AP Free Response Answers

Advanced Placement Chemistry

1983 Free Response Answers

Notes

[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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a) two points

n = 3.509 g ÷ 135.0 g/mol

P = (nRT) / V

= [(0.02600 mol) (0.08205 L atm mol¯¹ K¯¹) (375 K)] / 1.00 L

P = 0.800 atm

b) three points

P_SO₂Cl₂ = 0.800 atm - y atm

P_SO₂ = P_Cl₂ = y atm

P_tot = P_SO₂Cl₂ + P_SO₂ + P_Cl₂

1.43 atm = 0.800 atm - y + y + y (by substitution)

P_SO₂ = P_Cl₂ = y = 0.63 atm

P_SO₂Cl₂ = 0.800 - 0.63 = 0.17 atm

c) three points

K_p = (P_SO₂ x P_Cl₂) / P_SO₂Cl₂

K_p = (0.63 atm)² / 0.17 atm

K_p = 2.3 atm

Alternative approach in (b) and (c) is to determine the number of moles of each species, then molarity, and finally K_c.

n_SO₂Cl₂ = 0.0260 - z

n_{SO₂ = n_Cl₂ = z mol}

n_tot = n_SO₂Cl₂ + n_SO₂ + n_Cl₂

n_tot = PV / RT = (1.43 atm x 1.00 L) / (0.08205 L atm mol¯¹ K¯¹ x 375 K)

n_tot = 0.0465 mol = 0.0260 - z + z + z

n_Cl₂ = n_{SO₂ = z = 0.0205 mol}

n_SO₂Cl₂ = 0.0260 - 0.0205 = 0.00550 mol

K_c = ([SO₂] [Cl₂] ) / [SO₂Cl₂]

K_c = [0.0205]² / [0.0055]

K_c = 0.076 M

d) one point

An endothermic process absorbs heat during dissociation so K₅₀₀ > K₃₇₅ or a stress is placed upon the system and K increases in order to remove the stress.

A maximum of 1 point was deducted for math for unit errors or for reporting an unreasonable number of significant figures. This same procedure was also used for Problems 2 and 3.

a) two points

[delta]G° = [sigma] [delta]G_f° (prod) - [sigma] [delta]G_f° (react)

[delta]G° = - 166.2 - (- 137.3 + 2 (0))

[delta]G° = -28.9 KJ/mol

b) two points

[delta]G° = - RT ln K (or - 2.3 RT log K)

- 28.9 = - ( 8.31 x 10¯³) (298) ln K

ln K = 11.67

K = 1.17 x 10⁵

c) two points

[delta]G° = [delta]H° - T [delta]S°

- 28,900 = -128,100 - 298 [delta]S°

[delta]S° = - 99,200 / 298 = - 333 J / mol-K

d) three points

[delta]H_f° (H₂) = 0 and [delta]G_f° (H₂) = 0

deltaS° = [sigma]S° (prod) - [sigma]S° (react)

- 333 J / mol-K = 126.8 J / mol-K - 197.9 J / mol-K - 2 S° (H₂)

S° = 131 J / mol-K

a) three points

mol H₂C₂O₄^. 2H₂O = 1.2596 g H₂C₂O₄^. 2H₂O / 126.066 g/mol

= 9.9916 x 10¯³ mol

H₂C₂O₄ + 2NaOH ---> Na₂C₂O₄ + 2H₂O

mol NaOH = 9.9916 x 10¯³ mol H₂C₂O₄ x (2 mol NaOH / 1 mol H2C2O4 ) = 1.9983 x 10¯²

M_NaOH = 0.019983 mol / 0.04124 L = 0.4846

b) two points

mol HX = mol NaOH

0.03821 L x 0.4846 mol/L = 0.01852 mol HX

c) two points as follows: a logical argument plus a quantitative comparison = 2; a logical argument but no quantitative comparison = 1; one error in the quantitative comparison = 1

(0.01852 mol HX / 50.00 mL) x 250.00 mL = 0.09260 mol

MW = (15.126 g x 0.09260 mol) = 163.3 g/mol

d) two points

The calculated molecular weight is smaller than true value, because of the following:

Measured g H₂C₂O₄ is larger than true value.
Calculated mol H₂C₂O₄ is larger than true value.
Calculated mol NaOH is larger than true value.
Calculated M NaOH is larger than true value.
Calculated mol HX is larger than true value.

Therefore,

MW = gHX (true value) / mol HX (calculated, and too large)

One point was subtracted for an arithmetic error(s) or for a serious significant figure error.

a) three points

Plot ln k or log k vs 1/T

E_act = - R (slope) or - 2,303 R (slope)

For partial credit, if the 2-point equation is given for the activation evergy, the student may receive a point. A student may also receive a point if it is stated that k is plotted vs 1/T or if ln K or log k is plotted vs T.

b) five points

Plot ln P_A or log P_A vs time.

Plot 1/P_A vs time.

If ln P_A vs time is linear, the reaction is first order. If 1/P_A vs time is linear, the reaction is second order.

If first order, slope = - k₁ or - k₁ / 2.303.

If second order, slope = k₂.

a) three points: correct filling of a molecular orbital energy diagram for NO⁺, correct filling of a molecular orbital energy diagram for NO¯, correct labeling of both molecular orbital diagrams.

These MO diagrams will be posted when drawn (note added Feb 1997)

(b) four points

Bond order for NO⁺ = 3.

Bond order for NO¯ = 2.

Or bond order NO⁺ > bond order NO¯.

Bond length of NO⁺ is shorter than the bond length of NO¯.

Bond strength of NO⁺ is greater than the bond strength of NO¯.

NO¯ is paramagnetic since the anti-bonding pi orbitals are degenerate so the electrons will be unpaired in these orbitals.

(a) three points

A buffer solution resists changes in pH upon the addition of an acid or base.

Preparation:

Mix a weak acid + a salt of a weak acid.
Or mix a weak base + a salt of a weak base.
Or mix a weak acid with about half as many moles of strong base.
Or mix a weak base with about half as many moles of strong acid.
Or mix a weak acid and a weak base.

(b) five points

Carla has the correct procedure. She has mixed a weak base, NH₃, with the salt of a weak base, NH₄Cl.

Archie has buffer solution but it has a pH around 5.

Beula does not have a buffer solution, since her solution consists of a strong acid and a salt of a weak base.

Dexter does not have a buffer solution, since his solution consists of a weak base plus a strong base.

(a) four points

Representation of the cobalt complex as an octahedral complex.
Ethylenediamine shown as bidentate ligand.
Representation of the cis isomer.
Representation of the trans isomer.
If ethylenediamine is shown as a monodentate ligand, it is still possible to receive credit for the representation of both the cis and trans isomers.

Diagrams will be added when drawn (note added Feb 1997)

(b) two points

Selection of the cis isomer as the proper geometrical isomer.
Representation of the optical pair.

One mole of AgCl would be precipitated.

One chloride ion is outside the coordination sphere and thus available for precipitation, whereas the other two chloride ions are in the inner coordination sphere and are bonded directly to the Co³⁺ ion. These ions are not available for precipitation.

(a) three points

Ti³⁺ + H₂O ---> TiO²⁺ + 2 H⁺ + e¯

H⁺ + HOBr + 2e¯ ---> Br¯ + H₂O

2 Ti³⁺ + HOBr + H₂O ---> 2 TiO²⁺ + 3 H⁺ + Br¯

The third point was available for the correct addition of two half-reactions even if the half-reaction(s) were incorrect.

(b) one point

HOBr is the oxidizing agent and Ti³⁺ is the reducing agent.

The observed voltage will be greater than E° since E = E°- 0.059/2 (log [0.1]³).

Or 2 points for correctly discussing the relationships between cell potential and concentration for the balanced net equation that the student gave in part (a). One point could be achieved for discussing the cell potential and concentration relationship but omitting H⁺ or the exponets.

(d) two points

Identification of the property, [delta]G, K or pH.

Statement of the relationshop [delta]G° = - n F E° or E° = 0.059/n (log K).

Or 1 point was given if the student stated that reaction is spontaneous if E > 0.

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