Homework Solutions

November 5

15: 22  "a plot of 1/[A] versus time resulted in a straight line with a slope value of +3.60 x 10^-2 L mol^-1 s^-1."  This means the reaction is second order, k = slope.

a.    rate law:  rate = 3.60 x 10^-2 L mol^-1 s^-1[A]²

       integrated rate law:  1/[A] = 3.60 x 10^-2 L mol^-1 s^-1 t + 1/[A]o

      k = 3.60 x 10^-2 L mol^-1 s^-1

b.    t(1/2) = 1/k[A]o = 1/((3.60 x 10^-2 L mol^-1 s^-1)*(2.80 x 10^-3 mol L^-1))    =    9920 s = 165 min = 2.75 hours

c.    rearrange integrated rate law, set [A] = 7.00 x 10^-4 mol L^-1:

    t = (1/k)*(1/[A] - 1/[A]o) = 27.78*(1428 - 357) = 29748 s = 29700 s = 496 min = 8.26 hours

15.24 " a plot of [A] versus time resulted in a straight line" -- this means the reaction is zeroth order; slope = -k.

a.    rate law: rate = k = 4.00 x 10^-6 mol L^-1 s^-1

    integrated rate law:  [A] = -kt + [A]o = - 4.00 x 10^-6 mol L^-1 s^-1 * t + [A]o

    k = 4.00 x 10^-6 mol L^-1 s^-1

b.    t(1/2) = [A]o/2k = 1.25 x 10^-2 mol L^-1 / 2*4.00 x 10^-6 mol L^-1 s^-1 = 1562.5 s = 1560 s = 26.0 min

c.    Rearrange the integrated rate law with [A] = 0.  t = [A]o/k = 1.25 x 10^-2 mol L^-1/4.00 x 10^-6 mol L^-1 s^-1 = 2*t(1/2) = 52.0 min

15: 25  In Excel, plot [A] vs t, ln [A] vs t, and 1/[A] vs t to determine whether the reaction is zeroth, first, or second order, respectively.  The excel file is linked to the schedule page.