Answers to Final Exam Review June 2012
Thermochemistry:
1.
exothermic – reaction releases energy into the environment; products have less Pe than reactants.
endothermic – reaction absorbs energy from the environment; products have more Pe than reactants.
2.
89.7 J
3.
a) exothermic
b) 12.5 kJ
q = mcΔt
= (10.0 g)(0.897 J g-1 0C-1)(10.00C)
(5.10 g NH3)(1 mol / 17.04 g)(83.6 kJ / 2 mol NH3) = 12.5 kJ
4.
-86.3 kJ mol- KOH (exothermic reaction)
q = mcΔt = (250 g H2O)(4.18 J g-1 0C-1)(6.80C) = 7106 J = 7.11 kJ
n = (4.62 g KOH)(1 mol / 56.11 g) = 0.0823 mol
ΔH = q/n = 7.11 kJ / 0.0823 mol = 86.4 kJ mol-1 KOH
5.
ΔH = - 1 560.4 kJ
[2(-393.5) + 3(-285.8)] – [ (-84.7) + 7/2(0)] kJ = -1 560.4 kJ
Decrease in entropy….fewer moles of gas on the product side..less random
6.
a) positive
7.
ΔH = -802.5 kJ
8.
ΔH = - 2 757 kJ
9.
10.
11.
12.
A
[(-535 - -547) / (-547)] x 100% = 2.2%
D
D this reaction requires input of energy
B the NaBr bond is less stable ( releases less energy when it forms)
13.
-802.3 kJ
b) negative
[(-393.5) + 2(-241.8)] – [(-74.6) + 2(0)] = -802.5 kJ
[12(414) + 2(347) + 7(499)] – [8(799) + 12(460)] kJ = -2 757 kJ
reverse the second reaction using Hess law, makes its ΔH -118.7 kJ
Reaction Rates:
1.
no, it will be 5 times faster
2.
a) very rapid - no bonds to break, +/- attractions
b) very rapid - no bonds to break, +/- attractions
3.
4.
70 kJ
5.
6.
7.
a) N2O2 b) 2NO + O2 → 2NO2
Br
C
c) slow - two solids reacting, no +/- attractions
8.
C
9.
D
10.
a) ≈ 0.115 M
11.
D
12.
735 times faster at 500C
use exponential form of the Arrhenius equation:
k(323K) / k(283K) = Ae-125 000 J / (323K)(8.314J) / Ae-125 000 J / (283K)(8.314 J)
= e-46.5 / e-53.1 = e6.6 = 735 times faster
13.
EA = 7.2 kJ
use logarithmic form of the Arrhenius equation:
EA = -RT[ln(k/A)] = (-8.414 J)(473K)[ln(0.8/5)] = 7 200 J or 7.2 kJ
b) ≈ 21 min c) r ≈ -0.006 mol/L/min
(interpolate from your graph)
if reaction time is 20 min at 00C, then at 100C it will be twice as fast (10min)
at 200C it will be 5 min and at 300C 2.5 minutes
Equilibrium:
1.
Keq = [SO3] / [O2]3/2
or
2.
i) decrease
iii) increase
3.
C
4.
the higher the Keq, the more products at equilibrium
5.
as temperature decreases, the reaction will shift to the product side to create more
heat; more products will form, less reactants, and therefore the Keq increases.
6.
a) Keq = [BrF5]2 / [Br2][F2]5
7.
2.14 x 10-6
Keq = [CO][H2]3 / [CH4][H2O]
ii) decrease
Keq = [SO3]2 / [O2]3
iv) no effect
b) Keq = [Ca2+]3[PO43-]2 / 1
= (5.45 x 10-3)(2.1 x 10-3)3 / (2.97 x 10-3)( 7.94 x 10-3)
8.
i) towards Products
ii) no change iii) no change iv) towards Reactants
9.
calculate a trial value for K, using initial concentrations:
i) since Ktrial (5) is less than Keq, shift to form more products
ii) since Ktrial (13.3) is less than Keq, shift to form more products
10.
Product
11.
low pressure
12.
B
a very large Keq value indicates high concentration of products
this will favor the formation of more moles of gas
changes in temperature will change Keq values for equilibria
13.
B
complete ice box and solve for ‘x’
[N2]
[Initial]
0.96 M
[Change]
-x
[Equilibrium] 0.84
x = 0.12M
[H2]
0.72 M
-3x
0.36
[NH3]
---+2x
0.24 M
substitute the equilibrium values into the Keq expression:
Keq = [NH3]2 / [N2][H2]3 = (0.24)2 / (0.84)(0.36)3 = 1.46
14.
D
reading the graph shows a decrease in [X] at time t1, so [Y] and [Z] should also
decrease (LeChatelier’s principle) and looking at the graph, they do at time t1
15.
In a saturated solution, AgI(s)  Ag+(aq) + I-(aq)
Ksp = [Ag+][I-]
Assume [Ag+] = [I-]
1.5 x 10-16 = [Ag+]2
1.2 x 10-8 M = [Ag+]
16.
AgBr(s)   Ag+(aq) + Br-(aq)
Ksp = [Ag+][Br-]
6.5 x 10-13 = (2.5 x 10-8)[Br-]
2.6 x 10-5 M = max [Br-] in the solution
17.
Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq)
Ksp = [Mg2+][OH-]2
Let ‘s’ = solubility of magnesium hydroxide in mol/L, then [Mg2+] = ‘s’ and [OH-] = ‘2s’
4.0 x 10-5 = (s)(2s)2 = 4s3
2.2 x 10-2 mol/L = solubility of magnesium hydroxide
In g/L = (2.2 x 10-2 mol / L)( 58.33 g / mol) = 1.28 g /L
18.
Ksp = [Ca2+][OHl-]2
After dilution: [Ca2+] = (0.0322 M)(62.6 ml) / 93.9 ml = 0.0215 M
After dilution: [OH-] = (0.0145 M)(31.3 ml) / 93.9 ml = 0.00483 M
Qsp = (0.0215)(0.00483)2 = 5.02 x 10-7
Qsp < Ksp, no precipitate forms
19.
ScF3(s)  Sc3+(aq) + 3 F-(aq)
Ksp
= [Sc3+][F-]3
-18
4.2 x 10 = [Sc3+][0.076]3
9.6 x 10-15 = [Sc3+]
Oxidation-Reduction/Electrochemistry
1.
Zn + 2NO3- + 4H+ → 2NO2 + 2H2O + Zn2+
2.
5H+ + Cr2O72- + 3SO2 → 2Cr3+ + 3HSO4- + H2O
3.
Not spontaneous; the E0 value is -0.28v (diagonal rule using appendix 7)
4.
a) Zinc b) Bromine c) Br2
5.
a) K = 1, Mn = 7, O = -2; b) V = 3, O = -2; c) Cr = 6, O = -2; d) Fe = 8/3, O = -2
e) Fe = 2, S = -2
6.
Carbon is oxidized from 3 to 4; the Oxygen atoms in H2O2 are reduced from -1 to -2
7.
8.
9.
C
D
A
10.
from the Mg to the Cr
anode: Mg → Mg2+ + 2ecathode: Cr3+ + 3e- → Cr
3Mg + 2 Cr3+ → 2Cr + 3 Mg2+
11.
12.
C
A
13.
9.50 g
solution: use mass to mass stoichiometry from balanced equation:
(3.00 g Zn)(1 mol/65.4 g)(1 Pb/1 Zn)(207.2 g / 1 mol) = 9.50 g
14.
1.50v
Fe  Fe2+ + 2e- E0= 0.44v
2e + Br2  2BrE0 = 1.06v
-
15.
16.
C
C
this causes the Fe to attain a negative charge, and not to lose electrons
17.
902 Amp
solution: use faraday’s law g = A(s)(MM)/nF, therefore, A = gnF/sMM
A = (1.75molCu)(63.5 g/mol)(2e-)(96 500 A-s)/374.4s(63.5g) =
18.
10.8 g
solution: g = (1.00A)(9650s)(107.9 g)/ (1e-)(96500A-s) = 10.8 g
19.
the Mg rod will rust before the iron in the steel will; thus it protects the iron from rusting
20.
A the cathode in an electroplating experiment must be the object that is being plated.
ions of the metal plating over top of the object must be present in the electrolytic cell
Acid/Base
1.
pH = 5.66 acidic
2.
H2O + HNO3
base2
acid 1
3.
0.018 M
[H3O+] = (1 x 10-14)/(4.5 x 10-9) = 2.2 x 10-6; pH = -log(2.2 x 10-6M)
↔
NO3base 1
+
H3O+
acid 2
HNO3 + KOH → H2O + KNO3 use MAVA = MBVB
(0.0300M)(15.0 ml) = (MB)(25.0ml) MB = 0.018M
4.
[H3O+] = 2.24 x 10-5M ; [OH-] = 4.47 x 10-10M
[H3O+] = 10-4.65
[OH-] = (1 x 10-14) / (2.24 x 10-5M)
5.
1.69 x 10-3
6.
if HNO3 to this equilibrium system, HNO2 + H2O ↔ NO2- + H3O+
it will increase the [H3O+]; this will cause the equilibrium to shift to the reactant side,
decreasing the ionization of the nitrous acid.
7.
3 H2SO4(aq) + 2 Al(OH)3(aq) ↔ 6 H2O(l) + Al2(SO4)3(aq)
8.
a) ↔ SO42- + H2O
9.
HCl has a very large ionization constant (Ka) it therefore produces a large amount of ions
acetic acid has a relatively small ionization constant – produces only a few ions
10.
3.62 x 10-3M
11.
this titration curve will show buffering at the beginning, and will have an equivalence point
less than pH of 7; methyl red is one good choice for an indicator
12.
KA = [H3O+]2 / [HA] = (0.0260)2 / (0.400)
b) ↔ Br- + H3PO4
KA = [H3O+][NO2-] / [HNO2]
[H3O+] = (7.24 x 10-4)(0.15)/ (0.0300)
14
buffering
pH
- ≈5.0 at eq. pt.
0
the buffer must be able to react with both acids and bases , therefore it requires both a
weak acid and its conjugate base ; ex. CH3COONa / CH3COOH
13.
14.
C
6.67 x 10-9 ; 10.2
for conjugates, KW = KAKB
KA = 1 x 10-14 / 1.5 x 10-6
since pH = -log(H+) we need to find [H+] in the solution
N2H4 + H2O  OH- + N2H5+
KB = [OH-]2 / [N2H4]
[OH-] = √(1.5 x 10-6)(0.02) = 1.77 x 10-4; [H+] = 1 x 10-14 / 1.77 x 10-4 = 5.77 x 10-11
pH = -log(5.77 x 10-11) = 10.2
15.
B
KA = [H+]2 / [HA]
16.
D
2 HCl + Ca(OH)2 → CaCl2 + 2H2O
(0.250 M)(15.0 ml) = 2(MB)(22.5 ml)
17.
pH = 5.14
18.
19.
HCO3- is the best choice. Its pKa of 10.33 is closest to the desired pH of 10.5
B
note that after the equivalence point is reached, as more acid is added, the pH stays at about 5
use approximation [H+] ≈ √(2.9 x 10-8)(0.85) = 1.57 x 10-4
use MAVA = 2MBVB
MB = 0.0833M
use Henderson-hasselbach equation: pH = pKa + log[(A-]/(HA)]
pKa for CH3COOH = -log(1.8 x 10-5) = 4.74
pH of buffer = 4.74 + log[1.0/0.4] = 4.74 + 0.40 = 5.14
Solutions/Solubility
1.
a) K3PO4 soluble b) CuBr2 soluble c) NiSO4 soluble
2.
mass required is 4.23 grams ; g = cvMM = (0.0450 mol L-1)(0.250L)(375.7 g mol-1)
3.
I2 is nonpolar solid – it will form strong attractive forces to nonpolar solvents such
as benzene and alcohol which is partly non-polar.
4.
5.
6.
7.
8.
9.
a) 6.04 x 10-5 M b) 3.53 ppm c) 3 533 ppb d) 3.53 x 10-4 %
c = n/v = (5.300 g)(1 mol / 58.5 g) / 1 500 L
ppm = (5.300 g NaCl / 1.500 x 106 g water) X 1 x 106 = 3.53 ppm
ppb = ( 5.300 g NaCl / 1.500 x 106 g water) X 1 x 109 = 3 533 ppb
% wt = (g solute / g solution) x 100% = (5.300 g / 1.500 x 106 g) x 100% = 3.533 x 10-4%
A
C
c = n/v = (11.7 g)(1 mol / 58.5 g) / 0.250 L
A
B
0.208L or 208 mL
V1 = C2V2/C1 = (0.500 mol L-1)(5.00L) / (12.0 mol L-1) = 0.208L
10.
Pb2+(aq) + 2 I-(aq) →
11.
C1(2.00L) = (0.00400 M)(0.0400 L)
C1 = 8.00 x 10-5 M
12.
(2.00 M)(V1) = (0.800 M)(0.500 L)
V1 = 0.200 L or 200 ml
PbI2(s)