 2 ]
[HI]
1 point
b.
From the graph, [HI] eq is 0.80 M
1 point
c.
2HI(g) -->
H2(g) +
I2(g)
init
1.0 M
0M
0M
chang -0.20 M
+0.10 M
+0.10 M
final 0.80 M
0.10 M
0.10 M
[H2] = [I2] = 0.10 M
1 point for stoichiometric relationship between HI and H2 and I2
1 point for equilibrium concentrations of H2 and I2
d.
From the graph, [H2] eq is 0.10 M. The curve should have the following
characteristics: start at 0 M, increase to 0.10 M, reach equilibrium at the
same time [HI] reaches equilibrium
1 point for any two characteristics, 2 points for all three characteristics
[H ][I ] (0.10)(0.10)
 0.016
e.
i) K c  2 22 
[HI]
(0.80) 2
1 point for correct substitution which must agree with parts b and c
ii) KP = Kc = 0.016 because the number of moles of gaseous products is
equal to the number of moles of gaseous reactants
1 point
[H 2 ][I 2 ] (0.10)(0.50)
Q

 0.089
f.
[HI]2
(0.75) 2
Kc = 0.026, Q > Kc, therefore to establish equilibrium, the numerator must
decrease and the denominator must increase. Therefore, [HI] will
increase.
1 point for calculating Q and comparing to Kc
1 point for predicting correct change in [HI]
2004B #1
N2(g) + 3 H2(g)  2 NH3(g)
For the reaction represented above, the value of the equilibrium constant, Kp is 3.1 × 10-4
at 700 K.
a.
Write the expression for the equilibrium constant, Kp, for the reaction.
b.
Assume that the initial partial pressures of the gases are as follows:
P(N2) = 0.411 atm, P(H2) = 0.903 atm, and P(NH3) = 0.224 atm.
4
i)
Calculate the value of the reaction quotient, Q, at these initial conditions.
ii)
Predict the direction in which the reaction will proceed at 700. K if the
initial partial pressures are those given above. Justify your answer.
c.
Calculate the value of the equilibrium constant, Kc, given that the value of Kp for
the reaction at 700. K is 3.1 × 10-4.
d.
The value of Kp for the reaction represented below is 8.3 × 10-3 at 700. K.
NH3(g) + H2S(g)  NH4HS(g)
Calculate the value of Kp at 700. K for each of the reactions represented below.
i)
NH4HS(g)  NH3(g) + H2S(g)
ii)
2 H2S(g) + N2(g) + 3 H2(g)  2 NH4HS(g)
10 points
a.
b.
c.
d.
KP 
2
PNH
3
PN 2 PH3 2
1 point for pressure expression
1 point for correct substitution
2
PNH
(0.224) 2
3

 0.166
i) Q 
PN 2 PH3 2 (0.411)(0.903)3
1 point for calculation of Q with correct mass action expression consistent
with part a
ii) Since Q > Kp, the numerator must decrease and the denominator must
increase, so the reaction must proceed from right to left to establish
equilibrium
1 point for direction or for stating that Q > Kp
1 point for explanation
Kp = Kc (RT)Δn, with Δn = 2 – 4 = -2
3.1 × 10-4 = Kc (0.0821 L atm mol-1 K-1 × 700 K)-2
3.1 × 10-4 = Kc (57.5)-2 --> 3.1 × 10-4 = Kc(3.1 × 10-4 )
Kc = 1
1 point for calculating Δn
1 point for correct substitution and value of Kc
i) Kp = (8.3 × 10-3)-1 = 1.2 × 102
1 point
ii) 2 × [NH3 + H2S --> NH4HS]
Kp = (8.3 × 10-3)2
N2 + 3 H2 --> 2 NH3
Kp = 3.1 × 10-4
2 H2S + N2 + 3H2 --> 2 NH4HS
5
Kp = (8.3 × 10-3)2(3.1 × 10-4) = 2.1 × 10-8
1 point for squaring Kp for NH4HS or for multiplying Kp’s
1 point for correct Kp
1988 #6
NH4HS(s)  NH3(g) + H2S(g)
For this reaction, ΔH° = + 93 kilojoules. The equilibrium above is established by placing
solid NH4HS in an evacuated container at 25 °C. At equilibrium, some solid NH4HS
remains in the container. Predict and explain each of the following.
a.
The effect on the equilibrium partial pressure of NH3 gas when additional solid
NH4HS is introduced into the container.
b.
The effect on the equilibrium partial pressure of NH3 gas when additional H2S gas
is introduced into the container.
c.
The effect on the mass of solid NH4HS present when the volume of the container
is decreased.
d.
The effect on the mass of solid NH4HS present when the temperature is increased
Average score = 4.31
a) two points
The equilibrium pressure of NH3 gas would be unaffected Kp = (PNH3) (PH2S).
Thus the amount of solid NH4HS present does not affect the equilibrium.
b) two points
The equilibrium pressure of NH3 gas would decrease. In order for the pressure
equilibrium constant, Kp, to remain constant, the equilibrium pressure of NH3
must decrease when the pressure of H2S is increased.
Kp = (PNH3) (PH2S)
(A complete explanation based on Le Chatelier's principle is also acceptable.)
c) two points
The mass of NH4HS increases. A decrease in volume causes the pressure of each
gas to increase. To maintain the value of the pressure equilibrium constant, Kp,
the pressure of each of the gases must decrease. That decrease realized by the
formation of more solid NH4HS.
Kp = (PNH3) (PH2S)
(A complete explanation based on Le Chatelier's principle is also acceptable.)
d) two points
The mass of NH4HS decreases because the endothermic reaction absorbs heat
and goes nearer to completion (to the right) as the temperature increases. (One
6
point was assigned for each correct prediction and one point for each correct
explanation.)
1980 #6
NH4Cl(s)  NH3(g) + HCl(g) for this reaction, ΔH = +42.1 kilocalories
Suppose the substances in the reaction above are at equilibrium at 600 K in volume V and
at pressure P. State whether the partial pressure of NH3(g) will have increased, decreased,
or remained the same when equilibrium is reestablished after each of the following
disturbances of the original system. Some solid NH4Cl remains in the flask at all times.
Justify each answer with a one- or two-sentence explanation.
a.
A small quantity of NH4Cl is added.
b.
The temperature of the system is increased.
c.
The volume of the system is increased.
d.
A quantity of gaseous HCl is added.
e.
A quantity of gaseous NH3 is added.
a. Partial pressure of ammonia will not change, ammonium chloride is a solid and
does not effect equilibrium position. If partial pressure of ammonia would
increase the partial pressure so would partial pressure of HCl and Keq value
would change.
b. Increase the temp shifts equilibrium to the right to favor endothermic
so partial pressure of NH3 would increase.
process,
c. Increasing the volume (decrease the pressure) would not change the partial
pressure because both NH3 and HCl would decrease and cause a change in the
Keq value. Only temperature can cause this change.
d. The reaction would shift to the left so partial pressure of NH3 would decrease.
e. The partial pressure of NH3 would increase because the equilibrium would shift
to the left, amount of HCl decreases so amount of NH3 would need to increase to
keep Keq constant.
7
AP Chemistry Concept List – ACID - BASE
pH = - log [H+]
pOH = - log [OH-]
Kw = [H+] [OH-] = 1×10-14 at 25 oC
If you know one quantity, you know the other three
pH
[H+]
pOH
[OH-]
Definitions
Acid
Donates H+
Donates protons
Accepts e- pairs (AlCl3)
Base
Donates OHAccepts protons - {anions?}
Donates e- pairs (NH3)
Theory
Arrhenius
Bronsted – Lowry
Lewis
Conjugate Acid – Base Pairs
1.
HCl + H2O → H3O+ + Cl-
2.
NH3 + H2O  NH4+ + OH-
3.
HSO4- + H2O  H3O+ + SO42-
4.
CO32- + H3O+  HCO3- + H2O
A.
Ka
Ka 
Weak Acid
HCN  H+ + CN-
[H  ] [CN ]
 6.2  1010
[HCN]
What is the ph of a 0.5 M HCN solution?
B.
Kb
Kb 
Weak base NH3 + H2O  NH4+ + OH-
[ NH4 ] [OH ]
 1.8  10 5
[ NH3 ]
What is the pH of a 0.5 M NH2OH solution?
8
C.
Ksp
Insoluble Salts
MgF2(s)  Mg2+ + 2F-
Ksp = [Mg2+] [F-]2 = 6.6 × 10-9
What is the solubility of MgF2 in molarity?
D.
Buffers – a weak acid/base and its soluble salt (conjugate base or acid) mixture
pH  pK a  log
[base]
[acid ]
What is the pH of a 0.5 M HC2H3O2 in 2 M NaC2H3O2 solution? Ka = 1.8 × 10-5
E.
Salts of Weak Acids and Weak Bases
What is the pH of a 1 M NaC2H3O2 solution?
Titrations and Endpoints
At endpoint: acid moles = base moles or [H+] = [OH-]
Strong acid – strong base
endpoint pH = 7
Strong acid – weak base
endpoint pH < 7
Weak acid – strong base
endpoint pH > 7
The last two are important because of conjugate acid and base pairs
9
2007 #1
1.
HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
Ka = 7.2 × 10-4
Hydrofluoric acid, HF(aq), dissociates in water as represented by the equation above.
a.
Write the equilibrium constant expression for the dissociation of HF(aq) in water.
b.
Calculate the molar concentration of H3O+ in a 0.40 M HF(aq) solution.
HF(aq) reacts with NaOH(aq) according to the reaction represented below.
HF(aq) + OH-(aq)  H2O(l) + F-(aq)
A volume of 15 mL of 0.40 M NaOH(aq) is added to 25 mL of 0.40 M HF(aq) solution.
Assume volumes are additive.
c.
Calculate the number of moles of HF(aq) remaining in the solution.
d.
Calculate the molar concentration of F-(aq) in the solution.
e.
Calculate the pH of the solution.
9 points
a.
b.
c.
d.
e.
[ H3O ][ F  ]
1 point
[ HF ]
[ H3O ][ F  ]
( x)( x)
Ka 

 7.2 10 4
[ HF ]
0.40  x
Assume x << 0.40, then x2 = (0.40)(7.2 × 10-4), x = [H3O+] = 0.017 M
1 point is earned for the correct setup (or that consistent with part a)
1 point is earned for the correct concentration
mol HF = initial mol HF – mol NaOH added
= (0.025 L)(0.40 mol L-1)–(0.015 L)(0.40 mol L-1)=0.010–0.0060=0.0040
mol
1 point is earned for determining initial number of moles of HF and OH1 point is earned for setting up and doing correct subtraction
mol F- formed = mol NaOH added = 0.0060 mol F0.0060 mol F- / (0.015 + 0.025) L of solution = 0.15 M F1 point is earned for determining the number of moles of F1 point is earned for dividing the number of moles of F- by the correct
total volume
[HF] = 0.004 mol HF / 0.040 L = 0.10 M HF
[ H O ][ F  ]
[ HF ]  K a
0.10  7.2  104

Ka  3


[
H
O
]

 4.8  10 4
3

[ HF ]
[F ]
0.15
-4
pH = - log (4.8 × 10 ) = 3.32
Ka 
10
OR
Henderson – Hasselbach equation
1 point is earned for indicating that the resulting solution is a buffer (by
showing a ration of [F-] to [HF] or moles of F- to HF)
1 point is earned for the correct calculation of pH
2005B #1 Ka
Ka 
[H3O ][OCl ]
 3.2  108
[HOCl]
Hypochlorous acid, HOCl, is a weak acid in water. The Ka expression for HOCl is shown
above.
a.
Write a chemical equation showing how HOCl behaves as an acid in water.
b.
Calculate the pH of a 0.175 M solution of HOCl.
c.
Write the net ionic equation for the reaction between the weak acid HOCl(aq) and
the strong base NaOH(aq)
d.
In an experiment, 20.00 mL of 0.175 M HOCl(aq) is placed in a flask and titrated
with 6.55 mL of 0.435 M NaOH(aq).
i)
Calculate the number of moles of NaOH(aq) added.
ii)
Calculate [H3O+] in the flask after the NaOH(aq) has been added.
iii)
Calculate [OH-] in the flask after the NaOH(aq) has been added.
10 points
a.
HOCl + H2O --> OCl- + H3O+
b.
0.175
0
0
-x
x
x change
0.175 –x
x
x final
Ka 
1 point
initial
[H3O ][OCl ]
xx

 3.2  108
[HOCl]
0.175  x
Assume that x <<< 0.175, x = 7.5 × 10-5 M, pH = 4.13
1 point is earned for calculating the value of [H3O+]
1 point is earned for calculating the pH
c.
HOCl + OH- --> OCl- + H2O
1 point for both correct reactants
1 point for both correct products
11
d.
i) mol(NaOH)=6.55 mL(1L/1000 mL)(0.435 M)=2.85×10-31 point
ii) mol(HOCl) = 20.00 (1/1000) (0.175) = 3.50 ×10-3 mol 1 point
OH- is the limiting reactant, therefore all of it reacts
HOCl +
OH- -->
OCl-
0.00350
0.00285
0
-0.00285
-0.00285
0.00285
0.00065
0
0.00285
+ H2O
M(HOCl) = 0.00065 / 0.02655 = 0.0245 M
M(OCl-) = 0.00285 / 0.02655 = 0.107 M
HOCl + H2O -->
H3O+ +
0.0245
0
0.107
-x
x
x
0.0245
x
0.107 + x
Ka 
OCl-
1 point
[H3O ][OCl ] x (0.107  x)

 3.2  108
[HOCl]
(0.0245  x)
Assume 0.107 + x = 0.107 and 0.0245-x = 0.0245
x = 7.3 × 10-9 M
1 point
iii) [H3O+][OH-]= 10-14 = Kw
[OH-] = 10-14 / [H3O+] = 10-14 / 7.3 × 10-9 = 1.4 × 10-6 M 1 point
1999 #1 Kb
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
In aqueous solution, ammonia reacts as represented above. In 0.0180 M NH3(aq) at
25°C, the hydroxide ion concentration, [OH-] , is 5.60 × 10-4 M. In answering the
following, assume that temperature is constant at 25°C and that volumes are additive.
a.
Write the equilibrium-constant expression for the reaction represented above.
b.
Determine the pH of 0.0180 M NH3(aq).
12
c.
Determine the value of the base ionization constant, Kb, for NH3(aq).
d.
Determine the percent ionization of NH3 in 0.0180 M NH3(aq).
e.
In an experiment, a 20.0 mL sample of 0.0180 M NH3(aq) was placed in a
flask and titrated to the equivalence point and beyond using 0.0120 M
HCl(aq).
i. Determine the volume of 0.0120 M HCl(aq) that was added to
reach the equivalence point.
ii. Determine the pH of the solution in the flask after a total of 15.0
mL of 0.0120 M HCl(aq) was added.
iii. Determine the pH of the solution in the flask after a total of 40.0
mL of 0.0120 M HCl(aq) was added.
9 Points
(a)
[ NH 4 ][OH  ]
K
[ NH3 ]
(b)
pOH = 3.252
[OH ] = 5.60 × 10 → { or
[H+] = 1.79 × 10-11
(c)
Kb 
-
1 point
-4
} → pH = 10.748
(5.60  104 )2
 1.74  105 (or 1.80 ×10-5 )
4
0.0180  5.60  10
1 point
2 point
Note: 1st point for [NH4+] = [OH-] = 5.60 × 10-4; 2nd point for correct answer
5.60  10-4
 100% = 3.11% (or 0.0311)
0.0180
(d)
% ionization 
(e)
(i)
NH3 + H+ → NH4+
mol NH3 = 0.0180 M × 0.0200 L = 3.60 × 10-4 mol = mol H+ needed
3.60  10-4 mol
= 0.0300 L = 30.0 mL
1 point
vol HCl solution 
0.0120 M
mol H+ added = mol 0.0120 M × 0.0150 L = 1.80 × 10-4 mol H+ added
= 1.80 × 10-4 mol NH4+
produced
(ii)
1 point
1.80  104 mol
 0.00514 M  [ NH3 ]
1 point
0.0350 L
Note: Point earned for 1.80 × 10-4 mol, or 0.00514 M [NH3] or [NH4+] or
statement “halfway to equivalence point.”
NH 4 
13
[ NH 4 ][OH  ]
=[OH-] → pOH = 4.745 → pH = 9.255
[ NH 3 ]
(= 1.74 × 10-5)
(= 4.759)
(= 9.241) 1pt
(iii) 10.0 mL past equivalence point
0.0100 L × 0.0120 M = 1.20 × 10-4 mol H+ in 60.0 mL
[H+ ] = 0.000120 mol / 0.060 L = 0.00200 M
pH = − log (2.00 × 10-3) = 2.700
1 point
One point deduction for mathematical error (maximum once per question)
One point deduction for error in significant figures* (maximum once per question)
*number of significant figures must be correct within +/− one digit (except for pH: +/−
two digits)
Kb = 1.80 × 10-5 =
1996 A/B lab #6
A 0.500-gram sample of a weak, nonvolatile acid, HA, was dissolved in sufficient water
to make 50.0 milliliters of solution. The solution was then titrated with a standard NaOH
solution. Predict how the calculated molar mass of HA would be affected (too high, too
low, or not affected) by the following laboratory procedures. Explain each of your
answers.
a.
After rinsing the buret with distilled water, the buret is filled with the standard
NaOH solution; the weak acid HA is titrated to its equivalence point.
b.
Extra water is added to the 0.500-gram sample of HA.
c.
An indicator that changes color at pH 5 is used to signal the equivalence point.
for explanation point in 9 (a), (c), and (d), credit is earned at step indicted
in boldface type.
(a) two points
Calculated Mm(HA) too low
M(NaOH) => V(NaOH) => n(NaOH) => n(HA) => Mm(HA)
(M = n ÷ V) and (Mm = m÷ n)
(b) two points
Calculated Mm(HA) not affected
Any one of the following reasons. Water: does not change n(HA), changes
only M(HA) -- sense of dilution, is not a reactant
(c) two points
Calculated Mm(HA) too high
equivalence point => n(NaOH) => n(HA) => Mm(HA)
(expected pH higher)
Note: "no effect if NaOH standardized with same indicator" earns 2
points; no credit earned if pH=7 or neutral
(d) two points
14
Calculated Mm(HA) too low
V(NaOH) => n(NaOH) => n(HA) => Mm(HA)
Note: point earned for V(NaOH) only if:
(i) no explanation point is earned in (a)
(ii) explanation in (a) also includes V(NaOH)
2000 #8 A/B Lab
A volume of 30.0 mL of 0.10 M NH3(aq) is titrated with 0.20 M HCl(aq). The value of
the base-dissociation constant, Kb, for NH3 in water is 1.8 × 10-5 at 25 oC.
a.
Write the net-ionic equation for the reaction of NH3(aq) with HCl(aq).
b.
Using the axes provided below, sketch the titration curve that results when a total
of 40.0 mL of 0.20 M HCl(aq) is added dropwise to the 30.0 mL volume of 0.10
M NH3(aq).
c.
From the table below, select the most appropriate indicator for the titration.
Justify your choice
Indicator
Methyl Red
Bromothymol Blue
Phenolphthalein
d.
pKa
5.5
7.1
8.7
If equal volumes of 0.10 M NH3(aq) and 0.10 M NH4Cl(aq) are mixed, is the
resulting solution acidic, neutral, or basic? Explain
15
8 points
(a)
NH3(aq) + H+(aq) → NH4+(aq)
1 point
or
NH3(aq) + H3O+(aq) → NH4+(aq) + H2O(l)
Note: phase designations not required to earn point
(b)
Sketch of Titration Curve:
3 pnts
• 1st pt. → initial pH must be > 7 (calculated pH ≈ 11)
• 2nd pt. → equivalence point occurs at 15.0 mL ± 1 mL of HCl added
(equivalence point must be detectable from the shape of the curve or a mark on
the curve)
• 3rd pt. → pH at equivalence point must be < 7 (calculated pH ≈ 5).
Note: a maximum of 1 point earned for any of the following:
- a line without an equivalence point
- a random line that goes from high pH to low pH
- an upward line with increasing pH (equivalence point MUST be at 15.0 mL)
(c)
Methyl Red would be the best choice of indicator,
1 point
Because the pKa for Methyl Red is closest to pH atequivalence point.
1 point
Notes: • explanation must agree with equivalence point on graph
• alternative explanation that titration involves strong acid and weak base (with
product an acidic salt) earns the point
(d)
The resulting solution is basic.
1 point
Kb for NH3 (1.8 × 10-5) and Ka for NH4+ (5.6 × 10-10) indicate that NH3 is a
stronger base than NH4+ is an acid
or
[OH-] = Kb = 1.8 × 10-5 because of the equimolar and equivolume amounts of
ammonium and ammonia → cancellation in the buffer pH calculation. Thus pOH
 0.05 
≈ 5 and pH ≈ 9 (i.e., recognition of buffer, so that log 
  0 → pOH = pKb ≈
 0.05 
5 → pH = 14 – pOH ≈ 9)
Ksp 1998 #1
Solve the following problem related to the solubility equilibria of some metal hydroxides
in aqueous solution.
a.
The solubility of Cu(OH)2 is 1.72 × 10-6 gram per 100. mL of solution at 25 °C.
(i)
Write the balanced chemical equation for the dissociation of
Cu(OH)2(s) in aqueous solution.
(ii)
Calculate the solubility (in moles per liter) of Cu(OH)2 at 25 °C.
(iii)
Calculate the value of the solubility-product constant, Ksp, for
Cu(OH)2 at 25 °C.
16
b.
The value of the solubility-product constant, Ksp, for Zn(OH)2 is 7.7 × 10-17 at
25°C.
(i)
Calculate the solubility (in moles per liter) of Zn(OH)2 at 25°C in a
solution with a pH of 9.35.
(ii)
At 25°C, 50.0 mL of 0.100-molar Zn(NO3)2 is mixed with 50.0 mL of
0.300-molar NaOH. Calculate the molar concentration of Zn2+(aq) in
the resulting solution once equilibrium has been established. Assume
that volumes are additive.
9 points
a i.) Cu(OH)2(s) → Cu2+(aq) + 2 OH-(aq)
Correct stoichiometry and charges (but not phases) necessary
No credit earned if water as a reactant or product
ii)
1.72 10 6 g
 1.763 10 8 mol Cu(OH) 2
1
97.57 g mol
1 point
1 point
1.763 108 mol Cu(OH) 2
 1.76 107 mol L1
0.100 L
One point earned for conversion of mass to moles (need not be computed
explicitly)
One point earned for calculation of moles per liter
iii)
[Cu2+] = 1.76 × 10-7 M
[OH-] = 2 × (1.76 X 10-7 M) = 3.52 × 10-7 M
1 point
Ksp = [Cu2+][OH-] 2 = (1.76 × 10-7)( 3.52 × 10-7)2 = 2.18 × 10-20
1 point
2+
One point earned for correct [Cu ] and [OH ]
One point for correct substitution into Ksp expression and answer
Response need not include explicit statement of [OH-] if Ksp expression is written
with correct values of [Cu2+] and [OH-]
pH = 9.35 → pOH = 4.65 → [OH-] = 2.24 × 10-5 M
1 point
17
K sp
7.7  10
[ Zn 2 ] 

 1.5  10 7 M
 2
5 2
[OH ]
(2.24  10 )
One point earned for correct determination of [OH-]
One point for correct answer (assume [Zn2+] equals solubility in moles per liter)
No points earned if [OH-] is assumed equal to twice [Zn2+]
bi)
ii)
Zn2+ + 2OH-  Zn(OH)2(s)
initial
final
Zn2+
0.0050 mol
0 mol
OH0.0150 mol
0.0050 mol
Zn(OH)2 (s)
0 mol
0.0050 mol
17
or
0.0050 mol
 0.050 M
0.100 L
One point earned if precipitation reaction is clearly indicated and moles or
concentration of OH- is calculated correctly
Zn(OH)2

Zn2+ +
2 OHx
(0.050 + 2x)
-17
2+
- 2
Ksp = 7.7 × 10 = [Zn ][OH ] = (x) (0.050 + 2x)2 = (x)(0.050)2  [Zn2+] = x = 3.1
× 10-14 M 1 point
OR
Zn(OH)2

Zn2+ +
2 OH(0.050-x)
(0.150 - 2x)
[OH  ] 
Ksp = 7.7× 10-17 = [Zn2+][OH-] 2 = (0.050-x)(0.150-2x)2
Solve for x, then subtract x from 0.050 M to obtain [Zn2+]
1 point
1 point
2001 Ksp
Answer the following questions relating to the solubility of the chlorides of silver and
lead.
a.
b.
At 10 oC, 8.9 × 10-5 g of AgCl(s) will dissolve in 100. mL of water.
i.
Write the equation for the dissociation of AgCl(s) in water.
ii.
Calculate the solubility, in mol L-1, of AgCl(s) in water at 10 oC.
iii.
Calculate the value of the solubility-product constant, Ksp, for AgCl(s) at
10 oC.
At 25 oC, the value of Ksp for PbCl2(s) is 1.6 × 10-5 and the value of Ksp for
AgCl(s) is 1.8 × 10-10.
i.
If 60.0 mL of 0.0400 M NaCl(aq) is added to 60.0 mL of 0.0300 M
Pb(NO3)2(aq), will a precipitate form? Assume that volumes are additive.
Show calculations to support your answer.
ii.
Calculate the equilibrium value of [Pb2+(aq)] in 1.00 L of saturated PbCl2
solution to which 0.250 mole of NaCl(s) has been added. Assume that no
volume change occurs.
iii.
If 0.100 M NaCl(aq) is added slowly to a beaker containing both 0.120 M
AgNO3(aq) and 0.150 M Pb(NO3)2(aq) at 25 oC, which will precipitate
first, AgCl(s) or PbCl2(s)? Show calculations to support your answer.
(10 points)
18
(b)
(a)(i) AgCl(s) → Ag+(aq) + Cl-(aq)
1 point
· Correct charges needed to earn credit.
· Phases not necessary to earn credit.
8.9  105 g
(ii)
1 point
 6.2  10 7 mol (in 100 mL)
143.32 g/mol
Note: The first point is earned for the correct number of moles; the second point
is earned for the conversion from moles to molarity.
(iii) Ksp = [Ag+][Cl-] = (6.2 × 10-6)2 = 3.8 × 10-11
1 point
Note: Students earn one point for squaring their result for molarity in (a) (ii).
(i)
n(Cl-) = (0.060 L) (0.040 mol/L) = 0.0024 mol
1 point
[Cl ] = (0.0024 mol)/(0.120 L) = 0.020 mol/L = 0.020 M
n (Pb2+) = (0.060 L) (0.030 mol/L) = 0.0018 mol
[Pb2+] = (0.0018 mol)/(0.120 L) = 0.015 mol/L = 0.015 M
Q = [Pb2+][Cl-] 2 = (0.015)(0.020)2 = 6.0 × 10-6
1 point
Q < Ksp , therefore no precipitate forms
1 point
Note: One point is earned for calculating the correct molarities; one point is
earned for calculating Q ; one point is earned for determining whether or not a
precipitate will form.
(ii)
[Pb2+] = Ksp / [Cl- ] 2 = 1.6 × 10-5 / (0.25)2 = 2.6 × 10-4 M
1 point
AgCl
10
K
1.8  10
(iii) for AgCl solution: [Cl-] = sp  
1 point
 1.5  10 9 M
[ Ag ]
0.120
for PbCl2 solution: [Cl-] =
K spPbCl2


1.6  10 5
 1.0  10 2 M
0.150
[ Pb2 ]
The [Cl-] will reach a concentration of 1.5 × 10-9 M before it reaches a
concentration of 1.0 × 10-2 M, (or 1.5 × 10-9 << 1.0 × 10-2), therefore AgCl(s)
will precipitate first.
Note: One point is earned for calculating [Cl-] in saturated solutions with the
appropriate Ag+ and Pb2+ concentrations; one point is earned for concluding
which salt will precipitate first, based on the student’s calculations.
19
AP Chemistry Concepts List - KINETICS
1.
Rate definition
2.
Rate Law – differential versus integrated
3.
Factors affecting rate
a.
b.
c.
d.
e.
[C]
ΔT
catalysis
surface area
nature of reactants – distinguish between homo- and heterogenous
i. solids
ii. Liquids
iii. gases
iv. Ions (solutions)
4.
Collision theory – orientation and energy
5.
Mechanism – relationship between ΔT, ΔS, ΔH – catalysis
6.
Energy of activation (Ea) – Arrhenius equation – differentiate from ΔH
k  E  1 1 
E 1 1
ln  2   a      a   
R  T2 T1 
 k1  R  T1 T2 
7.
Order
a.
determined by
i.
experimental comparison
ii.
graphing
b.
zero, first, second – determining % remaining and/or % reacted
ex.
Ln (x2/x1) = kt
8.
Rate constants with units (units change with reaction order)
a.
unsuccessful versus effective collisions
b.
orientation and energy
20
1997 #4
2A+B→C+D
The following results were obtained when the reaction represented above was studied at
25 °C
Initial Rate
Initial
Initial
Experiment
of Formation
[A]
[B]
of C (mol L-1 min-1)
1
0.25
0.75
4.3 × 10-4
2
0.75
0.75
1.3 × 10-3
3
1.50
1.50
5.3 × 10-3
4
1.75
??
8.0 × 10-3
a.
Determine the order of the reaction with respect to A and B. Justify your answer.
b.
Write the rate law for the reaction. Calculate the value of the rate constant,
specifying units.
c.
Determine the initial rate of change of [A] in Experiment 3.
d.
Determine the initial value of [B] in Experiment 4.
e.
Identify which of the reaction mechanisms represented below is consistent with
the rate law developed in part (b). Justify your choice.
A+B→C+M
Fast
M+A→D
Slow
2
B <===> M
M+A→C+X
A+X→D
Fast equilibrium
Slow
Fast
3
A + B <===> M
M+A→C+X
X→D
Fast equilibrium
Slow
Fast
1
(a) three points
1.3 x 10-3 / 4.3 x 10¯4 = k (0.75)x (0.75)y / k (0.25)x (0.75)y leads to 3 = (3)x leads
to x = 1, first order in A
5.3 x 10¯3 / 1.3 x 10¯4 = k(1.50)(1.50)y/ k(0.75)(0.75)y => 4= 2(2)y => y=1 =>
First order in B
21
Notes; Verbal descriptions accepted, but no point earned for just "if A doubles,
the rate doubles". If A given as second order, 2 points can be earned for showing
that B must be zero order.
(b) two points
rate = k[A][B] (equation must be consistent with part (a))
k= 4.3 x 10¯4M min¯1 / (0.25M) (0.75M) = 2.3 x 10¯3 M¯ 1 min¯1
Note; Units must be correct to earn second point. If no part (a) shown, 1 point
can be earned for a reasonable (first or second order) rate law.
(c) one point
[A] / t = -2 (5.3 x 10¯3 M¯1 min¯1) = - 1.06 x 10¯2 M¯1 min¯ 1
Note; Units ignored; no penalty for ( ¯ ) sign.
(d) one point
8.0 x 10 ¯3 M¯1 min¯1 = (2.3 x 10¯2 M¯1 min¯1) (1.75 M) [B]
[B] = 2.0 M
Note; No penalty if answer is consistent with wrong part (b).
(e) two points
Mechanism 2 is consistent
rate proportional to [M][A] and [M] proportional to [B] => rate proportional to
[A][B]
Notes; Verbal discussion accepted for second point. Mechanism must be
consistent with rate law in part (b). Showing that mechanisms 1 and 3 are
inconsistent is not required.
1999 # 3
2 NO(g) + Br2(g) → 2 NOBr(g)
A rate study of the reaction represented above was conducted at 25°C. The data that were
obtained are shown in the table below.
Initial [NO]
Initial [Br2]
Initial Rate of Appearance
Experiment
-1
-1
(mol L )
(mol L )
of NOBr (mol L-1 s-1)
1
0.0160
0.0120
3.24 × 10-4
2
0.0160
0.0240
6.38 × 10-4
3
0.0320
0.0060
6.42 × 10-4
a.
Calculate the initial rate of disappearance of Br2(g) in experiment 1.
b.
Determine the order of the reaction with respect to each reactant, Br2(g) and
NO(g). In each case, explain your reasoning.
c.
For the reaction,
i.
write the rate law that is consistent with the data, and
22
ii.
d.
calculate the value of the specific rate constant, k, and specify units.
The following mechanism was proposed for the reaction:
Br2(g) + NO(g) → NOBr2(g) slow
NOBr2(g) + NO(g) → 2 NOBr(g) fast
Is this mechanism consistent with the given experimental observations? Justify
your answer.
9 points:
(a)
Rate of Br2(g) loss occurs at ½ the rate of NOBr(g) formation.
3.24  10 4 M
Rate of Br2(g) loss =
= 1.62 × 10-4 M sec-1 (or mol L-1 sec-1)
2 sec
1 point
Note: No penalty for missing units; ignore + or − signs
(b)
Comparing experiments 1 and 2, [NO] remains constant, [Br2] doubles,
and rate doubles; therefore, rate α [Br2] 1 → reaction is first-order with
respect to [Br2].
1 point
6.38  104
k[ NO]x [ Br2 ] k (0.0160) x (0.0240)  1 

1


   4 1
6.42  10 4
k[ NO]x [ Br2 ] k (0.0320) x (0.0240)  2 
x
x
1 1
    x = 2 → reaction is second-order with respect to [NO]
4 2
2 points
Note: One point earned for a proper set-up, comparing experiments 2 and 3 (as is
shown here) or experiments 1 and 3. Second point earned for solving the ratios
correctly and determining that the exponent = 2. Also, credit can be earned for a
non-mathematical approach (e.g., one point for describing the change in [Br2]
and subsequent effect on rate, another point ford escribing the change in [NO]
and subsequent effect on rate).
(c)
(i) Rate = k[NO] 2[Br2]
1 point
Note: Point earned for an expression that is consistent with answer in part (b)
Rate
3.24  104 M sec 1

(ii) k =
= 105 M -2 sec-1 (or 105 L2 mol-2 sec[ NO]2 [ Br2 ] (0.0160) 2 (0.0120) M 2
1
)
2 points
-4
-1
-2
(Using rate of Br2(g) loss = 1.62 × 10 M sec → k = 52.7 M sec-1 is correct.)
Note: One point for solving for the value of the rate constant consistent with the
rate-law expression found in (b) or stated in part (c); one point for the correct
units consistent with the rate-law expression found in part (b) or stated in (c).
(d)
No, it is not consistent with the given experimental observations. 1 point
23
This mechanism gives a reaction that is first-order in [NO], and first-order in
[Br2], as those are the two reactants in the rate-determining step. Kinetic data
show the reaction is second-order in [NO] (and first-order in [Br2]), so this
cannot be the mechanism.
1 point
Note: One point earned for “No” [or for “Yes” if rate = k[NO][Br2] in part (b)].
One point earned for justifying why this mechanism is inconsistent with the
observed rate-law [or consistent with rate law stated earlier in response].
One point deduction for mathematical error (maximum once per question)
One point deduction for error in significant figures* (maximum once per question)
*number of significant figures must be correct within +/− one digit
1996 # 8
The reaction between NO and H2 is believed to occur in the following three-step process.
NO + NO <===> N2O2 (fast)
N2O2 + H2 → N2O + H2O (slow)
N2O + H2 → N2 + H2O (fast)
a.
Write a balanced equation for the overall reaction.
b.
Identify the intermediates in the reaction. Explain your reasoning.
c.
From the mechanism represented above, a student correctly deduces that the rate
law for the reaction is rate = k[NO]2[H2]. The student then concludes that (1) the
reaction is third-order and (2) the mechanism involves the simultaneous collision
of two NO molecules and an H2 molecule. Are conclusions (1) and (2) correct?
Explain.
d.
Explain why an increase in temperature increases the rate constant, k, given the
rate law in part c.
(a) one point
2 NO + 2 H2 ---> N2 + 2 H2O
(b) two points
N2O2 and N2O are intermediates
because they appear in the mechanism but not in the overall products (or
reactants)
(c) three points; one for each half of conclusion (1) answer, third for conclusion
(2) answer
Student indicates conclusion (1) is correct,
because the sum of the exponents in rate law is 2 + 1 = 3
Student indicates conclusion (2) is incorrect,
because no step involves two NO molecules and a H2 molecule
24
(d) two points; T goes up therefore k goes up:
because increasing number of collisions between reactants
are occuring with sufficient energy to form an activated complex
OR
T goes up therefore rate goes up
because no change in concentration of reactants, therefore k must increase
OR
from Arrhenius equation (not required in AP Chemistry curriculum, but noted in
some student responses):
as T goes up, k goes up
1998 #6
Answer the following questions regarding the kinetics of chemical reactions.
a.
The diagram below at right shows the energy pathway for the reaction O3 + NO
→ NO2 + O2.
Clearly label the following directly on the diagram.
i.
ii.
b.
The activation energy (Ea) for the forward reaction
The enthalpy change (ΔH) for the reaction
The reaction 2 N2O5 → 4 NO2 + O2 is first order with respect to N2O5.
i.
Using the axes at right, complete the graph that represents the change in
[N2O5] over time as the reaction proceeds.
ii.
Describe how the graph in part i could be used to find the reaction rate
at a given time, t.
iii
Considering the rate law and the graph in part i, describe how the
value of the rate constant, k, could be determined.
iv.
If more N2O5 were added to the reaction mixture at constant
temperature, what would be the effect on the rate constant, k? Explain.
25
Data for the chemical reaction 2A → B + C were collected by measuring the
concentration of A at 10-minute intervals for 80 minutes. The following graphs
were generated from analysis of data.
c.
Use the information in the graphs above to answer the following.
i. Write the rate-law expression for the reaction. Justify your answer.
ii. Describe how to determine the value of the rate constant for the reaction.
8 point
a)
Response must clearly indicate (and distinguish between) Eact and ΔHrxn on graph.
Each earns one point (2 total)
26
bi).
Response shows a softly curving line that approaches the time axis
and whose slope changes continually
No penalty if curve crosses time
Curve must drop initially and
[N2O5] increases.
axis or levels out above time axis.
continually. No credit earned if
ii.
Reaction Rate is the slope of the line tangent to any point on the curve.
Rate must be tied somehow to slope of graph.
Answer may be indicated directly on the graph.
Instantaneous rate must be indicated rather than the average rate
iii) Since rate = slope = k[N205], the value of k can be determined algebraically from the
slope at a known value of [N 205].
No penalty for Rate =2k[N205], as reaction stoichiometry could suggest this
answer.
Point can be earned for rate constant = slope of graph of ln[N2O5] vs. time since
reaction is first order.
Use of half-life or integrated rate law to solve for k can be accepted.
iv. The value of the rate constant is independent of the reactant concentrations, so adding
more reactant will not affect the value of k.
No point earned for simply stating that value of k will not change.
Response must distinguish between rate. and rate constant.
bi.
Rate = k[A] or ln ([A]/[A]0) = -kt. Since graph of ln [A] vs. time is linear, it must
be a first-order reaction.
Either form of the rate law is acceptable, and both the equation and the brief justification
are required to earn the point
No point earned if response indicates first order because the first graph is not linear.
ii.
Determine the slope of the second graph and set “k= - slope”
Response must indicate both the negative sign and the slope.
27
AP Chemistry Concepts - ELECTROCHEMISTRY
1.
oxidation / reduction – balancing equations (review)
2.
galvanic cells – {positive, Red Cat}
3.
electrolytic cells
4.
cathode
5.
anode
6.
current, charge, Faradays, (voltage / EMF) (amps, coulombs and volts – unit
problem)
7.
cell notation
8.
salt bridge – “balance of charge” not electron balance
9.
Eo and spontaneity
10.
ΔGo = - n F Eo
11.
E = Eo – (0.059 / n) log Kc
28
2007 #3.
An external direct-current power supply is connected to two platinum electrodes
immersed in a beaker containing 1.0 M CuSO4(aq) at 25oC, as shown in the diagram
above. As the cell operates, copper metal is deposited onto one electrode and O2(g) is
produced at the other electrode. The two reduction half-reactions for the overall reaction
that occurs in the cell are shown in the table below.
half-reaction
O2(g) + 4 H+(aq) + 4 e-  2H2O(l)
Cu2+(aq) + 2e-  Cu(s)
Eo(V)
+1.23
+0.34
a.
On the diagram, indicate the direction of electron flow in the wire.
b.
Write a balanced net ionic equation for the electrolysis reaction that occurs in the
cell.
c.
Predict the algebraic sign of ΔGo for the reaction. Justify your prediction.
d.
Calculate the value of ΔGo for the reaction.
An electric current of 1.50 amps passes through the cell for 40.0 minutes.
e.
Calculate the mass, in grams, of the Cu(s) that is deposited on the electrode.
f.
Calculate the dry volume, in liters measured at 25oC and 1.16 atm, of the O2(g)
that is produced.
10 points
a.
b.
Electron flow in the wire is from the right to the left (counterclockwise)
1 point
2 H2O + 2 Cu2+ → 4 H+ + 2 Cu + O2
1 point is earned for all three products
1 point is earned for balancing the equation
29
c.
d.
e.
f.
The sign of ΔGo would be positive because the reaction is NOT
spontaneous.
1 point is earned for indicating that ΔGo is greater than 0 and supplying a
correct explanation
Eo = -1.23 V + 0.34 V = -0.89 V = -0.89 J C-1
ΔGo = -nFEo = -4 (96500) (-0.89) = 340,000 J = 340 kJ
1 point is earned for calculating Eo
1 point is earned for calculating ΔGo (consistent with Eo)
q = (1.50 C s-1) (40.0 min) (60 s / 1 min) = 3600 C
1 point
mass Cu = (3600 C) (1 mole / 96500 C) (1 mol Cu / mol e-) (63.55 g Cu /
moll) = 1.19 g Cu
1 point
OR can be calculated in one step (2 points)
n(O2) = (1.19 g Cu) (1 mol Cu / 63.55 g Cu) (1 mol O2 / 2 mol Cu) =
0.00936 mol O2
1 point
V = nRT / P = (0.00936) (0.0821) (298) / (1.16) = 0.197 L 1 point
1997 #3
In an electrolytic cell, a current of 0.250 ampere is passed through a solution of a chloride
of iron, producing Fe(s) and Cl2(g).
a.
Write the equation for the reaction that occurs at the anode.
b.
When the cell operates for 2.00 hours, 0.521 gram of iron is deposited at one
electrode. Determine the formula of the chloride of iron in the original solution.
c.
Write the balanced equation for the overall reaction that occurs in the cell.
d.
How many liters of Cl2(g), measured at 25 °C and 750 mmHg, are produced when
the cell operates as described in part (b)?
e.
Calculate the current that would produce chlorine gas at a rate of 3.00 grams per
hour.
(a) 1 point
2Cl¯ ---> Cl2 + 2e¯ (equation need not be balanced)
(b) three points
(0.250 coul / sec x 7,2000 sec) / (96,500 coul / mol e¯) = 1,800 coul / (96,500
coul / mol e¯) = 0.01865 mol e¯
mol Fe = 0.521 g Fe / (55.85 g / mol Fe) = 0.00933 mol Fe
mol e¯ / mol Fe = 1.865 x 10¯2 mol e¯ / 9.33 x 10¯3 approx. equals 2 e¯ per Fe
atom ---> FeCl2
(c) 1 point
Fe2+ + 2 Cl¯ ---> Fe + Cl2
Notes: "FeCl2(aq)" accepted for reactants. Any balanced equation corresponding
to answer in part (b) earns 1 point.
30
(d) 1 point
moles Fe2+ = moles Cl2 = 9.33 x 10¯3 mol Cl2
V = nRT / P = (0.00933 mol Cl2 x 0.0821 L.atm.mol¯1.K¯1 x 298 K) / (750 / 760)
atm = 0.231 L (or 231 mL)
(e) two points
(3.00 g Cl2 / 71 g mol¯1) / 3,600 sec = 0.0423 mol Cl2 / 3,600 sec = 1.17 x 10¯5
mol Cl2 / sec
current (in amperes) = (2 mol e¯ / mol Cl2) x (1.17 x 10¯5 mol Cl2 / sec) x (96,500
coul / 1 mol e¯) = 2.27 amp (or coul /sec)
alternate solution: 0.00933 mole Cl2 / 2 hrs = 0.662 g Cl2 / 2 hrs = 0.331 g Cl2 /
hr
0.20 amp / 0.331 g Cl2 = x / 3.00 g Cl2
x = (3.00 g x 0.250 amp) / 0.331 g = 2.27 amp
2000 # 2
2.
Answer the following questions that relate to electrochemical reactions.
a.
Under standard conditions at 25 oC, Zn(s) reacts with Co2+(aq) to produce
Co(s)
b.
i)
Write the balanced equation for the oxidation half reaction.
ii)
Write the balanced net-ionic equation for the overall reaction.
iii)
Calculate the standard potential, Eo, for the overall reaction at 25 oC.
At 25 oC, H2O2 decomposes according to the following equation.
2 H2O2(aq)  2 H2O(l) + O2(g)
Eo = 0.55 V
i)
Determine the value of the standard free energy change, ΔGo, for the
reaction at 25 oC.
ii)
Determine the value of the equilibrium constant, Keq, for the reaction at 25
o
C.
iii)
The standard reduction potential, Eo, for the half reaction O2(g) + 4 H+(aq)
+ 4e-  2 H2O(l) has a value of 1.23 V. Using this information in
addition to the information given above, determine the value of the
standard reduction potential, Eo, for the half reaction below.
O2(g) + 2 H+(aq) + 2e-  H2O2(aq)
31
c.
In an electrolytic cell, Cu(s) is produced by the electrolysis of CuSO4(aq).
Calculate the maximum mass of Cu(s) that can be deposited by a direct current of
100. amperes passed through 5.00 L of 2.00 M CuSO4(aq) for a period of 1.00
hour.
10 points
(a)
(i)
Zn(s) → Zn2+(aq) + 2 e1 point
2+
(ii)
Co (aq) + Zn(s) → Co(s) + Zn2+(aq)
1 point
(iii) 0.76 V + (−0.28 V) = 0.48 V
1 point
Note: phase designations not required in part (i) or part (ii)
(b)
(i) ΔG° = − nFE° = − 2(96,500)(0.55V)
2 pnts
= − 1.1 × 105 J or − 1.1 ×102 kJ
• First point earned for n = 2 (consistent use of n = 4 also accepted)
• Second point earned for negative sign, correct number (2 ± 1 sig. figs.),
and appropriate units (kJ or J or kJ/mole or J/mole)
(ii) ΔG° = – RT ln(K)
1 point
-1.1 × 105 J = – [8.31 J mol-1 K-1][298 K][ln (K)]
K = 2.0 × 1019
(full credit also for correct use of log K = nE/ 0.0592)
(iii) O2 + 2 H2O → 2 H2O2
-0.55 V
O2 + 4 H+ + 4 e- → 2 H2O
1.23 V
______________________________________________________
2 O2 + 4 H+ + 4 e- → 2 H2O2
0.68 V
2 points
→ O2 + 2 H+ + 2 e- → H2O2
0.68 V (not required)
• Two points earned for correct voltage with supporting numbers (chemical
equations not necessary)
• One point earned for correct chemical equations with incorrect voltage
• Two points earned for correct answer (3 ± 1 sig. figs.)
• One point earned for any two of these steps: (amp)(sec) → coulombs, coulombs
→ mol e-, mol e- → mol Cu, mol Cu → g Cu
1996 #7
Sr(s) + Mg2+ <===> Sr2+ + Mg(s)
Consider the reaction represented above that occurs at 25°C. All reactants and products
are in their standard states. The value of the equilibrium constant, Keq, for the reaction is
4.2 × 1017 at 25°C.
a.
Predict the sign of the standard cell potential, E°, for a cell based on the reaction.
Explain your prediction.
b.
Identify the oxidizing agent for the spontaneous reaction.
32
c.
If the reaction were carried out at 60°C instead of 25°C, how would the cell
potential change? Justify your answer.
d.
How would the cell potential change if the reaction were carried out at 25°C with
a 1.0 M solution of Mg(NO3)2 and a 0.10 M solution of Sr(NO3)2? Explain.
e.
When the cell reaction in part d reaches equilibrium, what is the cell potential?
(a) two points
The sign of the cell potential will be positive
because (any one is sufficient):
K is greater than 1
the reaction is spontaneous (occurs)
E° for Sr2+ is more positive
Standard reduction potential for Sr more negative
E° = + 0.52 V
Note: only 1 point earned for just E° positive because Keq positive.
(b) one point
The oxidizing agent is Mg2+
(c) two point
The cell potential would increase
Since all ions are at 1 M, Q for the system is 1 and E° = (RT/nF) ln K
so as T increases, so should E°
Note: no credit lost if student recognizes Keq dependence on T. For temperature
change in this problem, decrease in ln K term is small relative to the term RT/nF
OR
No change, because in the Nernst equation Ecell = E° - (RT/nF) ln Q
ln Q = 0, and Ecell = E°
Note: this second approach earns 1 point only
(d) two points
Ecell will increase
In the equation Ecell = E° - (0.0592 / n) log Q
Q = 0.1 therefore log Q is negative therefore term after E° is positive therefore
Ecell increases
OR
with the concentration of Mg2+ larger than that of Sr2+, Le Chatelier's principle
predicts the reaction will have a larger driving force to the right and a more
positive Ecell
(e) one point
33
At equilibrium, Ecell = 0
Note: "balanced", "neutral", or "no net reaction" not accepted
1998 #8
Answer the following questions regarding the electrochemical cell shown above.
a. Write the balanced net-ionic equation for the spontaneous reaction that occurs as
the cell operates, and determine the cell voltage.
b. In which direction do anions flow in the salt bridge as the cell operates? Justify
your answer.
c. If 10.0 mL of 3.0-molar AgNO3 solution is added to the half-cell on the right,
what will happen to the cell voltage? Explain.
d. If 1.0 grams of solid NaCl is added to each half-cell, what will happen to the cell
voltage? Explain.
e. If 20.0 mL of distilled water is added to both half-cells, the cell voltage decreases.
Explain.
8 points
a)
2 Ag+(aq) + Cd(s) ~ 2 Ag(s) + Cd2+(aq)
1 point
34
equation must be balanced and net ionic, phases not necessary
reaction direction and ion charges must be correct
0.80 - (-0040) = 1.20 V
evidence of where numbers came from should be present; if equation is exactly reversed,
-1.20 V earns the point
1 point
b)
Anions (or N03- ions) will flow to the Cd2+ solution or from the Ag+ solution to
balance the charges
OR
Anions will flow to the left to balance the positive charge of the new Cd 2+ ions
both the correct direction and justification needed to earn this point
direction may be indicated by arrow marked on diagram
1 point
c)
The cell voltage will increase.
1 point
Ag+ is a reactant, so increasing [Ag+] will increase the driving force (stress) for the
forward (spontaneous) reaction and the potential will increase
OR
1 point
Since Q = [Cd2+]/[Ag+]2, increasing [Ag+] will decrease Q. According to the Nernst
equation, E = E0 - (0.0592 log Q / n , if Q decreases, then voltage increases.
d.)
The cell voltage will decrease.
1 point
Adding NaCI will have no effect on the Cd cell, but will cause AgCI to precipitate
in the Ag cell (Ag+ + Cl-  AgCl). Thus [Ag+] decreases, and since Ag+ is a reactant,
decreasing [Ag+] causes a decrease in voltage.
1 point
+
Credit earned for decreasing [Ag ] results in decreased voltage or opposite of part c
e)
Since Q = [Cd2+]/ [Ag+]2 , diluting both solutions by the same amount will
increase the value of Q. According to the Nernst equation, E = E0 - (0.0592 log Q )/n , if
Q increases, then voltage decreases. ,
No credit earned for "since the solutions are diluted, the voltage will decrease"
2002 # 2
Answer parts (a) through (e) below, which relate to reactions involving silver ion, Ag+.
The reaction between silver ion and solid zinc is represented by the following equation
2 Ag+(aq) + Zn(s)  Zn2+(aq) + 2 Ag(s)
a.
A 1.50 g sample of Zn is combined with 250. mL of 0.110 M AgNO3 at 25 oC.
i.
Identify the limiting reactant. Show calculations to support your answer.
35
ii.
b.
On the basis of the limiting reactant that you identified in part (i),
determine the value of [Zn2+] after the reaction is complete. Assume that
volume change is negligible.
Determine the value of the standard potential, Eo, for a galvanic cell based on the
reaction between AgNO3(aq) and solid Zn at 25 oC.
Another galvanic cell is based on the reaction between Ag+(aq) and Cu(s), represented by
the equation below. At 25 oC, the standard potential, Eo, for the cell is 0.46 V.
2 Ag+(aq) + Cu(s)  Cu2+(aq) + 2 Ag(s)
c.
Determine the value of the standard free-energy change, Go, for the reaction
between Ag+(aq) and Cu(s) at 25 oC.
d.
The cell is constructed so that [Cu2+] is 0.045 M and [Ag+] is 0.010 M. Calculate
the value of the potential, E, for the cell.
e.
Under the conditions specified in part (d), is the reaction in the cell spontaneous?
Justify your answer.
Total Score 10 points
 1 mol Zn 
 = 2.29 × 10-2 mol Zn
i. n(Zn) = 1.50 g Zn 
65.4
g
Zn



 0.110 mol Ag 
 = 2.75 × 10-2 mol Ag+
n(Ag+) = 0.250 L 
1L


a.)
 1 mol Zn  2 mol Ag  
 = 4.59 × 10-2 mol Ag+ required

n(Ag+) = 1.50 g Zn 
2 
65.4
g
Zn
1
mol
Zn



-2
+
+
Since only 2.75 × 10 mol Ag available, Ag is the limiting reactant.
OR
 0.110 mol Ag  
+
 = 2.75 × 10-2 mol Ag+
n(Ag ) = 0.250 L 
1L


 1 mol Zn 
 = 1.38 × 10-2 mol Zn required
n(Zn) = 2.75 × 10-2 mol Ag+ 
 
2
mol
Ag


-2
Since 2.29 × 10 mol Zn are available, more is available than required, so Zn
is in excess and Ag+ is limiting.
(Correct solutions other than shown above earn both points.)
•1 point earned for the moles of one reactant and the proper stoichiometry
•1 point earned for the limiting reactant and the supporting calculation or
explanation
36
 1 mol Zn 2  
 = 1.38 × 10-2 mol Zn2+
ii. n(Zn2+) = 2.75 × 10-2 mol Ag+ 
 
2
mol
Ag


-2
2
1.38  10 mol Zn
= 0.0550 M Zn2+
0.250 L
OR
[Ag+] initial = 0.110 M , therefore [Zn2+] = (1/2) (0.110 M) = 0.0550 M
1 point earned for mol Zn2+
1 point earned for [Zn2+]
OR
2 points earned for [Zn2+]
**********************************************************
If the student concludes Zn is the limiting reactant, then
 1 mol Zn   1 mol Zn 2 
 = 2.29 × 10-2 mol Zn2+ formed
 
1.50 g Zn 
65.4
g
Zn
1
mol
Zn



2.29  10-2 mol Zn 2
= 0.0916 M Zn2+
0.250 L
1 point earned for mol Zn2+
1 point earned for [Zn2+]
b.)
E°cell = E°(reduction) − E°(reduction)
= (0.80 V) − (−0.76 V) = 1.56 V
2 Ag+(aq) + Zn(s) → Zn2+(aq) + 2 Ag(s) +1.56 V
OR
Ag+(aq) + e- → Ag(s) +0.80 V
Zn(s) → Zn2+(aq) + 2 e- +0.76 V
2 Ag+(aq) + Zn(s) → Zn2+(aq) + 2 Ag(s) +1.56 V
1 point earned for correct E°
c)
ΔG° = –nFE°
ΔG° = (–2 mol e-)(96,500 J V-1 mol-1)(+0.46 V)
ΔG° = – 89,000 J or – 89 kJ (units required)
1 point earned for n and E° in the correct equation
1 point earned for correct value and sign of ΔG°
d)
Ecell=E°–(RT/nF) lnQ =E°–(RT/nF) ln([Cu2+]/[Ag+] 2) = Eo−(0.0592/n)
log([Cu2+]/[Ag+] 2)
Note: Q must include only ion concentrations
8.314 J mol -1 · K -1 · 298 K  0.045 
ln 
Ecell = +0.46 V –

2 mol e- · 96500 J V-1 · mol -1  [0.010]2 
Ecell = +0.46 V – 0.0128 V ln 450 = +0.46 V – 0.0128 V · 6.11 = +0.46 V –
0.0782 V
Ecell = +0.38 V
1 point earned for correct substitution
37
1 point earned for correct answer
e)
Ecell = +0.38 V. The cell potential under the non-standard conditions in
part (d) is positive. Therefore the reaction is spontaneous under the conditions
stated in part (d). A correct reference (from answer in part (d)) to a negative ΔG
(not ΔG°) is acceptable. If no answer to (d) is given, students must make an
assumption or a general statement about Ecell, not E°.
1 point earned for correct answer and correct explanation
2001 # 7
Answer the following questions that refer to the galvanic cell shown in the diagram
below. (A table of standard reduction potentials is printed on the green insert and on
page 4 of the booklet with the pink cover.)
a.
Identify the anode of the cell and write the half-reaction that occurs there.
b.
Write the net ionic equation for the overall reaction that occurs as the cell operates
and calculate the value of the standard cell potential, Eocell.
c.
Indicate how the value of Ecell would be affected if the concentration of
Ni(NO3)2(aq) was changed from 1.0 M to 0.10 M and the concentration of
Zn(NO3)2(aq) remained at 1.0 M. Justify your answer.
d.
Specify whether the value of Keq for the cell reaction is less than 1, greater than 1,
or equal to 1. Justify your answer.
(8 points)
(a)
The anode is the electrode on the right (Zn is the anode)
· Point is also earned if the student points to the Zn cell in the diagram.
The half-reaction is Zn → Zn2+ + 2 e(b)
Zn + Ni2+ → Zn2+ + Ni
1 point
1 point
1 point
38
Eocell = (-0.25 V) - (-0.76 V) = 0.51 V
1 point
· Some work must be shown to support the answer.
(c)
Ecell would decrease
1 point
2+
Since Ni is a reactant, a decrease in its concentration decreases the driving
force for the forward reaction
1 point
or
[Zn 2  ]
Ecell = E° - (RT/n) ln Q , where Q =
[Ni2  ]
Decreasing [Ni2+] would increase the value of Q, so a larger number would be
subtracted from E°, thus decreasing the value of Ecell.
(d)
K>1
1 point
E° is positive, so K > 1
1 point
Note: The student’s score in part (d) is based on the sign of Eocell calculated in
part (b).
Note on Overall Question: If in part (a) a student incorrectly identifies Ni as
being oxidized, partial credit is earned if subsequent parts are followed through
consistently.
2003B # 6
Answer the following questions about electrochemistry.
a.
Several different electrochemical cells can be constructed using the materials
shown below. Write the balanced net-ionic equation for the reaction that occurs
in the cell that would have the greatest positive value of Ecello.
b.
Calculate the standard cell potential, Ecello, for the reaction written in part a.
c.
A cell is constructed based on the reaction in part a above. Label the metal used
for the anode on the cell shown in the figure below.
39
d.
Of the compounds, NaOH, CuS, and NaNO3, which one is appropriate to use in a
salt bridge? Briefly explain your answer, and for each of the other compounds,
include a reason why it is not appropriate.
e.
Another standard cell is based on the following reaction.
Zn + Pb2+  Zn2+ + Pb
If the concentration of Zn2+ is decreased from 1.0 M to 0.25 M, what effect does
this have on the cell potential? Justify your answer.
9 points
a.
b.
c.
d.
e.
Al(s) → Al3+ + 3 eCu2+ + 2 e- → Cu(s)
2 Al + 3 Cu2+ → 2 Al3+ + 3 Cu
1 point for selection of correct two redox couples
1 point for correctly balanced net ionic equation
Al3+ + 3 e- → Al
Eo = -1.66 V
Cu2+ + 2 e- → Cu
Eo = +0.34 V
Ecell = Ecathode – Eanode = +0.34 V – (-1.66 V) = +2.00 V
1 point
The metal is aluminum solid
1 point
NaOH is not appropriate. The anion, OH-, would migrate towards the
anode. The OH- would react with the Al3+ ion in solution
CuS is not appropriate. It is insoluble in water, so no ions would be
available to migrate to the anode and cathode compartment to balance the
charge.
NaNO3 is appropriate. It is soluble in water, and neither the cation or the
anion will react with the ions in the anode or cathode compartment.
1 point for correctly indicating whether each compound is appropriate,
along with an explanation (3 points total)
Ecell = Ecello – 0.059 ln ([Zn2+] / [Pb2+])
40
If [Zn2+] is reduced, then the ratio is < 1, therefore ln (ratio) <0, and Ecell
increases (becomes more positive)
1 point for correctly indicating how Ecell is affected
1 point for explanation in term of Nernst equation and Q
41
AP Chemistry Concepts - THERMODYNAMICS
1.
∆H0 rxn = ∑ ∆ Hf 0Products - ∑∆ Hf0 Reactants
= ∑ Bond Energy Reactants - ∑ Bond energy Products
∆Hrxn - exothermic
2.
∆S0 rxn = ∑ Sf0 Products - ∑ Sf0 Reactants
∆S0rxn - ordered
3.
∆Hrxn + endothermic
∆S0rxn
+ disordered
∆G0 rxn = ∆H0rxn - T ∆S 0rxn
∆G0rxn - spontaneous
4.
∆G0rxn = - RT ln Q
5.
∆G0 rxn = - nF E0
Q = Keq
∆G0rxn
+ nonspontaneous
free energy and equilibrium
free energy and electrochemistry
F = 96,500 coulombs / mole electrons
Faraday’s constant
6.
Phase diagrams
7.
∆H rxn = q = m ( c ) ( ∆T )
42
1998 # 3
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according
to the equation above, 64.98 kilojoules of heat is released. Use the information in the
table below to answer the questions that follow.
Substance
C(graphite)
CO2(g)
H2(g)
H2O(l)
O2(g)
C6H5OH(s)
Standard Heat of Formation,
ΔH°f, at 25°C (kJ/mol)
0.00
-395.5
0.00
-285.85
0.00
?
Absolute Entropy, S°, at 25°C (J/molK)
5.69
213.6
130.6
69.91
205.0
144.0
a.
Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C.
b.
Calculate the standard heat of formation, ΔH°f, of phenol in kilojoules per mole at
25°C.
c.
Calculate the value of the standard free-energy change, ΔG° for the combustion
of phenol at 25°C.
d.
If the volume of the combustion container is 10.0 liters, calculate the final
pressure in the container when the temperature is changed to 110°C. (Assume no
oxygen remains unreacted and that all products are gaseous.)
9 points
1 mol
 0.02125 mol phenol
1 point
94.113 g
64.98 kJ
 3,058 kJ mol 1
Heat released per mole 
0.02125 mol
Or, ΔHcomb = -3058 kJ mol-1
1 point
Units not necessary
b)
ΔHcomb = -3058 kJ mol-1
1 point
o
-3058 kJ = [6(-395.5) + 3(-285.85)] – [ΔHf (phenol)]
1 point
ΔHfo (phenol) = -161 kJ
1 point
One point earned for correct sign of heat of combustion, one point for correct use
of moles / coefficients, and one point for correct substitution
c)
ΔSo = [3(69.91) + 6(213.6)] [7(205.0) + 144.0] = -87.67 J!K
1 point
o
o
o
-1
ΔG = ΔH - TΔS = 3058 kJ – (298 K)(-0.08767 kJ K ) = -3032 kJ
1 point
a)
2.000 g 
43
Units not necessary; no penalty if correct except for wrong ΔH comb for part a
d)
moles gas = 9 × [moles from part a] = 9 (0.02125 mol) = 0.1913 moles gas
nRT (0.1913 mol)(0.0821 L atm mol 1 K 1 )(383 K )
P

 0.601 atm 1 point
V
10.0 L
Units necessary; no penalty for using Celcius temperature if also lost point in part c for
same error
1996 #3
C2H2(g) + 2 H2(g) → C2H6(g)
Information about the substances
Substance
S° (J/mol K)
(kJ/mol)
ΔH°f (kJ/mol)
Bond
Bond Energy
C2H2(g)
H2(g)
C2H6(g)
226.7
0
-84.7
C-C
C=C
C-H
H-H
347
611
414
436
200.9
130.7
--------
a.
If the value of the standard entropy change, ΔS°, for the reaction is -232.7 joules
per mole Kelvin, calculate the standard molar entropy, S°, of C2H6 gas.
b.
Calculate the value of the standard free-energy change, ΔG°, for the reaction.
What does the sign of ΔG° indicate about the reaction above?
c.
Calculate the value of the equilibrium constant, K, for the reaction at 298 K.
d.
Calculate the value of the C C bond energy in C2H2 in kilojoules per mole.
(a) two points; one for line of answer
- 232.7 J/K = S° (C2H6) - (261.4 + 200.9) J./K
S° (C2H6) = 229.6 J/K
units ignored; 1 point earned for 98.9 J/K; 1 point lost if stoichiometry is not
implied in process
(b) three points total; one point each portion; any value for T (e.g., 273 K or 298
K) is allowable:
ΔH° = (- 84.7 kJ) - (226.7 kJ) = -311.4kJ
= - 311.4kJ - (298 K) (- 0.2327kJ/K)
= - 311.4 kJ + 69.3 kJ
= - 242.1 kJ
Negative ΔG° therefore reaction is spontaneous, or Keq > 1 therefore reaction is
spontaneous, or products are favored at equilibrium.
44
(c) two points
ln K = 242.1 ÷ [(8.31 x 10¯3) (298)] = 97.7
K = 3 x 1042 (1,2,or 3 significant figures acceptable)
(d) two points; first point earned for correct substitution and correct number of
bonds, second point earned for setting equal to ΔHrxn and correct calculation of
answer; no points earned for "extrapolation" techniques to find carbon-carbon
triple bond energy; E* is the energy of the carbon-carbon triple bond.
- 311.4 kJ = [(2) (436) + E* + (2) (414)] - [(347) + (6) (414)]
E* = 820 kJ
1997 #7
For the gaseous equilibrium represented below, it is observed that greater amounts of
PCl3 and Cl2 are produced as the temperature is increased.
PCl5(g)  PCl3(g) + Cl2(g)
a.
What is the sign of ΔS° for the reaction? Explain.
b.
What change, if any, will occur in ΔG° for the reaction as the temperature is
increased. Explain your reasoning in terms of thermodynamic principles.
c.
If He gas is added to the original reaction mixture at constant volume and
temperature, what will happen to the partial pressure of Cl2? Explain.
d.
If the volume of the original reaction is decreased at constant temperature to half
the original volume, what will happen to the number of moles of Cl2 in the
reaction vessel? Explain.
(a) S° is positive (or "+", or ">0") 1 point
Moles products > moles reactants 1 point
Note; all species are gaseous, so (g) need not be indicated. To earn credit,
number of particles (moles) must be discussed. No explanation point earned for
just nothing that disorder increases, or that PCl5 is decomposing or dissociating.
(b) G° will decrease (or become more negative, or become smaller). 1 point
G° = H° - TS°
and since S° is positive, TS° is positive ( > 0). Thus increasing T will result in
a larger term being subtracted from H°, or, G° = -RT ln K and K is going up
in value since T is increasing.)
Note: full credit earned for part (b) if:
S° < 0 in part (a) which leads to G° is increasing because TS° is added to
H°, or,
S° = 0 in part (a) which leads to no change in G°
(c) no change (one point)
45
PHe is not part of the a) reaction (He is not involved) or, b) law of mass action or,
c) reaction quotient or, d) equilibrium constant expression; one point
hence altering PHe has no effect on the position at equilibrium
(d) moles of Cl2 will decrease (one point)
The decrease in volume leads to an increase in pressure (concentration),
therefore the reaction shifts to the left because:
(one point for any of the following)
Q > Ksp (Q > Kc, or,
the rate of the reverse reaction increase more than the rate of the forward
reaction, or,
the reaction shifts toward the lesser moles of gas.
Note: "LeChatelier's principle" alone is not sufficient to earn the explanation
point. If response suggests that the number of moles of Cl2 is halved because the
system is "cut" in half, only one point is earned.
1999 # 6
Answer the following questions in terms of thermodynamic principles and concepts of
kinetic molecular theory.
a.
Consider the reaction represented below, which is spontaneous at 298 K.
CO2(g) + 2 NH3(g) → CO(NH2)2(s) + H2O(l); ΔH°298 = -134 kJ
i. For the reaction, indicate whether the standard entropy change, ΔS°298, is
positive, or negative, or zero. Justify your answer.
ii. Which factor, the change in enthalpy, ΔH°298, or the change in entropy,
ΔS°298, provides the principal driving force for the reaction at 298 K? Explain.
iii. For the reaction, how is the value of the standard free energy change, ΔG°,
affected by an increase in temperature? Explain.
b.
Some reactions that are predicted by their sign of ΔG° to be spontaneous at room
temperature do not proceed at a measurable rate at room temperature.
i.
Account for this apparent contradiction.
ii.
A suitable catalyst increases the rate of such a reaction. What effect does
the catalyst have on ΔG° for the reaction? Explain.
(a)(i) ΔS° is negative (−) OR ΔS° < 0 OR entropy is decreasing.
1 point
3 moles of gaseous particles converted to 2 moles of solid/liquid. 1 point
• One point earned for correct identification of (−) sign of ΔS°
46
• One point earned for correct explanation (mention of phases is crucial
for point)
• No point earned if incorrect ΔS° sign is obtained from the presumed
value of ΔG°
(ii)
ΔH° drives the reaction.
1 point
The decrease in entropy (ΔS° < 0) cannot drive the reaction, so the
decrease in enthalpy (ΔH° < 0) MUST drive the reaction.
OR
ΔG° = ΔH° − TΔS°; for a spontaneous reaction ΔG° < 0, and a negative
value of ΔS° causes a positive ΔG°.
1 point
• One point earned for identifying ΔH° as the principal driving force for
the reaction
• One point earned for correct justification
• Justification point earned by mentioning the effects of changes in entropy
and enthalpy on the spontaneity of the reaction OR by a mathematical
argument using the Gibbs-Helmholtz equation and some implication about
the comparison between the effects of ΔS° and ΔH°
(iii) Given that ΔG° = ΔH° − TΔS° and ΔS° < 0, an increase in temperature
causes an increase in the value of ΔG° (ΔG° becomes less negative).
1 point
• One point earned for the description of the effect of an increase in
temperature on ΔS° and consequently on ΔG°
• No point earned for an argument based on Le Châtelier.s principle
(b)(i) The reaction rate depends on the reaction kinetics, which is determined by
the value of the activation energy, Eact. If the activation energy is large, a
reaction that is thermodynamically spontaneous may proceed very slowly
(if at all).
1 point
• One point earned for linking the rate of the reaction to the activation
energy, which may be explained verbally or using a reaction profile
diagram
(ii)
The catalyst has no effect on the value of ΔG°.
1 point
The catalyst reduces the value of Eact, increasing the rate of reaction, but
has no effect on the values of ΔH° and ΔS°, so it cannot affect the
thermodynamics of the reaction.
1 point
• One point earned for indicating no change in the value of ΔG°
• One point earned for indicating (verbally, or with a reaction-profile
diagram) that the catalyst affects the activation energy
2002 # 8
Carbon (graphite), carbon dioxide, and carbon monoxide form an equilibrium mixture, as
represented by the equation
C(s) + CO2(g)  2CO(g)
47
a.
Predict the sign for the change in entropy, S, for the reaction. Justify your
prediction.
b.
In the table below are data that show the percent of CO in the equilibrium mixture
at two different temperatures. Predict the sign for the change in enthalpy, H, for
the reaction. Justify your prediction.
Temperature
700 oC
850 oC
% CO
60
94
c.
Appropriate complete the potential energy diagram for the reaction by finishing
the curve on the graph below. Also, clearly indicate H for the reaction on the
graph.
d.
If the initial amount of C(s) were doubled, what would be the effect on the percent
of CO in the equilibrium mixture? Justify your answer.
a)
ΔS = +; There is more disorder in a gas than in a solid, so the product is
more disordered than the reactants. The change in entropy is therefore positive.
OR
There is 1 mole of gas in the reactants and 2 moles of gas in the product.
1 point earned for indicating that ΔS is positive
1 point earned for explanation
b)
ΔH = +; More CO at the higher temperature indicates that the reaction
shifts to the right with increasing temperature. For this to occur, the reaction
must be endothermic.
48
1 point earned for indicating that ΔH is positive
1 point earned for explanation
c)
1 point earned for completing the graph according to the information in
part (b)
1 point earned for appropriately labeling ΔHrxn for the reaction as drawn
d)
An increase in the amount of C(s) has no effect. Solids do not appear in
the equilibrium expression, so adding more C(s) will not affect the percent of CO
in the equilibrium
mixture.
1 point earned for indicating no effect
1 point earned for explanation
Note: Since the question asks about “percent of CO” a student might think of %
by mass or % by mole.
Adding carbon will not shift the equilibrium, so P(CO) and P(CO2) stay the same.
The % CO then decreases, because now there are more total moles in the system:
% CO = nCO/(nCO + nCO2 + nC). As nC is raised, the denominator increases,
and % CO decreases.
49
AP Chemistry Concepts - ATOMIC THEORY, BONDING AND
INTERMOLECULAR FORCES
1.
Quantum Numbers ,electron configurations, Hund’s rule, orbital diagrams
2.
ionic bonds
3.
Covalent bonds, Lewis structures, geometric shapes, bond polarity, molecular
polarity, resonance, hybridization
4.
Trends of the periodic table - a) size for atoms b) size of ions, c) IE, EA, EN
5.
Effective nuclear charge (Zeff ) increases as more protons added to same
energy level Zeff is a comparison tool.
6.
Effective nuclear charge (Zeff ) decreases as more shielding electrons are
present.
7.
Intermolecular Forces (IMF) are between molecules and help explain
differences in FP, BP, solids, liquids, gases, and solubilities.
a. ion – ion
b. dipole – dipole with H bonding
c. dipole – dipole
d. London dispersion forces ( LDF )
8.
When students talk about EN differences they are talking about bonds (within a
molecule) , we need them to talk about IMF (between molecules )
9.
Students often talk about atoms “wanting to gain/lose electrons”, being happy,
Full, rather than having a stable octet, complete energy level.
50
1996 # 9
Explain each of the following in terms of the electronic structure and/or bonding of the
compounds involved.
a.
At ordinary conditions, HF (normal boiling point = 20°C) is a liquid, whereas HCl
(normal boiling point = -114°C) is a gas.
b.
Molecules of AsF3 are polar, whereas molecules of AsF5 are nonpolar.
c.
The N-O bonds in the NO2¯ ion are equal in length, whereas they are unequal in
HNO2.
d.
For sulfur, the fluorides SF2, SF4, and SF6 are known to exist, whereas for oxygen
only OF2 is known to exist.
(a) two points
Hydrogen bonding (or dipole-dipole attraction) in HF is greater than it is in HCl
Note: only one point earned if simply stated that HF has greater intermolecular
forces than HCl
(b) two points
AsF3 has a trigonal pyramid shape and bond dipoles do NOT cancel (or
asymmetric molecule)
AsF5 has a trigonal bipyramid shape and bond dipoles cancel (or symmetric
shape)
Notes: explanation must refer to shape in order to earn point; one point earned if
only correct Lewis structures are given.
(c) two points
NO2¯ has resonance structures
HNO2 has no resonance structures
OR
one N-O single bond, one N=O double bond
Note: one point earned if only correct Lewis structures, including resonance for
NO2¯ given.
(d) two points
Sulfur uses d orbitals (or expanded octet), oxygen has no d orbitals in its valence
shell
OR
Sulfur is a larger atom, can accomodate more bonds.
51
1997 #5
Consider the molecules PF3 and PF5.
a.
Draw the Lewis electron-dot structures for PF3 and PF5 and predict the molecular
geometry of each.
b.
Is the PF3 molecular polar, or is it nonpolar? Explain.
c.
On the basis of bonding principles, predict whether each of the following
compounds exists. In each case, explain your prediction.
i.
ii.
NF5
AsF5
PF3
PF5
(Trigonal) pyramid(al)
(Trigonal) bipyramid(al)
1 point for each structure
Note ; One point (total) deducted if lone pairs not shown on F atoms in either
molecule.
(b) The PF3 molecule is polar
The three P-F dipoles do not cancel, or,
the lone pair on P leads to asymmetrical distribution of charge.
Note; "Molecule is not symmetrical" does not earn point. Both points can be
earned if answer is consistent with incorrect (a).
(c) NF5 does not exist because no 2d orbitals exist for use in bonding, or,
N is too small to accommodate 5 bonding pairs
AsF5 does exist because 4d orbitals are available for use in bonding, or,
As can accommodate an expanded octet using d orbitals
Note; Response with two correct predictions with no explanations earns one
point. Also, argument of "no expanded octet" vs. "expanded octet" alone does not
earn expalnation point
1999 # 8
Answer the following questions using principles of chemical bonding and molecular
structure.
a.
Consider the carbon dioxide molecule, CO2 , and the carbonate ion, CO32-.
i.
Draw the complete Lewis electron-dot structure for each species.
ii.
Account for the fact that the carbon-oxygen bond length in CO32- is greater
than the carbon-oxygen bond length in CO2.
52
b.
Consider the molecules CF4 and SF4.
i.
Draw the complete Lewis electron-dot structure for each molecule.
ii.
In terms of molecular geometry, account for the fact that the CF4 molecule
is nonpolar, whereas the SF4 molecule is polar.
1997 # 6
Explain each of the following observations using principles of atomic structure and/or
bonding.
a.
Potassium has a lower first-ionization energy than lithium.
b.
The ionic radius of N3- is larger than that of O2-.
c.
A calcium atom is larger than a zinc atom.
d.
Boron has a lower first-ionization energy than beryllium.
a) Response must contain a cogent discussion of the forces between the nucleus
and the outermost (or "ionized") electron. For example, a discussion of "the
outermost electron on K..." should include one of the following:
i. it is farther from nucleus than the outermost electron on Li
ii. it is more shielded from the nucleus (or "experiences a lower effective
nuclear charge") than the outermost electron on Li
iii. it is in a higher energy orbital (4s) than tne outermost electron on Li (2s)."
2 points for any one
Notes:"K is larger than Li" earns 1 point.
No points earned for "K electron is easier to remove" (or some other
restatement).
b) Nitrogen has one less proton than oxygen 1 point
Nitride and oxide ions are isoelectronic 1 point
or,
In nitride ion the electron/proton ratio is greater, causing more repulsion; thus,
nitride is the larger ion. 2 points
c) A Zn atom has more protons (10 more) than an atom of Ca 1 point
Electrons in d orbitals of Zn have a lower principal quantum number; thus, they
are not in orbitals that are farther from the nucleus. 1 point
d) Correct identification of the orbitals involved (2s versus 2p) 1 point
Clear statement that the two orbitals have different energies 1 point
Note: Arguments that "the 2p orbital is farther out than the 2s orbital", or that
"the Be atom has a filled subshell, which is a more stable configuration" earn no
explanation point.
General note:For all parts (a) through (d), discussions of position in the periodic
table earn no points.
53
2000 # 7
Answer the following questions about the element selenium, Se (atomic number 34).
a.
Samples of natural selenium contain six stable isotopes. In terms of atomic
structure, explain what these isotopes have in common, and how they differ.
b.
Write the complete electron configuration (e.g., 1s2 2s2 … etc.) for a selenium
atom in the ground state. Indicate the number of unpaired electrons in the
ground-state atom, and explain your reasoning.
c.
In terms of atomic structure, explain why the first ionization energy of selenium is
d.
i.
less than that of bromine (atomic number 35), and
ii.
greater than that of tellurium (atomic number 52).
Selenium reacts with fluorine to form SeF4. Draw the complete Lewis electrondot structure for SeF4 and sketch the molecular structure. Indicate whether the
molecule is polar or nonpolar, and justify your answer.
(8 points)
(a)
The isotopes have the same number (34) of protons,
1 point
but a different number of neutrons.
1 point
• No comment about the number of electrons is necessary
(b)
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4
or
1 point
2
2
6
2
6
10
2
4
1s 2s 2p 3s 3p 3d 4s 4p
• No point is earned for [Ar] 4s2 3d10 4p4, because the question specifically asks
for a complete electron configuration.
Since there are three different 4p orbitals, must be two unpaired electrons. 1 point
Notes: The second part should have some explanation of Hund’s rule, and may
include a diagram. The second point can still be earned even if the first point is
not IF the electron configuration is incorrect, but the answer for the second part
is consistent with the electron configuration given in the first part.
(c) (i) The ionized electrons in both Se and Br are in the same energy level, but Br
has more protons than Se, so the attraction to the nucleus is greater.
1 point
Note: There should be two arguments in an acceptable answer -- the electrons
removed are from the same (4p) orbital and Br has more protons (a greater
nuclear charge) than Se.
(ii) The electron removed from a Te atom is in a 5p orbital, while the electron
removed from an Se atom is in a 4p orbital. The 5p orbital is at a higher energy
than the 4p orbital, thus the removal of an electron in a 5p orbital requires less
energy.
1 point
(d)
Figure showing the see-saw structure of SF4.
1 point
54
Notes: One point earned for a correct Lewis diagram and a sketch. The Lewis
diagram and the molecular structure may be combined into one sketch if both
aspects (electron pairs and structure) are correct. Dots, lines, or a mixture of
both can be used in the Lewis diagram. The lone pair of electrons need not be
shown in the sketch -- just the atomic positions. No credit earned for just a verbal
description of molecular geometry (“see-saw”, “saw-horse”, or something
“distorted”), because the question clearly asks the student to “sketch the
molecular structure”.
The SeF4 molecule is polar, because the polarities induced by the bonds and the
lone pair of electrons do not cancel.
1 point
2003 # 8
Using the information in the table, answer the following questions about organic
compounds.
a.
b.
Compound Name
Compound Formula
ΔHvapo (kJ mol-1)
Propane
Propanone
1-propanol
CH3CH2CH3
CH3COCH3
CH3CH2CH2OH
19.0
32.0
47.3
For propanone,
i.
draw the complete structural formula (showing all atoms and
bonds)
ii.
predict the approximate carbon-to-carbon-to-carbon bond angle.
For each pair of compounds below, explain why they do not have the same value
for their standard heat of vaporization, ΔHvapo. (You must include specific
information about both compounds in each pair.)
i.
propane and propanone
ii.
propanone and 1-propanol
c.
Draw the complete structural formula for an isomer of the molecule you drew in
part a. i.
d.
Given the structural formula for propyne below,
55
8 points
a.
b.
c.
d.
i.
indicate the hybridization of the carbon atom indicated by the
arrow in the structure above;
ii.
indicate the total number of sigma (σ) bonds and the total number
of pi (π) bonds in the molecule.
i)1 point for complete and correct structural formula (3 carbon chain,
with ketone group on middle carbon)
ii)The C-C-C bond angle is 120o
1 point
i) The intermolecular attractive forces in propane are dispersion forces
only. The IMF’s in propanone are dispersion and dipole-dipole. Since
the IMF’s differ in the two substances, the enthalpy of vaporization will
differ
1 points for correctly identifying the IMF’s for each substance
ii) The IMF’s in 1-propanol are dispersion forces and hydrogen bonding.
The IMF’s in propanone are dispersion and dipole-dipole. Since the
IMF’s differ in the two substances, the enthalpy of vaporization will differ
1 point for correctly identifying the IMF’s for each substance
1 point for a correct, complete structural formula with a 3 carbon chain
and a COOH group at one end.
i) sp hybridization
1 point
ii) 6 sigma bonds, 2 pi bonds
1 point for correct number of sigma bonds, 1 point for correct number of
pi bonds
2007 B #6
First Ionization
Energy (kJ mol-1)
Element 1
Element 2
Element 3
Element 4
1251
496
738
1000
Second Ionization
Energy (kJ mol-1)
2300
4560
1450
2250
Third Ionization
Energy (kJ mol-1)
3820
6910
7730
3360
The table above shows the first three ionization energies for atoms of four elements from
the third period of the periodic table. The elements are numbered randomly. Use the
information in the table to answer the following questions.
56
a.
Which element is most metallic in character? Explain your reasoning.
b.
Identify element 3. Explain your reasoning.
c.
Write the complete electron configuration for an atom of element 3.
d.
What is the expected oxidation state for the most common ion of element 2?
e.
What is the chemical symbol for element 2?
f.
A neutral atom of which of the four elements has the smallest radius?
8 points
a.
b.
c.
d.
e.
f.
Element 2. It has the lowest first ionization energy. Metallic elements
lose electron(s) when they become ions, and element 2 requires the least
amount of energy to remove an electron
1 point for the identification, 1 point for the justification
Magnesium. Element 3 has low first and second ionization energies
relative to the third ionization energy, indicating that the element has two
valence electrons, which is true for magnesium. (The third ionization of
element 3 is dramatically higher, indicating the removal of an electron
from a noble gas core)
1 point for the identification, 1 point for the justification
1s2 2s2 2p6 3s2; 1 point for the correct electron configuration
1+
1 point
Na
1 point
Element 1
1 point
57
AP Chemistry Concept List – CONCENTRATION UNITS OF
SOLUTIONS / COLLIGATIVE PROPERTIES
1.
Molarity M = mole of solute/ L of solution
2.
molality m = mole of solute / Kg of solvent
3.
% by volume = volume of solute / total volume of solution
4.
% by weight = weight of solute / total weight of solution
5.
mole fraction = xa = mole of a /total moles in solution
Colligative Properties
1.
∆ FP ↓ = (kf ) ( m ) ( i ) freezing point depression
2.
∆ BP ↑ = ( kb ) (m ) ( i ) boiling point elevation
3.
∏ = ( M ) ( R ) ( T ) ( i ) osmotic pressure
4.
Vapor Pressure Lowering = VPL = (x solvent) VP pure solvent
The main use of colligative properties is to find the molecular weight of an unknown
compound, thus it is related to problems in earlier chapters about empirical or molecular
formulas.
i = Van’t Hoff factor
for organic solutes nonelectrolytes
i=1
for electrolytes i = 2,3,4… NaCl i = 2
AlCl3 i = 4
H2SO4 i = 3
58
1998# 2
An unknown compound contains only the three elements C,H, and O. A pure sample of
the compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass.
a.
Determine the empirical formula of the compound.
b.
A solution of 1.570 grams of the compound in 16.08 grams of camphor is
observed to freeze at a temperature 15.2 Celsius below the normal freezing
point of pure camphor. Determine the molar mass and apparent molecular
formula of the compound. (The molal freezing-point depression constant, kf,
for camphor is 40.0 kg-K-mol-1.)
c.
When 1.570 grams of the compound is vaporized at 300 °C and 1.00
atmosphere, the gas occupies a volume of 577 milliliters. What is the molar
mass of the compound based on this result?
d.
Briefly describe what occurs in solution that accounts for the difference
between the results obtained in parts (b) and (c).
Assume a 100 – gram sample (not necessary for credit):
1 mol C
65 g C 
 5.462 mol C
12.01 g C
1 mol H
9.44 g H 
 9.366 mol H
1.0079 g H
Mass O = [100 – (65.60 + 9.44)] = 24.96 g O
1 mol O
24.96 g O 
 1.560 mol O
15.9994 g O
C5.462H9.366O1.560 --> C3.5H6.0O1.0 --> C7H12O2
One point earned for determining moles of C and moles of H
One point earned for determining moles of O
One point earned for correct empirical formula
15.2 o C
T
m

 0.380 mol kg 1
b)
1
Kf
40.0 K kg mol
0.380 mol
0.01608 kg 
 0.00611 mol
1 kg
1.570 g
molar mass 
 257 g mol 1
0.00611 mol
One point earned for determination of molality
One point earned for conversion of molality to molar mass
OR
T  kg solvent
moles solute 
 0.00611 mol
Kf
a)
1 point
1 point
1 point
1 point
1 point
1 point
59
molar mass 
1.570 g
 257 g mol 1
0.00611 mol
1 point
OR
molar mass  mass 
Kf
T  kg solvent
 257 g mol 1
2 pts
Empirical mass of C7H12O2 = 7(12) + 12(1) + 2(1) = 128 g mol-1
128 g mol-1 = ½ molar mass --> molecular formula = 2 × (empirical
formula)--> molecular formula = C14H24O4
1 point
One point earned if molecular formula is wrong but is consistent with
empirical formula and molar mass
No penalty for simply ignoring the van’t Hoff factor
Only one point earned for part b if response indicates that
ΔT = (15.2 + 273) = 288 K and molar mass = 13.6 g mol-1
1 atm0.577 L 
PV
n

 0.0123 mol
c)
1 point
RT 0.0821 L atm mol 1 K 1 573 K 
mass of sample
1.570 g
molar mass 

 128 g mol 1
1 point
moles of sample 0.0123 mol
Only one point can be earned for part c if wrong value of R is used and / or T
is not converted from C to K
d)
The compound must form a dimer in solution, because the molar mass
in solution is twice that it is in the gas phase,
OR,
The compound must dissociate in the gas phase (A(g) --> 2 B(g)) because the
molar mass in the gas phase is half that it is in solution
One point earned for a reference to either or both the ideas of dimerization
and dissociation
No point earned for a “non-ideal behavior” argument
1999 # 7
Answer the following questions, which refer to the 100 mL samples of aqueous solutions
at 25°C in the stoppered flasks shown below.
60
a.
Which solution has the lowest electrical conductivity? Explain.
b.
Which solution has the lowest freezing point? Explain.
c.
Above which solution is the pressure of water vapor greatest? Explain.
d.
Which solution has the highest pH? Explain.
2001 # 5
Answer the questions below that related to the five aqueous solutions at 25 oC shown
below.
a.
Which solution has the highest boiling point? Explain.
b.
Which solution has the highest pH? Explain.
c.
Identify a pair of the solutions that would produce a precipitate when mixed
together. Write the formula of the precipitate.
d.
Which solution could be used to oxidize the Cl-(aq) ion? Identify the product of
the oxidation.
e.
Which solution would be the least effective conductor of electricity? Explain.
(10 points)
61
In each part, one point is earned for the correct solution or solutions, and one
point is earned for the correct explanation (in parts a, b, and e), precipitate (in
part c), or product (in part d).
(a)
Pb(NO3)2 (Solution 1)
1 point
Pb(NO3)2 has the largest value of i , the van’t Hoff factor, so the solution has the
highest number of solute particles (it dissociates into the most particles). 1 point
· Student must address the relative number of particles.
(b)
KC2H3O2 (Solution 5)
1 point
The acetate ion is the conjugate base of a weak acid, so it is a weak base (or
KC2H3O2 is the salt of a strong base and a weak acid, so the solution is basic).
1 point
(c)
Pb(NO3)2 and NaCl (Solutions 1 and 2)
1 point
PbCl2 would precipitate
1 point
· Points can also be earned for KMnO4 plus one of the other solutions
(with the precipitation of MnO2).
· Points can also be earned for KMnO4 plus Pb(NO3)2
(with the precipitation of PbO2, or MnO2).
(d)
KMnO4 (Solution 3)
1 point
The product of the oxidation is Cl2
1 point
(e)
C2H5OH (Solution 4)
1 point
Ethanol is the only nonelectrolyte given. It does not readily dissociate into ions,
so it would not produce charged species that would conduct a current.
1 point
· One point can also be earned for explanations using i, the van’t Hoff factor.
62
AP Chemistry – LABORATORY QUESTIONS
2007 #5
5.
5 Fe2+(aq) + MnO4-(aq) + 8H+(aq)  5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
The mass percent of iron in a soluble iron(II) compound is measured using a titration
based on the balanced equation above.
a.
What is the oxidation number of manganese in the permanganate ion, MnO4-(aq)?
b.
Identify the reducing agent in the reaction represented above.
The mass of a sample of the iron(II) compound is carefully measured before the sample is
dissolved in distilled water. The resulting solution is acidified with H2SO4(aq). The
solution is then titrated with MnO4-(aq) until the end point is reached.
c.
Describe the color change that occurs in the flask when the end point of the
titration has been reached. Explain why the color of the solution changes at the
end point.
d.
Let the variables g, M, and V be defined as follows:
g = the mass, in grams, of the sample of the iron(II) compound
M= the molarity of the MnO4-(aq) used as the titrant
V= the volume, in liters, of MnO4-(aq) added to reach the end point
In terms of these variables, the number of moles of MnO4-(aq) added to reach the
end point of the titration is expressed as M x V. Using the variables defined
above, the molar mass of iron (55.85 g mol-1), and the coefficients in the balanced
chemical equation, write the expression for each of the following quantities.
e.
i.
The number of moles of iron in the sample
ii.
The mass of iron in the sample, in grams
iii.
The mass percent of iron in the compound
What effect will adding too much titrant have on the experimentally determined
value of the mass percent of iron in the compound? Justify your answer.
9 points
a.
+7
1 point
63
b.
c.
d.
e.
Fe2+(aq)
1 point
The solution in the flask changes from colorless to faint purple pink at the
endpoint of the titration. At the endpoint there is no Fe2+(aq) left in the
flask to reduce the colored permanganate ion, so when a small amount of
permanganate ion is added after the endpoint, the unreacted
permanganate ion present in the solution colors the solution faint purple /
pink.
1 point is earned for stating that a faint pink color appears (unless
indication of acid-base reaction)
1 point is earned for a correct explanation involving excess MnO4- after
all Fe2+ has reacted
i) mol Fe2+ = 5 × M × V
1 point
2+
2+
OR mol Fe = (5 mol Fe /1 mol MnO4 ) M V
ii) mass Fe = 5 × M × V × 55.85
1 point
2+
OR mass Fe = mol Fe × 55.85
iii) mass % Fe = (5 × M × V × 55.85) / g × 100
2 point
OR mass % Fe = mass Fe / g × 100
The experimentally determined mass percent of iron in the compound will
be too large. V is too large --> expression in d)iii) above is too large
1 point
2007 B #5
Answer the following questions about laboratory situations involving acids, bases, and
buffer solutions.
a.
Lactic acid, HC3H5O3, reacts with water to produce an acidic solution. Shown
below are the complete Lewis structures of the reactants.
In the space provided above, complete the equation by drawing the complete
Lewis structures of the reaction products.
b.
Choosing from the chemicals and equipment listed below, describe how to
prepare 100.00 mL of a 1.00 M aqueous solution of NH4Cl (molar mass 53.5 g
mol-1). Include specific amounts and equipment where appropriate.
NH4Cl(s)
50 mL buret
Distilled water 100 mL beaker
100 mL graduated cylinder
100 mL volumetric flask
100 mL pipet
Balance
64
c.
Two buffer solutions, each containing acetic acid and sodium acetate, are
prepared. A student adds 0.10 mol of HCl to 1.00 L of each of these buffer
solutions and to 1.0 L of distilled water. The table below shows the pH
measurements made before and after the 0.10 mol of HCl is added.
Distilled water
Buffer 1
Buffer 2
pH before
HCl added
pH after
HCl added
7.0
4.7
4.7
1.0
2.7
4.3
i.
Write the balanced net ionic equation for the reaction that takes place
when the HCl is added to buffer 1 or buffer 2.
ii.
Explain why the pH of buffer 1 is different from the pH of buffer 2 after
0.10 mol of HCl is added.
iii.
Explain why the pH of buffer 1 is the same as the pH of buffer 2 before
0.10 mol of HCl is added.
8 points
a.
b.
c.
1 point is earned for each correct structure. (2 total)
mass of NH4Cl = (0.100 L) (1.00 mol L-1) (53.5 g mol-1) = 5.35 g NH4Cl
- measure out 5.35 g NH4Cl using the balance
- use the 100 mL graduated cylinder to transfer approximately 25 mL of
distilled water to the 100 mL volumetric flask
- transfer the 5.35 g NH4Cl to the 100 mL volumetric flask
- continue to add distilled water to the volumetric flask while swirling the
flask to dissolve the NH4Cl and remove all NH4Cl particles adhered to the
walls
- carefully add distilled water to the 100 mL volumetric flask until the
bottom of the meniscus of the solution reaches the etched mark on the
flask
1 point is earned for the mass
1 point is earned for using a volumetric flask
1 point is earned for diluting to the mark
i) C2H3O2- + H3O+ --> HC2H3O2 + H2O
1 point
ii) Before the HCl was added, each buffer had the same pH and thus had
the same [H+]. Because Ka for acetic acid is a constant, the ratio of [H+]
to Ka must also be constant; this means that the ratio of [HC2H3O2] to
[C2H3O2-] is the same for both buffers, as shown by the following
equation, derived from the equilibrium constant expression for the
dissociation of acetic acid.
[HC 2 H3O2 ] [H 3O  ]

[C 2 H3O 2 ]
Ka
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After the addition of the H+, the ratio in buffer 1 must have been greater
than the corresponding ratio in buffer 2, as evidenced by their respective
pH values. Thus a greater proportion of the C2H3O2- in buffer 1 must
have reacted with the added H+ compared to the proportion that reacted
in buffer 2. The difference between these proportions means that the
original concentrations of HC2H3O2 and C2H3O2- had to be smaller in
buffer 1 than in buffer 2.
1 point is earned for a correct answer involving better buffering capacity
or relative amount of base (acetate ion)
iii) Both buffer solutions have the same acid to conjugate base mole ratio
in the formula below. Therefore, the buffers have the same [H+] and pH.
[ HC 2 H 3O2 ]
[H  ]  Ka
[C2 H 3O2 ]
1 point is earned for the correct answer involving the ratio of acid to base
in the buffer.
1997 # 9
An experiment is to be performed to determine the mass percent of sulfate in an unknown
soluble sulfate salt. The equipment shown above is available for the experiment. A drying
oven is also available.
a.
Briefly list the steps needed to carry out this experiment.
b.
What experimental data need to be collected to calculate the mass percent of
sulfate in the unknown?
c.
List the calculations necessary to determine the mass percent of sulfate in the
unknown.
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d.
Would 0.20 M MgCl2 be an acceptable substitute for the BaCl2 solution provided
for this experiment? Explain.
a) 2 points Mix unknown and BaCl2(aq) as reactants
Collect precipitate / set up filtration
b) 2 points Mass of unknown salt as reactant
(sulfate="salt"=unknown salt, unless otherwise specified)
Mass BaSO4 (must be specified) as dried precipitate/product
Note: "Dried" must appear to earn all 4 points for (a) and (b)
c) 2 points
Mass BaSO4 --> moles SO42¯ --> mass SO42¯ (to be used in) -----> mass SO4 2¯ /
mass unknown
Notes: A list alone is acceptable. Method, if correct, acceptable as list. Response
must clearly distinguish between SO42¯, BaSO4, and unknown sulfate. Only one of
two points earned if mass SO42¯ incorrect but fraction for percent clearly
indicates part (of original salt) / whole (of original salt).
d) 2 points
MgCl2 is NOT an acceptable substitute for BaCl2.
MgCl2 is too soluble.
Note: 1 point earned if response indicates MgCl2 is acceptable and reason given
is that Mg2+ behaves like Ba2+ to form an insoluble SO42¯ precipitate (response
must previously specify BaSO4 as product)
1998 # 5
An approximately 0.1 M solution of NaOH is to be standardized by titration. Assume that
the following materials are available.
Clean, dry 50 mL buret
250 mL Erlenmeyer flask
Wash bottle filled with distilled water
Analytical balance
Phenolphthalein indicator solution
Potassium hydrogen phthalate, KHP, a pure solid
monoprotic acid (to be used as the primary standard)
a.
Briefly describe the steps you would take, using materials listed above, to
standardize the NaOH solution.
b.
Describe (i.e., set up) the calculations necessary to determine the concentration of
the NaOH solution.
c.
After the NaOH solution has been standardized, it is used to titrate a weak
monoprotic acid, HX. The equivalence point is reached when 25.0 mL of NaOH
solution has been added. In the space provided at the right, sketch the titration
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curve, showing the pH changes that occur as the volume of NaOH solution added
increases from 0 to 35.0 mL. Clearly label the equivalence point on the curve.
d.
Describe how the value of the acid-dissociation constant, Ka, for the weak acid
HX could be determined from the titration curve in part (c).
e.
The graph below shows the results obtained by titrating a different weak acid,
H2Y, with the standardized NaOH solution. Identify the negative ion that is
present in the highest concentration at the point in the titration represented by the
letter A on the curve.
8 points
a.
1)
2)
3)
4)
4 essential steps
weigh KHP
fill buret with NaOH solution
add indicator (phenolphthalein)
titrate to endpoint (color change)
2 points
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Two points earned for all 4 steps; one point earned for 2 or 3 steps
Titration of acid into base accepted if described correctly
mass KHP
moles KHP 
b.
molar mass KHP
1 point
moles OH 
1 point
 [OH  ]
liters NaOH
Acceptable if some parts of part b appear in a
c.
Curve should have 3 important features
2 points
1) Curve begins above pH 1, but below 7
2) Equivalence point at 25 mL
3) Equivalence point above pH 7
Both points earned for all 3 features
One point earned for any 2 of the 3 features
d.
At the half-way point in the titration, pH = pKa
1 point
e.
At point A in the titration, the anion in highest concentration is Y2- 1 point
Also accepted: Y2-, Y--, Y=, and specific anions such as SO42-, SO32HY-, Y- and “Y ion” not accepted
moles KHP = moles OH- at equivalence and
1999 # 5
A student performs an experiment to determine the molar mass of an unknown gas. A
small amount of the pure gas is released from a pressurized container and collected in a
graduated tube over water at room temperature, as shown in the diagram above. The
collection tube containing the gas is allowed to stand for several minutes, and its depth is
adjusted until the water levels inside and outside the tube are the same. Assume that:
 the gas is not appreciably soluble in water
 the gas collected in the graduated tube and the water are in thermal equilibrium
69

a barometer, a thermometer, an analytical balance, and a table of the equilibrium
vapor pressure of water at various temperatures are also available.
a.
Write the equation(s) needed to calculate the molar mass of the gas.
b.
List the measurements that must be made in order to calculate the molar mass of
the gas.
c.
Explain the purpose of equalizing the water levels inside and outside the gas
collection tube.
d.
The student determines the molar mass of the gas to be 64 g mol-1. Write the
expression (set-up) for calculating the percent error in the experimental value,
assuming that the unknown gas is butane (molar mass 58 g mol-1). Calculations
are not required.
e.
If the student fails to use information from the table of the equilibrium vapor
pressures of water in the calculation, the calculated value for the molar mass of
the unknown gas will be smaller than the actual value. Explain.
2000 # 5
The molar mass of an unknown solid, which is nonvolatile and a nonelectrolyte, is to be
determined by the freezing-point depression method. The pure solvent used in the
experiment freezes at 10 oC and has a known molal freezing-point depression constant, kf.
Assume that the following materials are also available.
Test tubes
Beaker
a.
stirrer
stopwatch
pipet
graph paper
thermometer
hot-water bath
balance
ice
Using the two sets of axes provided below, sketch cooling curves for (i) the pure
solvent and for (ii) the solution as each is cooled for 20 oC to 0.0 oC.
70
b.
Information from these graphs may be used to determine the molar mass of the
unknown solid.
i)
Describe the measurements that must be made to determine the molar
mass of the unknown solid by this method.
ii)
Show the setup(s) for the calculation(s) that must be performed to
determine the molar mass of the unknown solid from the experimental
data.
iii)
Explain how the difference(s) between the two graphs in part a) can be
used to obtain information needed to calculate the molar mass of the
unknown solid.
c.
Suppose that during the experiment a significant but unknown amount of solvent
evaporates from the test tube. What effect would this have on the calculated
value of the molar mass of the solid (i.e., too large, too small, or no effect)?
Justify your answer.
d.
Show the setup for the calculation of the percentage error in a student’s result if
the student obtains a value of 126 g mol-1 for the molar mass of the solid when the
actual value is 120. g mol-1.
(10 points)
(a)
2 points - Cooling curve graphs for pure material and solutionNotes: One
point is earned for each correct graph. The first graph should show a line that
drops to 10°C, holds steady at 10°C, and then falls steadily to 0°C. There must be
a discernable plateau at 10°C to earn this point. The second graph should show a
71
(c)
(d)
line that drops to below 10°C, levels off (or slants down a bit), and then falls
more sharply to 0°C.
(b)
(i) Measure mass of solute, mass of solvent, mass of solution
1 point
(two of three must be shown)
Measure the ΔTfp (or the freezing point of the solution)
1 point
• Volume of solution (without density), molality, or number of moles do not earn
points
(ii) Given:
ΔT = iKf m (or ΔT = Kf m)
2 pnts
m = (mol solute)/(kg solvent)
moles = g/(molar mass)
Combine to get: molar mass = (i)(Kf)(g solute)/(ΔT)(kg solvent)
Notes: One point is earned for any two equations, and two points are earned
for all three equations. “Solute” and “solvent” must be clearly identified in
the equations.
(iii) the difference in the vertical position of the horizontal portions of the graphs
is equal to ΔTfp , the change in freezing point due to the addition of the solute.
1 point
The molar mass is too small.
1 point
If some of the solvent evaporates, then the (kg solvent) term used in the equation
in (b) (ii) is larger than the actual value. If the (kg solvent) term used is too large,
then the value calculated for the molar mass will be too small.
1 point
or
If some of the solvent evaporates, then the concentration (molality) of the solute
will be greater than we think it is. More moles of solute results in a smaller molar
mass (or since ΔT = iKf m, then the ΔTobs would be greater than it should be).
Since the molar mass of the unknown solute is inversely proportional to ΔT, an
erroneously high value for ΔT implies an erroneously low value for the molar
mass (calculated molar mass would be too small).
1 point
1
1
126 g mol  120 g mol
% error =
1 point
 100%
120 g mol 1
or
6 g mol 1
% error =
 100%
120 g mol 1
2005 # 5
Answer the following questions that relate to laboratory observations and procedures.
a.
An unknown gas is one of three possible gases: nitrogen, hydrogen, or oxygen.
For each of the three possibilities, describe the result expected when the gas is
tested using a glowing splint (a wooden stick with one end that has been ignited
and extinguished, but still contains hot, glowing, partially burned wood).
b.
The following three mixtures have been prepared: CaO plus water, SiO2 plus
water, and CO2 plus water. For each mixture, predict whether the pH is less than
7, equal to 7, or greater than 7. Justify your answers.
72
c.
Each of three beakers contains a 0.1 M solution of one of the following solutes:
potassium chloride, silver nitrate, or sodium sulfide. The three beakers are
labeled randomly as solution 1, solution 2, and solution 3. Shown below is a
partially completed table of observations made of the results of combining small
amounts of different pairs of the solutions.
Solution 1
Solution 1
Solution 2
Solution 3
black precipitate
Solution 2
no reaction
Solution 3
i)
Write the chemical formula of the black precipitate.
ii)
Describe the expected results of mixing solution 1 with solution 3.
iii)
Identify each of the solutions 1, 2, and 3.
9 points
a.
Nitrogen: When the glowing splint is inserted into the gas sample, the
glowing splint will be extinguished.
Hydrogen: When the glowing splint is inserted into the gas sample, a
popping sound (explosion) can be heard.
Oxygen: When the glowing splint is inserted into the gas sample, the
splint will glow brighter or reignite.
1 point is earned for each description
b.
CaO plus water: The pH of the solution will be greater than 7. CaO is
water forms the base Ca(OH)2 (or metal oxides are basic, or basic
anhydrides)
SiO2 plus water: The pH of the solution will be equal to 7. SiO2 is
insoluble in water, so there would not be a change in the pH of the
mixture.
CO2 plus water: The pH of the solution will be less than 7. CO2 in water
forms the acid H2CO3 (or nonmetal oxides are acidic, or acidic anhydride)
1 point is earned for each description
c.
i) The black precipitate is Ag2S.
1 point
73
ii) A precipitate will be produced when the two solutions are mixed.
1 point
iii) Solution 1 is silver nitrate. Solution 2 is sodium sulfide. Solution 3 is
potassium chloride.
1 point is earned for the correct identification of all three solutions.
2005B #5
2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 22 H2O(l)  2 KAl(SO4)2·12 H2O + 3 H2(g)
In an experiment, a student synthesizes alum, KAl(SO4)2·12H2O(s), by reacting
aluminum metal with potassium hydroxide and sulfuric acid, as represented in the
balanced equation above.
a.
In order to synthesize alum, the student must prepare a 5.0 M solution of sulfuric
acid. Describe the procedure for preparing 50.0 mL of 5.0 M H2SO4 using any of
the chemicals and equipment listed below. Indicate specific amounts and
equipment where appropriate.
10.0 M H2SO4
Distilled water
100 mL graduated cylinder
100 mL beaker
50.0 mL volumetric flask
50.0 mL buret
25.0 mL pipet
50 mL beaker
b.
Calculate the minimum volume of 5.0 M H2SO4 that the student must use to react
completely with 2.7 g of aluminum metal.
c.
As the reaction solution cools, alum crystals precipitate. The student filters the
mixture and dries the crystals, then measures their mass.
d.
i)
If the student weighs the crystals before they are completely dry, would
the calculated percent yield be greater than, less than, or equal to the
actual percent yield? Explain.
ii)
Cooling the reaction solution in an ice bath improves the percent yield
obtained. Explain.
The student heats crystals of pure alum, KAl(SO4)2·12 H2O(s), in an open
crucible to a constant mass. The mass of the sample after heating is less than the
mass before heating. Explain.
10 points
a.
(50 mL) (1 L / 1000 mL) (5.0 mol H2SO4 / 1 L) = 0.25 mol H2SO4
(0.25 mol H2SO4) (1 L/ 10.0 mol H2SO4) (1000 mL / 1 L) = 25.0 mL
74
b.
c.
d.
Put on goggles. Measure approximately 20 mL of distilled water using the
100 mL graduated cylinder, and add the distilled water to the 50.0 mL
volumetric flask. Measure 25.0 mL of the 10.0 M H2SO4 using the 25.0
mL pipet, and transfer the concentrated acid slowly, with occasional
swirling, to the 50.0 mL volumetric flask containing the distilled water.
After adding all the concentrated acid, carefully add distilled water until
the meniscus of the solution is at the 50.0 mL mark on the neck of the flask
at 20 oC.
1 point is earned for the volume of the 10.0 M H2SO4
1 point is earned for using a volumetric flask and the pipet
1 point is earned for adding the acid to the water
1 point is earned for filling to the mark with water
1 mol Al 4 mol H 2SO 4
1L
 0.040 L
V(H2SO4) = (2.7 g Al)
27.0 g Al 2 mol Al 5.0 mol H 2SO 4
1 point is earned for the number of moles of Al
1 point is earned for the correct stoichiometry
1 point is earned for the answer
i) If the KAl(SO4)2•12H2O(s) crystals have not been properly dried, there
will be excess water present, making the mass of the product greater than
it should be and the calculated percent too high. Therefore, the calculated
percent yield will be greater than the actual percent yield.
1 point is earned for the prediction and a correct explanation
ii) If the solubility of KAl(SO4)2•12H2O(s) decreases with decreasing
temperature, cooling the reaction solution would result in the
precipitation of more KAl(SO4)2•12H2O(s)
1 point is earned for the correct explanation
KAl(SO4)2•12H2O(s) is a hydrate. For the mass of the sample to be less
after heating, the water of hydration must be lost. Heating the sample of
KAl(SO4)2•12H2O(s) crystals will drive off the water first, decreasing the
mass of the sample
1 point for the correct explanation
75
AP Chemistry – NUCLEAR QUESTIONS
2003B # 8
The decay of the radioisotope I-131 was studied in a laboratory. I-131 is known to decay
by beta (-10e) emission.
a.
Write a balanced nuclear equation for the decay of I-131.
b.
What is the source of the beta particle emitted from the nucleus?
The radioactivity of a sample of I-131 was measured. The data collected are plotted on
the graph below.
c.
Determine the half-life, t1/2, of I-131 using the graph above.
d.
The data can be used to show that the decay of I-131 is a first-order reaction, as
indicated on the graph below.
76
e.
i.
Label the vertical axis of the graph above.
ii.
What are the units of the rate constant, k, for the decay reaction?
iii.
Explain how the half-life of I-131 can be calculated using the slope of the
line plotted on the graph.
Compare the value of the half-life of I-131 at 25 oC to its value at 50 oC
8 points
a.
b.
c.
d.
e.
0
I131
1 point
54 Xe  1 e
A neutron spontaneously decays to an electron and a proton.
1 point for identifying a neutron as the source of the beta emission
The half life is 8 days. That is the time required for the disintegration rate
to fall from 16,000 to one half its initial value, 8000
1 point
i) The label on the y axis should be ln or log of one of the following:
disintegrations or moles or atoms of [I-131] or disintegration rate 1 point
ii) From the graph, the units on the rate constant are days-1 (Units of
time-1 are acceptable)
1 point
iii) The slope of the line is –k. The slope is negative, so k is a positive
number. The half life can then be calculated using the relationship, t1/2 =
0.693 / k.
1 point for indicating slope is k
1 point for half life equation
The half life will be the same at the different temperatures. The half life of
a nuclear decay process is independent of temperature
1 point
131
53
77