ln(k) = ( E a /R) (1/T) + ln(a) Copyright 2018 Dan Dill 1

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1 TP For A B, is 0.003/min at 25 o C and 0.025/min at 35. For C D, is 0.003/min at 25 and 0.035/min at 35. Compared to the activation energy of A B, the activation energy of C D is 1. smaller 2. the same 3. larger 4. More information needed Lecture 36 CH102 A2 (MWF 11:15 am) Friday, April 27, 2018 Complete: Rate versus temperature: and. Catalysis Half life Half life calculations Next lecture: First law, second law, equilibrium, and kinetics 1 TP For A B, is 0.003/min at 25 o C and 0.025/min at 35. For C D, is 0.003/min at 25 and 0.035/min at 35. Compared to the activation energy of A B, the activation energy of C D is 1. smaller 2. the same 3. larger 4. More information needed ln(k) = ( E a /R) (1/T) + ln(a) For A B, is 0.003/min at 25 and 0.025/min at 35. slope / K / / For C D, is 0.003/min at 25 and 0.035/min at 35. slope / K / / The larger, the greater the temperature dependence. Response Counter

2 ln(k) = ( E a /R) (1/T) + ln(a) An example to try: /min, /min at 25 Catalyst: New path with lower barrier bigger k slope K E a slope 100. kj/mol 8 11 Catalyst: New path with lower barrier bigger k Only molecules with energy greater than can react Catalyst: New path with lower barrier bigger k The lower, the more molecules with energy at least

3 Catalyst: New path with lower barrier bigger k Quiz Compared to temperature dependence of an uncatalyzed reaction, when a catalyst is used, increasing temperature would speed the reaction up 1. more 2. the same amount 3. less 4. More information needed 14 Response Counter ln(k) = ( E a /R) (1/T) + ln(a) The larger, the greater the temperature dependence. This means that a catalyzed reaction will be less sensitive to temperature than its non catalyzed version. First order reactions have the special property that the time for reduction of the reactant by half,, does not change as the amount of material present changes due to transformation into products.. Let s see why this is true

4 For process A... that is first order in A, rate A A/ da/a Add successive time intervals of length covering the total elapsed time,, Add corresponding successive ratios A/A, covering to total elapsed time, /0 A/A 1 da/a ln / This sum is the definition of the natural logarithm. Therefore, the relation for a process A... that is first order in A, rate A A/ da/a means that for the total elapsed time, ln A / A

5 Since half life is how long it takes for the amount to drop by half, ln A / A ln 1/2 ln 2 the half life is ln 2 / That is, first order reactions have the special property that the time for reduction of the reactant by half,, does not change no matter the amount of material present. Since half life is the same no matter how much material we have, after halflife A 0 is reduced to A A A and so, the fraction remaining tells us the number of half lives, This is true even if is not an integer! Types of first order decay calculations Types of first-order decay calculations The essential equations of half life are ln Let s see how to use these to carry out various kinds of half life calculations

6 Types of first-order decay calculations Given amount consumed in time, how many half lives have elapsed? Given how many half lives have elapsed in time, what is? Given, how much remains after time? Given amount consumed in time, what is the rate constant? Types of first-order decay calculations Given amount consumed in time, how many half lives have elapsed? Given how many half lives have elapsed in time, what is? Given, how much remains after time? Given amount consumed in time, what is the rate constant? In half lives, A 0 is reduced to A A A So, the fraction remaining tells us the number of half lives, This is true even if is not an integer! Finding the half-lives elapsed, n In a process, a 60.% of a substance has decayed. This means A /A The number of half lives elapsed is log0.40/log1/2 log 0.40/log