Advanced Placement Chemistry
1995 Free Response Questions
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1) Solve the following problem.
H2(g) + CO2(g) <===> H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation
above. In one experiment, the following equilibrium concentrations were measured.
[H2] = 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L
(a) What is the mole fraction of CO(g) in the equilibrium mixture?
(b) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium
constant for the reaction.
(c) Determine Kp, in terms of Kc for this system.
(d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is
converted back to CO2(g). Calculate the value of Kc at this lower temperature.
(e) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter
reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO(g)
at this temperature.
2) Propane, C3H8, is a hydrocarbon that is commonly used as fuel for cooking.
(a) Write a balanced equation for the complete combustion of propane gas, which yields CO2(g)
and H2O(l)
(b) Calculate the volume of air at 30°C and 1.00 atmosphere that is needed to burn completely
10.0 grams of propane. Assume that air is 21.0 percent O2 by volume.
(c) The heat of combustion of propane is -2,220.1 kJ/mol. Calculate the heat of formation, H°f,
of propane given that H°f of H2O(l) = -285.3 kJ/mol and H°f of CO2(g) = -393.5 kJ/mol.
(d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00
kilograms of water (specific heat = 4.18 J/g ⋅ K), calculate the increase in temperature of the
water.
3) A sample of dolomitic limestone containing only CaCO3 and MgCO3 was analyzed.
(a) When a 0.2800 gram sample of this limestone was decomposed by heating, 75.0 milliliters of
CO2 at 750 mm Hg, and 20°C were evolved. How many grams of CO2 were produced?
(b) Write equations for the decomposition of both carbonates described above.
(c) It was also determined that the initial sample contained 0.0448 gram of calcium. What
percent of the limestone by mass was CaCO3?
(d) How many grams of the magnesium-containing product were present in the sample in (a)
after it had been heated?
4) Give the formulas to show the reactants and the products for FIVE of the following chemical
reactions. Each of the reactions occurs in aqueous solution unless otherwise indicated. Represent
substances in solution as ions if the substance is extensively ionized. Omit formulas for any ions
or molecules that are unchanged by the reaction. In all cases a reaction occurs. You need not
balance.
Example: A strip of magnesium is added to a solution of silver nitrate.
Mg + Ag+ ---> Mg2+ + Ag
(a) Ethanol is burned in oxygen.
(b) Solid barium oxide is added to distilled water.
(c) Chlorine gas is bubbled into a cold, dilute solution of potassium hydroxide.
(d) A solution of iron(II) nitrate is exposed to air for an extended period of time.
(e) Excess concentrated sulfuric acid is added to solid calcium phosphate.
(f) Hydrogen sulfide gas is bubbled into a solution of mercury(II) chloride.
(g) Solid calcium hydride is added to distilled water.
(h) A bar of zinc metal is immersed in a solution of copper(II) sulfate.
5) The conductivity of several substances was tested using the apparatus represented by the
diagram below. The results of the tests are summarized in the following data table.
AgNO3
Sucrose
Na
H2SO4
212°
185°
99°
Liquid at
Room Temp.
Liquid
(fused)
++
-
++
+
Water
solution
++
-
++(1)
++(2)
Solid
-
-
++
Not Tested
Melting
Point
(°C)
Key:
++
Good conductor
of flammable gas
+
Poor conductor
added slowly and carefully to water
Nonconductor
(1) Dissolves, accompanied by evolution
(2) Conduction increases as the acid is
Using models of chemical bonding and atomic or molecular structure, account for the differences
in conductivity between the two samples in each of the following pairs.
(a) Sucrose solution and silver nitrate solution
(b) Solid silver nitrate and solid sodium metal
(c) Liquid (fused) sucrose and liquid (fused) silver nitrate
(d) Liquid (concentrated) sulfuric acid and sulfuric acid solution
6)
The phase diagram for a pure substance is shown above. Use this diagram and your knowledge
about changes of phase to answer the following questions.
(a) What does point V represent? What characteristics are specific to the system only at point V?
(b) What does each point on the curve between V and W represent?
(c) Describe the changes that the system undergoes as the temperature slowly increases from X
to Y to Z at 1.0 atmosphere.
(d) In a solid-liquid mixture of this substance, will the solid float or sink? Explain.
7) Explain the following in terms of the electronic structure and bonding of the compounds
considered.
(a) Liquid oxygen is attracted to a strong magnet, whereas liquid nitrogen is not.
(b) The SO2 molecule has a dipole moment, whereas the CO2 molecule has no dipole moment.
Include the Lewis (electron-dot) structures in your explanation.
(c) Halides of cobalt(II) are colored, whereas halides of zinc(II) are colorless.
(d) A crystal of high purity silicon is a poor conductor of electricity; however, the conductivity
increases when a small amount of arsenic is incorporated (doped) into the crystal.
8) Lead iodide is a dense, golden yellow, slightly soluble solid. At 25°C, lead iodide dissolves in
water forming a system represented by the following equation.
PbI2(s) <===> Pb2+ + 2 I¯ H = +46.5 kilojoules
The solubility-product constant, Ksp, for PbI2 is 7.1 x 10¯9 at 25°C.
(a) How does the entropy of the system PbI2(s) + H2O(l) change as PbI2(s) dissolves in water at
25°C? Explain.
(b) If the temperature of the system were lowered from 25°C to 15°C, what would be the effect
on the value of Ksp? Explain.
(c) If additional solid PbI2 were added to the system at equilibrium, what would be the effect on
the concentration of I¯ in the solution? Explain.
(d) At equilibrium, G = O. What is the initial effect on the value of G of adding a small
amount of Pb(NO3)2 to the system at equilibrium? Explain.
9)
(I) A2 + B2 ---> 2 AB
(II) X2 + Y2 ---> 2 XY
Two reactions are represented above. The potential-energy diagram for reaction I is shown
below. The potential energy of the reactants in reaction II is also indicated on the diagram.
Reaction II is endothermic, and the activation energy of reaction I is greater than that of reaction
II.
(a) Complete the potential-energy diagram for reaction II on the graph above.
(b) For reaction I, predict how each of the following is affected as the temperature is increased
by 20°C. Explain the basis for each prediction.
(i) Rate of reaction
(ii) Heat of reaction
(c) For reaction II, the form of the rate law is rate = k[X2]m[Y2]n. Briefly describe an experiment
that can be conducted in order to determine the values of m and n in the rate law for the reaction.
(d) From the information given, determine which reaction initially proceeds at the faster rate
under the same conditions of concentration and temperature. Justify your answer.
Advanced Placement Chemistry
1995 Free Response Answers
Notes
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[delta] and [sigma] are used to indicate the capital Greek letters.
[square root] applies to the numbers enclosed in parenthesis immediately following
All simplifying assumptions are justified within 5%.
One point deduction for a significant figure or math error, applied only once per problem.
No credit earned for numerical answer without justification.
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1)
a) two points
mole fraction for CO = moles CO ÷ moles (CO + CO2 + H2O + H2)
= 0.55 ÷ (0.55 + 0.55 + 0.20 + 0.30) = 0.55 ÷ 1.60 = 0.34 (or 34%)
Either mole fraction for CO definition above or numerator or denominator correct: one pt.
Numerator, denomiator and calculation correct or fraction 11/32: one pt.
One pt. deducted if units included with answer.
b) two points
Kc = ([H2O] [ CO]) ÷ ([H2] [CO2]) = [ (0.55) (0.55) ] ÷ [ (0.20) (0.30) ] = 5.0
Correct set up and/or correct substitution: one pt.
Calculated answer: one pt.
One pt. deducted if expression(s) inverted but otherwise correct.
One pt. deducted if H2O missing.
c) one point
Kp = Kc (RT)[delta]n ; [delta]n = 0 therefore Kp = Kc (or, Kp = 5.0)
The above earns one pt. Note: consistent [delta]n calculation is allowed. Numerical value same
as part (b) allowed.
d) two points
moles CO reacting = moles H2O reacting = moles CO2 formed = 0.16 moles
equilibrium [ ]'s at lower T:
[H2] = 0.36 mol/L
[CO2] = 0.46 mol/L
[H2O] = [CO] = 0.39 mol/L
Kc = (0.39)2 ÷ [ (0.46)(0.36) ] = 0.92
e) two points
x = number of moles that react
[H2]
[CO2]
[H2O]
[CO]
Initial
0.50/3.0
0.50/3.0
0
0
Change
-x/3.0
-x/3.0
x/3.0
x/3.0
x/3.0
x/3.0
Equilibrium (0.50-x)/3.0 (0.50-x)/3.0
[x2 / 9] ÷ [(0.50-x) / 3.0]2 = 5.0
x ÷ (0.50 - x) = 2.2
x = 0.34 mol
[CO] = 0.34 mol ÷ 3.0 L = 0.11 mol/L
Note: without final correct [CO], one point awarded if:
all 3.0's in denominators missing, or
all four [ ]'s substituted correctly, or
any two [ ]'s and proper K substituted correctly
2)
a) one point
C3H8 + 5 O2 --> 3 CO2 + 4 H2O
ignore phases (even when wrong)
multiples are OK
if balanced wrong, parts b and c should be consistent
b) four points
10.0 g C3H8 x (1 mol C3H8 / 44.1 g C3H8) = 0.227 mol C3H8
0.227 mol C3H8 x (5 mol O2 / 1 mol C3H8) = 1.13 mol O2
V = [ (1.13 mol O2) (0.0821 L atm mol¯1 K¯1) (303K) ] ÷ 1.00 atm = 28.1 L O2
28.1 L O2 x (100 L air / 21.0 L O2) = 134 L air
Note: answer must be consistent with part a
c) two points
[delta]H°rxn = [sigma] [delta]H°f (products) - [sigma] [delta]H°f (reactants)
-2,220.1 kJ = [ 4 (-285.3kJ) + 3 (-393.5 kJ)] - [5 (0 kJ) + [delta]H°f (C3H8)]
-2,220.1 kJ = -1,141.2 kJ - 1,180.5 kJ - [delta]H°f (C3H8)
-2,220.1 kJ = -2,321.7 kJ - [delta]H°f (C3H8)
-101.6 kJ = [delta]H°f (C3H8)
answer should be consistent with part a
1 point deducted if negative sign missing from answer
1 point deducted if -2,220.1 kJ substituted for [delta]H°f (C3H8)
no points earned if coefficients are inconsistent and not set equal to [delta]H°
d) two points
30.0 g C3H8 x (1 mol C3H8 / 44.1 g C3H8) x (2,220.1 kJ / 1 mol C3H8) = 1.51 x 103 kJ
1.51 x 103 kJ = 1.51 x 106 J = (8,000 g) (4.18 J g¯1 K¯1) ([delta]T)
45.1 K (or °C) = [delta]T
must correctly substitute into q = mCp[delta]T for 1 point
1 point earned if q value wrong but [delta]T consistent
3)
a) three points
75.0 mL = 0.0750 L = V
n = PV / RT = [ (750/760) atm x 0.0750L ] / [ (0.0821 L atm mol¯1 K¯1) (20 °C + 273) ] =
0.00308 mol CO2
= 3.08 x 10¯3 mol CO2
3.08 x 10¯3 mol CO2 x (44.0 g CO2 / 1 mol CO2) = 0.136 g CO2 = 1.36 x 10¯1 g CO2
b) two points
CaCO3(s) ---> CO2(g) + CaO(s)
MgCO3(s) ---> CO2(g) + MgO(s)
c) two points
0.0448 g Ca x (1 mol Ca / 40.08 gCa) x (1 mol CaCO3 / 1 mol Ca) x (100.0 g CaCO3 / 1 mol
CaCO3) = 0.112 g CaCO3
(0.112 g CaCO3 / 0.2800 g sample) x 100% = 40.0%
d) two points
0.2800 g sample x 60% = 0.168 g MgCO3
0.168 g MgCO3 x (1 mol MgCO3 / 84.31 g MgCO3) x (1 mol MgO / 1 mol MgCO3) x (40.30 g
MgO / 1 mol MgO) = 0.0802 g MgO
4) General Guidelines:
3 points per problem; 1 for both correct reactants, 2 for important reaction products
Balanced equation NOT necessary (ignore coefficients)
1 point penalty for including extraneous spectator ions
a) C2H5OH + O2 --> CO2 + H2O
CO acceptable as oxidized form of carbon
C acceptable as product if accompanied be CO or CO2
b) BaO + H2O --> Ba2+ + OH¯
Only 1 product point awarded for Ba(OH)2
c) Cl2 + OH¯ --> Cl¯ + ClO¯ (+ H2O)
both an oxidized and a reduced form of Cl necessary for 2 product points
H2O not necessary as product
both ClO¯ and ClO2¯ acceptable as oxidized forms of Cl 1 point deducted if acidic products
shown (e.g., H+, HClO, HCl)
d) Fe2+ + O2 (+ H2O) --> Fe2O3 or FeO(OH) or Fe(OH)3
Fe3+, Fe3+ + OH¯, and FeO2 or Fe3O4 awarded only 1 product point
e) H2SO4 + Ca3(PO4)2 --> H3PO4 + CaSO4
no ionized form of H3PO4 acceptable
CaSO4 may appear as Ca2+ + SO42¯, or Ca2+ + HSO4¯, or Ca(HSO4)2
f) H2S + Hg2+ --> HgS + H+
OR
H2S + HgCl2 --> HgS + H+ + Cl¯
If reactant is HgCl2, products must include Cl¯
g) CaH2 + H2O --> Ca2+ + OH¯ (or Ca(OH)2) + H2
no ionized form of CaH2 is acceptable; no H¯ as a reactant
h) Zn + Cu2+ --> Zn2+ + Cu
5) Two points are alloted for each of the four parts. One point is given for naming of the charge
carriers which must be explicitly stated as mobile or free to move. The second point is given for
the structure or bonding of the substance. A complete explanation of the conductivity of one
sample in each pair is sufficient for credit providing that some minimal reference is subsequently
made to the second substance of the pair.
The entire answer is scored holistically--that is, a principle once stated in one part need only be
referenced later for credit to be awarded in that latter part. For example, a mobility reference in
(a) or (c) is relevant to the other.
Ideas which contribute to correct responses:
(a) Sucrose Solution "-"
covalent bonding
Molecules in water
no ions formed therefore no charge flow
AgNO3 Solution "++"
ionic bonding
ions in water are free to move therefore conduct
(b) Solid Na "+"
Na+ in a sea of electrons
Solid AgNO3 "-"
ionic bonding
ions fixed in lattice therefore cannot move
therefore no conductivity
conducts since electrons can move
(c) Sucrose Fused "-"
covalent bonding
fused but still molecular
no ions present; no motion; no conductivity
AgNO3 Fused "++"
ionic bonding
fused alows ions to move therefore conduct
ionic bonds broken when melted
(d) Conc. H2SO4 "+"
not 100% dissociated
forms ion pairs
Diluted H2SO4 "++"
100% dissociated or more ions
H2SO4 + H2O --> 2H3O+ + SO42¯
water addition allows production of many more
ions and greater conductivity
Note SO42¯ alone is insufficient
fewer ions therefore lower conductivity
[mobility not checked]
Key Words
For 'conduction' point:
Good
move
free
transfer
For 'structure' point:
Poor
Good
conducts [merely repeats question] ionic bonds, ionic
separate
ions
line up
molecular
flow
mobile
carry
dissolve
break up
dissociate
travel
pass through (?)
sea of electrons
transport or transfer of e¯ in a
solution
molecules
crystal lattice
sea of electrons
metallic bonding (not just
"metal")
covalent
spread out
delocalized
conduction band, band
theory
Irrelevant erroneous statements do not reduce credit. Relevant erroneous statements have more
effect on a potential 2 point part score that on a potential 1 point part score.
6)
a) two points
V is the triple point (or point where 3 phases coexist).
Solid, liquid, and vapor (or 3 phases) are in equilibrium.
b) two points
Each point on the curve represents the temperature and pressure where the liquid and vapor (or 2
phases) coexist.
At these temperatures and pressures, the two phases are in equilibrium.
OR
The points represent the vapor pressure of the liquid s a function of temperature.
OR
The points represent the boiling points of the liquid as a function of (applied) pressure.
c) two points
Changes: sublimation of change between two pgases, or energy, or density, or entropy change
Point Y: change in phase occurs specifically at Y
d) two points
The solid will sink.
The positive slope of the solid/liquid equilibrium curbe indicates that the solid is more dense
than the liquid.
Notes:
If the phase diagram is labelled and if it is done incorrectly, 1 point is deducted from the total for
parts b), c), and d).
If the response for part c) indicates that a phase change from a less condensed phase to a more
condensed phase (e.g., gas to solid), and additional point is reduced.
7)
a) one point
Oxygen has unpaired electrons and is paramagnetic; thus it is attracted to a magnetic field.
b) three points
Lewis structures:
The CO2 molecule is linear, thus bond dipoles cancel; the SO2 molecule is bent, resulting in a net
dipole.
c) two points
Co2+ has a partially filled d subshell, whereas the d subshell in Zn2+ is filled.
As a result, an electron can be excited between d orbitals in Co2+, causing visible light
absorption; this cannot happen in Zn2+.
d) two points
Si has all of its valence electrons localized in covalent bonds (in a network lattice), hence it is a
poor conductor.
Introduction of As atoms with their "extra" electrons into the Si lattice allows for an increase in
conductivity.
8)
a) two points
[delta]S increases
Dissolving converts highly organized solid to less organized hydrated ions
OR
dissociates, break down, etc.
OR
less particles => more particles
OR
[delta]G negative, [delta]H positive => - T[delta]S negative => [delta]S positive
b) two points
Ksp decreases
lowering T decreases the solubility since the reaction is endothermic
OR
if T decreases, - T[delta]S vecomes less negative or [delta]G becomes more positive, i.e., less
soluble; Ksp decreases
c) two points
There is no effect on [I¯]
PbI2 is a solid; its concentration does NOT change on addition of more PbI2
OR
PbI2(s) is NOT included in the expression for Ksp or Q
OR
the solution is saturated => no more solid will dissolve
d) two points
[delta]G becomes more positive (increases, gets larger, etc.)
Due to the common ion effect; the increase in [Pb2+] shifts the equilibrium to the left.
OR
the reverse reaction becomes more spontaneous as [Pb2+] increases due to the common ion effect
9)
a) two points
Sketch must show start at X2 + Y2, rise to Ea that is less than that for reaction I, then drop to 2
XY, which must be at a higher PE than X2 + Y2 to show an endothermic process.
b) two points
The rate increases because more molecules have energy greater than Ea at higher temperature
and/or the higher speed of the molecules generates a greater collision frequency.
[delta]H for the reaction is changed only slightly, or not at all with small changes in temperature.
c) two points
Hold one reactant concentration constant, vary the other, and measure initial rate of reactant
disappearance or product formation. Take ratios of rates equal to ([ ]1 / [ ]2)n; repeat for the other
reactant.
OR
Follow concentrations of both reactants as functions of time, plot and test for order of each (1st,
2nd, etc.) or do integral law calculations.
OR
Add large concentrations of one reactant, follow the concentration of the other as a function of
time; repeat for the other reactant.
Note: 1 point awarded for less complete description
d) two points
Reaction II proceeds faster.
The reaction with the lower Ea, under the same conditions of concentration and temperature, will
have a faster rate.
OR
It is not possible to determine which reaction has the faster rate without knowledge of other
(preexponential) factors. It cannot be assumed these factors will be the same for X2, Y2, as for
A2, B2, or that a similar mechanism is involved.