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Advanced Placement Chemistry
1996 Free Response Questions
Question 1 is question 4 in previous years, question 2 is question 1 in previous years and
questions 3&4 are questions 2&3 in previous years.
students are now allowed 10 minutes to answer question 1, after which they must seal
that portion of the test.
Go to Answers
Return to Additional Materials Menu
1) 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
Example: A strip of magnesium is added to a solution of silver nitrate.
Mg + Ag+ ---> Mg2+ + Ag
(a) Solid calcium carbonate is strongly heated.
(b) A piece of nickel metal is immersed in a solution of copper(II) sulfate.
(c) Equal volumes of equimolar solutions of disodium hydrogen phosphate and hydrochloric acid
are mixed.
(d) Chlorine gas is bubbled into a solution of sodium bromide.
(e) Ammonia gas is bubbled into a solution of ethanoic (acetic) acid.
(f) Solid ammonium carbonate is added to a saturated solution of barium hydroxide.
(g) Drops of liquid dinitrogen trioxide are added to distilled water.
(h) Solutions of potassium permanganate and sodium oxalate are mixed.
HOCl <===> OCl¯ + H+
Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The aciddissociation constant, Ka, for the reaction represented above is 3.2 x 10¯8.
(a) Calculate the [H+] of a 0.14-molar solution of HOCl.
(b) Write the correctly balanced net ionic equation for the reaction that occurs NaOCl is
dissolved in water and calculate the numerical value of the equilibrium constant for the reaction.
(c) Calculate the pH of a solution made by combining 40.0 milliliters of 0.14-molar HOCl and
10.0 milliliters of 0.56-molar NaOH.
(d) How many millimoles of solid NaOH must be added to 50.0 milliliters of 0.20-molar HOCl
to obtain a buffer solution that has a pH of 7.49? Assume that the addition of the solid NaOH
results in a negligible change in volume.
(e) Household bleach is made by dissolving chlorine gas in water, as represented below.
Cl2(g) + H2O --> H+ + Cl¯ + HOCl(aq)
Calculate the pH of such a solution if the concentration of HOCl in the solution is 0.065 molar.
C2H2(g) + 2 H2(g) ---> C2H6(g)
Information about the substances
S° (J/mol K)
H°f (kJ/mol)
Bond Energy (kJ/mol)
(a) If the value of the standard entropy change, S°, for the reaction is -232.7 joules per mole
Kelvin, calculate the standard molar entropy, S°, of C2H6 gas.
(b) Calculate the value of the standard free-energy change, G°, for the reaction. What does the
sign of G° indicate about the reaction above?
(c) Calculate the value of the equilibrium constant, K, for the reaction at 298 K.
(d) Calculate the value of the C [triple bond] C bond energy in C2H2 in kilojoules per mole.
4) Concentrated sulfuric acid (18.4-molar H2SO4) has a density of 1.84 grams per milliliter. After
dilution with water to 5.20-molar, the solution has a density of 1.38 grams per milliliter and can
be used as an electrolyte in lead storage batteries for automobiles.
(a) Calculate the volume of concentrated acid required to prepare 1.00 liter of 5.20-molar H2SO4.
(b) Determine the mass percent of H2SO4 in the original concentrated solution.
(c) Calculate the volume of 5.20-molar H2SO4 that can be completely neutralized with 10.5
grams of sodium bicarbonate NaHCO3.
(d) What is the molality of the 5.20-molar H2SO4 ?
Represented above are five identical balloons, each filled to the same volume at 25°C and 1.0
atmosphere pressure with the pure gas indicated.
(a) Which balloon contains the greatest mass of gas? Explain.
(b) Compare the average kinetic energies of the gas molecules in the balloons. Explain.
(c) Which balloon contains the gas that would be expected to deviate most from the behavior of
an ideal gas? Explain.
(d) Twelve hours after being filled, all the balloons have decreased in size. Predict which balloon
will be the smallest. Explain your reasoning.
6) A 0.500-gram sample of a weak, nonvolatile acid, HA, was dissolved in sufficient water to
make 50.0 milliliters of solution. The solution was then titrated with a standard NaOH solution.
Predict how the calculated molar mass of HA would be affected (too high, too low, or not
affected) by the following laboratory procedures. Explain each of your answers.
(a) After rinsing the buret with distilled water, the buret is filled with the standards NaOH
solution; the weak acid HA is titrated to its equivalence point.
(b) Extra water is added to the 0.500-gram sample of HA.
(c) An indicator that changes color at pH 5 is used to signal the equivalence point.
(d) An air bubble passes unnoticed through the tip of the buret during the titration.
Sr(s) + Mg2+ <===> Sr2+ + Mg(s)
Consider the reaction represented above that occurs at 25°C. All reactants and products are in
their standard states. The value of the equilibrium constant, Keq, for the reaction is 4.2 x 1017 at
(a) Predict the sign of the standard cell potential, E°, for a cell based on the reaction. Explain
your prediction.
(b) Identify the oxidizing agent for the spontaneous reaction.
(c) If the reaction were carried out at 60°C instead of 25°C, how would the cell potential change?
Justify your answer.
(d) How would the cell potential change if the reaction were carried out at 25°C with a 1.0-molar
solution of Mg(NO3)2 and a 0.10-molar solution of Sr(NO3)2? Explain.
(e) When the cell reaction in (d) reaches equilibrium, what is the cell potential?
8) The reaction between NO and H2 is believed to occur in the following three-step process.
NO + NO <===> N2O2 (fast)
N2O2 + H2 ---> N2O + H2O (slow)
N2O + H2 ---> N2 + H2O (fast)
(a) Write a balanced equation for the overall reaction.
(b) Identify the intermediates in the reaction. Explain your reasoning.
(c) From the mechanism represented above, a student correctly deduces that the rate law for the
reaction is rate = k[NO]2[H2]. The student then concludes that (1) the reaction is third-order and
(2) the mechanism involves the simultaneous collision of two NO molecules and an H2 molecule.
Are conclusions (1) and (2) correct? Explain.
(d) Explain why an increase in temperature increases the rate constant, k, given the rate law in
9) Explain each of the following in terms of the electronic structure and/or bonding of the
compounds involved.
(a) At ordinary conditions, HF (normal boiling point = 20°C) is a liquid, whereas HCl (normal
boiling point = -114°C) is a gas.
(b) Molecules of AsF3 are polar, whereas molecules of AsF5 are nonpolar.
(c) The N-O bonds in the NO2¯ ion are equal in length, whereas they are unequal in HNO2.
(d) For sulfur, the fluorides SF2, SF4, and SF6 are known to exist, whereas for oxygen only OF2 is
known to exist.
Advanced Placement Chemistry
1996 Free Response Answers
Question 1 is question 4 in previous years, question 2 is question 1 in previous years and
questions 3&4 are questions 2&3 in previous years.
students are now allowed 10 minutes to answer question 1, after which they must seal
that portion of the test.
[delta] is used to indicate the capital Greek letter.
[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.
Return to Questions
Return to Additional Materials Menu
1) Reaction question
(a) CaCO3 ---> CaO + CO2
(b) Ni + Cu2+ ---> Ni2+ + Cu
hydrated ions acceptable with correct charge
1 point for Ni(OH)2 as product
(c) HPO42¯ + H+ ---> H2P04¯
incorrect charge on H2P04¯ when only one product occurs, 1 point only
1 product point for transfer if H+ from an ionic reactant to product when a phosphate species is
incorrectly but consistently written.
(d) Cl2 + Br- ---> Cl¯ + Br2
no credit for monatomic Cl as reactant or Br as product
(e) NH3 + HC2H3O2 ---> C2H3O2¯ + NH4+
1 product point for NH4C2H3O2
1 point for NH3 + H+ ---> NH4
(f) (NH4)2CO3 + Ba2+ + OH¯ ---> NH3 + BaCO3 + H2O
1 product point for either NH3 or BaCO3
2 product points for all three species correct
(g) N2O3 + H2O ---> HNO2
1 product point for H+ + NO2¯
(h) MnO4¯ + C2O42¯ ---> MnO2 + CO2
no penalty for OH¯ or H2O in equation
no point earned for Mn2+ as product
a) two points total ; one point for correct substitutions; one point for computation
[H+] = [OCl¯] = square root (0.14 x 3.2 x 10¯8) = 6.7 x 10¯5 M
since Ka
= ( [H+][OCl¯] ) / [HOCl]
= [H+]2 / cHOCl
(b) two points total: one point each
OCl¯ + H2O <===> HOCl + OH¯ (or NaOCl + H2O ---> Na+ + HOCl + OH¯)
Kb = Kw / Ka = 1 x 10¯14 ÷ 3.2 x 10¯8 = 3.1 x 10¯7
(c) two points total; one for concentrations and one for pH calc.
Concentrations before reaction:
[HOCl] = [(0.0400) (0.14)] / 0.050 = 0.11 M
[OH¯] = [(0.0100) (0.56)] / 0.050 = 0.11 M
Thus reaction is essentially complete and exactly equals a solution of NaOCl and [OCl¯] = 0.11
M (or reaction is at equivalence point).
[OH¯] = [HOCl]
Kb = [OH¯]2 / 0.11 = 3.1 x 10¯7
[OH¯] = square root [(0.11) (3.1 x 10¯7)] = 1.8 x 10¯4
pOH = 3.73
pH = 14 - 3.73 = 10.27
(d) two points; one for half-neutralized; one for mmol calcs.
pH = 7.49 therefore [H+] = 3.2 x 10¯8
pH = pKa , or [H+] = Ka.
So [OCl¯] / [HOCl] = 1 , or solution must be half neutralized.
initial mmol HOCl = 50.0 x 0.20 = 10.0 mmol
mmol NaOH required = 10.0 ÷ 2 = 5.0 mmol
(e) one point
From equation, 1 mol H+ produced for each 1 mole of HOCl produced, thus [H+] = [HOCl] =
0.065 therefore pH = 1.19
(a) two points; one for line of answer
- 232.7 J/K = S° (C2H6) - (261.4 + 200.9) J./K
S° (C2H6) = 229.6 J/K
units ignored; 1 point earned for 98.9 J/K; 1 point lost if stoichiometry is not implied in process
(b) three points total; one point each portion; any value for T (e.g., 273 K or 298 K) is allowable:
[delta]H° = (- 84.7 kJ) - (226.7 kJ) = -311.4kJ
= - 311.4kJ - (298 K) (- 0.2327kJ/K)
= - 311.4 kJ + 69.3 kJ
= - 242.1 kJ
Negative [delta]G° therefore reaction is spontaneous, or Keq > 1 therefore reaction is
spontaneous, or products are favored at equilibrium.
(c) two points
ln K = 242.1 ÷ [(8.31 x 10¯3) (298)] = 97.7
K = 3 x 1042 (1,2,or 3 significant figures acceptable)
(d) two points; first point earned for correct substitution and correct number of bonds, second
point earned for setting equal to [delta]Hrxn and correct calculation of answer; no points earned
for "extrapolation" techniques to find carbon-carbon triple bond energy; E* is the energy of the
carbon-carbon triple bond.
- 311.4 kJ = [(2) (436) + E* + (2) (414)] - [(347) + (6) (414)]
E* = 820 kJ
(a) one point
V1M1 = V2M2
(1.00L) (5.20 mol/L) = (x) (18.4 mol/L)
x = 5.2 mol / (18.4 mol/L) = 0.283 L (or 283 mL)
(b) two points
mass 1 liter of concentrated H2SO4 = 1 L x (1.84 g/mL) x (1,000 ml/L) = 1,840 g H2SO4
18.4 mol H2SO4 x 98.1 g/mol = 1,805 g H2SO4
mass percent H2SO4 = (1,805 g / 1,840 g) x 100 = 98.1%
(c) three points
Stoichiometric ratio of NaHCO3 to H2SO4 = 2:1
10.5 g NaHCO3 x (1 mol NaHCO3 / 84.0 g NaHCO3) = 0.125 mol NaHCO3
Since 1 mol H2SO4 reacts with 2 mol NaHCO3, 0.125 mol NaHCO3 reacts with 0.0625 mol
0.0625 mol H2SO4 = V x M = (V) (5.20 M)
V = 0.0625 mol / (5.20 mol/L) = 0.0120 L (or 12.0 mL)
(d) three points
molality = moles solute / 1,000 g solvent = moles solute / 1 kg solvent
mass of 1 L of 5.20 M H2SO4 = 1 L x (1,000 mL / 1 L) x (1.38 g 1 mL) = 1,380 g
mass of H2SO4 in 1 L = (5.20 mol/L) (98.1 g.mol) = 510 g
mass of H2O in 1 L = 1,380 - 510 = 870 g
molality = (5.20 mol H2SO4 / 870 g) x (1,000 g / 1 kg) = 5.98 m
Note: no credit earned for 5.20 mol / 1.38 kg = 5.77 m
(a) two points
because all contain same number of molecules (moles), and CO2 molecules are the heaviest
Note: total of 1 point earned if CO2 not chosen but same number of molecules (moles) is
(b) two points
All are equal
because same temperature, therefore same average kinetic energy
Note: just restatement of "same conditions, etc." does not earn second point
(c) two points
either one:
it has the most electrons, hence is the most polarizable
it has the strongest intermolecular (London) forces
Note: also allowable are "polar bonds", "inelastic collisions"; claiming larger size or larger
molecular volume does not earn second point
(d) two points
Any one:
greatest movement through the balloom wall
smallest size
greatest molecular speed
most rapid effusion (Graham's law)
6) for explanation point in 9 (a), (c), and (d), credit is earned at step indicted in boldface type.
(a) two points
Calculated Mm(HA) too low
M(NaOH) => V(NaOH) => n(NaOH) => n(HA) => Mm(HA)
(M = n ÷ V) and (Mm = m÷ n)
(b) two points
Calculated Mm(HA) not affected
Any one of the following reasons. Water:
does not change n(HA)
changes only M(HA) -- sense of dilution
is not a reactant
(c) two points
Calculated Mm(HA) too high
equivalence point => n(NaOH) => n(HA) => Mm(HA)
(expected pH higher)
Note: "no effect if NaOH standardized with same indicator" earns 2 points; no credit earned if
pH=7 or neutral
(d) two points
Calculated Mm(HA) too low
V(NaOH) => n(NaOH) => n(HA) => Mm(HA)
Note: point earned for V(NaOH) only if:
(i) no explanation point is earned in (a)
(ii) explanation in (a) also includes V(NaOH)
(a) two points
The sign of the cell potential will be positive
because (any one is sufficient):
K is greater than 1
the reaction is spontaneous (occurs)
E° for Sr2+ is more positive
Standard reduction potential for Sr more negative
E° = + 0.52 V
Note: only 1 point earned for just E° positive because Keq positive.
(b) one point
The oxidizing agent is Mg2+
(c) two point
The cell potential would increase
Since all ions are at 1 M, Q for the system is 1 and E° = (RT/nF) ln K
so as T increases, so should E°
Note: no credit lost if student recognizes Keq dependence on T. For temperature change in this
problem, decrease in ln K term is small relative to the term RT/nF
No change, because in the Nernst equation Ecell = E° - (RT/nF) ln Q
ln Q = 0, and Ecell = E°
Note: this second approach earns 1 point only
(d) two points
Ecell will increase
In the equation Ecell = E° - (0.0592 / n) log Q
Q = 0.1 therefore log Q is negative therefore term after E° is positive therefore Ecell increases
with the concentration of Mg2+ larger than that of Sr2+, Le Chatelier's principle predicts the
reaction will have a larger driving force to the right and a more positive Ecell
(e) one point
At equilibrium, Ecell = 0
Note: "balanced", "neutral", or "no net reaction" not accepted
(a) one point
2 NO + 2 H2 ---> N2 + 2 H2O
(b) two points
N2O2 and N2O are intermediates
because they appear in the mechanism but not in the overall products (or reactants)
(c) three points; one for each half of conclusion (1) answer, third for conclusion (2) answer
Student indicates conclusion (1) is correct,
because the sum of the exponents in rate law is 2 + 1 = 3
Student indicates conclusion (2) is incorrect,
because no step involves two NO molecules and a H2 molecule
(d) two points; T goes up therefore k goes up:
because increasing number of collisions between reactants
are occuring with sufficient energy to form an activated complex
T goes up therefore rate goes up
because no change in concentration of reactants, therefore k must increase
from Arrhenius equation (not required in AP Chemistry curriculum, but noted in some student
as T goes up, k goes up
graph as below with proper explanation
(a) two points
Hydrogen bonding (or dipole-dipole attraction) in HF is greater than it is in HCl
Note: only one point earned if simply stated that HF has greater intermolecular forces than HCl
(b) two points
AsF3 has a trigonal pyramid shape and bond dipoles do NOT cancel (or asymmetric molecule)
AsF5 has a trigonal bipyramid shape and bond dipoles cancel (or symmetric shape)
Notes: explanation must refer to shape in order to earn point; one point earned if only correct
Lewis structures are given.
(c) two points
NO2¯ has resonance structures
HNO2 has no resonance structures
one N-O single bond, one N=O double bond
Note: one point earned if only correct Lewis structures, including resonance for NO2¯ given.
(d) two points
Sulfur uses d orbitals (or expanded octet), oxygen has no d orbitals in its valence shell
Sulfur is a larger atom, can accomodate more bonds.