Advanced Placement Chemistry

1996 Free Response Answers

Notes

• 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.
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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 K_a

= ( [H⁺][OCl¯] ) / [HOCl]
= [H⁺]² / c_HOCl

(b) two points total: one point each

OCl¯ + H₂O <===> HOCl + OH¯ (or NaOCl + H₂O ---> 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 half-neutralized; 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° (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 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 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]H_rxn 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 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 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 K_eq 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 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 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 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 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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