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AP® Chemistry 2006 Free-Response Questions

The College Board: Connecting Students to College Success

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睿途教育北美出国长线规划专家 www.oneplusone.cn

GO ON TO THE NEXT PAGE. 2

INFORMATION IN THE TABLE BELOW AND IN THE TABLES ON PAGES 3-5 MAY BE USEFUL IN ANSWERING THE QUESTIONS IN THIS SECTION OF THE EXAMINATION.



STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25°C

Half-reaction E °( )V

2F ( ) 2g e-+ Æ 2 F- 2.87

3+Co e-+ Æ 2Co + 1.82 3+Au 3 e-+ Æ Au( )s 1.50 2Cl ( ) 2g e

-+ Æ 2 Cl- 1.36

+2O ( ) 4 H 4g e-+ + Æ 22 H O( )l 1.23

2Br ( ) 2l e-+ Æ 2 Br- 1.07

2+2 Hg 2 e-+ Æ 2+2Hg 0.92

2+Hg 2 e-+ Æ Hg( )l 0.85

+Ag e-+ Æ Ag( )s 0.80

2+2Hg 2 e-+ Æ 2 Hg( )l 0.79

3+Fe e-+ Æ 2+Fe 0.77

2I ( ) 2s e-+ Æ 2 I- 0.53

+Cu e-+ Æ Cu( )s 0.52

2+Cu 2 e-+ Æ Cu( )s 0.34 2+Cu e-+ Æ +Cu 0.15 4+Sn 2 e-+ Æ 2+Sn 0.15 +S( ) 2 H 2s e-+ + Æ 2H S( )g 0.14 +2 H 2 e-+ Æ 2H ( )g 0.00 2+Pb 2 e-+ Æ Pb( )s – 0.13 2+Sn 2 e-+ Æ Sn( )s – 0.14 2+Ni 2 e-+ Æ Ni( )s – 0.25 2+Co 2 e-+ Æ Co( )s – 0.28 2+Cd 2 e-+ Æ Cd( )s – 0.40 3+Cr e-+ Æ 2+Cr – 0.41 2+Fe 2 e-+ Æ Fe( )s – 0.44 3+Cr 3 e-+ Æ Cr( )s – 0.74 2+Zn 2 e-+ Æ Zn( )s – 0.76 22 H O( ) 2l e

-+ Æ 2H ( ) + 2 OHg- – 0.83

2+Mn 2 e-+ Æ Mn( )s – 1.18 3+Al 3 e-+ Æ Al( )s – 1.66 2+Be 2 e-+ Æ Be( )s – 1.70 2+Mg 2 e-+ Æ Mg( )s – 2.37 +Na e-+ Æ Na( )s – 2.71 2+Ca 2 e-+ Æ Ca( )s – 2.87 2+Sr 2 e-+ Æ Sr( )s – 2.89 2+Ba 2 e-+ Æ Ba( )s – 2.90 +Rb e-+ Æ Rb( )s – 2.92 +K e-+ Æ K( )s – 2.92 +Cs e-+ Æ Cs( )s – 2.92 +Li e-+ Æ Li( )s – 3.05


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ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS

E

v n

m

p

= == == ==

energy velocity

frequency principal quantum number

wavelength mass

momentum

u

l

Speed of light, 3.0 10 m s

Planck’s constant, 6.63 10 J s

Boltzmann’s constant, 1.38 10 J K

Avogadro’s number 6.022 10 mol

Electron charge, 1.602 10 coulomb

1 electron volt per atom 96.5 kJ mo

8 1

34

23 1

23 1

19

1

l

c

h

k

e

= ¥

= ¥

= ¥

= ¥

= - ¥

=

-

-

- -

-

-

-

Equilibrium Constants

(weak acid)

(weak base)

(water)

(gas pressure)

(molar concentrations)

K

K

K

K

K

a

b

w

p

c

S

H

G

E

T

n

m

q

c

C

E

k

A

p

a

=

=

=

=======

=

standard entropy

standard enthalpy

standard free energy

standard reduction potential

temperature

moles

mass

heat

specific heat capacity

molar heat capacity at constant pressure

activation energy

= rate constant

= frequency factor

Faraday's constant, coulombs per mole

of electrons

Gas constant, J mol K

L atm mol K

volt coulomb mol K

� =

=

=

=

- -

- -

- -

96 500

8 31

0 0821

8 31

1 1

1 1

1 1

,

.

.

.

R

ATOMIC STRUCTURE

E hv c v

p m

En

n

= =

=

= - ¥-

l

lu

u= hm

2 178 10 18

2.

joule

EQUILIBRIUM

K

K

K

a

b

b

=

=

=

+ −

− +

− + −

+ −

−

+

= ×= ×

= − = −= +

= +

= +

= − = −

=

= −

[ ] [ ][ ]

[ ] [ ][ ]

[ ] [ ] .

log [ ], log [ ]

log[ ][ ]

log[ ]

[ ]log , log

( ) ,

H AHA

OH HBB

OH H @ 25 C

pH H pOH OH

pH pOH

pH pAHA

pOH pHB

Bp p

where moles product gas moles reactant gas

K

K

K

K

K K K K

K K RT

n

w

a

a

b

a a b b

p cn

1 0 10

14

14

D

D

THERMOCHEMISTRY/KINETICS

D

D D D

D D D

D D D

D D D

D

DD

S S S

H H H

G G G

G H T S

RT K RT K

n E

G G RT Q G RT Q

q mc T

CHT

f f

f f

p

= -

-

= -

-= - = -

= -

= + = +=

=

Â ÂÂ ÂÂ Â

=

=

products reactants

products reactants

products reactants

ln . log

ln . log

2 303

2 303

�

ln lnA A

A A

t

t

kt

kt

- = -

- =

0

0

1 1

ln lnkER T

Aa= - +1e j



GASES, LIQUIDS, AND SOLUTIONS

1 1 2 2

1 2

1 2

2 1

2

2

2

,

( )

moles Awheretotal moles

...

K C 273

3 3

1 per molecule23 per mole2

molarity, moles solu

u

Ê ˆÁ ˜Ë ¯

= ¥

=

=

=

+ - =

=

= + + +

=

= +

=

= =

=

=

=

M

M

M

M

A total A A

total A B C

rms

P P

PV nRT

n aP V nb nRTV

X X

P P P P

mn

PV P V

T T

mDV

kT RTum

KE m

KE RT

r

r

M te per liter solution

molality moles solute per kilogram solvent

molality

molality

D

D

p

== ¥= ¥==

f f

b b

T iK

T iK

iMRT

A abc

OXIDATION-REDUCTION; ELECTROCHEMISTRY

Q a b c d

Iqt

E ERTn

Q En

Q

KnE

c d

= + Æ +

=

= - = -

=

[ ] [ ]

[ ] [ ]

ln.

log @

log.

,C D

A Bwhere A B C D

C

cell cell cell

a b

�0 0592

25

0 0592

P

V

T

n

D

m

=======

pressure

volume

temperature

number of moles

density

mass

velocityu

u

KE

r

i

K

K

Q

I

q

t

E

K

rms

f

b

A

a

b

c

=======

=

====

==

====

root-mean-square speed

kinetic energy

rate of effusion

molar mass

osmotic pressure

van't Hoff factor

molal freezing-point depression constant

molal boiling-point elevation constant

reaction quotient

current (amperes)

charge (coulombs)

time (seconds)


equilibrium constant

absorbance

molar absorptivity

path length

concentration

M

p


L atm mol K

volt coulomb mol K

Boltzmann's constant, J K

for H O K kg mol

for H O K kg mol

STP C and atm


of electrons

atm mm Hg

torr

R

k

K

K

f

b

=

=

=

= ¥

=

=

=

- -

- -

- -

- -

-

-

==

=

8 31

0 0821

8 31

1 38 10

1 86

0 512

0 000 1 000

96 500

1 1

1 1

1 1

23 1

21

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1 760

760

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.

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,�


2006 AP® CHEMISTRY FREE-RESPONSE QUESTIONS

© 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).


CHEMISTRY Section II

(Total time—90 minutes)

Part A Time—40 minutes

YOU MAY USE YOUR CALCULATOR FOR PART A.

CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. Attention should be paid to significant figures. Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the pink cover. Do NOT write your answers on the green insert. Answer Question 1 below. The Section II score weighting for this question is 20 percent. 1. Answer the following questions that relate to solubility of salts of lead and barium.

(a) A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0 L of solution at

25°C. The concentration of Pb2+(aq) in the saturated solution is found to be 1.3 × 10 − 3 M . The chemical equation for the dissolution of PbI2(s) in water is shown below.

PbI2(s) →← Pb2+(aq) + 2 I−(aq)

(i) Write the equilibrium-constant expression for the equation.

(ii) Calculate the molar concentration of I−(aq) in the solution.

(iii) Calculate the value of the equilibrium constant, Ksp .

(b) A saturated solution is prepared by adding PbI2(s) to distilled water to form 2.0 L of solution at 25°C. What are the molar concentrations of Pb2+(aq) and I−(aq) in the solution? Justify your answer.

(c) Solid NaI is added to a saturated solution of PbI2 at 25°C. Assuming that the volume of the solution does not change, does the molar concentration of Pb2+(aq) in the solution increase, decrease, or remain the same? Justify your answer.

(d) The value of Ksp for the salt BaCrO4 is 1.2 × 10−10. When a 500. mL sample of 8.2 × 10− 6 M Ba(NO3)2 is added to 500. mL of 8.2 × 10− 6 M Na2CrO4 , no precipitate is observed.

(i) Assuming that volumes are additive, calculate the molar concentrations of Ba2+(aq) and CrO42−(aq)

in the 1.00 L of solution.

(ii) Use the molar concentrations of Ba2+(aq) ions and CrO42−(aq) ions as determined above to show

why a precipitate does not form. You must include a calculation as part of your answer.





Answer EITHER Question 2 below OR Question 3 printed on page 8. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 20 percent.

CO(g) + 12

O2(g) → CO2(g)

2. The combustion of carbon monoxide is represented by the equation above.

(a) Determine the value of the standard enthalpy change, rxnHD , for the combustion of CO(g) at 298 K using

the following information.

C(s) + 12

O2(g) → CO(g) 298HD = − 110.5 kJ mol−1

C(s) + O2(g) → CO2(g) 298HD = − 393.5 kJ mol−1

(b) Determine the value of the standard entropy change, rxnSD , for the combustion of CO(g) at 298 K using

the information in the following table.

Substance 298

S

(J mol−1 K−1)

CO(g) 197.7

CO2(g) 213.7

O2(g) 205.1

(c) Determine the standard free energy change, rxnGD , for the reaction at 298 K. Include units with

your answer.

(d) Is the reaction spontaneous under standard conditions at 298 K ? Justify your answer.

(e) Calculate the value of the equilibrium constant, Keq , for the reaction at 298 K.




8

3. Answer the following questions that relate to the analysis of chemical compounds.

(a) A compound containing the elements C , H , N , and O is analyzed. When a 1.2359 g sample is burned

in excess oxygen, 2.241 g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H.

(i) Determine the mass, in grams, of C in the 1.2359 g sample of the compound.

(ii) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound.

(iii) Determine the mass, in grams, of O in the original 1.2359 g sample of the compound.

(iv) Determine the empirical formula of the compound.

(b) A different compound, which has the empirical formula CH2Br , has a vapor density of 6.00 g L− 1 at 375 K

and 0.983 atm. Using these data, determine the following.

(i) The molar mass of the compound

(ii) The molecular formula of the compound

STOP

If you finish before time is called, you may check your work on this part only. Do not turn to the other part of the test until you are told to do so.





CHEMISTRY

Part B Time—50 minutes

NO CALCULATORS MAY BE USED FOR PART B.

Answer Question 4 below. The Section II score weighting for this question is 15 percent.

4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described below. No more than five choices will 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.

Example: A strip of magnesium is added to a solution of silver nitrate.

(a) Solid potassium chlorate is strongly heated.

(b) Solid silver chloride is added to a solution of concentrated hydrochloric acid.

(c) A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide.

(d) Ammonia gas is bubbled into a solution of hydrofluoric acid.

(e) Zinc metal is placed in a solution of copper(II) sulfate.

(f) Hydrogen phosphide (phosphine) gas is added to boron trichloride gas.

(g) A solution of nickel(II) bromide is added to a solution of potassium hydroxide.

(h) Hexane is combusted in air.





Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses. Answer BOTH Question 5 below AND Question 6 printed on pages 11-12. Both of these questions will be graded. The Section II score weighting for these questions is 30 percent (15 percent each). 5. Three pure, solid compounds labeled X , Y , and Z are placed on a lab bench with the objective of identifying

each one. It is known that the compounds (listed in random order) are KCl , Na2CO3 , and MgSO4 . A student performs several tests on the compounds; the results are summarized in the table below.

Compound pH of an Aqueous

Solution of the Compound

Result of Adding 1.0 M NaOH to a


Result of Adding 1.0 M HCl Dropwise to

the Solid Compound

X > 7 No observed reaction Evolution of a gas

Y 7 No observed reaction No observed reaction

Z 7 Formation of a white precipitate

No observed reaction

(a) Identify each compound based on the observations recorded in the table.

Compound X ______________________

Compound Y ______________________

Compound Z ______________________

(b) Write the chemical formula for the precipitate produced when 1.0 M NaOH is added to a solution of compound Z .

(c) Explain why an aqueous solution of compound X has a pH value greater than 7. Write an equation as part of your explanation.

(d) One of the testing solutions used was 1.0 M NaOH . Describe the steps for preparing 100. mL of 1.0 M NaOH from a stock solution of 3.0 M NaOH using a 50 mL buret, a 100 mL volumetric flask, distilled water, and a small dropper.

(e) Describe a simple laboratory test that you could use to distinguish between Na2CO3(s) and CaCO3(s). In your description, specify how the results of the test would enable you to determine which compound was Na2CO3(s) and which compound was CaCO3(s) .





6. Answer each of the following in terms of principles of molecular behavior and chemical concepts.

(a) The structures for glucose, C6H12O6 , and cyclohexane, C6H12 , are shown below.

Identify the type(s) of intermolecular attractive forces in

(i) pure glucose

(ii) pure cyclohexane

(b) Glucose is soluble in water but cyclohexane is not soluble in water. Explain.

(c) Consider the two processes represented below.

Process 1: H2O(l) → H2O(g) ∆H ° = + 44.0 kJ mol−1

Process 2: H2O(l) → H2(g) + 21 O ( )2

g ∆H ° = + 286 kJ mol−1

(i) For each of the two processes, identify the type(s) of intermolecular or intramolecular attractive forces

that must be overcome for the process to occur.

(ii) Indicate whether you agree or disagree with the statement in the box below. Support your answer with a short explanation.

When water boils, H2O molecules break apart to form hydrogen molecules and oxygen molecules.





(d) Consider the four reaction-energy profile diagrams shown below.

(i) Identify the two diagrams that could represent a catalyzed and an uncatalyzed reaction pathway for the

same reaction. Indicate which of the two diagrams represents the catalyzed reaction pathway for the reaction.


Adding a catalyst to a reaction mixture adds energy that causes the reaction to proceed more quickly.





Answer EITHER Question 7 below OR Question 8 printed on page 14. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 15 percent. 7. Answer the following questions about the structures of ions that contain only sulfur and fluorine.

(a) The compounds SF4 and BF3 react to form an ionic compound according to the following equation.

SF4 + BF3 → SF3BF4

(i) Draw a complete Lewis structure for the SF3+ cation in SF3BF4.

(ii) Identify the type of hybridization exhibited by sulfur in the SF3+ cation.

(iii) Identify the geometry of the SF3+ cation that is consistent with the Lewis structure drawn

in part (a)(i).

(iv) Predict whether the F–S–F bond angle in the SF3+ cation is larger than, equal to, or smaller

than 109.5°. Justify your answer.

(b) The compounds SF4 and CsF react to form an ionic compound according to the following equation.

SF4 + CsF → CsSF5

(i) Draw a complete Lewis structure for the SF5− anion in CsSF5.

(ii) Identify the type of hybridization exhibited by sulfur in the SF5− anion.

(iii) Identify the geometry of the SF5− anion that is consistent with the Lewis structure drawn

in part (b)(i).

(iv) Identify the oxidation number of sulfur in the compound CsSF5.




14

8. Suppose that a stable element with atomic number 119, symbol Q , has been discovered.

(a) Write the ground-state electron configuration for Q , showing only the valence-shell electrons.

(b) Would Q be a metal or a nonmetal? Explain in terms of electron configuration.

(c) On the basis of periodic trends, would Q have the largest atomic radius in its group or would it have the smallest? Explain in terms of electronic structure.

(d) What would be the most likely charge of the Q ion in stable ionic compounds?

(e) Write a balanced equation that would represent the reaction of Q with water.

(f) Assume that Q reacts to form a carbonate compound.

(i) Write the formula for the compound formed between Q and the carbonate ion, CO32− .

(ii) Predict whether or not the compound would be soluble in water. Explain your reasoning.

STOP

END OF EXAM


AP® Chemistry 2006 Scoring Guidelines





AP® CHEMISTRY 2006 SCORING GUIDELINES


2

Question 1 1. Answer the following questions that relate to solubility of salts of lead and barium.

(a) A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0 L of solution

at 25°C. The concentration of Pb2+(aq) in the saturated solution is found to be 1.3 × 10 −3 M . The chemical equation for the dissolution of PbI2(s) in water is shown below.

PbI2(s) →← Pb2+(aq) + 2 I−(aq)

(i) Write the equilibrium-constant expression for the equation.

22+[Pb ][I ]spK−=

One point is earned for the correct expression.

(ii) Calculate the molar concentration of I−(aq) in the solution.

By stoichiometry, [I−] = 2 × [Pb2+] , thus [I− ] = 2 × (1.3 × 10 −3) = 2.6 × 10 −3 M

One point is earned for the correct concentration.

(iii) Calculate the value of the equilibrium constant, Ksp .

Ksp = [Pb2+][I−]2 = (1.3 × 10−3)(2.6 × 10−3) 2

= 8.8 × 10−9

One point is earned for a value of Ksp that is consistent with the answers in parts (a)(i) and (a)(ii).

(b) A saturated solution is prepared by adding PbI2(s) to distilled water to form 2.0 L of solution at 25°C. What are the molar concentrations of Pb2+(aq) and I−(aq) in the solution? Justify your answer.

The molar concentrations of Pb2+(aq) and I−(aq)

would be the same as in the 1.0 L solution in part (a)

(i.e., 1.3 × 10 −3 M and 2.6 × 10 −3 M , respectively). The concentrations of solute particles in a saturated

solution are a function of the constant, Ksp , which is independent of volume.

One point is earned for the concentrations (or stating they are the same as in the solution

described in part (a)) and justification.




3

Question 1 (continued)

(c) Solid NaI is added to a saturated solution of PbI2 at 25°C. Assuming that the volume of the solution does not change, does the molar concentration of Pb2+(aq) in the solution increase, decrease, or remain

the same? Justify your answer.

[Pb2+] will decrease.

The NaI(s) will dissolve, increasing [I− ] ; more I−(aq) then combines

with Pb2+(aq) to precipitate PbI2(s) so that the ion product [Pb2+][I− ] 2

will once again attain the value of 8.8 × 10−9 (Ksp at 25°C).

One point is earned for stating that [Pb2+] will

decrease.

One point is earned for justification (can involve a

Le Chatelier argument).

(d) The value of Ksp for the salt BaCrO4 is 1.2 × 10−10. When a 500. mL sample of 8.2 × 10− 6 M Ba(NO3)2 is added to 500. mL of 8.2 × 10− 6 M Na2CrO4 , no precipitate is observed.

(i) Assuming that volumes are additive, calculate the molar concentrations of Ba2+(aq) and CrO4

2−(aq) in the 1.00 L of solution.

New volume = 500. mL + 500. mL = 1.000 L , therefore [Ba2+] in 1.000 L is one-half its initial value:

[Ba2+] = 500. mL

1,000. mL × (8.2 × 10− 6 M ) = 4.1 × 10− 6 M

[CrO42−] =

500. mL

1,000. mL × (8.2 × 10− 6 M ) = 4.1 × 10− 6 M

One point is earned for the correct concentration.

(ii) Use the molar concentrations of Ba2+(aq) ions and CrO42−(aq) ions as determined above to show

why a precipitate does not form. You must include a calculation as part of your answer.

The product Q = [Ba2+][CrO42−]

= (4.1 × 10− 6 M)(4.1 × 10− 6 M)

= 1.7 × 10−11

Because Q = 1.7 × 10−11 < 1.2 × 10−10 = Ksp , no precipitate forms.

One point is earned for calculating a value of Q that is consistent with the concentration

values in part (d)(i).

One point is earned for using Q to explain why no precipitate forms.




4

Question 2

CO(g) + 12

O2(g) → CO2(g)

2. The combustion of carbon monoxide is represented by the equation above.

(a) Determine the value of the standard enthalpy change, rxnH∆ , for the combustion of CO(g) at 298 K using the following information.

C(s) + 12

O2(g) → CO(g) 298H∆ = − 110.5 kJ mol−1

C(s) + O2(g) → CO2(g) 298H∆ = − 393.5 kJ mol−1

Reverse the first equation and add it to the second equation to obtain the third equation.

CO(g) → 12

O2(g) + C(s) 298H∆ = + 110.5 kJ mol−1

+ C(s) + O2(g) → CO2(g) 298H∆ = − 393.5 kJ mol−1

_______________________________

CO(g) + 12

O2(g) → CO2(g) rxnH∆ = 110.5 + (− 393.5)

= −283.0 kJ mol−1

OR

rxnH∆ = fH∆ of CO2(g) − fH∆ of CO(g)

= −393.5 kJ mol−1 − (−110.5 kJ mol−1) = − 283.0 kJ mol−1

One point is earned for reversing the first equation.

One point is earned for the correct answer (with sign).

OR

Two points are earned for

determining rxnH∆ from the enthalpies of formation.

(If sign is incorrect, only one point is earned.)

(b) Determine the value of the standard entropy change, rxnS∆ , for the combustion of CO(g) at 298 K using the information in the following table.

Substance

298S

(J mol−1 K−1)

CO(g) 197.7

CO2(g) 213.7

O2(g) 205.1




5


rxnS∆ = 213.7 J mol−1 K−1 − (197.7 J mol−1 K−1 + 12

(205.1 J mol−1 K−1))

= − 86.5 J mol−1 K−1

One point is earned for

taking one-half of 298S

for O2(g).

One point is earned for the answer (with sign).

(c) Determine the standard free energy change, rxnG∆ , for the reaction at 298 K. Include units with your answer.

rxnG∆ = rxnH∆ − T rxnS∆

= −283.0 kJ mol−1 − (298 K)(− 0.0865 kJ mol−1 K−1)

rxnG∆ = −257.2 kJ mol−1

One point is earned for substitutingthe values from parts (a) and (b)

into the equation.

One point is earned for the answer (with sign and units).

(d) Is the reaction spontaneous under standard conditions at 298 K ? Justify your answer.

Yes, the reaction is spontaneous because the value

of rxnG∆ for the reaction is negative (−257.2 kJ mol−1).

One point is earned for an answer with justification (consistent with the

answer in part (c)).

(e) Calculate the value of the equilibrium constant, Keq , for the reaction at 298 K.

rxnG∆ = −R T ln Keq ⇒ rxnG

RT∆− = ln Keq

1

1 1

257,200 Jmol

(8.31 J mol K )(298 K)

-

- --

- = ln Keq ⇒ Keq = 1.28 × 1045

One point is earned for correct substitution into the equation.

One point is earned for the answer.




6

Question 3 3. Answer the following questions that relate to the analysis of chemical compounds.

(a) A compound containing the elements C , H , N , and O is analyzed. When a 1.2359 g sample is burned

in excess oxygen, 2.241 g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H.

(i) Determine the mass, in grams, of C in the 1.2359 g sample of the compound.

2.241 g CO2(g) × 2

2

1 molCO44.01g CO

× 2

1 mol C1 mol CO

× 12.011g C1 mol C

= 0.6116 g C

One point is earned for the correct answer.

(ii) When the compound is analyzed for N content only, the mass percent of N is found

to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound.

1.2359 g sample × 0.2884 = 0.3564 g N One point is earned for the correct answer.

(iii) Determine the mass, in grams, of O in the original 1.2359 g sample of the compound.

Because the compound contains only C , H , N , and O ,

mass of O = g sample − ( g H + g C + g N ) = 1.2359 − (0.0648 + 0.6116 + 0.3564) = 0.2031 g

One point is earned for the answer consistent with the

answers in parts (a)(i) and (a)(ii).

(iv) Determine the empirical formula of the compound.

Converting all masses to moles,

0.6116 g C × 1 mol C12.011g C

= 0.05092 mol C

0.0648 g H × 1 mol H1.0079 g H

= 0.06429 mol H

0.3564 g N × 1 mol N14.007 g N

= 0.02544 mol N

0.2031 g O × 1 mol O16.00 g O

= 0.01269 mol O

One point is earned for all masses converted to moles.

Note: Moles of C may be shown in part (a)(i).




7


Divide all mole quantities by the smallest number of moles: 0.05092 mol ÷ 0.01269 mol = 4.013 0.06429 mol ÷ 0.01269 mol = 5.066 0.02544 mol ÷ 0.01269 mol = 2.005 0.01269 mol ÷ 0.01269 mol = 1.000 ⇒ Empirical formula is C4H5N2O

One point is earned for dividing by the smallest number of moles.

One point is earned for the empirical formula consistent with the ratio of moles

calculated.

(b) A different compound, which has the empirical formula CH2Br , has a vapor density of 6.00 g L−1 at 375 K and 0.983 atm. Using these data, determine the following.

(i) The molar mass of the compound

PV = nRT ⇒ PVRT

= n

1 1

(0.983 atm)(1.00 L)

(0.0821 L atm mol K )(375 K)- - = 0.0319 mol

molar mass of gas (M) = 6.00 g

0.0319 mol = 188 g mol−1

OR

M = DRT

P =

1 1 16.00 g L 0.0821 L atm mol K 375 K

0.983 atm

− − −× ×

= 188 g mol−1

One point is earned for applying the gas law to calculate n.

One point is earned for calculating the molar mass.

OR

Two points are earned for calculating

the molar mass using M = DRT

P

(ii) The molecular formula of the compound

Each CH2Br unit has mass of 12.011 + 2(1.0079) + 79.90 = 93.9 g,

and 188 g

93.9 g = 2.00, so there must be two CH2Br units per molecule.

Therefore, the molecular formula of the compound is C2H4Br2 .

One point is earned for the molecular formula that is

consistent with the molar mass calculated in part (b)(i).




8

Question 4 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 reaction: 1 point for correct reactant(s) and 2 points for correct product(s). Designation of physical states is not required.

(a) Solid potassium chlorate is strongly heated.

KClO3 → KCl + O2

(b) Solid silver chloride is added to a solution of concentrated hydrochloric acid.

AgCl + Cl− → [AgCl2 ]− (c) A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide.

HC2H3O2 + OH− → H2O + C2H3O2−

(d) Ammonia gas is bubbled into a solution of hydrofluoric acid.

NH3 + HF → NH4+ + F − (e) Zinc metal is placed in a solution of copper(II) sulfate.

Zn + Cu2+ → Zn2+ + Cu (f) Hydrogen phosphide (phosphine) gas is added to boron trichloride gas.

PH3 + BCl 3 → H3PBCl 3 Note: PH3BCl3 also acceptable as a product. (g) A solution of nickel(II) bromide is added to a solution of potassium hydroxide.

Ni2+ + OH− → Ni(OH)2 (h) Hexane is combusted in air.

C6H14 + O2 → CO2 + H2O




9

Question 5

5. Three pure, solid compounds labeled X , Y , and Z are placed on a lab bench with the objective of

identifying each one. It is known that the compounds (listed in random order) are KCl , Na2CO3 , and MgSO4 . A student performs several tests on the compounds; the results are summarized in the table below.

Compound pH of an Aqueous


Result of Adding 1.0 M NaOH to a


Result of Adding 1.0 M HCl Dropwise to

the Solid Compound

X > 7 No observed reaction Evolution of a gas

Y 7 No observed reaction No observed reaction

Z 7 Formation of a white

precipitate No observed reaction

(a) Identify each compound based on the observations recorded in the table.

Compound X ______ Na2CO3 _______

Compound Y ______ KCl _________

Compound Z _____ MgSO4 _________

One point is earned for one correct identification, and a second point is earned for a second correct identification.

(No points are earned if all three identifications are the same compound; no second point is earned if two identifications are the same compound.)

(b) Write the chemical formula for the precipitate produced when 1.0 M NaOH is added to a solution of compound Z .

Mg(OH)2 One point is earned for the correct formula.

(c) Explain why an aqueous solution of compound X has a pH value greater than 7. Write an equation as part of your explanation.

CO32− reacts with water to form OH−.

CO3

2−(aq) + H2O(l) → OH −(aq) + HCO3−(aq)

One point is earned for identifying CO32−

as a base.

One point is earned for a correct equation.




10


(d) One of the testing solutions used was 1.0 M NaOH . Describe the steps for preparing 100. mL of 1.0 M NaOH from a stock solution of 3.0 M NaOH using a 50 mL buret, a 100 mL volumetric flask, distilled water, and a small dropper.

1,000 mL of 1.0 M NaOH contains 1.0 mol NaOH; therefore, 100. mL of 1.0 M NaOH contains 0.10 mol NaOH (i.e., 0.10 mol NaOH is needed for the solution)

volume of 3.0 M NaOH needed = 0.10 mol NaOH × 1,000 mL

3.0 mol NaOH

= 33 mL

Step 1: Use the buret to deliver 33 mL of the 3.0 M NaOH stock solution into the clean 100 mL volumetric flask.

Step 2: Add distilled water to the flask until the liquid level is just below the calibration line in the neck of the flask; swirl gently to mix.

Step 3: Use the small dropper to add the last amount of distilled water, drop by drop, until the bottom of the meniscus in the flask neck is level with the calibration line. Insert the stopper, and invert the flask to mix.


using the buret to dispense 33 mL of

NaOH(aq).


adding distilled water to the calibration

mark.

(e) Describe a simple laboratory test that you could use to distinguish between Na2CO3(s) and CaCO3(s).

In your description, specify how the results of the test would enable you to determine which compound

was Na2CO3(s) and which compound was CaCO3(s) .

A water solubility test would work. Put a small amount of one substance in a beaker of distilled water. If the substance dissolves readily when stirred, then it is Na2CO3 ; if it does not dissolve, it is CaCO3 .

OR

A flame test would work. Dip a moistened wire into a sample of one of the substances and place the wire in the flame of a bunsen burner. If a bright orange-yellow color is observed, then the sample is Na2CO3 ; if a brick red color is observed, it is CaCO3 . Note: The student does NOT have to perform a confirmatory test on

the other substance if one has already been identified with a test.

One point is earned for any reasonable test.

One point is earned for interpreting the results that

will identify one compound.




11

Question 6

6. Answer each of the following in terms of principles of molecular behavior and chemical concepts.

(a) The structures for glucose, C6H12O6 , and cyclohexane, C6H12 , are shown below.

Identify the type(s) of intermolecular attractive forces in

(i) pure glucose

Hydrogen bonding OR dipole-dipole interactions OR van der Waals interactions (London dispersion forces may also be mentioned.)

One point is earned for a correct answer.

(ii) pure cyclohexane

London dispersion forces

One point is earned for London dispersion forces.

(b) Glucose is soluble in water but cyclohexane is not soluble in water. Explain.

The hydroxyl groups in glucose molecules can form strong hydrogen bonds with the solvent (water) molecules, so glucose is soluble in water. In contrast, cyclohexane is not capable of forming strong intermolecular attractions with water (no hydrogen bonding), so the water-cyclohexane interactions are not as energetically favorable as the interactions that already exist among polar water molecules.

OR • Glucose is polar and cyclohexane is nonpolar. • Polar solutes (such as glucose) are generally soluble in

polar solvents such as water. • Nonpolar solutes (such as cyclohexane) are not soluble in

the polar solvent.

One point is earned for explaining the solubility of glucose in terms of hydrogen bonding or dipole-dipole

interactions with water.

One point is earned for explaining the difference in the polarity of

cyclohexane and water.

OR

One point is earned for any one of the three concepts; two points are earned

for any two of the three concepts.




12


(c) Consider the two processes represented below.

Process 1: H2O(l) → H2O(g) ∆H ° = + 44.0 kJ mol−1

Process 2: H2O(l) → H2(g) + 21 O ( )2

g ∆H ° = + 286 kJ mol−1

(i) For each of the two processes, identify the type(s) of intermolecular or intramolecular attractive

forces that must be overcome for the process to occur.

In process 1, hydrogen bonds (or dipole-dipole interactions) in liquid water are overcome to produce distinct water molecules in the vapor phase.

In process 2, covalent bonds (or sigma bonds, or electron-pair bonds) within water molecules must be broken to allow the atoms to recombine into molecular hydrogen and oxygen.

One point is earned for identifying the type of intermolecular force involved

in process 1.

One point is earned for identifying the type of intramolecular bonding

involved in process 2.


When water boils, H2O molecules break apart to form hydrogen molecules and oxygen molecules.

I disagree with the statement. Boiling is simply Process 1, in which only intermolecular forces are broken and the water molecules stay intact. No intramolecular or covalent bonds break in this process.

One point is earned for disagreeing with the statement and providing a

correct explanation.




13


(d) Consider the four reaction-energy profile diagrams shown below.

(i) Identify the two diagrams that could represent a catalyzed and an uncatalyzed reaction pathway for the same reaction. Indicate which of the two diagrams represents the catalyzed reaction pathway for the reaction.

Diagram 1 represents a catalyzed pathway and diagram 2 represents an uncatalyzed pathway for the same reaction.

One point is earned for identifying the correct graphs and indicating which

represents which pathway.


Adding a catalyst to a reaction mixture adds energy that causes the reaction to proceed more quickly.

I disagree with the statement. A catalyst does not add energy, but provides an alternate reaction pathway with a lower activation energy.

One point is earned for disagreeing with the statement and providing

an explanation.




14

Question 7

7. Answer the following questions about the structures of ions that contain only sulfur and fluorine.

(a) The compounds SF4 and BF3 react to form an ionic compound according to the following equation.

SF4 + BF3 → SF3BF4

(i) Draw a complete Lewis structure for the SF3+ cation in SF3BF4.

One point is earned for the correct Lewis structure (the structure must include lone pairs of electrons, which may be represented as dashes).

(ii) Identify the type of hybridization exhibited by sulfur in the SF3+ cation.

sp3

One point is earned for the correct hybridization.

(iii) Identify the geometry of the SF3+ cation that is consistent with the Lewis structure drawn

in part (a)(i).

Trigonal pyramidal

One point is earned for the correct shape.

(iv) Predict whether the F–S–F bond angle in the SF3+ cation is larger than, equal to, or smaller

than 109.5°. Justify your answer.

The F–S –F bond angle in the SF3+ cation is expected to be slightly

smaller than 109.5° because the repulsion between the nonbonding pair of electrons and the S –F bonding pairs of electrons “squeezes” the F–S –F bond angles together slightly.

One point is earned for stating that the angle is

smaller, with justification.




15


(b) The compounds SF4 and CsF react to form an ionic compound according to the following equation.

SF4 + CsF → CsSF5

(i) Draw a complete Lewis structure for the SF5− anion in CsSF5.

One point is earned for the correct Lewis structure (the structure must include lone pairs of electrons, which

may be represented as dashes).

(ii) Identify the type of hybridization exhibited by sulfur in the SF5− anion.

sp3d2

One point is earned for the correct hybridization.

(iii) Identify the geometry of the SF5− anion that is consistent with the Lewis structure drawn

in part (b)(i).

Square pyramidal

One point is earned for the correct shape.

(iv) Identify the oxidation number of sulfur in the compound CsSF5.

+ 4

One point is earned for the correct oxidation number.




16

Question 8

8. Suppose that a stable element with atomic number 119, symbol Q , has been discovered.

(a) Write the ground-state electron configuration for Q , showing only the valence-shell electrons.

8s 1

One point is earned for the electron configuration.

(b) Would Q be a metal or a nonmetal? Explain in terms of electron configuration.

It would be a metal (OR an alkali metal). The valence electron would be held only loosely.

One point is earned for the correct answer and explanation, which must include reference to the

valence electron.

(c) On the basis of periodic trends, would Q have the largest atomic radius in its group or would it have the smallest? Explain in terms of electronic structure.

It would have the largest atomic radius in its group because its valence electron is in a higher principal shell.

One point is earned for the correct answer and explanation; the size must refer to number of

electron shells.

(d) What would be the most likely charge of the Q ion in stable ionic compounds?

+ 1 One point is earned for the correct charge.

(Must be consistent with configuration in part (a).)

(e) Write a balanced equation that would represent the reaction of Q with water.

2 Q(s) + 2 H2O(l) → 2 Q+ (aq) + 2 OH −(aq) + H2(g)

One point is earned for H2 as a product.

One point is earned for balancing the equation.

(f) Assume that Q reacts to form a carbonate compound.

(i) Write the formula for the compound formed between Q and the carbonate ion, CO32− .

Q2CO3 One point is earned for the formula consistent

with the charge given in part (d).

(ii) Predict whether or not the compound would be soluble in water. Explain your reasoning.

It would be soluble in water because all alkali metal carbonates are soluble.

One point is earned for the answer consistent with the identification of Q.


AP® Chemistry 2006 Free-Response Questions

Form B






INFORMATION IN THE TABLE BELOW AND IN THE TABLES ON PAGES 3-5 MAY BE USEFUL IN ANSWERING THE QUESTIONS IN THIS SECTION OF THE EXAMINATION.



STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25°C

Half-reaction E °( )V

2F ( ) 2g e-+ Æ 2 F- 2.87

3+Co e-+ Æ 2Co + 1.82 3+Au 3 e-+ Æ Au( )s 1.50 2Cl ( ) 2g e

-+ Æ 2 Cl- 1.36

+2O ( ) 4 H 4g e-+ + Æ 22 H O( )l 1.23

2Br ( ) 2l e-+ Æ 2 Br- 1.07

2+2 Hg 2 e-+ Æ 2+2Hg 0.92

2+Hg 2 e-+ Æ Hg( )l 0.85

+Ag e-+ Æ Ag( )s 0.80

2+2Hg 2 e-+ Æ 2 Hg( )l 0.79

3+Fe e-+ Æ 2+Fe 0.77

2I ( ) 2s e-+ Æ 2 I- 0.53

+Cu e-+ Æ Cu( )s 0.52

2+Cu 2 e-+ Æ Cu( )s 0.34 2+Cu e-+ Æ +Cu 0.15 4+Sn 2 e-+ Æ 2+Sn 0.15 +S( ) 2 H 2s e-+ + Æ 2H S( )g 0.14

+2 H 2 e-+ Æ 2H ( )g 0.00 2+Pb 2 e-+ Æ Pb( )s – 0.13 2+Sn 2 e-+ Æ Sn( )s – 0.14 2+Ni 2 e-+ Æ Ni( )s – 0.25 2+Co 2 e-+ Æ Co( )s – 0.28 2+Cd 2 e-+ Æ Cd( )s – 0.40 3+Cr e-+ Æ 2+Cr – 0.41 2+Fe 2 e-+ Æ Fe( )s – 0.44 3+Cr 3 e-+ Æ Cr( )s – 0.74 2+Zn 2 e-+ Æ Zn( )s – 0.76 22 H O( ) 2l e

-+ Æ 2H ( ) + 2 OHg- – 0.83

2+Mn 2 e-+ Æ Mn( )s – 1.18 3+Al 3 e-+ Æ Al( )s – 1.66 2+Be 2 e-+ Æ Be( )s – 1.70 2+Mg 2 e-+ Æ Mg( )s – 2.37 +Na e-+ Æ Na( )s – 2.71 2+Ca 2 e-+ Æ Ca( )s – 2.87 2+Sr 2 e-+ Æ Sr( )s – 2.89 2+Ba 2 e-+ Æ Ba( )s – 2.90 +Rb e-+ Æ Rb( )s – 2.92 +K e-+ Æ K( )s – 2.92 +Cs e-+ Æ Cs( )s – 2.92 +Li e-+ Æ Li( )s – 3.05



ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS

E

v n

m

p

= == == ==

energy velocity

frequency principal quantum number

wavelength mass

momentum

u

l

Speed of light, 3.0 10 m s

Planck’s constant, 6.63 10 J s

Boltzmann’s constant, 1.38 10 J K

Avogadro’s number 6.022 10 mol

Electron charge, 1.602 10 coulomb

1 electron volt per atom 96.5 kJ mo

8 1

34

23 1

23 1

19

1

l

c

h

k

e

= ¥

= ¥

= ¥

= ¥

= - ¥

=

-

-

- -

-

-

-

Equilibrium Constants

(weak acid)

(weak base)

(water)

(gas pressure)

(molar concentrations)

K

K

K

K

K

a

b

w

p

c

S

H

G

E

T

n

m

q

c

C

E

k

A

p

a

=

=

=

=======

=

standard entropy

standard enthalpy

standard free energy


temperature

moles

mass

heat

specific heat capacity

molar heat capacity at constant pressure

activation energy

= rate constant

= frequency factor


of electrons


L atm mol K

volt coulomb mol K

� =

=

=

=

- -

- -

- -

96 500

8 31

0 0821

8 31

1 1

1 1

1 1

,

.

.

.

R

ATOMIC STRUCTURE

E hv c v

p m

En

n

= =

=

= - ¥-

l

lu

u= hm

2 178 10 18

2.

joule

EQUILIBRIUM

K

K

K

a

b

b

=

=

=

+ −

− +

− + −

+ −

−

+

= ×= ×

= − = −= +

= +

= +

= − = −

=

= −

[ ] [ ][ ]

[ ] [ ][ ]

[ ] [ ] .

log [ ], log [ ]

log[ ][ ]

log[ ]

[ ]log , log

( ) ,

H AHA

OH HBB

OH H @ 25 C

pH H pOH OH

pH pOH

pH pAHA

pOH pHB

Bp p

where moles product gas moles reactant gas

K

K

K

K

K K K K

K K RT

n

w

a

a

b

a a b b

p cn

1 0 10

14

14

D

D

THERMOCHEMISTRY/KINETICS

D

D D D

D D D

D D D

D D D

D

DD

S S S

H H H

G G G

G H T S

RT K RT K

n E

G G RT Q G RT Q

q mc T

CHT

f f

f f

p

= -

-

= -

-= - = -

= -

= + = +=

=

Â ÂÂ ÂÂ Â

=

=

products reactants

products reactants

products reactants

ln . log

ln . log

2 303

2 303

�

ln lnA A

A A

t

t

kt

kt

- = -

- =

0

0

1 1

ln lnkER T

Aa= - +1e j



GASES, LIQUIDS, AND SOLUTIONS

1 1 2 2

1 2

1 2

2 1

2

2

2

,

( )

moles Awheretotal moles

...

K C 273

3 3

1 per molecule23 per mole2

molarity, moles solu

u

Ê ˆÁ ˜Ë ¯

= ¥

=

=

=

+ - =

=

= + + +

=

= +

=

= =

=

=

=

M

M

M

M

A total A A

total A B C

rms

P P

PV nRT

n aP V nb nRTV

X X

P P P P

mn

PV P V

T T

mDV

kT RTum

KE m

KE RT

r

r

M te per liter solution

molality moles solute per kilogram solvent

molality

molality

D

D

p

== ¥= ¥==

f f

b b

T iK

T iK

iMRT

A abc

OXIDATION-REDUCTION; ELECTROCHEMISTRY

Q a b c d

Iqt

E ERTn

Q En

Q

KnE

c d

= + Æ +

=

= - = -

=

[ ] [ ]

[ ] [ ]

ln.

log @

log.

,C D

A Bwhere A B C D

C

cell cell cell

a b

�0 0592

25

0 0592

P

V

T

n

D

m

=======

pressure

volume

temperature

number of moles

density

mass

velocityu

u

KE

r

i

K

K

Q

I

q

t

E

K

rms

f

b

A

a

b

c

=======

=

====

==

====

root-mean-square speed

kinetic energy

rate of effusion

molar mass

osmotic pressure

van't Hoff factor

molal freezing-point depression constant

molal boiling-point elevation constant

reaction quotient

current (amperes)

charge (coulombs)

time (seconds)


equilibrium constant

absorbance

molar absorptivity

path length

concentration

M

p


L atm mol K

volt coulomb mol K

Boltzmann's constant, J K

for H O K kg mol

for H O K kg mol

STP C and atm


of electrons

atm mm Hg

torr

R

k

K

K

f

b

=

=

=

= ¥

=

=

=

- -

- -

- -

- -

-

-

==

=

8 31

0 0821

8 31

1 38 10

1 86

0 512

0 000 1 000

96 500

1 1

1 1

1 1

23 1

21

21

1 760

760

.

.

.

.

.

.

. .

,�


2006 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)



CHEMISTRY Section II

(Total time—90 minutes)

Part A Time—40 minutes

YOU MAY USE YOUR CALCULATOR FOR PART A.

CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if you do not. Attention should be paid to significant figures. Be sure to write all your answers to the questions on the lined pages following each question in the goldenrod booklet. Do NOT write your answers on the lavender insert. Answer Question 1 below. The Section II score weighting for this question is 20 percent.

C6H5COOH(s) →← C6H5COO– (aq) + H+(aq) Ka = 6.46 × 10–5

1. Benzoic acid, C6H5COOH, dissociates in water as shown in the equation above. A 25.0 mL sample of an aqueous solution of pure benzoic acid is titrated using standardized 0.150 M NaOH.

(a) After addition of 15.0 mL of the 0.150 M NaOH, the pH of the resulting solution is 4.37. Calculate each of the following.

(i) [H+] in the solution

(ii) [OH–] in the solution

(iii) The number of moles of NaOH added

(iv) The number of moles of C6H5COO– (aq) in the solution

(v) The number of moles of C6H5COOH in the solution

(b) State whether the solution at the equivalence point of the titration is acidic, basic, or neutral. Explain your reasoning.

In a different titration, a 0.7529 g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in

water and titrated with 0.150 M NaOH. The equivalence point is reached when 24.78 mL of the base solution is added.

(c) Calculate each of the following.

(i) The mass, in grams, of benzoic acid in the solid sample

(ii) The mass percentage of benzoic acid in the solid sample




7

Answer EITHER Question 2 OR Question 3 below. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 20 percent. 2. Answer the following questions about voltaic cells.

(a) A voltaic cell is set up using Al /Al3+ as one half-cell and Sn /Sn2+ as the other half-cell. The half-cells contain equal volumes of solutions and are at standard conditions.

(i) Write the balanced net-ionic equation for the spontaneous cell reaction.

(ii) Determine the value, in volts, of the standard potential, E°, for the spontaneous cell reaction.

(iii) Calculate the value of the standard free-energy change, ∆G°, for the spontaneous cell reaction. Include units with your answer.

(iv) If the cell operates until [Al3+] is 1.08 M in the Al /Al3+ half-cell, what is [Sn2+] in the Sn /Sn2+ half-cell?

(b) In another voltaic cell with Al /Al3+ and Sn /Sn2+ half-cells, [Sn2+] is 0.010 M and [Al3+] is 1.00 M. Calculate the value, in volts, of the cell potential, Ecell , at 25°C.

3. Answer the following questions about the thermodynamics of the reactions represented below.

Reaction X: 12

I2(s) + 12

Cl2(g) →← ICl(g) fHD = 18 kJ mol–1, 298SD = 78 J K–1 mol–1

Reaction Y: 12

I2(s) + 12

Br2(l) →← IBr(g) fHD = 41 kJ mol–1, 298SD = 124 J K–1 mol–1

(a) Is reaction X , represented above, spontaneous under standard conditions? Justify your answer with a calculation.

(b) Calculate the value of the equilibrium constant, Keq , for reaction X at 25°C.

(c) What effect will an increase in temperature have on the equilibrium constant for reaction X ? Explain your answer.

(d) Explain why the standard entropy change is greater for reaction Y than for reaction X .

(e) Above what temperature will the value of the equilibrium constant for reaction Y be greater than 1.0 ? Justify your answer with calculations.

(f) For the vaporization of solid iodine, I2(s) → I2(g) , the value of 298HD is 62 kJ mol–1. Using this

information, calculate the value of 298HD for the reaction represented below.

I2(g) + Cl2(g) →← 2 ICl(g)

S T O P If you finish before time is called, you may check your work on this part only.

Do not turn to the other part of the test until you are told to do so.





CHEMISTRY Part B

Time—50 minutes NO CALCULATORS MAY BE USED FOR PART B.

Answer Question 4 below. The Section II score weighting for this question is 15 percent. 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.

Example: A strip of magnesium is added to a solution of silver nitrate.

(a) Solid calcium carbonate is strongly heated.

(b) A strip of magnesium metal is placed in a solution of iron(II) chloride.

(c) Boron trifluoride gas is mixed with ammonia gas.

(d) Excess concentrated hydrochloric acid is added to a solution of nickel(II) nitrate.

(e) Solid ammonium chloride is added to a solution of potassium hydroxide.

(f) Propanal is burned in air.

(g) A strip of aluminum foil is placed in liquid bromine.

(h) Solid copper(II) sulfide is strongly heated in air.





Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy and relevance of the information cited. Explanations should be clear and well organized. Examples and equations may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses. Answer BOTH Question 5 below AND Question 6 printed on page 11. Both of these questions will be graded. The Section II score weighting for these questions is 30 percent (15 percent each). 5. A student carries out an experiment to determine the equilibrium constant for a reaction by colorimetric

(spectrophotometric) analysis. The production of the red-colored species FeSCN2+(aq) is monitored.

(a) The optimum wavelength for the measurement of [FeSCN2+] must first be determined. The plot of absorbance, A , versus wavelength, λ , for FeSCN2+(aq) is given below. What is the optimum wavelength for this experiment? Justify your answer.

(b) A calibration plot for the concentration of FeSCN2+(aq) is prepared at the optimum wavelength. The data below give the absorbances measured for a set of solutions of known concentration of FeSCN2+(aq).

Concentration

( mol L− 1 ) Absorbance

1.1 × 10 – 4 0.030 3.0 × 10 – 4 0.065 8.0 × 10 – 4 0.160

12 × 10 – 4 0.239 18 × 10 – 4 0.340





(i) Draw a Beer’s law calibration plot of all the data on the grid below. Indicate the scale on the horizontal axis by labeling it with appropriate values.

(ii) An FeSCN2+(aq) solution of unknown concentration has an absorbance of 0.300. Use the plot you drew in part (i) to determine the concentration, in moles per liter, of this solution.

(c) The purpose of the experiment is to determine the equilibrium constant for the reaction represented below.

Fe3+(aq) + SCN–(aq) →← FeSCN2+(aq)

(i) Write the equilibrium-constant expression for Kc .

(ii) The student combines solutions of Fe(NO3)3 and KSCN to produce a solution in which the initial

concentrations of Fe3+(aq) and SCN–(aq) are both 6.0 × 10 –3 M . The absorbance of this solution is measured, and the equilibrium FeSCN2+(aq) concentration is found to be 1.0 × 10 –3 M. Determine the value of Kc .

(d) If the student’s equilibrium FeSCN2+(aq) solution of unknown concentration fades to a lighter color before the student measures its absorbance, will the calculated value of Kc be too high, too low, or unaffected? Justify your answer.





GeCl4 SeCl4 ICl4– ICl4

+

6. The species represented above all have the same number of chlorine atoms attached to the central atom.

(a) Draw the Lewis structure (electron-dot diagram) of each of the four species. Show all valence electrons in your structures.

(b) On the basis of the Lewis structures drawn in part (a), answer the following questions about the particular species indicated.

(i) What is the Cl – Ge – Cl bond angle in GeCl4 ?

(ii) Is SeCl4 polar? Explain.

(iii) What is the hybridization of the I atom in ICl4– ?

(iv) What is the geometric shape formed by the atoms in ICl4+ ?




12

Answer EITHER Question 7 OR Question 8 below. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 15 percent. 7. Account for each of the following observations in terms of atomic theory and/or quantum theory.

(a) Atomic size decreases from Na to Cl in the periodic table.

(b) Boron commonly forms molecules of the type BX3. These molecules have a trigonal planar structure.

(c) The first ionization energy of K is less than that of Na.

(d) Each element displays a unique gas-phase emission spectrum.

8. Use chemical and physical principles to account for each of the following.

(a) An aluminum container filled with an aqueous solution of CuSO4 eventually developed a leak. Include a chemical equation with your answer.

(b) The inside of a metal container was cleaned with steam and immediately sealed. Later, the container imploded.

(c) Skin feels cooler after rubbing alcohol has been applied to it.

(d) The redness and itching of the skin caused by ant bites (injections of methanoic acid, HCO2H) can be relieved by applying a paste made from water and baking soda (solid sodium hydrogen carbonate). Include a chemical equation with your answer.

STOP

END OF EXAM


AP® Chemistry 2006 Scoring Guidelines

Form B





AP® CHEMISTRY 2006 SCORING GUIDELINES (Form B)


2

Question 1

C6H5COOH(s) →← C6H5COO– (aq) + H+(aq) Ka = 6.46 × 10 – 5

1. Benzoic acid, C6H5COOH, dissociates in water as shown in the equation above. A 25.0 mL sample of an

aqueous solution of pure benzoic acid is titrated using standardized 0.150 M NaOH.

(a) After addition of 15.0 mL of the 0.150 M NaOH, the pH of the resulting solution is 4.37. Calculate each of the following.

(i) [H+] in the solution

[H+] = 10− 4.37 M = 4.3 × 10− 5 M


(ii) [OH −] in the solution

[OH −] = [ H ]

wK+ =

14 2

5

1.0 10

4.3 10

M

M

−

−××

= 2.3 × 10−10 M


(iii) The number of moles of NaOH added

mol OH − = 0.0150 L × 0.150 mol L−1 = 2.25 × 10− 3 mol


(iv) The number of moles of C6H5COO– (aq) in the solution

mol OH − added = mol C6H5COO−(aq) generated, thus

mol C6H5COO−(aq) in solution = 2.25 × 10− 3 mol


(v) The number of moles of C6H5COOH in the solution

Ka = 6 5

6 5

[H ][C H COO ][C H COOH]

+ - ⇒ [C6H5COOH] =

6 5[H ][C H COO ]

aK

+ -

[C6H5COOH] =

35

5

2.25 10 mol(4.3 10 )

0.040 L

6.46 10

M-

-

-

��

� = 3.7 × 10−2 M

thus, mol C6H5COOH = (0.040 L)( 3.7 × 10−2 M) = 1.5 × 10−3 mol

One point is earned for the correct molarity.


the correct answer.




3


Alternative solution for part (a)(v):

pH = pKa + log 6 5

6 5

[C H COO ][C H COOH]

−

⇒ pH − pKa = log [C6H5COO−] − log [C6H5COOH]

⇒ log [C6H5COOH] = log [C6H5COO−] − ( pH − pKa )

= log (32.25 10 mol

0.040 L

−× ) − (4.37 − 4.190)

= −1.25 − 0.18 = −1.43

⇒ [C6H5COOH] = 10−1.43 = 3.7 × 10−2 M

thus, mol C6H5COOH = ( 0.040 L )( 3.7 × 10−2 M ) = 1.5 × 10−3 mol

(b) State whether the solution at the equivalence point of the titration is acidic, basic, or neutral. Explain your reasoning.

At the equivalence point the solution is basic due to the presence

of C6H5COO− (the conjugate base of the weak acid) that

hydrolyzes to produce a basic solution as represented below.

C6H5COO− + H2O →← C6H5COOH + OH−

One point is earned for the prediction and the explanation.

In a different titration, a 0.7529 g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved in

water and titrated with 0.150 M NaOH. The equivalence point is reached when 24.78 mL of the base solution is added.

(c) Calculate each of the following.

(i) The mass, in grams, of benzoic acid in the solid sample

mol C6H5COOH = (0.02478 L) × (0.150 mol OH− L−1) × 6 51 mol C H COOH

1 mol OH −

= 3.72 × 10− 3 mol C6H5COOH

mass C6H5COOH = 3.72 × 10− 3 mol C6H5COOH × 6 5

6 5

122 g C H COOH1 mol C H COOH

= 0.453 g C6H5COOH

One point is earned for the

correct answer.




4


(ii) The mass percentage of benzoic acid in the solid sample

mass % C6H5COOH = 6 50.453 g C H COOH

0.7529 g× 100

= 60.2%





5

Question 2

2. Answer the following questions about voltaic cells.

(a) A voltaic cell is set up using Al /Al3+ as one half-cell and Sn /Sn2+ as the other half-cell. The half-cells contain equal volumes of solutions and are at standard conditions.

(i) Write the balanced net-ionic equation for the spontaneous cell reaction.

3 Sn2+ + 2 Al → 3 Sn + 2 Al 3+ One point is earned for the correct direction.

One point is earned for the balanced net-ionic equation.

(ii) Determine the value, in volts, of the standard potential, E°, for the spontaneous cell reaction.

E° = − 0.14 V − (−1.66 V) = 1.52 V (or, 1.52 J C−1)


(Potential must be positive.)

(iii) Calculate the value of the standard free-energy change, ∆G°, for the spontaneous cell reaction. Include units with your answer.

∆G° = − n F E° = − 6 mol1 mol

e− × 96,500 C

1 mol e− × (1.52 J C−1)

= − 8.80 × 105 J mol−1 (or – 880 kJ mol−1)

One point is earned for indicating the correct mol e− to mol reaction ratio.

One point is earned for the correct answer with correct units.

(iv) If the cell operates until [Al3+] is 1.08 M in the Al /Al3+ half-cell, what is [Sn2+] in the Sn /Sn2+ half-cell?

change in [Sn2+] = 3+0.08 mol Al

1 L ×

2+

3+

3 mol Sn

2 mol Al =

2+0.12 mol Sn1 L

[Sn2+] = 1.00 mol L−1 – 0.12 mol L−1 = 0.88 mol L−1


(b) In another voltaic cell with Al /Al3+ and Sn /Sn2+ half-cells, [Sn2+] is 0.010 M and [Al3+] is 1.00 M. Calculate the value, in volts, of the cell potential, Ecell , at 25°C.

Ecell = 1.52 V −

0.0592

6 log

2

3(1.00)

(0.010)

= 1.52 V − 0.0592 V = 1.46 V

Answers must be consistent with part (a)(i). One point is earned for the proper exponents.

One point is earned for the correct substitution of concentrations.





6

Question 3

3. Answer the following questions about the thermodynamics of the reactions represented below.

Reaction X: 12

I2(s) + 12

Cl2(g) →← ICl(g) fH∆ = 18 kJ mol–1, 298S∆ = 78 J K–1 mol–1

Reaction Y: 12

I2(s) + 12

Br2(l) →← IBr(g) fH∆ = 41 kJ mol–1, 298S∆ = 124 J K–1 mol–1

(a) Is reaction X

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