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.
1) Reaction question
(a) CaCO_{3} > CaO + CO_{2}
(b) Ni + Cu^{2+} > Ni^{2+} + Cu
hydrated ions acceptable with correct charge
1 point for Ni(OH)_{2} as product
(c) HPO_{4}^{2}¯ + H^{+} > H_{2}P0_{4}¯
incorrect charge on H_{2}P0_{4}¯ 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) Cl_{2} + Br > Cl¯ + Br_{2}
no credit for monatomic Cl as reactant or Br as product
(e) NH_{3} + HC_{2}H_{3}O_{2} > C_{2}H_{3}O_{2}¯ + NH_{4}^{+}
1 product point for NH_{4}C_{2}H_{3}O_{2}
1 point for NH_{3} + H^{+} > NH_{4}
(f) (NH_{4})_{2}CO_{3} + Ba^{2+} + OH¯ > NH_{3} + BaCO_{3} + H_{2}O
1 product point for either NH_{3} or BaCO_{3}
2 product points for all three species correct
(g) N_{2}O_{3} + H_{2}O > HNO_{2}
1 product point for H^{+} + NO_{2}¯
(h) MnO_{4}¯ + C_{2}O_{4}^{2}¯ > MnO_{2} + CO_{2}
no penalty for OH¯ or H_{2}O in equation
no point earned for Mn^{2+} as product
2)
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 K_{a}
= ( [H^{+}][OCl¯] ) / [HOCl]
= [H^{+}]^{2} / c_{HOCl}
(b) two points total: one point each
OCl¯ + H_{2}O <===> HOCl + OH¯ (or NaOCl + H_{2}O > Na^{+} + HOCl + OH¯)
K_{b} = K_{w} / K_{a} = 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).
Then
[OH¯] = [HOCl]
K_{b} = [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 halfneutralized; one for mmol calcs.
pH = 7.49 therefore [H^{+}] = 3.2 x 10¯^{8}
pH = pK_{a} , or [H^{+}] = K_{a}.
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
3)
(a) two points; one for line of answer
 232.7 J/K = S° (C_{2}H_{6})  (261.4 + 200.9) J./K
S° (C_{2}H_{6}) = 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 K_{eq} > 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 10^{42} (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]H_{rxn} and correct calculation of answer; no points earned for "extrapolation" techniques to find carboncarbon triple bond energy; E* is the energy of the carboncarbon triple bond.
 311.4 kJ = [(2) (436) + E* + (2) (414)]  [(347) + (6) (414)]
E* = 820 kJ
4)
(a) one point
V_{1}M_{1} = V_{2}M_{2}
(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 H_{2}SO_{4} = 1 L x (1.84 g/mL) x (1,000 ml/L) = 1,840 g H_{2}SO_{4}
18.4 mol H_{2}SO_{4} x 98.1 g/mol = 1,805 g H_{2}SO_{4}
mass percent H_{2}SO_{4} = (1,805 g / 1,840 g) x 100 = 98.1%
(c) three points
Stoichiometric ratio of NaHCO_{3} to H_{2}SO_{4} = 2:1
10.5 g NaHCO_{3} x (1 mol NaHCO_{3} / 84.0 g NaHCO_{3}) = 0.125 mol NaHCO_{3}
Since 1 mol H_{2}SO_{4} reacts with 2 mol NaHCO_{3}, 0.125 mol NaHCO_{3} reacts with 0.0625 mol H_{2}SO_{4}
0.0625 mol H_{2}SO_{4} = 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 H_{2}SO_{4} = 1 L x (1,000 mL / 1 L) x (1.38 g 1 mL) = 1,380 g
mass of H_{2}SO_{4} in 1 L = (5.20 mol/L) (98.1 g.mol) = 510 g
mass of H_{2}O in 1 L = 1,380  510 = 870 g
molality = (5.20 mol H_{2}SO_{4} / 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
5) (a) two points
CO_{2}
because all contain same number of molecules (moles), and CO_{2} molecules are the heaviest
Note: total of 1 point earned if CO_{2} not chosen but same number of molecules (moles) is specified
(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
CO_{2}
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
He
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 M_{m}(HA) too low
M(NaOH) => V(NaOH) => n(NaOH) => n(HA) => Mm(HA)
(M = n ÷ V) and (M_{m} = m÷ n)
(b) two points
Calculated M_{m}(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 M_{m}(HA) too high
equivalence point => n(NaOH) => n(HA) => M_{m}(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 M_{m}(HA) too low
V(NaOH) => n(NaOH) => n(HA) => M_{m}(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)
7)
(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 Sr^{2+} is more positive
Standard reduction potential for Sr more negative
E° = + 0.52 V
Note: only 1 point earned for just E° positive because K_{eq} positive.
(b) one point
The oxidizing agent is Mg^{2+}
(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 K_{eq} dependence on T. For temperature change in this problem, decrease in ln K term is small relative to the term RT/nF
OR
No change, because in the Nernst equation E_{cell} = E°  (RT/nF) ln Q
ln Q = 0, and E_{cell} = E°
Note: this second approach earns 1 point only
(d) two points
E_{cell} will increase
In the equation E_{cell} = E°  (0.0592 / n) log Q
Q = 0.1 therefore log Q is negative therefore term after E° is positive therefore E_{cell} increases
OR
with the concentration of Mg^{2+} larger than that of Sr^{2+}, Le Chatelier's principle predicts the reaction will have a larger driving force to the right and a more positive E_{cell}
(e) one point
At equilibrium, E_{cell} = 0
Note: "balanced", "neutral", or "no net reaction" not accepted
8)
(a) one point
2 NO + 2 H_{2} > N_{2} + 2 H_{2}O
(b) two points
N_{2}O_{2} and N_{2}O 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 H_{2} 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
OR
T goes up therefore rate goes up
because no change in concentration of reactants, therefore k must increase
OR
from Arrhenius equation (not required in AP Chemistry curriculum, but noted in some student responses):
as T goes up, k goes up
OR
graph as below with proper explanation
9)
(a) two points
Hydrogen bonding (or dipoledipole 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
AsF_{3} has a trigonal pyramid shape and bond dipoles do NOT cancel (or asymmetric molecule)
AsF_{5} 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
NO_{2}¯ has resonance structures
HNO_{2} has no resonance structures
OR
one NO single bond, one N=O double bond
Note: one point earned if only correct Lewis structures, including resonance for NO_{2}¯ given.
(d) two points
Sulfur uses d orbitals (or expanded octet), oxygen has no d orbitals in its valence shell
OR
Sulfur is a larger atom, can accomodate more bonds. 