Question

Determine which of these data sets represents a zero-order, which represents a first order, and which represents a second order reaction. For each compound, A, B and C:

  

Time (min)	0.0	10.0	20.0	30.0	40.0	50.0	60.0	80.0	100.0	150.0	200.0
[A], M	2.5	2.25	2.07	1.94	1.79	1.69	1.61	1.49	1.28	1.06
[B], M	1.51	1.22	1.06	0.95	0.8	0.73	0.69	0.58	0.48	0.32	0.2
[C], M	3.25	3.05	2.85	2.65	2.45	2.24	2.11	1.78	1.43	0.72	0.05

Write the rate law and integrated rate law.

Determine the value of the rate constant, use appropriate units? Justify.

Calculate the half-life.

Determine the concentration of A, B, and C at 400 minutes.

Answer 1

Calculate slope for each A, B and C

For zero-order,

slope = -(final - initial) concentration/(final - initial) time = constant

slope = rate constant (k)

For C,

from Ist and IInd data set, slope = -(3.05 - 3.25)/10 = 0.02

from IInd and IIIrd data set, slope = -(2.85 - 3.05)/10 = 0.02

rate constant = 0.02 M-1.min-1

So,

[C] (M) vs time is a zero-order reaction

rate law,

rate = k = constant

independent of initial concentration

half-life, t1/2 = [A]o/2k

                     = 3.25/2 x 0.02

                     = 81.25 min

concentration of C after 400 min,

[C] (at t = 400 min) = 3.25 - 0.02 x 400

                               = -4.75 M

So technically all of [C] is consumed at this stage.

For a first order reaction,

slope = -(ln(final concentration) - ln(initial concentration))/(final - initial) time

         = rate constant

For A,

from Ist and IInd data set, slope = -(ln(2.25) - ln(2.5))/10 = 0.01

from IVth and Vth data set, slope = -(ln(1.79) - ln(1.94))/10 = 0.01

rate constant k = 0.01 min-1

So,

[A] (M) vs time is a first-order reaction

rate law,

rate = k[A]

dependent of initial concentration

half-life, t1/2 = ln(2)/k

                     = ln(2)/0.01

                     = 69.31 min

concentration of A after 400 min,

ln[A] (at t = 400 min) = ln(2.5) - 0.01 x 400

[A] = 0.046 M

Fors second-order reaction,

slope = -(1/[A]t - 1/[A]o)/(final - initial) time

From IInd and IIIrd data set,

slope = 0.012

From Vth and VIth data set,

slope = 0.012

rate constant = slope = 0.012 M-1.min-1

So,

[B] is second-order reaction,

rate law = k[B]^2

half-life = 1/0.012 x 1.51

            = 55.19 min

after 400 min

1/[B] = 1/1.51 + 0.012 x 400

concentration of [B] = 0.18 M