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Unformatted text preview: CHEM111 Practice Final Exam Key Fall 2014 Name Key Please ignore all point values on this Practice exam as they are all incorrect. Constants and equations: Speed of light: c = 2.997x108ms-‐1 Plank’s constant: h=6.626x10-‐34 J·∙s 2 -‐2
1 J = 1 kg·∙m ·∙s charge of an e-‐: e = 1.602x10-‐19 C mass of e-‐: me=9.11x10-‐31 kg mass of p+: mp=1.67x10-‐27 kg ! ∆! = ! Avogadro’s constant: NA = 6.022x1023 mol-‐1 1st order rate law: ln[A] = -‐kt ln[A]0 2nd order rate law: 1/[A]= kt + 1/[A]0 Ea/RT
k=Ae-‐ F = 96,500 C/mol e-‐ 1st Law of Thermodynamics: ΔEtotal = ΔEσ+ ΔEθ + ΔEwt + . . . =0 2nd Law of Thermodynamics: ΔStotal = ΔSσ + ΔSθ ≥ 0 and ΔSθ = ΔEθ/Tθ and ΔSwt = 0 3rd Law of Thermodynamics: Sº0K = 0 for pure, crystalline substances ΔG=ΔH-‐TΔS ΔrG = ΔrG° + RTlnQ E=Eo-‐(0.0592V mol e-‐/n)logQ ΔG=-‐nFE ΔrG°=-‐RTlnK R=0.08206 L atm mol-‐1 K-‐1= 8.314 J mol-‐1 K-‐1 ! ∆! !"# = !!"
!! = ! CHEM111 Practice Final Exam Key Fall 2014 Multiple Choice: 1. The degree of agreement among several measurements of the same quantity is called __________. It reflects the reproducibility of a given type of measurement. a) accuracy b) error c) precision d) significance e) certainty 2. A species with 12 protons and 10 electrons is a) Ne2+ b) Ti2+ c) Mg2+ d) Mg e) Ne2– 3. Which of the following atomic symbols is incorrect? a) 146 C b) 37 17 Cl
c) 15 P d) 39 19 K
4. Consider the element indium, atomic number 49, atomic mass 114.8 g. The nucleus of an atom of indium-‐112 contains a) 49 protons, 63 neutrons, 49 electrons b) 49 protons, 49 neutrons c) 49 protons, 49 alpha particles d) 49 protons, 63 neutrons e) 49 protons, 112 neutrons 5. Phosphorus has the molecular formula P4, and sulfur has the molecular formula S8. How many grams of phosphorus contain the same number of molecules as 7.88 g of sulfur? a) 3.80 g b) 0.263 g c) 7.61 g d) 7.88 g e) none of these 6. You have exposed electrodes of a light bulb in a solution of H2SO4 such that the light bulb is on. You add a dilute solution and the bulb grows dim. Which of the following could be in the solution? a) Ba(OH)2 b) NaNO3 c) K2SO4 d) Cu(NO3)2 e) none of these CHEM111 Practice Final Exam Key Fall 2014 7) The cation M2+ reacts with NH3 to form a series of complex ions as follows: M2+ + NH3 M(NH3)2+ K1 = 102 M(NH3)2+ + NH3 M(NH3)22+ + NH3 M(NH3)22+ M(NH3)32+ –3 K2 = 103 K3 = 102 –5 A 1.0 x 10 mol sample of M(NO3)2 is added to 1.0 L of 15.0 M NH3 (Kb = 1.8 x 10 ). Choose the dominant species in this solution a) M2+ b) M(NH3)2+ c) M(NH3)22+ d) M(NH3)32+ e) M(NO3)2 8. All of the following reactions 2Al(s) + 3Br2(l) → 2AlBr3(s) 2Ag2O(s) → 4Ag(s) + O2(g) CH4(l) + 2O2(g) → CO2(g) + 2H2O(g) can be classified as a) oxidation-‐reduction reactions b) acid-‐base reactions c) precipitation reactions d) A and B e) A and C Four identical 1.0-‐L flasks contain the gases He, Cl2, CH4, and NH3, each at 0°C and 1 atm pressure. 9. Which gas has the highest density? a) He b) Cl2 c) CH4 d) NH3 e) all gases the same 10. For which gas do the molecules have the smallest average kinetic energy? a) He b) Cl2 c) CH4 d) NH3 e) all gases the same 11. If the volume of the container cut in half, which gas will have the highest pressure? a) He b) Cl2 c) CH4 d) NH3 e) all gases the same CHEM111 Practice Final Exam Key Fall 2014 12. For a 3rd order reaction, the units for the rate constant (k) are a) M s-‐1 b) M-‐1s-‐1 c) M-‐1s-‐2 d) M-‐2s-‐1 e) M-‐3s-‐1 13. The catalyzed pathway in a reaction mechanism has a __________ activation energy and thus causes a __________ reaction rate. a) higher, lower b) higher, higher c) lower, higher d) lower, steady e) higher, steady 14. The decomposition of ozone may occur through the two-‐step mechanism shown: O3 ! O2 + O step 1 O3 + O ! 2O2 step 2 The oxygen atom is considered to be a(n) a) reactant b) product c) catalyst d) reaction intermediate e) activated complex 15. The reaction 2A + 5B → products is second order in A and fourth order in B. What is the rate law for this reaction? a) rate = k[A]2[B]5
b) rate = k[A]4[B]2 c) rate = k[A]2[B]4 d) rate = k[A]5[B]2 e) rate = k[A]2/7[B]5/7 16. Which of the following is the strongest oxidizing agent? MnO4– + 4H+ + 3e– ! MnO2 + 2H2O ℰ° = 1.68 V I2 + 2e– ! 2I– 2+ – Zn + 2e ! Zn – a) MnO4 b) I2 c) Zn2+ d) Zn e) MnO2 ℰ° = 0.54 V ℰ° = –0.76 V CHEM111 Practice Final Exam Key Fall 2014 17. What is the cell reaction for the voltaic cell Cr(s) | Cr3+(aq) || Br–(aq) | Br2(g) | Pt? a) Cr(s) + 2Br–(aq) Br2(g) + Cr3+(aq) b) 2Cr3+(aq) + 6Br–(aq) 2Cr(s) + 3Br2(g) c) Cr(s) + 3Br2(g) Cr3+(s) + 2Br–(aq) d) 2Cr(s) + 3Br2(g) 2Cr3+(aq) + 6Br–(aq) e) none of these 18. At a given temperature, K = 0.033 for the equilibrium: PCl5(g) PCl3(g) + Cl2(g) What is K for: Cl2(g) + PCl3(g) a) b) c) d) e) PCl5(g)? 0.033 30. 0.00109 33 920 19. For a particular system at a particular temperature there ______ equilibrium constant(s) and there _______ equilibrium position(s). a) are infinite; is one b) is one; are infinite c) is one; is one d) are infinite; are infinite e) none of these 20. A weak acid, HF, is in solution with dissolved sodium fluoride, NaF. If HCl is added, which ion will react with the extra hydrogen ions from the HCl to keep the pH from changing to a large extent? a) OH– b) Na+ c) F– d) Na– e) none of these 21. Consider a thermodynamics universe consisting of an exothermic reaction in a Styrofoam cup that is open to the atmosphere that does not involve the evolution of a gas. Which are the correct signs of the following terms: a) ∆H = (-‐),∆!"! =(-‐), q=(+), ∆!"! =0, ∆!" =0, ∆!"## =(+) b) ∆H = (-‐),∆ =(-‐), q=(-‐), ∆ =0, ∆ =0, ∆ =(+) c) ∆H = (-‐),∆!"! =(-‐), q=(+), ∆!"! =0, ∆!" =0, ∆!"## =(-‐) d) a) ∆H = (-‐),∆!"! =(-‐), q=(-‐), ∆!"! =0, ∆!" =0, ∆!"## =(-‐) e) none of these CHEM111 Practice Final Exam Key Fall 2014 22. Which of the following salts shows the lowest solubility in water? (Ksp values: Ag2S = 1.6 ´ 10–49; Bi2S3 = 1.0 ´ 10–72; HgS = 1.6 ´ 10–54; Mg(OH)2 = 8.9 ´ 10–12; MnS = 2.3 ´ 10–13) a) Bi2S3 b) Ag2S c) MnS d) HgS e) Mg(OH)2 23) Two metals of equal mass with different heat capacities are subjected to the same amount of heat. Which undergoes the smallest change in temperature? a) The metal with the higher heat capacity. b) The metal with the lower heat capacity. c) Both undergo the same change in temperature. d) You need to know the initial temperatures of the metals. e) You need to know which metals you have. 24) Suppose a buffer solution is made from formic acid, HCHO2, and sodium formate, NaCHO2. What is the net ionic equation for the reaction that occurs when a small amount of sodium hydroxide is added to the buffer? a) NaOH(aq) + H3O+(aq) ® Na+(aq) + 2H2O(l) b) H3O+(aq) + OH–(aq) ® 2H2O(l) c) OH–(aq) + HCHO2(aq) ® CHO2–(aq) + H2O(l) d) NaOH(aq) + HCHO2(aq) ® NaCHO2(aq) + H2O(l) e) Na+(aq) + HCHO2(aq) ® NaH(aq) + HCO2+(aq) 25) In titrating 0.20 M hydrochloric acid, HCl, with 0.20 M NaOH at 25°C, the solution at the equivalence point is a) 0.20 M NaCl b) very acidic c) slightly acidic d) 0.10 M HCl and 0.20 M NaOH e) 0.10 M NaCl CHEM111 Practice Final Exam Key Fall 2014 26. Write balanced net ionic equations for the following reactions and identify what type of reaction it is: A. The neutralization of hydrochloric acid with sodium hydroxide. H+(aq) + OH-‐(aq) ! H2O(l) Acid-‐Base reaction B. The reaction of iron copper metal with silver (I) ions to produce copper (II) ions and silver metal. Cu(s) + 2Ag+(aq) !Cu2+(aq) + 2Ag Redox reaction C. The mixture of a 20 mL of a 2 M solution of lead (II) nitrate with 15 mL of a 1.5 M solution of sodium sulfate. Pb2+(aq) + SO42-‐(aq) ! PbSO4(s) Precipitation reaction D. A mixture is made of 50 mL of 0.5 M sodium chloride and 20 mL of 0.15 M potassium nitrate. NR CHEM111 Practice Final Exam Key Fall 2014 27. For the reaction: 2SO3(g) ⇌ 2SO2(g) + O2(g) a) Write the equilibrium constant, Kc, for the reaction. ! = ! ! ! ! ! b) At equilibrium the concentration of SO3(g) is 0.15 M, SO2(g) is 0.75M and O2(g) is 0.30 M, what is the value of KP at 250C? ! ! = ! !
! ! ! = ∆! = ! [0.30] 0.75
0.15 ! ! = 7.5 ∆ = 1 so ! = ! ∆! = 7.58 0.08206 298 = 183.4 c) What is the partial pressure of SO3(g) at equilibrium? PV=nRT P = (n/V)RT = [SO3]RT = 3.66 atm What will happen to the numbers of moles of SO3 in equilibrium with SO2 and O2 in the reaction d) Oxygen is added. Increased (shift toward reactants) e) The reaction is increased by decreasing the volume of the reaction container. Increase (less moles of gas on reactants side, thus shift that way.) f) In a rigid reaction container, the pressure is increased by adding argon gas. No effect. Partial pressure of sulfur trioxide, oxygen and sulfur dioxide are unchanged. g) The temperature is decreased (the reaction is endothermic). Increase (endothermic reaction shifts towards reactants when heat is decreased) h) Gaseous sulfur dioxide is removed. Decrease (shift toward products). CHEM111 Practice Final Exam Key Fall 2014 28) The following initial rate data were found for the reaction 2MnO4– + 5H2C2O4 + 6H+ ! 2Mn2+ + 10CO2 + 8H2O [MnO4–]0 [H
[H+]0 Initial Rate (M/s) 2C2O4]0 –3 –3 1 x 10
1.0 2 x 10–4 2 x 10–3 1 x 10–3 1.0 8 x 10–4 2 x 10–3 2 x 10–3 1.0 1.6 x 10–3 2 x 10–3 2 x 10–3 2.0 1.6 x 10–3 a. What is the rate law obtained from this data? b. What is the value of the rate constant? c. Would you expect the mechanism for this reaction of have a slow first elementary step? Why or why now? d. What species is oxidized during this reaction? What species is reduced? How many electrons are transferred? CHEM111 Practice Final Exam Key Fall 2014 29. Elemental analysis is done on a liquid compound containing only carbon, hydrogen and oxygen. The compound is found to be 52.2 % C and 13% H. The molar mass of the compound is between 40 and 50 g/mol. A. What is the molecular formula of the compound. 1
= 4.35 ×
= 2.00 1 12 2.175 1
= 13 ×
= 5.98 1 1 2.175 1 1
% = 100 − 52.2 − 13 = 34.8 100 ×
= 2.175 ×
= 1 16 2.175 EF = C2H6O, MM(EF) = 12*2+6+16 = 46 (within range) MF = C2H6O B. Write a balanced chemical reaction of the species above reacting with oxygen molecules to produce carbon dioxide and water. 52.2 × C2H6O(l) + 3O2(g) ! 2CO2(g) + 3H2O(g) C. If 57.0 g of the compound is burned in a sealed vessel containing 168.0 g of oxygen molecules, how many grams carbon dioxide will be produced? 57.0 ! ! ×
168.0 ! × 1 2 !
= 109 ! 46 1 ! ! 1 !
1 2 ! 44 !
= 154 ! 32 3 ! 1 ! 109 g CO2 D. How many grams of excess reagent will be left after the reaction is finished? grams CO2 not produced= 154 g – 109 g = 45 g CO2 45 ! × 1 ! 3 !
= 49.1 ! 44 ! 2 ! 1 There will be 49.1 grams of O2 left unreacted. CHEM111 Practice Final Exam Key Fall 2014 30. A solution is prepared by adding 0.10 mol Ni(NH3)6Cl2 to 0.50 L of 3.0 M NH3. Calculate [Ni(NH3)62+] and [Ni2+] in this solution. Koverall for Ni(NH3)62+ is 5.5x108. That is !! ! = 5.5×10! = ! ! !! ! ! The overall reaction is Ni2+(aq) + 6NH3(aq) ⇌ Ni(NH3)62+(aq) CHEM111 Practice Final Exam Key Fall 2014 31. a. A 60 mL volume of acetic acid is titrated with sodium hydroxide requires 30 mL of 3.0 M NaOH to reach the end point and has a pH of 4.80 at the half-‐equivalence point. What is the Ka of acetic acid? At ½ equivalence point, pH=pKa, so Ka = 1.59x10-‐5 b. What is the starting concentration of acetic acid? 0.030 L x 3.0 mol/L = 0.09 mol acetic acid Concentration – 0.09 mol/0.060L = 1.5M c. What is the pH of the starting solution of acetic acid? If you could not solve for the Ka or the initial concentration, assume they are 2.0x10-‐5 and 1.25 M, respectively. HA(aq) + H2O(l) ⇌ A-‐(aq) + H3O+(aq) I 1.5 M 0 0 C -‐x +x +x E 1.5-‐x x x ! ! ! !
= 2.0×10!! 1.5 1.5
x=[H3O+] = 5.5x10-‐3 M pH = -‐log(H3O+) = -‐log(5.5x10-‐3) = 2.26 d. What is true at the equivalence point of the titration? What kind of problem would you solve to determine the pH at this point? At the equivalence point, the moles of initial acid are equal to the moles of OH-‐ added, so all of the initial HA has been converted into A-‐. You would solve a weak base problem. e. In-‐between the initial point and the equivalence point we can solve a buffer problem. Why is this true? What is true about the solution at this time? As we add OH-‐ it reacts to produce A-‐, so we have a mixture of HA and A-‐, which is a buffer. CHEM111 Practice Final Exam Key Fall 2014 32) (20 points) Solubility Equilibrium Calcium carbonate (CaCO3(s)) is an important component in the shells of marine organisms and is a common substance found in rocks. (a) Calcium carbonate has a Ksp value of 5.0x10-‐9 at 25oC. What is the molar solubility of Calcium carbonate at 25oC? Ksp=5.0x10-‐9=[Ca2+][CO32-‐]=x2 x=molar solubility = 7.07x10-‐5M (b) A major concern to the rising of CO2(g) levels is ocean acidification. What will increased levels of dissolved CO2(aq) do to the pH of ocean water. Use chemical reactions to assist your answer. CO2(aq) + H2O(aq) ⇌ H2CO3(aq) H2CO3(aq) ⇌ H+(aq) + HCO3-‐(aq) HCO3-‐(aq) ⇌ H+(aq) + CO32-‐(aq) Carbon dioxide forms carbonic acid in water and will lead to a decrease in pH. (c) As more CO2(g) dissolves in the ocean, will calcium carbonate become more or less soluble? Why? An increase in pH will drive the final reaction towards the reactants and remove CO32-‐(aq) from the solution. This will in turn drive the solubility equilibrium in 9(a) toward the products, increasing solubility. (d) Solid calcium carbonate can go through the following chemical equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) Which has Kp=1.16 at 800.oC. If a 21.1-‐gram sample of CaCO3 is put into a 10.0-‐L container and heated to 800.°C. Fill in the boxes with symbols only (no values) in the possible route to determine the percent of CaCO3(s) that would react to reach equilibrium. Molar mass % !"#$% !"# !"# nCO2 !"#$%!. n CaCO3 PCO2 ! → ! → % CHEM111 Practice Final Exam Key Fall 2014 33. The following reaction occurs in an electrochemical cell:
H3AsO3(aq) + Mn3+(aq) ! As(s) + MnO2(s)
The reduction potential for H3AsO3 ! As is 0.248 V.
a. Balance this redox reaction in ACIDIC conditions and circle the reducing agent. H3AsO3(aq) + 3Mn3+ (aq)+ 3H2O(l) ! As(s)+3MnO2 (s)+ 9H+(aq)
b. The reaction is not spontaneous and requires an input of current above the cell potential of 0.702V to
proceed under standard conditions. With this in mind, what is the reduction potential of the manganese
containing half-reaction? Reduction potential for manganese is:
4 points c. Calculate the value of ΔGo for this reaction as written. Does your value support a nonspontaneous
ΔGo=-nFEo=-3 mol e- x 9.649x104 C mol-1x-0.702V = 203 kJ/mol
Positive value, so yes, nonspontaneous.
4 points d. Calculate the equilibrium constant for this reaction at 298K.
4 points e. For the electrolytic process, fill in the following values:
__0___ 4 points __-___ __-___ __+___ CHEM111 Practice Final Exam Key Fall 2014 34. (a) For the Haber process, N2(g) + 3H2(g)!2NH3(g) , an exothermic process, predict the signs of
each of the following components at 298K. The reaction is spontaneous. ΔH ΔE ΔS ΔS ΔEtot ΔStot ∆Ωtot 7 pts θ -‐ + σ θ - + 0 + + –––– –––– –––– –––– –––– –––– –––– (b) Calculate a value for ΔGo for this reaction. Δ H fo So (kJ/mol) (j/molK) H2(g) 0.0 130.68 N2(g) 0.0 191.61 NH3(g) -46.11 192.45 ΔH= ΣνPHfo(P)-‐ ΣνRHfo(R)=2(-‐46.11kJ/mol)-‐0=-‐92.220 kJ/mol
ΔS o= ΣνPSo(P)-‐ ΣνRSo(R)= 2(192.45 j/mol K)–3(130.68 J/molK)-‐(191.61 j/molK)=-‐198.75 J/molK
σ ΔGo=ΔHo-‐TΔS= -‐92.220 kJ/mol – 298 K * -‐0.19875 kJ/mol K = -‐32.99 kJ/mol 6pts
(c) At what temperature will the Haber process transfer from being spontaneous to nonspontaneous?
σ =(-‐92220j/mol)/-‐198.75 j/molK)=464.0K 6 pts ...
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