Notes
a) 5 points
Kp = (PNH3) (PH2S)
PNH3 = PH2S = (0.659 atm / 2) = 0.330 atm
Kp = (0.330 atm) (0.330 atm) = 0.109 atm2
b) 5 points
PNH3 = 2 PH2S
(2x) (x) = 0.109
x = 0.233 atm = PH2S
2x = 0.466 atm = PNH3
c) 5 points
equilibrium pressure of NH3 = equilibrium pressure of H2S = 0.330 atm
PNH3 that has reacted = PH2S
= 0.500 - 0.330 = 0.170 atm
n = PV / RT = (0.170 x 1.00) / (0.08205 x 298) = 6.95 x 10¯3
If a student calculated Kc rather than Kp, one point was deducted.
a) 6 points
rate = k [A]m [B]n
m = 2
n = 1
b) 3 points
Substitution of any one set of data from experiments 1-5 into the rate equation.
For example, experiment #1:
8.00 mol L¯1 hr¯1 = k [0.240 mol / L]2 [0.480 mol / L]
k = 289 L2 mol¯2 hr¯1
c) 2 points
rate = 289 L2 mol¯2 hr¯1 (0.0140 mol / L)2 (1.35 mol / L)
rate = 0.0766 mol L¯1 hr¯1
d) 4 points
acknowledgement of a limiting reagent
identification of B as the limiting reagent
correct mole ratio
[C] = (3C / 2B) x 0.120 mol / L of B
a) 5 points
MnO4¯ + 8 H+ + 5e¯ ---> Mn2+ + 4 H2O
or MnO4¯ (+7) ---> Mn2+ (with an indication of 5 e¯ exchanged)
H2C2O4 ---> CO2 + 2 H+ + 2e¯
or H2C2O4 (+3) ---> 2 CO2 (+4) (with an indication of 2 e¯ exchanged)
MnO4¯ + 5 H2C2O4 + 6 H+ ---> 10 CO2 + 8 H2O + 2 Mn2+
MnO4¯ (or Mn) oxidizing agent
H2C2O4 (or C) reducing agent
b) 4 points
moles MnO4¯ = 0.03563 L x 0.1092 mol/L = 3.890 x 10¯3 mol
moles H2C2O4 = 3.890 x 10¯3 mol MnO4¯ x (5 mol H2C2O4 / 2 mol MnO4¯) = 9.724 x 10¯3 mol
c) 2 points
# mol CaCO3 = # mol H2C2O4
therefore 9.72 x 10¯3 mol CaCO3 in sample
d) 4 points
g CaCO3 = 9.72 x 10¯3CaCO3 x (100.1 g CaCO3 / mol CaCO3) = 0.973 g CaCO3
% CaCO3 = (0.973 g CaCO3 / 1.2516 g sample) x 100 = 77.7%
a) Mg + N2 ---> Mg3N2
b) SO2 + CaO ---> CaSO3 (Ca2+ SO32¯ or Ca2+ + SO32¯)
c) Pb + Ag+ ---> Pb2+ + Ag
d) NH4+ + SO42¯ + Ba2+ OH¯ ---> BaSO4 + NH3 + H2O (or NH4OH)
e) CH3COOH + HCO3¯ ---> CH3COO¯ + CO2 + H2O (or H2CO3)
(Somewhat suprising to the ChemTeam, the answer sheet indicated that HAc or HOAc were accepted for acetic acid and Ac¯: or OAc¯ for acetate. Hmmmmm.)
f) Na2Cr2O7 (or Cr2O72¯ + I¯ ---> Cr3+ + I2 (I3¯ or IO3¯)
g) Fe3+ + SCN¯ ---> [FeSCN]2+ (or [Fe(SCN)x](3-x)+ where x = 2-6)
or
[Fe(H2O)6]3+ + SCN¯ ---> [Fe(H2O)5(SCN)]2+ + H2O
h) C2H5OH + O2 ---> CO2 + H2O (C2H6O or EtOH and CO if O2 is not in excess)
a) 2 points
Cu2+ + 2 e¯ ---> Cu
b) 2 points
2 H2O ---> O2 + 4 H+ + 4 e¯
c) 1 point
any reasonable sketch with correct labels
d) 3 points
measure mass of cathode before and after experiment
measure current
measure time
(Also acceptable: measure volume of O2, temperature, pressure, current, time.)
a) 3 points
Per 100 g of compound
85.7 g C x (1 mole C / 12.0 g C) = 7.14 mol C
14.3 g H x (1 mol H / 1.01 g H) = 14.2 mol H
emperical formula is CH2
molecular formula is C4H8
or
(56 g / mol) x 0.857 = 48 g / mol C
(56 g / mol) x 0.143 = 8.0 g / mol H
48 / 12 = C4 and 8.0 / 1.0 = H8
b) 4 points, one point for each correct structure
c) one point
a) 4 points
a solution of Al(NO3)3 is acidic
a solution of K2CO3 is basic
a solution of NaHSO4¯ is acidic
a solution of NH4Cl is acidic
b) 4 points
Al3+ + H2O ---> Al(OH)2+ + H+
or
Al(H2O)63+ + H2O ---> [Al(H2O)5OH]2+ + H+
Al3+ + 3 H2O ---> Al(OH)3 + 3 H+
CO32¯ + H2O ---> HCO3- + OH¯
HSO4¯ + H2O ---> SO42¯ + H3O+
NH4+ + H2O ---> NH3 + H3O+
Equations needed to be balanced and ions must show correct charges. Molecular formulas were acceptable if acids and bases formed were marked strong and weak.
a) 1 point
Quantized energy levels or discrete energies or wave properties of electron produce discrete energy states in a gas.
b) 2 points
The excited state atoms can relax to several lower energy states (see diagram in c).
Each final state energy level produces a separate series.
c) 2 points
d) 3 points
Emission spectra are photons emitted from excited state systems as they drop to lower energy states.
Absorption spectra result from the absorption of electromagnetic radiation. Electrons are excited to a highter energy state.
Hydrogen atoms are in the lowest electronic energy state at 25 °C (n = 1) so absorptions will be n = 1 to n = 2,3,4, etc.
Four effects and 4 explanations valued at 1 point each.
a) No, the reactant is favored. Treaction is endothermic and products are at highter energy than the reactant.
b) Yes, products are favored. [delta]S > O as 2 moles of gas are produced form 1 mole of gas.
c) Right, products are favored. For the system at constant pressure: absorption of heat favors the products. An argument using LeChatelier's principle can be used with heat considered as a reactant; or T[delta]S favors products as [delta]G becomes more negative. For the system at constant volume: products are favored but increase in the yield of products results in an increase in pressure which would drive the reaction to the left.
No points were deducted for students who mentioned both the thermodynamic effect and the counteractiong pressure effect at constant volume but were uncertain as to which effect is larger. Students who considered a system at constant volume and stated only that the increase in pressure would drive the reaction to the left received half credit.
d) Left, the reactant is favored. A decrease in the volume would increase the pressure and the strain is relieved by the reverse reaction which produces 1 mole of gas from 2 moles of gas.
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