Download CHEMSTRY FREE-RESPONSE QUESTIONS (Form B)

CHEMSTRY FREE-RESPONSE QUESTIONS (Form B)
1. Answer the following problems about gases.
a) The average atomic mass of naturally occurring neon is 20.18 amu. There are two
common isotopes of naturally occurring Neon as indicated in the table below.
Isotope
Ne-20
Ne-22
Mass (amu)
19.99
21.99
(i)
Using the information above, calculate the percent abundance of each
isotope.
Let x represent the natural abundance of Ne-20.
19.99 x  21.99(1  x)  20.18
19.99 x  21.99  21.99 x  20.18
2 x  1.81
x  0.905
Therefore, percent abundances are: Ne-20: 90.5%
Ne-22: 1 – 90.5% = 9.5%
(ii)
Calculate the number of N2-22 atoms in a 12.55 g sample of naturally
occurring neon.
12.55 g Ne 
1 mol Ne 0.095 mol Ne -22 6.022  1023 Ne-22 atoms


 3.6  1022 Ne-22 atoms
20.18 g Ne
1 mol Ne
1 mol Ne - 22
b) A major line in the emission spectrum of neon corresponds to a frequency of 1.34
x 10 14 s1. Calculate the wavelength, in nanometers, of light that corresponds to
this line.
c
c     

3.0  10 m/s
 6.9  107 m = 690 nm
14
4.34  10 Hz
c) In the upper atmosphere, ozone molecules decompose as they absorb ultraviolet
(UV) radiation, as shown by the equation below, Ozone serves to block harmful
ultraviolet radiation that comes from the sun.

8
UV
O3(g)
→
O2(g) + O(g)
A molecule of O3(g) absorbs a photon with a frequency of 1.00 x 1015 s-1.
(i)
How much energy, in joules, does the O3(g) molecule absorb per photon?
E  h   6.63  1034 J  s 1.00  1015 s-1   6.63  1019 J per photon
(ii)
The minimum energy needed to break and oxygen-oxygen bond in ozone
is 387 kJ mol-1. Does a photon with frequency of 1.00 x 1015 s-1 have
enough energy to break this bond? Support your answer with a calculation.
6.63  1019 J 6.022  1023 photons 1 kJ

 3  399 kJ  mol-1  387 kJ  mol-1
1 photon
1 mol
10 J
Therefore, the bond can be broken.
2.
The following questions relate to the synthesis reaction represented by the chemical
equation in the box above.
(a) Calculate the value of the standard free energy change, ΔSo298, for the reaction.
G298  H 298  T S298

 264 kJ  mol-1   298 K  0.278 kJ  mol-1  K 1

 181 kJ  mol
(b) Determine the temperature at which the equilibrium constant Keq, for the reaction is
equal to 1 .00 (Assume that ΔHo and ΔSo are independent of temperature).
When Keq = 1, then GT  RT ln Keq  RT ln1  0
Therefore,
GT  HT  T ST  0
-1
HT
264 kJ  mol-1

 950 K
ST 0.278 kJ  mol-1  K 1
(c) Calculate the standard enthalpy change, ΔHo, That occurs when 0.256 mol sample of
NF3(g) is formed from N2(g) and F2(g) at 1.0 atm and 298K.
From the chemical equation, -264 kJ standard enthalpy change occurs when 2 moles of
NF3(g) are produced at 1.0 atm and 298 K, so
264 kJ
0.256 mol NF3 ( g ) 
 33.8 kJ
2.00 mol NF3 ( g )
T
The enthalpy change in a chemical reaction is the difference between energy absorbed in
breaking bonds in the reactants and energy released by bond formation, in the products.
(d) How many bonds are formed when two molecules of NF3 are produced according to
the equation in the box above?
1 each molecule of NF3, there are 3 N-F bonds.
Therefore, there are 6 N-F bonds when two molecules of NF3 are produced.
(e) Use both the information in the box above and the table of average bond enthalpies
below to calculate the average enthalpy of the F – F bond.
Bond
Average Bond Enthalpy
NN
N-F
F-F
946
272
?
In the above chemical reaction, 1 N=N bond and 3 F-F bonds are broken. Meanwhile, 6
N-F bonds are formed.
H 298   Ebonds broken  Ebonds formed  264 kJ  mol-1
  BEN  N  3BEF  F    6 BEN  F 

 
 946 kJ  mol-1  3BEF  F  6 272 kJ  mol-1
3BEF  F   264  946  1632  kJ  mol

-1
BEF  F  141 kJ  mol-1
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 C02(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.
All C in CO2 comes from the C in the sample.
Since 2.241 g of CO2 (g) is produced, the moles of C in CO2 is then
2.241 g / 44.01 g/mol = 0.051 mol CO2 is produced.
Therefore, there are 0.051 mol of C in CO2. that is, there are 0.051 mol of C in the
sample.
The mass of C in the sample is
0.051 mol * 12.011 g/mol = 0.6116 g C.
1 mol CO2
1 mol C
12.011 g C
2.241 g CO2 ( g ) 


 0.6116 g C
44.01 g CO2 1 mol CO 2
1 mol C
(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.
Since N is 28.84% in mass, the mass in the sample is
1.2359 g sample  28.84% = 0.3564 g N.
(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 = mass of sample – (Mass of H + C + N)
= 1.2359 – (0.0648 + 0.6116 + 0.3564)
= 0.2031 g.
(iv)
Determine the empirical formula of the compound.
First converting all masses to moles,
1 mol C
0.6116 g C 
 0.05092 mol C
12.011 g C
1 mol H
0.0648 g H 
 0.06429 mol H
1.0079 g H
1 mol N
0.3564 g N 
 0.02544 mol N
14.007 g N
1 mol O
0.2031 g O 
 0.01269 mol O
16.00 g O
Divide all mol 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
Therefore, the empirical formula is C4H5N2O
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.
Apply the ideal gas law,
PV  nRT
n
 0.983 atm 1.00 L 
PV

 0.0319 mol
RT
0.0821 L  atm  mol-1  K 1  375 K 


So the molar mass of the gas is
M = 6.00 g / 0.0319 mol = 188 g/mol.
(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 / 93.9 = 2, so there must be two CH2Br units in each molecule.
The molecular formula of the compound is C2H4Br2.
Answer question 4 below. The section II score weighting for this question is 10 percent.
4. For each of the following three reactions in part (i) write a balanced equation for the
reaction and in part (ii) answer the question about the reaction, In part (i), coefficients
should be in terms of lowest whole numbers. Assume that solutions are aqueous
unless otherwise indicated. Represent substances in solutions as ions if the substances
extremely ionized. Omit formulas for any ions or molecules chat are unchanged by
the reaction. You may use the empty space at the bottom of the ‘next page for scratch
work but only the equations that are written in the answer boxes provided will be
graded,
Example:
A strip of Magnesium is added to a solution of silver(1) nitrate
(i) Balanced equation
(ii) Which substance is oxidized in the reaction.
(a) A solution of sodium hydroxide is added to a solution of lead(II) nitrate.
(i) Balanced equation
2 OH- + Pb2+  Pb(OH)2
(ii)
If 1.0 L volumes of 1.0 M solutions of sodium hydroxide and lead(II)
nitrate are mixed together, how many moles of the product(s) will be
produced? Assume the reaction goes to completion.
______________________________________________________________________
The moles of NaOH is (1.0 L) * (1.0 M) = 1 mol.
From the balanced equation, we know that 2 moles of OH- produce 1 mole of Pb(OH)2.
Therefore, with 1 mol of NaOH, a total of 0.5 mo of Pb(OH)2 will be produced.
______________________________________________________________________
(b) Excess nitric acid is added to solid calcium carbonate.
(i) Balanced equation
2H+ + CaCO3  Ca2+ + H2O + CO2
(ii)
Briefly explain why statues made of marble (calcium carbonate) displayed
outdoors in urban areas are deteriorating. .
The H+ ions in the acid rain react with the marble statues and the soluble compounds of
Ca that are formed wash away.
______________________________________________________________________
(c) A solution containing silver(1) ion (un oxidizing agent) is mixed with a solution
containing iron(II) ion (a reducing agent).
(i) Balanced equation
Ag+ + Fe2+  Ag + Fe3+
(ii)
If the contents of the reaction mixture described above are filtered, what
substance(s), if any, would remain on the filter paper.
______________________________________________________________________
The precipitated solid silver will remain on the filter paper.
______________________________________________________________________
(a) Solid ammonium carbonate decomposes as it is heated.
(i) Balanced equation
(NH4)2CO3  2NH3 + CO2 + H2O
(ii)
Predict the algebraic sign ΔSo for the reaction. Explain your reasoning.
______________________________________________________________________
The algebraic sign of ΔSo for the reaction will be positive because one mole of solid
(with relatively low entropy) is converted into four moles of gas (with much greater
entropy).
______________________________________________________________________
(b) Chlorine gas, an oxidizing agent, is bubbled into a solution of potassium bromide.
(i) Balanced equation
Cl2  2 Br   2Cl   Br2
(ii)
What is the oxidation number of chlorine before the reaction occurs. What
is the oxidation number of chlorine after the reaction occurs.
___________________________________________________________________
The oxidation number of chlorine is 0 before the reaction since it exists as Cl2 and -1
after the reaction.
______________________________________________________________________
(c) A small piece of sodium is placed in a beaker of distilled water.
(i) Balanced equation
2Na + 2H2O  H2 + 2Na+ + 2OH-
(ii)
The reaction is exothermic, and sometimes small flames are observed as
the sodium reacts with the water. Identify the product of the reaction that
burns to produce the flames.
______________________________________________________________________
It is the H2 gas that burns.
______________________________________________________________________
Answer Question 5 and Question 6. The Section 11 score weighting for these questions
is 15 percent each.
Your responses to these Questions 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, diffusive responses.
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) —> 5Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
5. 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)
The oxidation number of O in MnO4- is -2. The total is -1
So assume Mn as x
x + 2 * (-4) = -1
x = +7
The oxidation number of Mn is +7 in the permanganate ion, MnO4-(aq).
(b) Identify the reducing agent in the reaction represented above.
The oxidation number of Fe changes from +2 to +3 during the reaction. Therefore, it
loses an electron.
So the reducing agent is Fe2+ (aq).
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.
The solution in the flask changes from colorless to faint purple-pink at the end point of
the titration.
At the endpoint, there is no Fe2+ (aq) left in the flask to reduce the colored permanganate
ion, MnO4-(aq), so when a small amount of permanganate ion is added after the endpoint,
the un-reacted permanganate ion present in the solution colors the solution to faint
purple/pink.
(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 a 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.
(i) The number of moles of iron in the sample
The moles of permanganate ion, MnO4-(aq) is M* V.
From the balanced equation, we know that 1 mole permanganate ion, MnO4-(aq) reacts
with 5 moles of Fe2+.
So the mole of Fe2+ in the sample is 5 * M * V.
(ii) The mass of iron in the sample, in grams
The molar mass of Fe is 55.85 g/mol.
So Mass of Fe = 5 * M * V * 55.85
(iii) The mass percent of iron in the compound
Mass of Fe
5  M  V  55.85
Mass % Fe =
 100=
 100
Mass of smaple
g
(e) 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.
The experimentally determined mass percent of iron in the compound will be too large.
When more titrant is added, V is too large, so expression in (d) (iii) above is too large.
6. Answer the fol1owing questions, which pertain to binary compounds.
(a)
In the box provided below draw a complete Lewis electron - dot diagram for the
IF3 molecule.
(b)
On the basis of the Lewis electron—dot diagram that you drew (a), predict the
molecular geometry of the IF3 molecule.
Trigonal planar electronic geometry or T-shaped.
(c)
In the SO2 molecule, both of the bonds between sulfur and oxygen have the same
length. Explain this observation, supporting your explanation by drawing in the
box below a Lewis electron—dot diagram (or diagrams) for the SO2, molecule.
O == S == O
The bonds are the same length because they are both double bonds.
(d) On the basis of your Lewis election—dot diagram(s) in part (c), identify the
hybridization of the sulfur atom, in the SO2 molecu1e.
2
sp
The reaction between SO2(g) and O2(g) to form SO3(g) is represented below
2 SO2(g) + O2(g)
↔
3 SO3(g)
The reaction is exothermic. The reaction is slow a 25oC; however, a catalyst will cause
the reaction to proceed faster.
(e) Using the axes provided in the next page draw the complete potential—energy
diagram for both the catalyzed and uncatalyzed reactions. Clearly label the curve that
represents the catalyzed reaction.
For the uncatalyzed reaction curve, Ea  0 and H  0 for this exothermic reaction.
For the catalyzed curve, Ea < uncatalyed Ea, and the catalyzed reaction begin and end at
the same energies as the uncatalyzed curve.
(f)
Predict how the ratio of the equilibrium pressures PSO2/PSO3, would
change when the temperature of the uncatalyzed reaction mixture is
increased. Justify your prediction.
The ratio PSO2/PSO3 would increase as the temperature increases. Because the
reaction is exothermic ( H  0 ), as the temperature is raised, the reaction shifts
to the left-hand side.
(g)
How would the presence of a catalyst affect the change in the ration
described in part (f)? Explain.
The catalyst would not affect the value of the two equilibrium ratios but would
increase the rate of the shifting of the system to the new equilibrium position. The
catalyst does this by providing an alternate path with a lower activation energy.
6. Answer each of the following in terms of principles of molecular behavior arid
chemical concepts.
(a) The structures of 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 interaction,
(ii)
pure cyclohexane
London dispersion forces
(b) Glucose is soluble in water but cyclohexane is not soluble in water. Explain.
Glucose is polar and cyclohexane is nonpolar.
Polar solutes are generally soluble in polar solvents such as water.
Nonpolar solutes are not soluble in polar solvents.
(c) Consider the processes represented below:
Process 1: H2O(l)
→
H2O(g)
ΔHo = +44.0 kJ mol-1
Process 2: H2O(l)
→
H2(g) + 1/2 O2 (g)
ΔHo = +286.0 kJ mol-1
(i) For each of the two processes, identify the type(s) of intermolecular or
intraamolecular attractive forces that must be overcome for the process to occur.
In process 1, hydrogen bonds in liquid water are overcome to produce distinct water
molecules in the vapor phase.
In process 2, covalent bonds within water molecules must be broken to allow the atoms to
recombine into molecular hydrogen and oxygen.
(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.
Disagree.
Boiling is simply process 1, in which only intermolecular forces are broken and the water
molecules stay intact. No covalent bonds break in this process.
(d)
Consider the four reaction-energy profile diagrams
(i)
Identify the two diagrams that could represent catalyzed and an
uncatalyzed reaction. Indicate which of the two diagrams
represents the catalyzed reaction pathway for the reaction.
For the catalyzed curve, Ea < uncatalyed Ea, and the catalyzed reaction begin and
end at the same energies as the uncatalyzed curve.
Therefore, Diagram 1 represents a catalyzed pathway and diagram 2 represents an
uncatalyzed pathway for the same reaction.
(ii)
Indicate whether you agree or disagree with the statement in the
box below. Support your answer with a short explanation.
Adding a catalyst to a reaction mixture adds energy that
causes the reaction to proceed more quickly.
Disagree.
A catalyst does not add energy, but provides an alternate reaction pathway with a lower
activation enrgy.
2007 AP CHEMISTRY FREE-RESPONSE QUESTIONS (FORM B)
Element 1
Element 2
Element 3
Element 4
First
Ionization Energy
kJ mol-1
1.251
496
738
1,000
Second
Ionization Energy
kJ mol-1
2,300
4,560
1,450
2,250
Third
Ionization Energy
kJ mol-1
3,820
6,910
7,730
3,360
6. 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
(a)
Which element is most metallic in character? Explain your reasoning.
Element 2.
It has the lower first-ionization energy. Metallic elements lose electrons when
they become ions, and element 2 requires the least amount of energy to remove an
electron.
(b)
Identify element 3. Explain your reasoning.
Magnesium Mg.
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 Mg.
(c)
Write the complete electron configuration for an atom of element 3.
2
2
1s 2s 2 p 6 3s 2
(d)
What is the expected oxidation state for the most common ion of element
2?
+1. The 2nd and 3rd ionization energy is dramatically higher. So the most common
oxidation state should be +1.
(e)
Na
What is the chemical symbol for element 2?
(f)
A neutral atom of which of the four elements has the smallest radius?
Element 1 because it has the largest 1st ionization energy.
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 arid fluorine.
(a)
The compounds SF4 arid BF3 react to form and ionic compound according
to the following equation.
SF4 + BF3 —> SF3 BF4
(i)
(ii)
Draw a complete Lewis structure for the SF3+ cation in SF3BF4
Identify the type of hybridization exhibited by sulfur in the SF3+
cation.
The hybridization is sp3
Identify the geometry o the SF3+ cation that is consistent with the
Lewis structure drawn in part (a)(i).
Trigonal pyramidal
(iii)
(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+ caton is expected to be slightly smaller
than 109.5° because the repulsion between the nonbonding pair of
electrons and the S-F bonding pairs electrons squeezes the F-S-F bond
angles together slightly.
(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 SF5anion.
sp3d 2
(iii)
Identify the geometry of the SF5- anion that is consistent with the
Lewis structure drawn in part (b)(i).
Square pyramidal
(iv)
Identify the oxidation number of sulfur in the compound CsSF5.
Cs: +1
F: -1
The total is 0
+1 + x + 5*(-1) = 0
x = +4
so the oxidation number of sulfur is +4 in the compound CsSF5.