BCIT Winter 2014 Chem 0012 Final Exam Name: ___________________ Attempt all questions in this exam. Read each question carefully and give a complete answer in the space provided. Part marks given for wrong answers with partially correct reasoning/calculations. Constants and equations are attached at the back. Data sheets and tables are provided. Total points = 50 Page 1 of 12
Section I: Multiple choice (25 points total, 1 point each) Choose the BEST answer to the following questions. 1. Consider the following reaction involving 1.0 g of powdered zinc: Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) Trial Temperature (°C) Concentration of HCl 1 40 3.0 2 20 3.0 3 40 6.0 The reaction rates , in order of fastest to slowest, are a. 1, 2, 3 b. 2, 1, 3 c. 3, 1, 2 d. 3, 2, 1 2. Consider the following reaction: C5H12(g) +8O2(g)  5CO2(g) + 6H2O(g) Which of the following explains, in terms of collision theory, why this reaction occurs in more than one step? a. a low C5H10 concentration
b. low temperature of reactant mixture
c. low probability of a multi‐particle collision
d. particles collide with insufficient kinetic energy
3. What would happen if the kinetic energy of the reactants was not enough to provide the needed activation energy? a. The products would be produced at a lower energy b. The activated complex would convert into products c. The reactants would re‐from d. The products would form at an unstable energy state Page 2 of 12
4. Consider the following reaction mechanism: Step 1 OClˉ + H2O  HOCl + OHˉ Step 2 Iˉ + HOCl  HOI + Clˉ Step 3 HOI + OHˉ  OIˉ + H2O The catalyst is a. OIˉ b. H2O c. OClˉ d. HOCl 5. A chemical equilibrium is described as “dynamic” because a. maximum randomness has been achieved. b. the pressure and temperature do not change. c. both reactants and products continue to form. d. the concentrations of chemical species remain constant. 6. Consider the following potential energy diagram for an equilibrium system: When the temperature of the system is increased, the equilibrium shifts to the a. left and Keq increases b. left and Keq decreases c. right and Keq increases d. right and Keq decreases Page 3 of 12
7. What is Keq expression for the following equilibrium? 3Fe(s) + 4H2O(g) ⇌Fe3O4 s 4H2 g a.
b.
c.
d.
8. Consider the reaction 2Li(s) + 2H2O()  2LiOH(aq) + H2(g) H = ‐433 J What will entropy and enthalpy factors favour? Entropy Enthalpy a. products products b. products reactants c. reactants products d. reactants reactants 9. Consider the following system at equilibrium: Keq = 1.2x104 2SO2(g) + O2(g) ⇌ 2SO3(g) If additional SO2 is added to the system, what happen to the equilibrium and the value of Keq? Equilibrium Keq a. shifts left decreases b. shifts right increases c. shifts right no change d. no change no change Page 4 of 12
10. Which of the following will dissolve in water to produce a molecular solution? a. CaCl2 b. NaOH c. CH3OH d. Sr(OH)2 11. What happens when 10.0 mL of 0.2 M KOH is added to 10.0 mL of 0.2 M CuSO4? a. No precipitate forms b. A precipitate of K2SO4 forms c. A precipitate of Cu(OH)2 forms d. Precipitates of K2SO4 and Cu(OH)2 form 12. The complete ionic equation for the reaction between MgS and Sr(OH)2 is a. MgS(aq) + Sr(OH)2(aq)  Mg(OH)2(s) + SrS(s) b. MgS(aq) + Sr(OH)2(aq)  Mg(OH)2(s) + SrS(aq) c. Mg2+(aq) + S2‐(aq) + Sr2+(aq) + 2OH‐(aq)  Mg2+(aq) + 2 OH‐(aq) + SrS(s) d. Mg2+(aq) + S2‐(aq) + Sr2+(aq) + 2OH‐(aq)  Mg(OH)2(s) + Sr2+(aq) + S2‐(aq) 13. Consider the following equilibrium: Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH‐(aq) Which of the following will cause the equilibrium to shift to the right? a. Adding KOH b. Adding Na2S c. Adding Fe(OH)2 d. Adding Fe(NO3)2 14. Which of the following chemical indicators has a Ka = 2.5x10‐5? a. methyl orange b. phenolphthalein c. thymolphthalein d. bromcresol green 15. The value of Kb for HPO42ˉ is a. 2.2x10‐13 b. 6.2x10‐8 c. 1.6x10‐7 d. 4.5x10‐2 Page 5 of 12
16. Which of the following 1.0 M solutions would have a pH greater than 7.00? a. HCN b. KNO3 c. NH4Cl d. NaCH3COO 17. What volume of 0.100 M KOH is required to completely neutralize 15.0 mL of 0.100 M H3PO4? a. 5.00 mL b. 15.0 mL c. 30.0 mL d. 45.0 mL 18. Which of the following represents a redox reaction? a. CaCO3  CaO + CO2 b. SiCl4 + 2Mg  Si + 2MgCl2 c. 2NaOH + H2SO4  2H2O + Na2SO4 d. AgBr + 2S2O32‐  Ag(S2O3)23‐ + Brˉ 19. Manganese has an oxidation number of +4 in a. MnO b. MnO2 c. Mn2O3 d. Mn2O7 20. Which of the following 1.0 M solutions will react spontaneously with lead? a. KCl b. CuCl2 c. ZnCl2 d. MgCl2 21. During the corrosion of magnesium, the anode reaction is a. Mg  Mg2+ + 2eˉ b. Mg2+ + 2eˉ  Mg c. 4OHˉ  O2 + 2H2O + 4eˉ d. O2 + 2H2O + 4eˉ  4OHˉ Page 6 of 12
22. Consider the following spontaneous reactions: 3Cd2+ + 2Np  3Cd + 2Np3+ Cd + Pd2+  Cd2+ + Pd Np3+ + Ce  Np + Ce3+ Which is the strongest oxidizing agent? a. Cd2+ b. Ce3+ c. Np3+ d. Pd2+ Use the following diagram to answer questions 23 and 24 23. As the cell operates, the cations move towards the a. the Pb electrode and the Pb electrode gains mass. b. the Pb electrode and the Pb electrode loses mass. c. the Cu electrode and the Cu electrode gains mass d. the Cu electrode and the Cu electrode loses mass 24. As the cell operates, the electrons move towards the a. Pb electrode and the cell voltage decreases over time b. Pb electrode and the cell voltage increases over time c. Cu electrode and the cell voltage decreases over time d. Cu electrode and the cell voltage increases over time Page 7 of 12
25. The process of applying an electric current through a cell to produce a chemical change is called a. corrosion b. ionization c. hydrolysis d. electrolysis Section II: Written problems (25 points total). 26. Use the potential energy diagram shown to answer the following: (3 points) a. Which letter shows the activation energy? b. Which letter shows the change in energy, H, for the overall reaction? ______d______ c. Which letter shows the activation energy using a catalyst? ______a______ _____e_______ 27. Consider the following equilibrium: 2NO(g) + Cl2(g) ⇌ 2NOCl(g) At equilibrium, the system contains 4.00 mole NO, 1.00 mole Cl2 and 0. 200 mole NOCl in a 2.0 L container. Calculate the value of Keq for the equilibrium. (2 points) 4.00
2.0
1.00
2.0
0.200
2.0
2.0 0.50 0.10 0.10
2.0 0.50
0.0050
5.0 10 Page 8 of 12
28. Tabulated below are initial rate data for the reaction. 2Fe(CN)63‐ + 2Iˉ  2Fe(CN)64‐ + I2 [Fe(CN)63‐]0 (M) 0.010 0.010 0.020 Run 1 2 3 [Iˉ]0 (M) 0.10 0.40 0.40 Initial Rate (M/s) 1.0 x 10‐5 4.0 x 10‐5 1.6 x 10‐4 a. What is the experimental rate law? (3 points) Rate = k[Fe(CN)63‐]x[Iˉ]y 4
4 ∴
4
4.0 10 1.0 10 0.010
0.010
0.40
0.10
0.020
0.010
0.40
0.40
0.40
0.10
4 1 1.6 10 4.0 10 2 ∴
2 →
Fe CN
0.020
0.010
2 ¯ b. Determine the rate constant k? (2 points) Fe CN
¯
1.0 10 0.010
0.10
1.0
29. a. What would be the solubility of PbCl2 in pure water at 25°C? (2 points). Let s = solubility of PbCl2 then [Pb2+] = s and [Clˉ] = 2s Ksp = [Pb2+] [Clˉ]2 = s(2s)2 = 4s³ = 1.2x10‐5 1.2 10
4
0.014
b. What would be the solubility of PbCl2 in 0.10 M CaCl2 at 25°C? (2 points) Let s = solubility of PbCl2 then [Pb2+] = s and [Clˉ] = 2s +0.20 Ksp = [Pb2+] [Clˉ]2 = 1.2x10‐5 = s(2s + 0.20)2 ≈ s(0.20)2 s = 1.2x10‐5/(0.20)2 = 3.0x10‐4 M assumption is OK Page 9 of 12
30. At 25°C, cocaine, C17H21O4N, is soluble in water to the extent of 1.7 g/L. The molar mass of cocaine is 303 g/mol. Kb for cocaine is 2.6x10‐6 at 25°C. What is the pH of a saturated solution of cocaine at 25°C? (5 points) 1.7 5.61 10 303 C17H21O4N(aq) + H2O() ⇌ C17H21O4NH+(aq) + OHˉ(aq) 0.00561 0 0 ‐x x x 0.00561‐x x x I C E ¯
= 2.6x10 =
2.6 10
5.61 10
=
x²
.
≈ x²
.
assume |x|<< 0.00561 1.2 10 assumption is OK [OHˉ] = x= 1.2x10‐4 M pOH = ‐log[OHˉ] = ‐log(1.2x10‐4) = 3.92 pH = 14.00 –OH = 14.00 – 3.92 = 10.08 Page 10 of 12
31. Refer to the galvanic cell below (the contents of each half‐cell are written beneath each compartment): Pt Pt 1.0 M MnO4‐ 1.0 M Mn2+ 1.0 M H+ 1.0 M Cr3+ 1.0 M Cr2O72‐ 1.0 M H+ a. What is the balanced reaction occurring in the cell on the left side? (1 point) MnO4ˉ(aq) + 8H+(aq) + 5eˉ  Mn2+(aq) + 4H2O() b. What is the balanced reaction occurring in the cell on the right hand side? (1 point) 2Cr3+(aq) + 7H2O()  Cr2O72ˉ (aq) + 14H=(aq) + 6eˉ c. What is the balanced overall reaction? (1 point) 6MnO4ˉ(aq) + 10Cr3+(aq) + 11H2O()  6Mn2+(aq) + 5Cr2O72ˉ + 22H+(aq) d. What is the value of E°cell? (1 point) E°cell = 1.51 V – 1.23 V = 0.28 V e. In which direction do electrons flow in the external circuit? (1 point) from the Cr half‐cell to the Mn half‐cell (i.e. from right to left) f. Which side is the anode? (1 point) the Cr half reaction side (i.e. the right side) Page 11 of 12
Equations and Constants PV = nRT T(K) = T(°C) + 273.15 R = 0.0820575 L atm mol‐1 K‐1 = 8.314 J mol‐1 K‐1 Δ
Kp = Kc(RT) n 1 atm = 760 mm Hg = 760 Torr = 101.325 kPa pOH = ‐log[OHˉ] pH = ‐log[H+] [OHˉ] = 10‐pOH [H+] = 10‐pH pX = ‐log(X) X = 10‐pX Kw = 1.0x10‐14 @ 25°C 14.00 = pH + pOH @ 25°C Kw = Ka Kb  [ A ] 
 pH  pK a  log
 [ HA] 
F = 96485 C/mol °
°
ln
.
log
at 25°C The solution to the quadratic equation ax² + bx +c = 0 is x
 b  b 2  4ac
2a
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