Principles of Chemistry I - Exam 4 Solutions | CH 302, Exams of Chemistry

Download Principles of Chemistry I - Exam 4 Solutions | CH 302 and more Exams Chemistry in PDF only on Docsity! Version 173 – Exam 4 302 – sutcliffe – (51060) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points While electroplating a spoon with silver (molar mass = 107.9 g/mol) from an Ag2+ so- lution, 19.3× 103 C of charge is used. By how much has the mass of the spoon increased? 1. 5.39 g 2. 107.9 g 3. 21.58 g 4. 10.79 g correct 5. 0 g 6. 53.9 g Explanation: This is a 2 electron process. g = Q(molar mass) nF = (19.3 × 103 C)(107.9 g/mol) (2)(9.65× 104 C/mol) = 10.79 g 002 10.0 points Note: when this question says ”approxi- mately” that doesn’t mean you ignore those concentrations!! Consider the voltaic cell Pt | Sn4+ (0.0010 M), Sn2+ (0.10 M) || Ag+ (0.010 M) | Ag Sn4+ + 2 e− → Sn2+ E0 = 0.15 V Ag+ + 1 e− → Ag(s) E0 = 0.8 V The experimental cell potential for the cell is approximately 1. 0.65 V. 2. 0.68 V. 3. 0.72 V. 4. 0.62 V. 5. 0.59 V. correct Explanation: The species in contact with the electrode surfaces are Sn2+ and Ag+. Note under standard conditions: Sn2+ → Sn4+ + 2 e− E0anode = −0.15 V Ag+ + 1 e− → Ag(s) E0cathode = 0.8 V E0cell = 0.65 V But these aren’t standard conditions, as the concentrations are not at 1 M. Use the Nernst equation to calculate E’s at these concentra- tions: E = E0 − 0.0592 V n log ( [Red]y [Ox]x ) For Sn4+ + 2 e− → Sn2+, E = 0.15 V − 0.0592 V 2 log ( 0.10 M 0.0010 M ) = 0.0908 V For Ag+ + 1 e− → Ag(s), E = 0.8 V − 0.0592 V 1 log ( 1 0.010 ) = 0.6816 V So, using the appropriate values of E, Sn2+ → Sn4+ + 2 e− E0anode = − 0.0908 V Ag+ + 1 e− → Ag(s) E0cathode = + 0.6816 V Ecell = 0.5908 V 003 10.0 points Consider the voltaic cell Pt | H2 (1 atm) | H + (? M) || Cl− (1 M) | AgCl(s) | Ag 2 H+ + 2 e− → H2 E 0 = 0.00 V AgCl + 1 e− → Ag + Cl− E0 = 0.222 V If the measured cell potential for the cell is 0.430 volts, what is the pH of the solution? 1. 0.253 2. 3.75 Version 173 – Exam 4 302 – sutcliffe – (51060) 2 3. 4.00 4. less than 1.00 5. 3.52 correct Explanation: 004 10.0 points The equilibrium constant for the reaction 2 Hg(ℓ) + 2 Cl−(aq) + Ni2+(aq) → Ni(s) + Hg2Cl2(s) is 5.6 × 10−20 at 25◦C. Calculate the value of E◦ for a cell utilizing this reaction. 1. − 1.14 V 2. + 0.57 V 3. − 0.57 V correct 4. − 0.25 V 5. + 1.14 V Explanation: 005 10.0 points If k = 2.7 × 10−6 M−1s−1 for the reaction A → B which of the following is the correct rate law? 1. rate = k [A]0 2. rate = k [A]0 [B]−1 3. rate = k [A]2 correct 4. rate = k [A]1 5. rate = k [A]2 [B]−1 Explanation: 006 10.0 points The principle of inhibiting the corrosion of iron by using a sacrificial anode is to allow 1. a metal with a lower oxidation potential to be sacrificed. 2. the iron to function as the anode. 3. a metal with a higher reduction potential to be sacrificed. 4. a metal that is more active than iron to function as a cathode. 5. a metal that is a better reducing agent to be sacrificed. correct Explanation: 007 10.0 points Note: use a = 1 in this question. A compound decomposes with a half-life of 8.0 s and the half-life is independent of the concentration. How long does it take for the concentration to decrease to one-ninth of its initial value? 1. 32 s 2. 3.6 s 3. 72 s 4. 25 s correct 5. 64 s Explanation: 008 10.0 points Consider the reaction 2 Cu+(aq) → Cu(s) + Cu2+(aq) . If the standard potentials of the Cu2+|Cu and Cu+|Cu couples are +0.34 and +0.52 V, respectively, calculate the value of E◦ for the given reaction. 1. − 0.18 V 2. − 0.70 V 3. + 0.70 V Version 173 – Exam 4 302 – sutcliffe – (51060) 5 2. ON−NO3(g) 3. NO3(g) correct 4. NO(g) 5. O2(g) Explanation: 015 10.0 points Consider the reaction 2 O3(g) → 3 O2(g) rate = k[O3] 2[O2] −1 . What is the overall order of the reaction and the order with respect to [O3]? 1. −1 and 3 2. 2 and 2 3. 3 and 2 4. 1 and 2 correct 5. 0 and 1 Explanation: 016 10.0 points Zinc Silver 1.56 V V Voltmeter e− e− 1 M Zn2+(aq) 1 M Ag+(aq) Salt bridge to carry ions In this electrochemical cell, what is the cathode? 1. the solid silver electrode correct 2. the Ag+(aq) ions in the 1 M solution 3. the solid zinc electrode 4. the Zn2+(aq) ions in the 1 M solution Explanation: Zn(s) + 2 Ag+(aq) → Zn2+(aq) + 2 Ag(s) Reduction occurs at the cathode. In this cell the reduction half reaction is Ag+(aq) + e− → Ag(s) . Ag+ cations are attracted to the solid Ag electrode where they are reduced to Ag(s). 017 10.0 points What half reaction occurs at the anode during the electrolysis of molten sodium bromide? 1. Na → Na+ + 1 e− 2. Na+ + 1 e− → Na 3. Br2 + 2 e − → 2 Br− 4. 2 Br− → Br2 + 2 e − correct 5. 2 H2O + 2 e − → 2 OH− + H2 Explanation: 018 10.0 points Consider the potential energy diagram shown below. A B Reaction progress E n er g y (k J ) 450 350 550 What is the activation energy Ea for the reaction A → B? 1. −100 kJ 2. 100 kJ correct 3. 550 kJ Version 173 – Exam 4 302 – sutcliffe – (51060) 6 4. 200 kJ 5. 450 kJ Explanation: The minimum energy required by A to react is represented by the height from A to the peak. 019 10.0 points In this question, n=8. This is because C starts as CH4 and becomes CO2. If the standard free energy change for com- bustion of 1 mole of CH4(g) is − 818 kJ · mol−1, calculate the standard voltage that could be obtained from a fuel cell using this reaction. 1. + 4.24 V 2. − 1.06 V 3. + 8.48 V 4. + 0.53 V 5. + 1.06 V correct Explanation: 020 10.0 points Catalysts work by 1. increasing the exothermicity of the reac- tion, which increases entropy in the surround- ings, making the reaction more favorable. 2. increasing the diffusion rates (collision rate) of molecules in solution. 3. increasing the kinetic energy of the reac- tant molecules. 4. creating a new reaction path with a lower activation barrier. correct Explanation: Definition 021 10.0 points This is a BONUS question: get it correct your max points are 250/240; get it wrong and your max points are 240/240. The overall reaction for the discharge of a nickel/cadmium cell is Cd + NiO2 + 2 H2O → Cd(OH)2 + Ni(OH)2 Which of the following statements is true? 1. Cadmium metal loses 2 electrons and is converted to Cd2+ at the cathode. 2. It is impossible to recharge a nickel/cadmium cell. 3. The nickel is oxidized during this reac- tion. 4. Insoluble Cd(OH)2 is deposited on the anode. correct Explanation: Looking at the reaction above, Cd → Cd2+ + 2 e− is an oxidation reaction that will occur at the anode. The nickel is not oxidized because we have shown that Cd is being oxidized. Nickel/cadmium batteries are rechargable. Insoluble CD(OH)2 is deposited on the an- ode since Cd is oxidized at the anode. 022 10.0 points A certain reaction is found to have a rate constant of 1.50 ×10−8 sec−1 at 0◦C and an activation energy of 45.0 kJ · mol−1. What will be its rate constant at 100◦C? 1. k = 68.4 s−1 2. None of the other answers is correct. 3. k = 7.37 × 10−11 s−1 4. k = 3.05 × 10−6 s−1 correct 5. k = 3.10 × 10−3 s−1 Explanation: Version 173 – Exam 4 302 – sutcliffe – (51060) 7 T1 = 273 K T2 = 373 K EA = 45.0 kJ k1 = 1.50 × 10 −8 sec−1 ln k2 k1 = ln k2 − ln k1 = EA R ( 1 T1 − 1 T2 ) ln k2 = ln k1 + EA R ( 1 T1 − 1 T2 ) = ln 1.50 × 10−8 + 45000 8.314 ( 1 273 − 1 373 ) = −12.699876 k2 = 3.05 × 10 −6 s−1 023 10.0 points Note: use a = 1 in this question. For the reaction cyclopropane → propene a plot of ln[cyclopropane] vs time in seconds gives a straight line with slope−4.1×10−3 s−1 at 550◦C. What is the rate constant for this reaction? 1. 1.8 × 10−3 s−1 2. 8.2 × 10−3 s−1 3. 4.1 × 10−3 s−1 correct 4. 3.9 × 10−2 s−1 5. 2.1 × 10−3 s−1 Explanation: 024 10.0 points Note: use a = 1 in this question. Consider the dimerization reaction 2 A → A2 rate = k [A] 2 . When the initial concentration of A is 2.0 M, it requires 30 min for 60% of A to react. Calculate the rate constant. 1. 1.1 × 10−3 M−1 · s−1 2. 3.2 × 10−4 M−1 · s−1 3. 5.0 × 10−4 M−1 · s−1 4. 4.2 × 10−4 M−1 · s−1 correct 5. 1.9 × 10−4 M−1 · s−1 Explanation: 025 10.0 points Given the standard potentials E0 ZnS(s) + 2 e− → Zn(s) + S2−(aq) −1.440 V Zn2+(aq) + 2 e− → Zn(s) −0.763 V calculate Ksp for ZnS at 298.15 K. R = 8.314 J molK 1 F = 96,485 Coulombs 1 volt = 1 J/Coulomb 1. 1.98 × 10−53 2. 1.14 × 10−10 3. 3.60 × 10−12 4. 6.13 × 10−20 5. 1.29 × 10−23 correct Explanation: