AP Free Response Answers

Advanced Placement Chemistry

1994 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

K_sp = [Mg²⁺][F¯]²

= (1.21 x 10¯³) (2 x 1.21 x 10¯³)²

= 7.09 x 10¯⁹

b) two points

K_sp = [Mg²⁺] (2x + 0.100)²

since 2x is much less than 0.100

= 7.09 x 10¯⁹ = [Mg²⁺] (0.010)²

[Mg²⁺] = (7.09 x 10¯⁹) / (10¯²)

= 7.09 x 10¯⁷ M

Note: OK if 0.102 is used for [F¯], then K_sp = 6.76 x 10¯⁷

c) three points (first point earned if both concentrations are correct; correct substitution and calculation of the wrong concentration values earns the second point [calc. of Q], but not the first

[Mg²⁺]:
100.0 x 3.00 x 10¯³ = 300.0 x [Mg²⁺]
[Mg²⁺] = 1.00 x 10¯³ M

[F¯]:
200.0 x 2.00 x 10¯³ = 300.0 x [F¯]
[F¯] = 1.33 x 10¯³ M

Q = Ion Product = [Mg²⁺] [F¯]²

= (1.00 x 10¯³) (1.33 x 10¯³)²

= 1.77 x 10¯⁹

Since Q < K_sp, no precipitate will form.

Note: conclusion must be consistent with Q value.

d) two points

Solubility of MgF₂ decreases with the increasing temperature, thus dissolution process is exothermic.

MgF₂ ---> Mg²⁺ + 2F¯ + Q (or H)

Reason:

i) Increased temperature puts a stree on the system (LeChâtelier). The system will reduce the stree by shifting the equilibrium in the endothermic (left) direction.

ii) A data supported argument such as comparing ion concentrations, calculating second K_sp and giving proper interpretations.

a) three points (point for each order must include justification)

From exps. 1 and 2: doubling [H₂] while keeping [NO] constant doubles the rate, therefore the reaction is first order in [H₂].

From exps. 3 and 4; doubling [NO] while keeping [H₂] constant quadruples the rate, therefore the reaction is second order in [NO].

Rate = k [H₂] [NO]²

Note: full credit is earned for the rate expression as long as it is consistent with orders described by student.

b) two points (one for value and one for units)

k= Rate / ([H₂] [NO]²)

From exp. 1: k = 1.8 x 10¯⁴ M/min / [(1.0 x 10¯³ M) (6.0 x 10¯³ M)²]

= 5.0 x 10³ M¯² min¯¹

Note: the unit is often written as L² mol¯² min¯¹

c) one point

Stoichiometry: NO : H₂ is 1:1

When 0.0010 mole of H₂ had reacted , it must have reacted with 0.0010 mole NO; thus [NO] remaining = 0.0060 - 0.0010 = 0.0050 M

d) three points

(i)

For I : K_eq = [N₂O₂] / [NO]²

For II: Rate= k[H₂] [N₂O₂]

[N₂O₂] = K_eq [NO]²

Rate = k' [H₂][NO]²

Note: there must be some clear algebraic manipulation showing that [N₂O₂] is proportional (NOT equal) to [NO]².

Step II is the rate determining step.

(ii)

I: NO + NO ---> N₂O₂
II: N₂O₂ + H₂ ---> H₂O + N₂O
III: N₂O + H₂ ---> N₂ + H₂O
I + II + III: 2NO + 2H₂ ---> N₂ + 2 H₂O

a) three points

n = PV ÷ RT = [(721) (0.090)] ÷ [(62.4) (298)]= 3.49 x 10¯³ mol H₂

Each worth one point:

25° to 298 K
745 - 24 = 721 mm Hg
calculation of moles H₂

Note: 62.4 is R in units of L torr per mol K

b) two points

[(23.8) (0.090)] ÷ [(62.4 ) (298)] = 1.15 x 10¯⁴ mol H₂O

(1.15 x 10¯⁴) (6.02 x 10²³) = 6.92 x 10¹⁹ molecules H₂O

c) two points (one for formula, one for calculation)

The average kinetic energies are equal, so:

(1/2 mv²)_H₂O = (1/2 mv²)_H₂

v_H₂ / v_H₂O = square root [MM_H₂O / MM_H₂]

= square root (18 / 2) = 3

Note: credit also given for correct use of v_rms = square root (3RT / M)

d) two points

H₂O deviates more from ideal behavior.

Explanation:

1) The volume of the H₂O molecule is larger than that of the H₂ molecule

2) The intermolecular forces among the H₂O molecules are stronger than those among H₂ molecules.

Guiding principles:

Each reaction is worth a total of three points.
Reactants are +1 point, products are +2 points.
Ignore balancing and states.
Inappropriate ionization = maximum one point penalty per equation.

a) CN¯ + Ag⁺ ---> Ag(CN)₂¯

Note: any complex ion of Ag⁺ with cyanide with consistent charge earns 3 points; AgCN given as product earns one product point.

b) Mn²⁺ + S²¯ ---> MnS

Note: If Mg is used instead of Mn, maximum possible score is two points.

c) P₄O₁₀ (or P₂O₅) + H₂O ---> H₃PO₄

Note: Acidic species (H⁺ or oxyacid of phosphorous) earns one product point; P in +5 oxidation state in oxyanion earns one product point; anions of oxyacids of phosphorous require H⁺ for full credit for products.

d) (NH₄)₂CO₃ ---> NH₃ + H₂O + CO₂

Note: any one product earns one point; all three earn two points. NH₄OH + CO₂ earns one product point. NH₃ + H₂CO₃ eans one product point.

e) CO₂ + OH¯ ---> HCO₃¯

Note: CO₃²¯ + H₂O as products earns two product points. CO₃²¯ alone as product earns one product point. HCO₃¯ + H₂O earns one product point.

f) H⁺ + Cl¯ + KMnO₄ ---> K⁺ + Mn²⁺ + Cl₂ + H₂O

Note: HCl and MnO₄¯ acceptable as reactants. Any valid redox product earns one point. All four product earns two points. K⁺ and/or H₂O only as products earns no credit. If both H⁺ and H₂O omitted, then maximum of two points possible.

g) Na + H₂O ---> H₂ + Na⁺ + OH¯

Note: all three products earn two product points. Any valid redox product earns one product point.

h) Cr₂O₇²¯ + Fe²⁺ + H⁺ ---> Cr³⁺ + Fe³⁺ + H₂O

Note: All three products earn two product points. Any valid redox product earns one product point. H₂O only earns no credit. If Cl¯ ---> Cl₂ instead of Fe²⁺ ---> Fe³⁺, then maximum of two points possible.

a) two points

Volume decreases in beaker A; the concentration decreases in beaker B (either obs. earns one point provided the other is not wrong.)

The vapor pressure of pure H₂O is greater than the vapor pressure of H₂O in solution.

The rate of evaporation of H₂O molecules from pure H₂O is greater than that from the sugar solution, while the condensation rates are the same.

b) two points

The water will begin to boil (or evaporate).

The external pressure on the water will become equal to the vapor pressure of the water, causing it to boil.

The drop in external pressure causes the boiling point to drop to the temperature of the water.

c) two points

Solid copper is deposited on the zinc strip; the zinc strip goes into solution. No reaction occurs with silver.

Zinc is a better reducing agent or a more active metal than copper and will be oxidized. Silver is a less reactive metal than copper is.

d) two points

Two layers will form, one of which is colored. Iodine is nonpolar and will dissolve in TTE. Water is polar and will not dissolve in TTE.

Placment of I₂ must be correctly indicated for second point.

6) Note: for parts (a), (b), and (c), just writing an equation is not sufficient for the 'explanation" point. To earn credit, the student must connect the equation to the issue to be explained.

a) two points

Statement that [delta]S° is negative

3 moles of gas ---> 2 moles of gas plus solid (3 moles ---> 2 moles earns no points),

2 gases ---> one gas + solid

use of [delta]G° = [delta]H° - T[delta]S° with [delta]G° = 0

b) two points

[delta]G° is less negative, goes to 0, goes positive, gets larger

Explanation using [delta]G° = [delta]H° - T[delta]S°

c) two points

K_eq decreases (exponent goes more negative) as T increases

K_eq goes from > 1, to 1, to < 1, as T increases

Correct explanation using the equation

[delta]G° = - RT ln K (or ln(k₁ / k₂) = ([delta]H° / R) (1/T₂ - 1/T₁)

higher T favors the reverse reaction (Le Châtelier) because the forward reaction is exothermic.

Note: if answer for (a) is that [delta]S° is positive, then statement that K_eq will decrease or increase depending on the relative magnitudes of T and [delta]G° change earns two points.

d) two points

Since [delta]G° = 0 at this point, the equation is T= [delta]H° / [delta]S°

([delta]G° = [delta]H° - T [delta]S°S is NOT sufficient without [delta]G° = 0.)

Prediction is not exact since [delta]H° and/or [delta]S° vary with T.

a) two points

PO₄³¯ + H⁺ ---> HPO₄²¯

HPO₄²¯ + H⁺ ---> H₂PO₄¯

Note: any proton transfer to any P_xO_y species earns one point.

b) two points

explicit 32 e¯

explicit 2¯ charge (somewhere)

not more than 1 double P-O bond

Note: HPO₄²¯ (formula only) or other P_xO_y species with correct diagram earns one point.

c) three points

Graph goes from upper left to the lower right. (pH decreases)

In either direction:

Two protons transferred
Two "buffers"
Two "equivalences"

Explain/correctly label at least one "buffer" or "equivalence" region

d) one point

H₂PO₄¯ + H⁺ <===> H₃PO₄

Note: other proton transfer earns one point if consistent with product in part (a)

a) two points

The addition of a solute lowers the freezing point of water.

A mole of NaCl contains (dissociates into) 2 moles of ions/particles, whereas a mole of CaCl₂ contains (dissociates into) 3 moles of ions. Therefore, CaCl₂ is more effective.

b) two points

Hydrogen bonding is the most important intermolecular attractive force between molecules of H₂O and between molecules of NH₃.

Water is a liquid because the hydrogen-bonding forces are stronger between the adjacent H₂O molecules than between adjacent NH₃ molecules.

c) two points

Graphite's structure consists of 2-dimensional sheets of covalently bonded carbon atoms.

The attractive forces between sheets (layers) are weak London (dispersion) forces, which allow the sheets to slide easily over one another. (Note: must indicate layers and sliding to earn point.)

Diamond consists of an extended 3-dimensional covalent network of carbon atoms. This makes diamond a very hard substance.

d) two points

Vinegar, a dilute solution of acetic acid, reacts with the white solid, which contains metal carbonates, in a neutralization reaction to form gaseous CO₂.

a) two points

Ca²⁺ has fewer electrons, thus it is smaller than Ca.

The outermost electron in Ca is in a 4s orbital, whereas the outermost electron in Ca²⁺ is in a 3p orbital.

Note: The first point is earned for indicating the loss of electrons, the second point for indicationg the outermost electrons are in different shells -- must account for the magnitude of the size difference between Ca and Ca²⁺.

b) two points

U for CaO is more negative than U for K₂O, so it is more difficult to break up the CaO lattices (stronger bonds in CaO). Ca²⁺ is smaller than K⁺, so internuclear separations (between cations and O²¯) are less.

Ca²⁺ is more highly charged than K⁺, thus cation-O2 bonds are stronger

Note: understanding what "lattice energy" is earns 1 point; size or charge explanation needed for the second point. Responses that use Lewis structures or otherwise indicate molecules rather than ionic lattice earn no points.

c) two points

i) Ca has ore protons and is smaller. The outermost electrons are more strongly held by the nuclear charge of Ca.

ii) The outermost electrons in Ca are in the 4s, which is a higher energy orbital (more shielded) than the 3p electrons in K.

Note: for (i), the idea of attraction between nucleus and electrons must be present; for (ii), a "noble-gas configuration" argument must be tied to an energy argument in order to earn credit.

d) two points

the highest energy (outermost) electrons in Al is in a 3p orbital, whereas that electron in Mg is in a 3s orbital.

The 3p electron in Al is of higher energy (is more shielded) than is the 3s electron in Mg.

Note: noting that different orbitals are involved earns the first point; a correct energy argument earns the second point.

Responses that attribute the greater stability of Ca over K (or K⁺ over Ca⁺, or Mg over Al) to the stability of a completely filled (vs. half or partially filled) orbital earn NO credit.

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