Notes
a) three points
Ka = ( [H+] [C3H5O2¯] ) ÷ [HC3H5O2]
1.3 x 10¯5 = x2 ÷ 0.20
x = [H+] = 1.6 x 10¯3
b) one point
% dissoc. = [H+] ÷ [HC3H5O2]
= 1.6 x 10¯3 ÷ 0.20 = 0.80%
c) two points
[H+] = antilog (- 5.20) = 6.3 x 10¯6
1.3 x 10¯5 = (6.3 x 10¯6) x ([C3H5O2¯] ÷ [HC3H5O2]
[C3H5O2¯] ÷ [HC3H5O2] = 1.3 x 10¯5 ÷ 6.3 x 10¯6 = 2.1
An alternate solution for (c) based on the Henderson-Hasselbalch equation.
pH = pKa + log ([base] ÷ [acid])
5.20 = 4.89 + log ([C3H5O2¯] ÷ [HC3H5O2])
log ([C3H5O2¯] ÷ [HC3H5O2]) = 0.31
[C3H5O2¯] ÷ [HC3H5O2] = 2.0
d) six points
0.10 L x 0.35 mol/L = 0.035 mol HC3H5O2
0.10 L x 0.50 mol/L = 0.050 mol C3H5O2¯
0.035 mol - 0.004 mol = 0.031 mol HC3H5O2
0.050 mol + 0.004 mol = 0.054 mol C3H5O2¯
1.3 x 10¯5 = [H+] x [(0.054 mol/0.1 L) ÷ (0.031 mol/0.1 L)]
Can use 0.54 and 0.31 instead.
[H+] = 7.5 x 10¯6
pH = 5.13
An alternate solution for (d) based on the Henderson-Hasselbalch equation.
use [ ]s or moles of HC3H5O2 and C3H5O2¯
pH = pKa + log (0.054 / 0.031)
= 4.89 + 0.24 = 5.13
a) three points
7.2 g H2O ÷ 18.0 g/mol = 0.40 mol H2O
0.40 mol H2O x (2 mol H / 1 mol H2O) = 0.80 mol H
7.2 L CO2 ÷ 22.4 L/mol = 0.32 mol CO2
0.32 mol CO2 x (1 mol C / 1 mol CO2) = 0.32 mol C
OR
n = PV ÷ RT = [(1 atm) (7.2 L)] ÷ [(0.0821 L atm mol¯1 K1) (273 K)] = 0.32 mol CO2
0.80 mol H ÷ 0.32 = 2.5
0.32 mol C ÷ 0.32 = 1
2.5 x 2 = 5 mol H
1 x 2 = 2 mol C
empirical formula = C2H5
b) two points
mol O2 for combustion = mol CO2 + 1/2 mol H2O = 0.32 + 0.20 = 0.52 mol O2
0.52 mol O2 x 32 g/mol = 17 g O2
alternate approach for mol O2 from balanced equation
C2H5 + 13/4 O2 ---> 2 CO2 + 5/2 H2O
other ratio examples:
1, 6.5 ---> 4, 5
0.25, 1.625 ---> 1, 1.25
mol O2 = 0.40 mol H2O x (13/4 mol O2 / 5/2 mol H2O) = 0.52 mol O2
Note: starting moles of C2H5 = 0.16 mol C2H5
c) three points
MM stands for molar mass.
DT = (Kf (g/MM)) / kg of solvent
0.5 °:C = ((4.68 °:C kg mol¯:1) x (0.60 g / MM)) / 0.1 kg
MM = (4.68 x 0.60) / (0.5 x 0.1) = 56 or 6 x 101
an alternate solution for (c)
molality = 0.5 °:C / (4.68 0.5 °:C/m) = 0.107 m
mol solute =( 0.107 mol / kg solvent) x 0.100 kg solvent = 0.0107 mol
MM = 0.60 g / 0.0107 mol = 56 or 6 x 101
d) one point
(56 g/mol of cmpd) / (29 g/mol of empirical formula) = 1.9 empirical formula per mol
OR
6 x 101 / 29 = 2.1
empirical formula times 2 equals molecular formula = C4H10
a) four points
rate = k [ClO2] [F2]
one point - rate equation form, k
one point - F2 order
two points - ClO2 order
b) two points
k = rate / ([ClO2] [F2])
= 2.4 x 10¯3 mol L¯1 sec¯1 / ((0.010 mol/L) (0.10 mol/L))
= 2.4 L mol¯1 sec¯1
one point - value consistent with equation in (a)
one point - units consistent with equation in (a)
c) one point
2 ClO2 + F2 ---> 2 ClO2F
- d[F2] / dt = 1/2 (d[ClO2F] / dt)
= 1/2 (9.6 x 10¯3)
= 4.8 x 10¯3 mol L¯1 sec¯1
d) two points
mechanism I
defense:
slow step is first order
three equations add to proper stoichiometry
Note: if ClO2 order in rate equation of part (a) is zero, mechanism II must be chosen to obtain credit.
a) Al2O3 + OH¯ ---> Al(OH)4¯
OR
Al2O3 + H2O ---> Al(OH)3
(Personal note by John Park: I think the H2O in the second equation above comes from the fact that the NaOH concentration was not specified in the original problem. For example, suppose [OH¯] were 10¯12 M. Then the second equation becomes the predominate one.)
b) CaO + SO3 ---> CaSO4
c) H+ + OH¯ ---> H2O
d) Ca + N2 ---> Ca3N2
e) CuS + O2 ---> Cu + SO2 (also CuO, Cu2O)
f) H+ + Cl¯ + MnO2 ---> Mn2+ + Cl2 + H2O (one pt. for either redox product, two pts. for all three products)
g) Zn2+ + NH3 ---> Zn(NH3)42+
OR
Zn2+ + NH3 + H2O ---> Zn(OH)2 + NH4+
h) Cu2+ + SO42¯ + Ba2+ + OH¯ ---> Cu(OH)2 + BaSO4
A rare double precipitation.
Partial credit was allowed for some alternate solutions, e.g.
Cu2+ + OH¯ ---> Cu(OH)2
Ba2+ + SO42¯ ---> BaSO4
a) two points
DS will be negative. The system becomes more ordered as two gases form a solid.
b) two points
DH must be negative. For the reaction to be spontaneous, DG must be negative, so DH must be more negative than -TDS is positive.
c) two points
As T increases, -TDS increases. Since DS is negative, the positive -TDS term will eventually exceed DH (which is negative), making DG positive. (In the absence of this, DG = DH - TDS and general discussion of the effect of T and DS gets 1 point.)
d) two points
The equilibrium constant is 1. The system is at equilibrium at this temperature with an equal tendancy to go in either direction.
OR
DG = 0 at equilibrium so K = 1 in DG = -RT ln K
(In the absence of these, DG = -RT ln K gets 1 point).
The above concludes the AP scoring standards published in 1991. The following is simply alternate ways of answering which the AP readers may or may not have given full credit to.
a) The amount of entropy goes down, DS is negative.
b) DG = DH - TDS. If DS is negative, then DH must also be negative to get a negative DG.
c) Let us say DG is positive when DH is positive and DS is positive. As T goes up - TDS becomes more negative until it makes DG (which equals DH - TDS) become negative.
d) At the temperature when the direction changes, the rate forward = the rate reverse. Since K = kf / kr, this equals 1.
a) two points
Mass of vaporized liquid (or liquid or substance)
two of three in parts a or b
atmosperic pressure
volume of flask
temperature of vapor (water)
b) three points
Procedure for:
Flask containing liquid is heated until liquid disappears
c) one point
mass ÷ mole where mole is determined from PV = nRT
d) two points
Molar mass is too high
because the non-volatile inpurity contribute additional mass (but insignificant volume).
a) two points
Cl¯ is more easily oxidized than H2O
H2O is more easily oxidized than Na+
no 2nd pt is awarded for H+ ---> H2 unless H2O is implied.
no 2nd pt for Na(s) + H2O ---> Na + OH¯ unless H2 is indicated.
b) two points
Fe2+ requires 2 Faraday / mol Fe (s) or 1Faraday ---> 1/ 2 mol Fe(s)
Fe3+ require 3 Faraday / mol Fe (s) or 1 Faraday ---> 1/3 mol Fe (s)
for equal numbers of Faraday (1/2 : 1/3 as 1.5 : 1) (Or inverse relationship is clear)
no 2nd point unless flow of e¯ to mass (moles) is clear and logically correct
c) two points
Le Châtlier's argument
if [Zn2+ ] goes down; reaction shifts right, i.e. cell potential goes up
if [Pb2+ ] goes down; reaction shifts left, i.e. cell potential goes down
OR
Nernst Equation argument
E = E° - RT ln Q with Q = [Zn2+ ] / [Pb2+ ]
if [Zn2+ ] goes down Q < 1, therefore E > E°
if [Pb2+ ] goes down Q > 1, therefore E < E°
reasoning must indicate correct usage of equation
d) two points
[Zn2+ ] / [Pb2+ ] does not change; regardless of values; i.e. E=E°
OR
[Zn2+ ] / [Pb2+ ] = 1 so ln Q = 0; i.e. E=E°
no pt is awarded for just stating concentrations are equal
ratio or proportion concept is required for 2nd point
a) three points
MgCl2 is ionic and SiCl4 is covalent. The elctrostatic, interionic forces in MgCl2 are much stronger than the intermolecular (dispersion) forces in SiCl4 and lead to a higher melting point. Molten MgCl2 contains mobile ions that conduct electricity whereas molten SiCl4 is molecular, not ionic, and has no conductivity.
b) two points
MgF2 has a higher melting point than MgCl2 because the smaller F¯ ions and smaller interionic distances in MgF2 cause stronger forces and higher melting point.
c) one point
The bond length in Br2 is larger than in F2 because the Br atom is larger than the F atom.
d) two points
The bond length in N2 is less than in F2 because the N-N bond is triple and the F-F is single. Triple bonds are stonger and therefore shorter than single bonds.
a) two points
When nucleons are combined in nuclei, some of their mass is converted to energy (binding energy) which is released and stabilizes the nucleus. (Key concepts: mass defect; binding energy)
b) two points
Alpha particles have a greater mass than beta particles. Thus their speed (penetrating potential) is less. (Alternate explanation could be based on charge.)
c) two points
The neutron/proton ratio in Sr-90 and Cs-137 is too large and they emit beta particles (converting neutrons to protons) to lower this ratio.
d) two points
Large amounts of energy are neded to initiate fusion reactions in order to overcome the repulsive forces between the positively charged nuclei. Large amounts of energy are not required to cause large nuclei to split.
Copyright © 1991 by College Entrance Examination Board and
Educational Testing Service. All rights reserved.