# Ch12b - 2 O 2 has reacted Half-Life • Time needed to 1/2...

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Kinetics 2 1) Integrated Rate Laws 2) Half-Life 3) Activation Energy

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Rate = k[A] 0 = k Integrated Rate Law: [A] t = -k t + [A] 0
Rate = k[A] Integrated Rate Law: ln [A] t = -k t + ln [A] 0

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Rate = k[A] 2 Integrated Rate Law: 1/[A] t = k t + 1/[A] 0
Order Rate Law Integrated Rate Law Straight Line Plot Slope of Plot 0 k[A] 0 [A] = -kt + [A] i [A] vs. t -k 1 k[A] 1 ln [A] = -kt + ln [A] i ln [A] vs. t -k 2 k[A] 2 1/[A] = kt + 1/[A] i 1/[A] vs. t k

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Using Integrated Rate Laws 2H 2 O 2 → 2H 2 O + O 2 Look for linearity Test plots [ H 2 O 2 ] vs. time ln [ H 2 O 2 ] vs. time 1/[ H 2 O 2 ] vs. time Time (min) [H 2 O 2 ] (mol/ L) 0 0.0200 200 0.0160 400 0.0131 600 0.0106 800 0.0086 1000 0.0069 1200 0.0054 1600 0.0037 2000 0.0024

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2 H 2 O 2    2 H 2 O + O 2 1) Rate law? 2) Integrated rate law? 3) [H 2 O 2 ] 0 ? 4) k value? 5)[H 2 O 2 ] after 500 min? 6) Time that [H 2 O 2 ]=0.0030? 7) Time when 50% of H

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Unformatted text preview: 2 O 2 has reacted? Half-Life • Time needed to 1/2 concentration ([A] is ½ [A] ) • For First Order Reactions: ln [A] t = -k t + ln [A] t 1/2 = 0.693 / k Decomposition of H 2 O 2 is first order. At 20°C and initial [H 2 O 2 ] of 0.30 mol/L, rate constant is 0.00106 min-1 . • Half-Life? • Concentration after 1000 min? • Time to get to 0.075 mol/L? Activation Energy, E a • Energy needed for reactants to collide successfully; reach transition state Energy Profile, Endothermic Energy Profile, Exothermic Draw the Energy Profile • Ea = +25 kJ • ∆E = -60 kJ • Exo or Endo? • Reverse Ea? • Reverse ∆E?...
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Ch12b - 2 O 2 has reacted Half-Life • Time needed to 1/2...

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