Integrated Rate Laws

This is a continuation of Rate and Order of Reactions

Skills to develop

Apply either the differential rate laws or integrated rate laws to solve chemical kinetic problems.

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 = A_o 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

For the polymerization of butadiene, 2 C₄H₆(g) = C₈H₁₂(g), the rate of decreasing of partial pressure of butadiene is 2 kPa/min. What is the rate of increasing of partial pressure of (octadiene)?

Discussion -
Consider the relationship
-d[C₄H₆]/dt = 2 d[C₈H₁₂]/dt
A cylinder containing pure butadiene has a pressure of 101 kPa. After 10 minutes, the total pressure drops to 95 kPa. What is the partial of C₈H₁₂ (octadiene) formed? 2 C₄H₆(g) = C₈H₁₂(g).

Discussion -
The drop in pressure is 101-95 = 6 kPa. Consider the reaction:
```
 2 C₄H₆(g) = C₈H₁₂(g).
  12 kPa      6 kPa
```
Since two moles of butadiene combine to give one mole octadiene, the difference in total pressure is the partial pressure of octadiene, 12 kPa of butadiene converted to 6 kPa of octadiene. What is the partial pressure of butadiene?
At some temperature, the total pressure of a butadiene cylinder drops from 101 kPa to 95 kPa in 10 min. The polymerization is known to be second order. What is the rate constant?

Discussion -
The reaction is:
2 C₄H₆(g) = C₈H₁₂(g). Since the total pressure drops from 101 to 95 kPa after 10 min, the partial pressure of C₄H₆ goes from 101 to [101 - 2*(101-95)] = 89 kPa. Using the integrated rate law:
1/89 - 1/101 = k * 10
k =1.33e-4
What is the total pressure when the reaction is completed?
Radioactive decay always follow first order kinetics. Carbon-11 is a radioactive isotope of carbon, and its half life is 20.3 min. What is the decay (or rate) constant?

Skill -
Evaluate half-life according to conditions. Apply the equation,
k * half-life = ln 2 in this case.