Unformatted text preview: [A] = [A]t at t = t 1 ! [A] d[A] = [A]0 [A]t t ! k dt
0 [A] " [A] % ln $ t ' = kt $ [A] ' # 0&
Temps Time ln[A]t = ln[A]0 kt CHM1311 Chemical Kinetics 32 First Order Reactions [A] vs t = CURVE ln [A] vs t = STRAIGHT LINE k = slope ln[A]t = ln[A]0 kt CHM1311 Chemical Kinetics 33 Example 1: First order reactions For the natural decomposition of sucrose, the rate law is rate = k [sucrose] where k = 0.21 hr 1. If [sucrose]o = 0.010 M, how long will it take for the concentration to drop by 90%? The concentration of sucrose has dropped by 90% this means there is 10% left! [sucrose]0 = 0.010M  Ao , initial concentration [sucrose]t = 0.10 x [sucrose]0 = 0.001 M At CHM1311 Chemical Kinetics 34 Example 1: First order reactions For the natural decomposition of sucrose, the rate law is rate = k [sucrose] where k = 0.21 hr 1. If [sucrose]o = 0.010 M, how long will it take for the concentration to drop by 90%? ln[A]t = ln[A]0 kt
ln (0.001) = (0.21 h 1) t + ln (0.010) t = 11 hours CHM1311 Chemical Kinetics 35 Example 2: First order reactions Using the data below, is the following reaction first order in N2O5? 2 N2O5(g) 4 NO2(g) + O2(g) Time (min) 0 1.0 2.0 5.0 [N2O5]0 (mol/L) ln l[N2O5] 5]0 n [N2O x 1.00 0.705 0.497 0.173 0 0 0.35 0.7 0.35 1.75 0.70 1.75 CHM1311 Chemical Kinetics 36 Example 2: First order reactions 2 N2O5(g) 4 NO2(g) + O2(g) 1.2 1 Zeroorder plot: [N O ] vs. time
2 5 Firstorder plot: ln [N O ] vs. time 2 5 0.5 0.8 0.6 0.4 0.2 0 0 1 2 3 4 Time (min) 5 6 ln [N O ] (mol/L) [N O ] (mol/L) 0 0.5 1 1.5 2 0 1 2 3 4 Time (min) 5 6 5 CHM1311 2 Chemical Kinetics 2 5 37 First Order HalfLife the halflife of a reaction, t1/2, is the time needed for the concentration of a reactant to decrease by half for a first order reaction: ln t= [A]o k 1 [A]o 2 t1 =
2 [A]t at t 1 [A]t =
2 therefore ln 2 k for first order reactions, the halflife is independent of the concentration CHM1311 Chemical Kinetics 38 Example: Halflife
of a first order reaction The halflife of a first order reaction is 84.1 min. Calculate the rate constant for the reaction. t1 =
2 ln2 k ! k= ln2 t1
2 k= ln 2 = 8.24 "10#3min #1 84.1 min or k= ln 2 (84.1 min)(60 s min ) = 1.37 "10#4s#1 CHM1311 Chemical Kinetics 39 Second Order Reactions the sum of exponents is two ( x + y + ... = 2) the graph of [A] as a function of time is a biexponential curve Usually have a fast component and then a slow component
[A]
o A products rate = k [A]2 [A] units of k = M1 s1
Temps Time CHM1311 Chemical Kinetics 40 Second Order Reactions Integrating the rate law from time = 0 to time = t : [A] o d[A] rate = = k[A]2 dt
[A] = [A]0 at t = 0 [A] [A] = [A]t at t = t 1 ! [A]2 d[A] = [A]0 [A]t t ! k dt
0 Temps Time 1 1 = + kt [A]t [A]0 CHM1311 Chemical Kinetics 41 Second Order Reactions [A] vs t = CURVE [A]
o ln [A] vs t = CURVE [A]
o 1/[A] vs t = STRAIGHT LINE [A] [A] k = slope Temps Time Temps Time Temps Time 1 1 = + kt [A]t [A]0 CHM1311 Chemical Kinetics 42 Second Order HalfLife the halflife of a reaction, t1/2, is the time...
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 Fall '08
 MAYER
 Reaction, Kinetics, Chemical reaction, Rate equation

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