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Unformatted text preview: AP® Chemistry
2005 Scoring Guidelines
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2005 SCORING GUIDELINES (Form B)
Question 1
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.
HOCl(aq) + H2O(l) → OCl−(aq) + H3O+(aq) One point is earned for the correct chemical equation. (b) Calculate the pH of a 0.175 M solution of HOCl.
→
HOCl(aq) + H2O(l) ← OCl−(aq) + H3O+(aq)
0.175
–
0
~0
−x
–
+x
+x
0.175 − x
–
+x
+x I
C
E
Ka = [H3O + ][OCl− ]
( x)( x)
=
[HOCl]
(0.175 − x) One point is earned for calculating the value of
[H3O+]. Assume that 0.175 − x ≈ 0.175
x2
3.2 × 10−8 =
0.175
2
x = (3.2 × 10−8 ) (0.175) = 5.6 × 10−9 One point is earned for calculating the pH. x = [H3O+ ] = 7.5 × 10−5 M
pH = – log [H3O+ ] = – log (7.5 × 10−5) = 4.13
(c) Write the net ionic equation for the reaction between the weak acid HOCl(aq) and the strong base
NaOH(aq).
HOCl(aq) + OH−(aq) → OCl−(aq) + H2O(l) One point is earned for both of the correct reactants.
One point is earned for both of the correct products. (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.
molNaOH = 6.55 mL × 1L
0.435 mol NaOH
×
1,000 mL
1L One point is earned for the correct number of
moles of NaOH. molNaOH = 2.85 × 10−3 mol NaOH Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 1 (continued) (ii) Calculate [H3O+] in the flask after the NaOH(aq) has been added. molHOCl = 20.00 mL × 1L
0.175 mol NaOH
×
= 3.50 × 10−3 mol
1,000 mL
1L − OH (aq) is the limiting reactant, therefore all of it reacts One point is earned for calculating
the initial number of moles
of HOCl. HOCl(aq) + OH−(aq) → OCl− (aq) + H2O(l)
I
0.00350
0.00285
0
–
C − 0.00285
– 0.00285
+ 0.00285
–
E 0.00065
0
0.00285
–
MHOCl =
MOCl = I
C
E Ka = 0.00065 mol
= 0.0245 M
0.02655 L
0.00285 mol
= 0.107 M
0.02655 L HOCl(aq) + H2O(l) → H3O+(aq) + OCl−(aq)
0.0245
–
~0
0.107
–
+x
+x
–x
0.0245 – x
–
+x
0.107 + x One point is earned for the
concentration or number of moles of
HOCl and OCl− after the
neutralization reaction. [H3O + ][OCl− ]
( x)(0.107 + x)
=
[HOCl]
(0.0245 − x) Assume that 0.107 + x ≈ 0.107 and that 0.0245 – x ≈ 0.0245
3.2 × 10−8 = ( x)(0.107)
(0.0245)
One point is earned for the
correct [H3O+ ]. x = [H3O+ ] = 7.3 × 10−9 M (iii) Calculate [OH−] in the flask after the NaOH(aq) has been added.
[H3O+][OH−] = 1.0 × 10−14 = Kw
[OH−] = 1.0 × 10 −14 + [H3O ] = 1.0 × 10 −14 7.3 × 10−9 = 1.4 × 10−6 M One point is earned for the correct concentration
of OH−. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 2 Water was electrolyzed, as shown in the diagram above, for 5.61 minutes using a constant current of
0.513 ampere. A small amount of nonreactive electrolyte was added to the container before the electrolysis began.
The temperature was 298 K and the atmospheric pressure was 1.00 atm.
(a) Write the balanced equation for the half reaction that took place at the anode.
2 H2O(l) → O2(g) + 4 H+(aq) + 4 e− One point is earned for the correct half reaction. (b) Calculate the amount of electric charge, in coulombs, that passed through the solution. 0.513 amp = 0.513 ( coul
sec electric charge = 0.513 One point is earned for
the setup. ) ⎛ 60 sec ⎞
coul
× (5.61 min) × ⎜
= 173 coulombs
sec
⎝ 1 min ⎟
⎠ One point is earned for the
answer. (c) Why is the volume of O2(g) collected different from the volume of H2(g) collected, as shown in the
diagram?
When water decomposes according to the balanced chemical
equation 2 H2O(l) → O2(g) + 2 H2(g) , twice as many moles
of hydrogen are produced than moles of oxygen. One point is earned for the correct
explanation based on the
stoichiometry of the
decomposition reaction. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 2 (continued) (d) Calculate the number of moles of H2(g) produced during the electrolysis.
The halfreaction that takes place at the cathode is:
2 H2O(l) + 2 e− → H2(g) + OH−(aq)
⎛
⎞
⎛ 1 mol H 2 ( g ) ⎞
1 mol e −
molH 2 = 173 coulombs × ⎜
⎟×⎜
⎟
⎝ 2 mol e − ⎠
⎝ 96,500 coulomb ⎠ One point is earned for the number
of coulombs.
One point is earned for recognizing
the 1 : 2 stoichiometry. molH 2 = 8.96 × 10− 4 mol
(e) Calculate the volume, in liters, at 298 K and 1.00 atm of dry H2(g) produced during the electrolysis. VH 2 = VH 2 nH 2 RT
P L atm ⎞
⎛
(8.96 × 10− 4 mol) × ⎜ 0.0821
⎟ × (298 K)
⎝
mol K ⎠
=
= 0.0219 L
1 atm One point is earned for the
substitution into the gas law
equation.
One point is earned for the
correct answer. (f) After the hydrolysis reaction was over, the vertical position of the tube containing the collected H2(g) was
adjusted until the water levels inside and outside the tube were the same, as shown in the diagram below.
The volume of gas in the tube was measured under these conditions of 298 K and 1.00 atm, and its volume
was greater than the volume calculated in part (e). Explain. Because the electrolysis of water occurs in water, there is some water vapor
in the tube of H2(g) that was collected. The volume calculated in part (e)
was the volume of only the H2(g) in the tube at the given temperature and
pressure. The presence of another gas (water vapor) results in a greater
volume at the given temperature and pressure. One point is earned for
recognizing that there is some
water vapor in the sample of
hydrogen gas. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 3 X → 2Y+Z
The decomposition of gas X to produce gases Y and Z is represented by the equation above. In a certain
experiment, the reaction took place in a 5.00 L flask at 428 K. Data from this experiment were used to produce the
information in the table below, which is plotted in the graphs that follow.
Time
(minutes)
0
10.
20.
30.
50.
70.
100. [X]
(mol L− 1)
0.00633
0.00520
0.00427
0.00349
0.00236
0.00160
0.000900 ln [X]
−5.062
−5.259
−5.456
−5.658
− 6.049
− 6.438
−7.013 [X] − 1
(L mol −1 )
158
192
234
287
424
625
1,110 (a) How many moles of X were initially in the flask?
[X] at 0 minutes = 0.00633, so
mol X
5.00 L × 0.00633
= 3.17 × 10−2 mol X
L One point is earned for correct number of moles
of X. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 3 (continued) (b) How many molecules of Y were produced in the first 20. minutes of the reaction?
After 20. minutes of reaction, the number of moles of X remaining
mol X
) = 2.14 × 10−2 mol X .
in the flask is (5.00 L) × (0.00427
L
Then the number of moles of X that reacted in the first 20 minutes One point is earned for the
number of moles of X that
react or for the correct
stoichiometry between X
and Y. is (3.17 × 10−2 mol X ) – (2.14 × 10−2 mol X) = 1.03 × 10−2 mol X.
Thus the number of molecules of Y produced in the first 20. minutes is ⎛ 2 mol Y produced ⎞ ⎛ 6.02 × 1023 molecules Y ⎞
(1.03 × 10−2 mol X) × ⎜
×
⎟
1 mol Y
⎝ 1 mol X reacted ⎟ ⎜
⎠⎝
⎠
= 1.24 × 1022 molecules Y produced One point is earned for the
number of molecules of Y
produced. (c) What is the order of this reaction with respect to X ? Justify your answer.
The reaction is first order with respect to X because a plot of
ln [X] versus time produces a straight line with a negative slope. One point is earned for the
correct order and an explanation. (d) Write the rate law for this reaction.
rate = k[X]1 One point is earned for the rate law consistent with part (c). (e) Calculate the specific rate constant for this reaction. Specify units. ln [X]t
= −kt
[X]0 ⎛ 0.00520 ⎞
= −k (10 min)
From the first two data points, ln ⎜
⎝ 0.00633 ⎟
⎠
⎛ ln 0.821⎞
k = −⎜
= 0.0197 min−1
⎝ 10 min ⎟
⎠ One point is earned for the
magnitude of the rate constant.
One point is earned for the units. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 3 (continued (f) Calculate the concentration of X in the flask after a total of 150. minutes of reaction. ln [X]t
= −kt means the same thing as ln [X]t − ln[X]0 = −kt
[X]0 ln [X]150 − ln (0.00633) = −(0.0197 min−1)(150 minutes) One point is earned for substituting
into the integrated rate law. ln [X]150 = −(0.0197 min−1 )(150 minutes) + ln (0.00633)
ln [X]150 = −(0.0197 min−1 )(150 minutes) + (−5.062)
ln [X]150 = −2.955 + (−5.062) = −8.017
eln[X]150 = e−8.017 = 3.30 × 10− 4 [X] at 150. minutes = 3.30 × 10−4 M One point is earned for the correct
concentration of X. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 4 Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described
below. Answers to more than five choices will not be graded. In all cases, a reaction occurs. Assume that
solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances
are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You
need not balance the equations.
General Scoring: Three points are earned for each: 1 point for correct reactant(s) and 2 points for correct
product(s).
(a) A solution of potassium carbonate is added a solution of strontium nitrate.
CO32− + Sr2+ → SrCO3
(b) Propene is burned in air.
C3H6 + O2 → CO2 + H2O
(c) Excess ammonia is added to a solution of zinc nitrate.
NH3 + Zn2+ → Zn(NH3)62+
(d) Ethanoic acid (acetic acid) is added to a solution of barium hydroxide.
OH− + HC2H3O2 → C2H3O2− + H2O
(e) A small piece of potassium is added to water.
K + H2O → OH− + H2 + K+
(f) Powdered iron metal is strongly heated with powdered sulfur.
Fe + S → FeS
Note: Fe2S3 also acceptable as a product
(g) A solution of sodium fluoride is added to a solution of hydrochloric acid.
H+ + F− → HF
(h) A strip of lead metal is added to a solution of silver nitrate.
Pb + Ag+ → Pb2+ + Ag
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2005 SCORING GUIDELINES (Form B)
Question 5 2 Al(s) + 2 KOH(aq) + 4 H2SO4(aq) + 22 H2O(l) → 2 KAl(SO4)2·12H2O(s) + 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 ⎛ 1 L ⎞ ⎛ 5.0 mol H 2SO 4 ⎞
(50.0 mL) ⎜
⎟ = 0.25 mol H2SO4
⎟⎜
1L
⎠
⎝ 1,000 mL ⎠ ⎝
⎛
⎞ ⎛ 1,000 mL ⎞
1L
(0.25 mol H2SO4) ⎜
⎟⎜
⎟ = 25.0 mL of 10.0 M H2SO4
⎝ 10.0 mol H 2SO 4 ⎠ ⎝ 1 L ⎠
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°C. One point is earned for
the volume of
10.0 M H2SO4.
One point is earned for
using a volumetric
flask and the pipet.
One point is earned for
adding the acid to the
water.
One point is earned for
filling to the mark
with water. (b) Calculate the minimum volume of 5.0 M H2SO4 that the student must use to react completely with
2.7 g aluminum metal.
⎞
⎛ 1 mol Al ⎞ ⎛ 4 mol H 2SO 4 ⎞ ⎛
1L
VH 2SO4 = (2.7 g Al) ⎜
⎟⎜
⎟ ⎜ 5.0 mol H SO ⎟
⎝ 27.0 g Al ⎠ ⎝ 2 mol Al ⎠ ⎝
2
4⎠
VH 2SO4 = 0.040 L One point is earned for the
number of moles of Al.
One point is earned for the
correct stoichiometry.
One point is earned for the
answer. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 5 (continued) (c) As the reaction solution cools, alum crystals precipitate. The student filters the mixture and dries the
crystals, then measures their mass.
(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.
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 yield too high. Therefore, the
calculated percent yield will be greater than the actual percent yield. One point is earned for the
prediction and a correct
explanation. (ii) Cooling the reaction solution in an ice bath improves the percent yield obtained. Explain.
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). One point is earned for the
correct explanation. (d) The student heats crystals of pure alum, KAl(SO4)2·12H2O(s), in an open crucible to a constant mass. The
mass of the sample after heating is less than the mass before heating. Explain.
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. One point is earned for the
correct explanation. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 6 Consider two containers of volume 1.0 L at 298 K, as shown above. One container holds 0.10 mol N2(g) and
the other holds 0.10 mol H2(g). The average kinetic energy of the N2(g) molecules is 6.2 × 10−21 J. Assume that
the N2(g) and the H2(g) exhibit ideal behavior.
(a) Is the pressure in the container holding the H2(g) less than, greater than, or equal to the pressure in the
container holding the N2(g) ? Justify your answer.
The pressure in the container holding the H2(g) is equal to the
pressure in the container holding the N2(g) because there is an
equal number of moles of both gases at the same temperature
RT
and volume (P = nK , where the constant K =
).
V One point is earned for the
correct choice.
One point is earned for the
correct explanation. (b) What is the average kinetic energy of the H2(g) molecules?
The average kinetic energy of the H2(g) molecules is
6.2 × 10−21 J because both gases are at the same temperature. One point is earned for the
correct energy. (c) The molecules of which gas, N2 or H2 , have the greater average speed? Justify your answer.
H2(g) molecules will have the greater average speed. Both
gases have the same average kinetic energy, but H2(g) has the
smaller molar mass. Therefore, the H2(g) molecules will have a
greater average speed because, at a given temperature , the
average (rootmeansquare) speed of gas molecules is inversely One point is earned for the correct
answer with an explanation. proportional to the square root of the molar mass of the gas:
urms = ( 3 RT ) 1
M Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 6 (continued) (d) What change could be made that would decrease the average kinetic energy of the molecules in the container?
The average kinetic energy of a gas particle depends on the
temperature of the gas sample. To decrease the average kinetic
energy of the gas particles in a gas sample, the temperature of
the N2(g) would have to be lowered. One point is earned for the correct
answer with an explanation. (e) If the volume of the container holding the H2(g) was decreased to 0.50 L at 298 K, what would be the
change in each of the following variables? In each case, justify your answer.
(i) The pressure within the container
The pressure would be doubled. PV is a constant when the
temperature and number of moles of gas are held constant.
Therefore, if the volume is halved the pressure is doubled.
P1 V1 = P2 V2 If V2 = () 1
⎛1 ⎞
1
V1 , then P1V1 = P2 ⎜ V1 ⎟ ⇒ P1 = P2
2
2
⎝2 ⎠ One point is earned for the
correct answer.
⇒ One point is earned for the
correct explanation. P2 = 2 P1 (ii) The average speed of the H2(g) molecules
The average speed is unchanged when the volume of the gas
sample is halved. Average speed depends on changes in
temperature, not changes in volume. One point is earned for the correct
answer with an explanation. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 7 Answer the following questions about thermodynamics.
Substance Combustion Reaction H2(g) H2(g) + Enthalpy of Combustion,
∆H comb , at 298 K (kJ mol−1) 1
O (g) → H2O(l)
22 −290 C(s) + O2(g) → CO2(g) C(s) −390 CH3OH(l) −730 (a) In the empty box in the table above, write a balanced chemical equation for the complete combustion of
one mole of CH3OH(l). Assume products are in their standard states at 298 K. Coefficients do not need
to be in whole numbers. CH3OH(l) + 3
O2(g) → CO2(g) + 2 H2O(l)
2 One point is earned for the correct products.
One point is earned for balancing the equation. (b) On the basis of your answer to part (a) and the information in the table, determine the enthalpy change for
the reaction C(s) + H2(g) + H2O(l) → CH3OH(l) .
Adding the following three equations,
–390 kJ mol−1 C(s) + O2(g) → CO2(g)
1
H2(g) + O2(g) → H2O(l)
2
CO2(g) + 2 H2O(l) → CH3OH(l) +
yields this equation: –290 kJ mol
3
O2(g)
2 C(s) + H2(g) + H2O(l) → CH3OH(l) −1 +730 kJ mol−1 One point is earned for
the correct equations.
One point is earned for
the correct value of ∆H°. +50 kJ mol−1 (c) Write the balanced chemical equation that shows the reaction that is used to determine the enthalpy of
formation for one mole of CH3OH(l).
1
C(s) + 2 H2(g) + 2 O2(g) → CH3OH(l) One point is earned for the correct equation. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 7 (continued) (d) Predict the sign of ∆S° for the combustion of H2(g). Explain your reasoning.
∆S° for the combustion of H2(g) is negative. Both reactants are
in the gas phase and the product is in the liquid phase. The
liquid phase is much more ordered than the gas phase, so the
product is more ordered compared to the reactants, meaning that
∆S° is negative.
(Note: There are fewer moles of products than reactants, which
also favors a more ordered condition in the products, but the
difference in phases is the more important factor.) One point is earned for the correct
sign of ∆S°.
One point is earned for a correct
explanation. (e) On the basis of bond energies, explain why the combustion of H2(g) is exothermic.
The combustion of H2(g) is exothermic ( ∆H° < 0 ) because
more energy is released during the formation of two moles of
O–H bonds than is required to break one mole of H–H bonds One point is earned for the correct
explanation. and onehalf of a mole of O–O bonds. Copyright © 2005 by College Board. All rights reserved.
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2005 SCORING GUIDELINES (Form B)
Question 8 Use principles of atomic structure, bonding, and intermolecular forces to answer the following questions. Your
responses must include specific information about all substances referred to in each part.
(a) Draw a complete Lewis electrondot structure for the CS2 molecule. Include all valence electrons in your
structure.
One point is earned for the correct structure. (b) The carbontosulfur bond length in CS2 is 160 picometers. Is the carbontoselenium bond length in CSe2
expected to be greater than, less than, or equal to this value? Justify your answer.
The carbontoselenium bond length in CSe2 is greater
than the carbontosulfur bond length in CS2. Because
the valence electrons in Se are in a higher shell (n = 4)
than the valence electrons in S (n = 3), Se has a
larger atomic radius than S has. Therefore, the carbontoselenium bond length is greater than the carbonto One point is earned for indicating that the
CSe bond length is greater than the CS
bond length.
One point is earned for indicating that Se is
larger than S. sulfur bond length. (c) The bond energy of the carbontosulfur bond in CS2 is 577 kJ mol−1. Is the bond energy of the carbontoselenium bond in CSe2 expected to be greater than, less than, or equal to this value? Justify your answer.
The carbontoselenium bond energy in CSe2 is less than
the carbontosulfur bond energy in CS2 because Se has a
larger atomic radius than S . Because Se is a larger
atom, the orbital overlap between the Se and C will be
smaller than the orbital overlap between S and C. One point is earned for indicating that the
CSe bond energy is less than the CS
bond energy.
One point is earned for the explanation. Copyright © 2005 by College Board. All rights reserved.
Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents). AP® CHEMISTRY
2005 SCORING GUIDELINES (Form B)
Question 8 (continued) (d) The complete structural formulas of propane, C3H8, and methanoic acid, HCOOH, are shown above. In the
table below, write the type(s) of intermolecular attractive forces(s) that occur in each substance.
Substance Boiling Point Propane 229 K Methanoic acid Intermolecular Attractive Force(s) 374 K Propane has dispersion forces.
Methanoic acid has dispersion forces and hydrogen bonding forces. One point is earned for IMFs in
propane.
One point is earned for IMFs in
methanoic acid. (e) Use principles of intermolecular attractive forces to explain why methanoic acid has a higher boiling point
than propane.
Hydrogen bonding IMFs among methanoic acid molecules are
much stronger than dispersion forces among propane molecules.
The stronger the IMFs , the more energy it takes to overcome them.
Therefore, methanoic acid has a higher boiling point than propane. One point is earned for comparing
the strengths of the IMFs. Copyright © 2005 by College Board. All rights reserved.
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This note was uploaded on 10/01/2009 for the course OC 9876 taught by Professor Dq during the Spring '09 term at UC Merced.
 Spring '09
 Dq

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