Download Answers 1998 Free Response

1998 AP Chemistry Scoring Guidelines
1.-9 points i.) Cu(OH)2(s)  Cu2+(aq) + 2 OH-(aq)
1 point
Correct stoichiometry and charges (but not phases) necessary
No credit earned if water as a reactant or product
ii)
1.72 10 6 g
 1.763 10 8 mol Cu(OH) 2
1
97.57 g mol
1 point
1.763 108 mol Cu(OH) 2
 1.76 107 mol L1
0.100 L
One point earned for conversion of mass to moles (need not be computed explicitly)
One point earned for calculation of moles per liter
iii)
[Cu2+] = 1.76 × 10-7 M
[OH-] = 2 × (1.76 X 10-7 M) = 3.52 × 10-7 M
1 point
2+
- 2
-7
-7 2
-20
Ksp = [Cu ][OH ] = (1.76 × 10 )( 3.52 × 10 ) = 2.18 × 10
1 point
One point earned for correct [Cu2+] and [OH-]
One point for correct substitution into Ksp expression and answer
Response need not include explicit statement of [OH-] if Ksp expression is written with
correct values of [Cu2+] and [OH-]
pH = 9.35 → pOH = 4.65 → [OH-] = 2.24 × 10-5 M
1 point
17
K
7.7  10
sp
[ Zn 2 ] 

 1.5  10 7 M
 2
5 2
[OH ]
(2.24  10 )
One point earned for correct determination of [OH-]
One point for correct answer (assume [Zn2+] equals solubility in moles per liter)
No points earned if [OH-] is assumed equal to twice [Zn2+]
b)
i)
ii)
Zn2+ + 2OH-  Zn(OH)2(s)
initial
final
Zn2+
0.0050 mol
0 mol
OH0.0150 mol
0.0050 mol
Zn(OH)2 (s)
0 mol
0.0050 mol
or
0.0050 mol
 0.050 M
0.100 L
One point earned if precipitation reaction is clearly indicated and moles or concentration of OHis calculated correctly
Zn(OH)2

Zn2+ +
2 OHx
(0.050 + 2x)
Ksp = 7.7 × 10-17 = [Zn2+][OH-]2 = (x) (0.050 + 2x)2 = (x)(0.050)2  [Zn2+] = x = 3.1 × 10-14 M 1 point
OR
[OH  ] 
Zn(OH)2

Zn2+ +
(0.050-x)
2 OH(0.150 - 2x)
Ksp = 7.7× 10-17 = [Zn2+][OH-]2 = (0.050-x)(0.150-2x)2
Solve for x, then subtract x from 0.050 M to obtain [Zn2+]
Question 2
a)
Assume a 100 – gram sample (not necessary for credit):
1 mol C
65 g C 
 5.462 mol C
12.01 g C
1 mol H
9.44 g H 
 9.366 mol H
1.0079 g H
Mass O = [100 – (65.60 + 9.44)] = 24.96 g O
1 mol O
24.96 g O 
 1.560 mol O
15.9994 g O
C5.462H9.366O1.560 --> C3.5H6.0O1.0 --> C7H12O2
One point earned for determining moles of C and moles of H
One point earned for determining moles of O
One point earned for correct empirical formula
15.2 o C
T
m

 0.380 mol kg 1
b)
Kf
40.0 K kg mol 1
0.380 mol
0.01608 kg 
 0.00611 mol
1 kg
1.570 g
molar mass 
 257 g mol 1
0.00611 mol
One point earned for determination of molality
One point earned for conversion of molality to molar mass
OR
T  kg solvent
moles solute 
 0.00611 mol
Kf
1.570 g
molar mass 
 257 g mol 1
0.00611 mol
OR
Kf
molar mass  mass 
 257 g mol 1
T  kg solvent
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
1 point
2 points
Empirical mass of C7H12O2 = 7(12) + 12(1) + 2(1) = 128 g mol-1
128 g mol-1 = ½ molar mass --> molecular formula = 2 × (empirical formula)
--> molecular formula = C14H24O4
1 points
One point earned if molecular formula is wrong but is consistent with empirical
formula and molar mass
c)
d)
No penalty for simply ignoring the van’t Hoff factor
Only one point earned for part b if response indicates that
ΔT = (15.2 + 273) = 288 K and molar mass = 13.6 g mol-1
1 atm0.577 L 
PV
n

 0.0123 mol
1 point
RT 0.0821 L atm mol 1 K 1 573 K 
mass of sample
1.570 g
molar mass 

 128 g mol 1
1 point
moles of sample 0.0123 mol
Only one point can be earned for part c if wrong value of R is used and / or T is not
converted from C to K
The compound must form a dimer in solution, because the molar mass in solution is
twice that it is in the gas phase,
OR,
The compound must dissociate in the gas phase (A(g) --> 2 B(g)) because the molar
mass in the gas phase is half that it is in solution
One point earned for a reference to either or both the ideas of dimerization and
dissociation
No point earned for a “non-ideal behavior” argument
Question 3 – 9 points
1 mol
 0.02125 mol phenol
1 point
94.113 g
64.98 kJ
 3,058 kJ mol 1
Heat released per mole 
0.02125 mol
Or, ΔHcomb = -3058 kJ mol-1
1 point
Units not necessary
b)
ΔHcomb = -3058 kJ mol-1
1 point
o
-3058 kJ = [6(-395.5) + 3(-285.85)] – [ΔHf (phenol)]
1 point
ΔHfo (phenol) = -161 kJ
1 point
One point earned for correct sign of heat of combustion, one point for correct use of
moles / coefficients, and one point for correct substitution
c)
ΔSo = [3(69.91) + 6(213.6)] [7(205.0) + 144.0] = -87.67 J!K
1 point
ΔGo = ΔHo - TΔSo = 3058 kJ – (298 K)(-0.08767 kJ K-1) = -3032 kJ
1 point
Units not necessary; no penalty if correct except for wrong ΔHcomb for part a
d)
moles gas = 9 × [moles from part a] = 9 (0.02125 mol) = 0.1913 moles gas
nRT (0.1913 mol)(0.0821 L atm mol 1 K 1 )(383 K )
P

 0.601 atm 1 point
V
10.0 L
Units necessary; no penalty for using Celcius temperature if also lost point in part c for same error
a)
2.000 g 
Question 4 – 15 points
a.
Sn2+ + Fe2+  Sn4+ + Fe2+
Two points earned if only error is wrong symbol for tin (eg, Ti)
b.
Co2+ + OH-  Co(OH)2
c.
C2H4 + O2  CO2 + H2O
No penalty for other oxidized forms of carbon as products (e.g. C, CO)
d.
H3PO4 + OH-  H2PO4- + H2O
One point earned for H+ + OH-  H2O
Two points earned for removal of H+ from any HxPyOz species and H2O as product
e.
CaSO3  CaO + SO2
Two points earned for CaSO4  CaO + SO3
f.
H+ + Cl- + [Ag(NH3)2]+  AgCl + NH4+
Cl- + [Ag(NH3)2]+  AgCl + NH3 (or NH4+) earns two points
H+ + [Ag(NH3)2]+  Ag+ + NH4+ earns two points
g.
Na2O + H2O  Na+ + OHTwo points earned if reactants correct but only product is NaOH
h.
Zn + H+  Zn2+ + H2
Two points earned for Zn + H+ + Br-  ZnBr2 + H2
Two points earned for Zn + HBr  Zn2+ + Br- + H2
Question 5 (8 points)
a.
4 essential steps
1)
weigh KHP
2)
fill buret with NaOH solution
3)
add indicator (phenolphthalein)
4)
titrate to endpoint (color change)
Two points earned for all 4 steps; one point earned for 2 or 3 steps
Titration of acid into base accepted if described correctly
mass KHP
moles KHP 
b.
molar mass KHP
moles OH 
moles KHP = moles OH- at equivalence and
 [OH  ]
liters NaOH
Acceptable if some pars of part b appear in a
c.
Curve should have 3 important features
1) Curve begins above pH 1, but below 7
2) Equivalence point at 25 mL
3) Equivalence point above pH 7
Both points earned for all 3 features
One point earned for any 2 of the 3 features
d.
At the half-way point in the titration, pH = pKa
e.
At point A in the titration, the anion in highest concentration is Y2Also accepted: Y2-, Y--, Y=, and specific anions such as SO42-, SO32HY-, Y- and “Y ion” not accepted
Question 6 – 8 point
a)
Response must clearly indicate (and distinguish between) Eact and ΔHrxn on graph.
Each earns one point (2 total)
b)
i.
Response shows a softly curving line that approaches the time axis
and whose slope changes continually
No penalty if curve crosses time axis or levels out above time axis.
Curve must drop initially and continually. No credit earned if [N2O5] increases.
2 points
1 point
1 point
2 points
1 point
1 point
ii.
Reaction Rate is the slope of the line tangent to any point on the curve.
Rate must be tied somehow to slope of graph .
Answer may be indicated directly on the graph .
Instantaneous rate must be indicated rather than the average rate
iii.) Since "rate = slope = k[N205J", the value of k can be determined algebraically from the slope at a known
value of [N20S].
No penalty for "Rate =2k[N205] ", as reaction stoichiometry could suggest this answer.
Point can be earned for rate constant = slope of graph of ln[N2O5] vs.. time since reaction is
first order.
Use of half-life or integrated rate law to solve for k can be accepted.
iv. The value of the rate constant is independent of the reactant concentrations, so adding more
reactant will not affect the value of k.
No point earned for simply stating that value of k will not change .
Response must distinguish between rate. and rate constant.
i.
Rate = k[A] or ln ([A]/[A]0) = -kt. Since graph of ln [A] vs. time is linear, it must be a first-order
reaction.
Either form of the rate law is acceptable, and both the equation and the brief justification are
required to earn the point
No point earned if response indicates first order because the first graph is not linear.
ii.
Determine the slope of the second graph and set “k= - slope”
Response must indicate both the negative sign and the slope.
Question 7 – 8 points
a
b.
c.
the number of moles of CO will decrease
1 point
because
adding H2 will make the reaction shift to the left
OR,
adding H2 will make the reaction quotient larger than K, thus the reaction shifts to the left
1 point
The number of moles of CO will increase
1 point
because
since the reaction is endothermic, addition of heat (as a reactant) will drive the reaction to
the right
1 point
The number of moles of CO will decrease
1 point
because
there are more moles of gas (2) on the right than on the left (1), thus decreasing the
volume which increases the pressure causing the reaction to shift to the left
1 point
d.
The number of moles of CO will stay the same
1 point
because
Solids are not involved in the equilibrium expression OR
Solids have no effect on the equilibrium
Question 8 (8 points)
a)
2 Ag+(aq) + Cd(s) ~ 2 Ag(s) + Cd2+(aq)
equation must be balanced and net ionic, phases not necessary
reaction direction and ion charges must be correct
1 point
1 point
0.80 - (-0040) = 1.20 V
evidence of where numbers came from should be present; if equation is exactly reversed, -1.20 V earns
the point
1 point
b)
Anions (or N03- ions) will flow to the Cd2+ solution or from the Ag+ solution to balance the
charges
OR
Anions will flow to the left to balance the positive charge of the new Cd:!+ ions
both the correct direction and justification1 needed to earn this point
direction may be indicated by arrow marked on diagram
1 point
c)
The cell voltage will increase.
1 point
Ag+ is a reactant, so increasing [Ag+] will increase the driving force (stress) for the forward (spontaneous)
reaction and the potential will increase
OR
1 point
2+
Since Q = [Cd ]/[Ag+]2, increasing [Ag+] will decrease Q. According to the Nernst equation, E = E0 (0.0592 log Q / n , if Q decreases, then voltage increases.
d.)
The cell voltage will decrease.
1 point
Adding NaCI will have no effect on the Cd cell, but will cause AgCI to precipitate
in the Ag cell (Ag+ + Cl-  AgCl). Thus [Ag+] decreases, and since Ag+ is a reactant, decreasing
[Ag+] causes a decrease in voltage.
1 point
Credit earned for" decreasing [Ag+] results in decreased voltage" or "opposite of part (c)"
e)
Since Q = [Cd2+]/ [Ag+/2 , diluting both solutions by the same amount will increase the value of Q.
According to the Nernst equation, E = E0 - (0.0592 log Q )/n , if Q increases, then voltage decreases. ,
• No credit earned for "since the solutions are diluted, the voltage will decrease"
Question 9 (8 points)
a)
At the higher altitude the ambient pressure is significantly less than 1.0 atm.
1 point
Under reduced pressure, water boils at less than 100 oC
1 point
Two points earned for "At the higher altitude water boils at less than JOO"C. and at the lower
temperature the chemical physical processes (the "cooking ") take longer
b)
Cu2+(aq) + 2 OH-(aq)  Cu(OH)2(s)
1 point
Response must indicate that an insoluble hydroxide forms. but equation is not necessary
Cu(OH)2(s) + 4NH3(aq)  [Cu(NH3)4]2+(aq)
1 point
Response must indicate some cationic ammine complex with a reasonable coordination number
c)
CH3CH2OH
1 point
Response must indicate a clear ethanol structure (Lewis diagram not necessary) 1 point
The hydroxyl group forms hydrogen bonds with water molecules.
Response must mention/indicate involvement of hydroxyl group
Point earned for "Ethanol is more polar than dimethyl ether", but no point earned for "dimethyl
ether is linear (or nonpolar)"
d.)
Au3+, Co3+, or F2
These oxidants are below the CI/Cl- reduction half-reactions, so they
would spontaneously oxidize CI to C12
1 point
Any species to the right of the arrow and above the C12/Cl reduction half-reaction on the standard
reduction potential table.
1 point
Identification and justification needed to earn each point - justification should minimally make
some reference to relative positions in the reduction potential table