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1442answerstest2 - 1442 Test 2 Chapters 16 17 Kinetics and...

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1442 Test 2 Chapters 16 & 17 Kinetics and Keq website: http://dipowell1.home.mindspring.com ( last item after powell is the number one) 1. [ratio M] a [ratio M] b = ratio of rates Using experiment 1 & 2 : [.1M/.1M] a [.2M/.3M] b = 0.004 M/s/0.0135 M/s Since the A concentrations are equal we can cancel out that factor and: [.2M/.3M] b = 0.004M/s / 0.0135 M/s To solve algebracially : b log 0.667 = log .296 b = 3 If you don’t understand this process, go to the website home page and under Chemistry Help Topics choose Rate Law: Kinetics. Using Exp. 2&4 : [.1M/.3M] b [.3M/.4M] a = 0.0135 M/s / .288 M/s Since b did not stay constant in this problem, we must now substitute the power of 3 for b, so the second factor becomes : [.3M/.4M] 3 [.1M/.3M] a [.3M/.4M] 3 = [0.0135 M/s / 0.288 M/s] [0.33] a [0.75 ] 3 = 0.0469 0.33 a = .111 a = 2 ( if you need help solving this see Rate Law: Kinetics) Answer: Rate = k[A] 2 [B] 3 2. For an elementary reaction ( occurs in a single event) just use the coefficients in the balanced equation for the powers. So, the rate law is: R = k[NO 2 ] 2 No calculations are required if the reaction is elementary 3. ln [A] 0 / [A] t = kt ln [1 atm]/ [0.803 atm] t = k (10,000s) 0.219 = 10000s k k = 2.2 X 10 -5 s -1 4. 1/ [A] t - 1/ [A] 0 = kt 1/[A] t = kt - 1/[A] 0 1/[A] t = (0.02 L mol -1 s -1 )(120s) + 1/[0.2] [A] t = 0.14 M

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5. To prove first order and second order plot the following: ln[A] 0 /[A] t = kt Plot the two things that change: ln[A] t vs t To prove second order: 1/[A] t -1/[A] 0
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