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chexam4 - Version 128 Exam 4 Holcombe(52460 This print-out...

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Version 128 – Exam 4 – Holcombe – (52460) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the data below: [CH 4 ] [O 2 ] initial rate M M M · s 1 Exp 1 0 . 5 1 . 3 2 . 5 × 10 3 Exp 2 0 . 5 2 . 6 1 . 0 × 10 2 Exp 3 1 . 0 1 . 3 5 . 0 × 10 3 Which of the following is a correct rate law for the reaction? 1. k [CH 4 ] 1 [O 2 ] 2. k [CH 4 ][O 2 ] 2 correct 3. k [CH 4 ][O 2 ] 4. k [CH 4 ] 1 [O 2 ] 2 5. k [CH 4 ] Explanation: rate 1 rate 2 = [O 2 ] 1 [O 2 ] 2 2 . 5 × 10 3 1 . 0 × 10 2 = parenleftbigg 1 . 3 2 . 6 parenrightbigg x x = 2 rate 3 rate 1 = [CH 4 ] 3 [CH 4 ] 1 5 . 0 × 10 3 2 . 5 × 10 3 = parenleftbigg 1 . 0 0 . 5 parenrightbigg y y = 1 002 10.0 points Magnesium metal is placed in contact with an underground steel storage tank to inhibit the corrosion of the tank because 1. magnesium has a lower oxidation poten- tial than iron. 2. magnesium is cheaper than gold. 3. magnesium has a higher reduction poten- tial than iron. 4. magnesium metal is a better reducing agent than iron. correct 5. magnesium functions as the cathode in the electrochemical cell instead of the iron. Explanation: 003 10.0 points Consider the reaction mechanism below: Step Reaction 1 Cl 2 + Pt -→ 2 Cl + Pt 2 Cl + CO + Pt -→ ClCO + Pt 3 Cl + ClCO -→ Cl 2 CO overall Cl 2 + CO -→ Cl 2 CO Which species is/are intermediates? 1. Pt, Cl 2. Pt, ClCO 3. ClCO 4. Cl, ClCO correct 5. Cl 6. Pt Explanation: Both Cl and ClO are produced in early steps and stiochiometrically consumed in sub- sequent steps and neither appear in the overall reaction. 004 10.0 points Consider the reaction 2 NOCl(g) 2 NO(g) + Cl 2 (g) with rate constant 0.0480 M 1 · s 1 when con- ducted at 200 C. The initial concentration of NOCl was 0.521 M. What is the concentration of NOCl after 0.268 minutes and at 200 C?
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Version 128 – Exam 4 – Holcombe – (52460) 2 1. 0.112 M 2. 0.372 M 3. 0.200 M 4. 0.111 M 5. 0.289 M correct 6. 0.514 M Explanation: k = 0 . 0480 M 1 · s 1 t = 0 . 268 min [A] 0 = 0.521 M The units on k indicates a second order reaction: 1 [A] t - 1 [A] 0 = akt 1 [A] t = 1 [A] 0 + akt = 1 0 . 521 M +2 ( 0 . 0480 M 1 · s 1 ) × (0 . 268 min) × 60 . 0 s min = 3 . 463 M 1 [A] t = 0 . 289 M 005 10.0 points What would be the E cell of an electrolytic cell made from the following two half reac- tions? Half reaction E AgCl( s ) + e -→ Ag( s ) + Cl ( aq ) +0 . 22 Al 3+ ( aq ) + 3 e -→ Al( s ) - 1 . 66 1. 1 . 88 2. - 1 . 44 3. 1 . 44 4. - 1 . 88 correct Explanation: E = E cathode - E anode = - 1 . 66 - 0 . 22 = - 1 . 88 006 10.0 points Consider the concentration-time dependence graph for a first-order reaction.
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