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Exam 4-solutions - Version 022 Exam 4 mccord(51625 This...

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Version 022 – Exam 4 – mccord – (51625) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. mccord-ch302 2pmclassonly-51625 K w = 1 . 0 × 10 14 1 F = 96485 C/mol e Standand Potentials at 25 C Cl 2 + 2 e -→ 2 Cl +1.36 V Ag + + e -→ Ag +0.80 V Hg 2+ 2 + 2 e -→ 2 Hg +0.79 V Cu 2+ + 2 e -→ Cu +0.34 V Hg 2 Cl 2 + 2e -→ 2Hg + 2Cl +0.27 V AgCl + e -→ Ag + Cl +0.22 V 2H + +2e -→ H 2 0.000V Ti 3+ + e -→ Ti 2+ -0.37 V Zn 2+ + 2 e -→ Zn -0.76 V Mn 2+ + 2 e -→ Mn -1.18 V Mn 3+ + e -→ Mn 2+ -1.51 V Ti 2+ + 2e -→ Ti -1.63 V Al 3+ + 3 e -→ Al -1.66 V Na + + e -→ Na -2.71 V Also,RememberinQuest... 1.23e-12 is the same as 1 . 23 × 10 12 5.2e5 is the same as 5 . 2 × 10 5 001 4.0points A catalyst 1. increases the fraction of the molecules that collide with enough energy to react. 2. provides energy to help the reactants achieve the required activation energy. 3. increases the activation energy of the re- action. 4. does not enter into a reaction in any way. 5. provides a different mechanism by which the reaction can occur. correct Explanation: A catalyst speeds up a chemical reaction by providing an alternate mechanism which re- quires a lower energy of activation. Although the catalyst takes part in the reaction it is not used up. Catalysts may be in solid, liquid, gaseous or aqueous phase and only a small amount is used. 002 4.0points For a zero th order reaction aA products what is the y -intercept when the concentra- tion of A is plotted versus time to obtain a straight line? 1. ln[A] 0 2. [A] 0 correct 3. - akt 4. 1 [A] 0 5. akt Explanation: The equation is [A] 0 - [A] = akt [A] = [A] 0 - akt ( y = c - mx ) 003 4.0points Determine the overall balanced equation for a reaction having the following proposed mech- anism Step1: B 2 + B 2 -→ E 3 + D slow Step2: E 3 + A -→ B 2 + C 2 fast and write an acceptable rate law. 1. B 2 + B 2 -→ E 3 + D; R = k [B 2 ] 2
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Version 022 – Exam 4 – mccord – (51625) 2 2. E 3 + A -→ B 2 + C 2 ; R = k [E 3 ] [A] 3. A + B 2 -→ C 2 + D; R = k [A][B 2 ] 4. A + B 2 -→ C 2 + D; R = k [B 2 ] 2 correct Explanation: Step 1: B 2 + B 2 -→ E 3 + D slow Step 2: E 3 + A -→ B 2 + C 2 fast balanced equation, rate law = ? A + B 2 -→ C 2 + D (from the 2 molecules of B 2 in the rate- determining step) 004 4.0points The potential ( E ) of the cell shown below is +0 . 35 V. Pt(s) | H 2 (g , 1 bar) | H + (pH =?) || Cl (aq , 1 mol / L) | Hg 2 Cl 2 (s) | Hg( ) What is the missing pH of this cell? 1. 1.35226 2. 1.52129 3. 3.38064 4. 1.85935 5. 5.91613 6. 4.39484 7. 3.21161 8. 2.70452 9. 1.01419 10. 3.88774 Correct answer: 1 . 35226. Explanation: At the cathode, Hg 2 Cl 2 (s) + 2 e 2 Hg( ) + 2 Cl (aq) E = +0 . 27 V At the anode, H 2 (g) 2 H + (aq) + 2 e - E = 0 V Combining the half reactions, H 2 (g) + Hg 2 Cl 2 (s) 2 H + (aq) + 2 Hg( ) + 2 Cl E cell = +0 . 27 V Using the Nernst equation, E = E - parenleftbigg 0 . 025693 n parenrightbigg ln parenleftbigg [H + ] 2 [Cl ] 2 P H 2 parenrightbigg 0 . 35 = 0 . 27 - parenleftbigg 0 . 025693 2 parenrightbigg ln parenleftbigg [H + ] 2 [1] 2 [1] parenrightbigg 0 . 08 = - (0 . 0128465) · ln[H + ] 2 ln[H + ] 2 = - 6 . 22738 [H + ] 2 = e 6 . 22738 = 0 . 00197462 [H + ] = 0 . 00197462 = 0 . 0444367 pH = - log [H + ] = - log(0 . 0444367) = 1 . 35226 .
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