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Integrated Rate Laws

This is a continuation of Rate and Order of Reactions

Skills to develop

Integrated Rate Laws

The differential rate laws and integrated rate laws are summarized in the table below to give you an overall view of reactions of these types.
\ Differential
rate law
Integrated
rate law
Linear plot Half life
first
order
- d[A]/dt = k [A] [A] = [A]o exp (-k t) ln [A] vs t
slope = - k
hl=ln(2)/k
second
order
- d[A]/dt = k [A]2
 1     1
--- - ---- = k t
[A]   [A]o
 1
--- vs t
[A]
slope = k
hl=1/k[A]o
In this module, more examples are given to show applications of integrated rate laws in problem solving.

Problem 1

The decomposition of A is first order, and [A] is monitored. The following data are recorded:
 t / min    0       1      2      4
 [A]/[M]    0.100 0.0905 0.0819 0.0670
Calculate k. (What is the rate constant?)
Calculate the half life. (What is the half life?)
Calculate [A] when t = 5 min.
Calculate t when [A] = 0.0100.
(Estimate the time required for 90% of A to decompose.)

Solution

A. We can calculate k from any two data points
 The integrated rate law for 1st order is
    A = Ao exp (- k t)
 Using the the first two points,
    0.0905 = 0.100 exp (- k * 1)
    - k = ln (0.0905/0.100)
        = ln(0.905) = -.0998 min-1
 Using the point when t = 2
    0.100 = 0.0819 exp (-k * 2)
    - k 2 = ln (0.0819/0.100)
        = ln (0.819)
        = -0.200
      k = 0.100 min-1
 Using the point when t = 4
   0.0670 = 0.100 exp (- k * 4)
   4 k = ln 1.49
       = 0.400
     k = 0.100 min-1
B. Two methods to evaluate half life are:
 (a) half life * k = ln 2
               = 0.693
     half life = 0.693 / 0.1 = 6.93 min
     (note the calculation of units)
 (b) Calculate the time t when [A] = 0.0500
     0.0500 = 0.100 exp (-0.100 t)
     0.100 * t = ln (0.100/0.0500)
     gives the same result.
C. When t = 5 min
     [A] = 0.100 exp (-0.100 *5)
         = 0.100 * 0.6065
         = 0.0607
D. When [A] is reduced by 90%, we have
   0.01 = 0.1 exp (-0.100 * t)
   0.100 * t = ln (10)
   t = 2.303/0.100 = 23.03 min
  check:
   [A] = 0.1 exp (-0.1 * 23.03)
       = 0.010

Problem 2

The dimerization reaction of butadiene is second order: 2 C4H6(g) = C8H12(g). The rate constant at some temperature is 0.100 /min. The initial concentration of butadiene [B] is 1 M. Calculate the concentration of butadiene when t is 1, 2, 5, 10, 20, 30, and 70.

Solution:

In fact, many of the following values can be evaluated without using a calculator.

  t/min   0   1     2     5     10   20    30    70.
  [B]/M  1.0 0.909 0.833 0.667 0.50 0.333 0.25  0.125

Discussion:
From the record shown as above,
Calculate k. (What is the rate constant?)
Calculate the half life for [B] = 1.0. (What is the half life for an initial concentration of 1.0?)
Calculate [B] when t = 40 min.
Calculate t when [B] = 0.100.
(Estimate the time required for 90% of B to polymerize.)
How does the total pressure change with time?
Getting the answers for the above is fun. Please try.

Confidence Building Questions

©cchieh@uwaterloo.ca