Chemical Kinetics © 2009, Prentice-Hall, Inc. First-Order Processes Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k. ln [A] t = -kt + ln [A] 0

Chemical Kinetics Ex The decomposition of certain insecticide in water follows first- order kinetics with rate constant of 1.45 /y at 12 c. A quantity of this insecticide is washed into alake on june 1, leading to concentration of 5.0 X 10-7 g/cm3. assume that the average tempreture of the lake is 12 A- what is the concentration of the insecticide on june of the following year B- how long will it take for the concentration of the insecicide to decrease to 3.0 x g/ cm3

Chemical Kinetics © 2009, Prentice-Hall, Inc. First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN

Chemical Kinetics © 2009, Prentice-Hall, Inc. First-Order Processes This data was collected for this reaction at °C. CH 3 NCCH 3 CN

Chemical Kinetics © 2009, Prentice-Hall, Inc. First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore, –The process is first-order. –k is the negative of the slope: 5.1  s −1.

Chemical Kinetics * The reaction 2A → B is first order in A with a rate costant of 2.8x10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 to 0.14 M, then calculate half life time (s) Solve: ln[A] = ln[A] 0 - kt, ln 0.14 = ln x10 -2 x t t = ln 0.14 – ln 0.88 / 2.8x10 -2 = 66 s Half life time ( t 1/2 ) = 0.693/ k = / 2.8x10 -2 = s.... * Given : CH 3 - N= N – CH 3 (g) → N 2(g) + C 2 H 6 (g) Time P N=N-CH3 mmHg ln P t = -kt + ln P ln 220 = -k x ln = - k x k = - ( ) /100 ______________ 2.55x10 -3 s -1 © 2009, Prentice-Hall, Inc.

Chemical Kinetics © 2009, Prentice-Hall, Inc. Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1 [A] t = kt + 1 [A] 0 also in the form y = mx + b

Chemical Kinetics © 2009, Prentice-Hall, Inc. Second-Order Processes So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k. Zero order reaction : Rate ( - ∆[A] / ∆t ) = k [A] 0 = k [A] = [A] 0 – k t, t 1/2 = [A] 0 /2 k 1 [A] t = kt + 1 [A] 0 1 [A]

Chemical Kinetics © 2009, Prentice-Hall, Inc. Second-Order Processes The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + O 2 (g) and yields data comparable to this: Time (s)[NO 2 ], M

Chemical Kinetics © 2009, Prentice-Hall, Inc. Second-Order Processes Plotting ln [NO 2 ] vs. t yields the graph at the right. Time (s)[NO 2 ], Mln [NO 2 ] − − − − −5.573 The plot is not a straight line, so the process is not first-order in [A].

Chemical Kinetics © 2009, Prentice-Hall, Inc. Second-Order Processes Graphing ln vs. t, however, gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] Because this is a straight line, the process is second- order in [A]. 1 [NO 2 ]

Chemical Kinetics © 2009, Prentice-Hall, Inc. Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0.

Chemical Kinetics © 2009, Prentice-Hall, Inc. Half-Life For a first-order process, this becomes 0.5 [A] 0 [A] 0 ln = −kt 1/2 ln 0.5 = − kt 1/2 −0.693 = − kt 1/2 = t 1/ k NOTE: For a first-order process, then, the half-life does not depend on [A] 0.

Chemical Kinetics © 2009, Prentice-Hall, Inc. Half-Life For a second-order process, [A] 0 = kt 1/2 + 1 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] 0 2 − 1 [A] 0 = kt 1/2 1 [A] 0 == t 1/2 1 k[A] 0

Chemical Kinetics Ex 14.9 The reaction of C4H9Cl with water is first order reaction. Figure shows How the concentration changes with time at particular temp. A- from the graph estamate the t1/2 B- use t1/2 and calculate the rate constant