Advanced Placement Chemistry: 1996 Free Response Answers



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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) 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

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 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).

Then

[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

3)


(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


4)


(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 H2SO4

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


5) (a) two points

CO2

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 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

CO2

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 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)

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 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
OR

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


OR

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

8)
(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
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 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




OR

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
OR

Sulfur is a larger atom, can accomodate more bonds.


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