Advanced Placement Chemistry

1990 Free Response Answers

Notes

1)

a) one point

Fe(OH)2 ------> Fe2+ + 2 OH¯

b) four points

Ksp = [Fe2+][OH¯]2

M.W. of Fe(OH)2 = 55.85 + 2(16.0) + 2(1.008) = 89.87

1.43 x 10¯3 g. x 1 mol / 89.87 g. = 1.59 x 10¯5 mole Fe(OH)2

this equals 1.59 x 10¯5 mole Fe2+ and 3.18 x 10¯5 mole OH¯

therefore the Ksp = (1.59 x 10¯5) (3.18 x 10¯5)2 = 1.6 x 10¯14

c) one point

[H+] = 1.0 x 10¯14 / [OH¯]
= 1.0 x 10¯14 / 3.18 x 10¯5
= 3.14 x 10¯10

pH = - log [H+] = 9.50

OR

[pOH] = - log [OH¯] = - log (3.18 x 10¯5) = 4.50

pH = 14 - pOH = 9.50

d) three points

50.0 mL of 3.0 x 10¯3 M Fe2+ diluted to 100 ml = 1.5 x 10¯3 M Fe2+

50.0 ml of 4.0 x 10¯6 M OH¯ diluted to 100 ml = 2.0 x 10¯6 M OH¯

Q = [Fe2+][OH¯]2 = (1.5 x 10¯3) (2.0 x 10¯6)2 = 6.0 x 10¯15

Precepitate will NOT form since Q < Ksp

2)

a) six points

2 H2 + O2 ---> 2 H2O

moles H2 = moles O2 initially but 2 moles of H2 react for every mole of O2. O2 is left.

Ptot = PH2 + PO2 + PH2O

1146 = PH2 + PO2 + 24 (v.p. of H2O = 24 mmHg)

PH2 + PO2 = 1122 mm Hg

1122 mm / 4 = PO2 left (1/2 of initial O2 which is 1/2 total)

PO2 = 280.5 mm

P1V1 / T1 = P2V2 / T2

((280.5) (0.5 L)) / (298) = ((760 mm)V2) / (273)

V2 = 0.169 L therefore

n = 0.169 L / 22.4 L mol¯1 = 7.55 x 10¯3 mol

OR

PV = nRT

((280.5 / 760)atm (0.5 L)) / ((0.0821 L atm/mol K) (298 K)) = n = 7.55 x 10¯3 mol

b) two points

PO2 (@ 90°C) / 363 = 280.5 / 298 = 342 mm Hg

OR

P = ((7.55 x 10¯3 mol) (0.0821) (363)) / 0.5 = 0.45 atm = 342 mm Hg

therefore

Ptot = PO2 + PH2O = 342 + 526 = 868 mm Hg

c) one point

VH2O (@ STP) = ((526 mm Hg) (0.5) (273)) / ((760) (363)) = 0.260 L

mol. of H2O = 0.260 L / 22.4 L mol¯1 = 0.0116 mol

OR

n = ((526 / 750)atm (0.50)) / ((0.0821) (363) = 0.0116 mol

3)

a) two points

[delta]H = bonds broken minus bonds formed.

C2H5Cl + Cl-Cl ---> C2H4Cl2 + HCl
[delta]H = (2794 + 243) minus (2757 + 431)

[delta]H = 3037 - 3188 = - 151 kJ mol¯1

OR

CH + Cl-Cl ---> C-Cl + HCl
[delta]H = (414 + 243) minus (377 + 431)

[delta]H = - 151 kJ mol¯1

b) four points

[delta]G = [[delta]G°f C2H4Cl2 + [delta]G°f HCl] - [[delta]G°f C2H5Cl + [delta]G°f Cl2]

= (- 80.3 - 95.3) - (- 60.5 + 0) = - 115 kJ

[delta]G = [delta]H - T[delta]S

[delta]S = ((- 151 kJ - (- 115 kJ)) / 298 = - 0.120 kJ mol¯1 K¯1

c) two points

[delta]G = - RT ln K

- ln K = - 11510 / (8.314 x 298)

ln K = 46.46

K = 1.50 x 1020

d) one point

Keq will decrease with an increase in T because the reverse (endothermic) reaction will be favored with addition of heat

OR

[delta]G will be less negative with an increase in temperature (from [delta]G = [delta]H - T[delta]S) which will cause K to decrease.

4)

a)* Pb2+ + 2I¯ -----> PbI2

b) Fe3+ + NH3 + H2O ----> Fe(OH)3 + NH4+ (NH4OH is OK in place of NH3 + H2O)

Fe2+ + others above-----> Fe(OH)2 + NH4+ (given two points)

c) H2O2 -----> H2O + O2

d)* Ag+ + CrO42¯ -----> Ag2CrO4

e) H2S + OH¯ -----> S2¯ (or HS¯) + H2O

f) N2O5 + H2O -----> H+ + NO3¯ (HNO3 = one point)

g) Bi + O2 -----> Bi2O3 (Bi2O5 = one point)

h) Cu + H+ + HSO4¯ (or SO42¯) -----> Cu2+ + SO2 + H2O (any two products is given one point; all three must be there for 3 points; omitting H+ and/or H2O is minus one point)

(*Correct equation but wrong charge on ion(s) = one point)

5)

a) two points

C2H4 has a multiple bond; C2H6 has a single bond.

Multiple bonds are stronger and therefore shorter than single bonds.

b) two points

NH3 has 3 bonding pairs and 1 lone pair of electrons.

Bond pairs are forced together because the repulsion between the lone pair and the bond pairs is greater than that netween bond pairs.

c) two points

The bonding in SO3 can be described as a combination of 3 resonance forms of 1 double and single bonds.

The actual structure is intermediate between the 3 resonance forms, having 3 bonds which are equal and stronger (therefore shorter) than a S-O single bond.

(Credit is given for three sulfur-oxygen double bonds @ 120° with explanation.)

d) two points

The central I atom has 3 lone pairs and 2 bond pairs around it. To minimize repulsion, the 3 lone pairs are arranged in a trigonal plane at right angles to the I-I-I axis.

6)

a) two points

Across the period from Li to Ne the number of protons is increasing in the nucleus hence the nuclear charge is increasing with a consequently stronger attraction for electrons and an increase in I.E.

b) two points

The electron ionized in the case of Be is a 2s electron wheras in the case of B it is a 2p electron. 2p electrons are higher in energy than 2s electrons because 2p electrons penetrate the core to a lesser degree.

c) two points

The electron ionized in O is paired with another electron in the same orbital, whereas in N the electron comes from a singly-occupied orbital. The ionization energy of the O electron is less because of the repulsion between two electrons in the same orbital.

d) two points

The ionization energy of Na will be less than those of both Li and Ne because the electron removed comes from an orbital which is farther from the nucleus, therefore less tightly held.

7)

a) two points

  1. the kinetic energy of the molecules
    (A certain minimum energy is required for a reaction to occur (activation energy))
  2. The orientation of the molecules relative to one another. Even very energetic collisions may not lead to a reaction if the molecules are not oriented properly.

b) two points

  1. A decrease in temperature would decrease the rate.
  2. Fewer molecules would have the energy necessary to react. (Fewer effective collisions)

c) two points

Rate = [A][B]2

d) two points

  1. Catalyst increases the rate by providing an alternate pathway which has a lower activation energy.
  2. The value of the equilibrium constant does not change as a catalyst does not affect the energies (or concentrations) of the reactants and products.

8)

a) two points

After the first H+ is lost from H2S, the remaining species, HS¯, has a negative charge. This increases the attraction of the S atom for the bonding electrons in HS¯. The bond is therefore stronger, H+ is harder to remove, and K2 is lower.

b) two points

Polar H2O can separate ionic NaOH into Na+ (aq) and OH¯ (aq) giving a basic solution. In HOCl, chlorine has a high attraction for electrons due to its greater charge density. This draws electrons in the H-O bond towards it and weakens the bond. H+ can be removed, making an acid solution.

c) two points

Water is a more basic solvent (greater attraction for H+) and removes H+ from Cl¯ and I¯ equally.

Acetic acid has little attraction for H+ but the H+ separates from the larger I¯ easier than from the smaller Cl¯.

d) two points

The bond between H and Cl is weaker than the bond between H and F. HCl is therefore a stronger acid.

9)

a) four points

  1. mass of Cu strip and compound minus mass of original clean Cu strip = mass of iodine
    mass of iodine / atomic mass of iodine (126.9 g/mol) = moles iodine
  2. mass of original clean Cu strip minus mass of strip after washing = mass of Cu
    mass of copper / atomic mass of copper (63.54 g/mol) = moles of Cu

b) one point

The empirical formula is the ratio of mol. iodine to mol. copper ( or alternatively, mol. copper to mol. iodine)

c) three points

  1. Unreacted iodine would make the apparent mass of compound and the iodine too high. Thus, the I:Cu ratio in the empirical formula would be too high.
  2. If some of the compound flaked off, the mass of compound (and the iodine) would be too low. Thus, the I:Cu ratio in the empirical formula would be too low.

In part (c) 2 points were awared for a correct response on either part (1) or part (2) that included the prediction (1 pt.) and reason (1 pt.) An additional one point was awarded for the correct prediction and reason in the other part.

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