# 4.11: Rate Laws - Integrated

This is a continuation of Rate and Order of Reactions.

Learning Objectives

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

In this module, more examples are given to show applications of integrated rate laws in problem solving.

## 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 Integrated Linear plot Half life rate law rate law $$\mathrm{-\dfrac{d[A]}{dt} = \mathit k [A]}$$ $$\mathrm{[A] = [A]_o} e^{-\large{k\, t}}$$ $$\mathrm{\ln [A]\: vs}\: t$$ $$\textrm{slope} = - k$$ $$t_{\frac{1}{2}}=\dfrac{\ln(2)}{k}$$ $$\mathrm{- \dfrac{d[A]}{dt} = \mathit k [A]^2}$$ $$\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t$$ $$\mathrm{\dfrac{1}{[A]}\: vs}\: t$$ $$\textrm{slope} = k$$ $$t_{\frac{1}{2}}=\dfrac{1}{k\mathrm{[A]_o}}$$

Example 1

The decomposition of $$\ce{A}$$ is first order, and $$\ce{[A]}$$ is monitored. The following data are recorded:

 t / min 0 1 2 3 [A]/[M] 0.1 0.0905 0.0819 0.67
1. Calculate k. (What is the rate constant?)
2. Calculate the half life. (What is the half life?)
3. Calculate $$\ce{[A]}$$ when t = 5 min.
4. Calculate t when $$\mathrm{[A] = 0.0100}$$ (i.e., estimate the time required for 90% of $$\ce{A}$$ to decompose.)

Solution

1. We can calculate k from any two data points.

The integrated rate law for 1st order is

$$\mathrm{A = A_o}\, e^{- \large{k\, t}}$$

Using the the first two points,

$$0.0905 = 0.100\, e^{(- \large k \times 1)}$$

\begin{align*} - k &= \ln \left(\dfrac{0.0905}{0.100}\right)\\ &= \mathrm{\ln(0.905) = -.0998\: min^{-1}} \end{align*}

Using the point when t = 2

$$0.100 = 0.0819\, e^{(-\large k \times 2)}$$

\begin{align*} - k\, 2 &= \ln \left(\dfrac{0.0819}{0.100}\right)\\ &= \ln (0.819)\\ &= -0.200\\ k &= \mathrm{0.100\: min^{-1}} \end{align*}

Using the point when t = 4

$$0.0670 = 0.100\, e^{(- \large k \times 4)}$$

\begin{align*} 4\, k &= \ln 1.49\\ &= 0.400\\ k &= \mathrm{0.100\: min^{-1}} \end{align*}

2. Two methods to evaluate half life are:
1. $$t_{\frac{1}{2}} \times k = \ln 2$$

$$= 0.693$$

$$t_{\frac{1}{2}} = \dfrac{0.693}{0.1} = \mathrm{6.93\: min}$$

(note the calculation of units)

2. Calculate the time t when $$\mathrm{[A] = 0.0500}$$
$$0.0500 = 0.100\, e^{(\large{-0.100\: t})}$$
$$0.100 \times t = \ln \left(\dfrac{0.100}{0.0500}\right)$$
gives the same result.
3. When t = 5 min

\begin{align*} \mathrm{[A]} &= 0.100\, e^{(\large{-0.100 \times5})}\\ &= 0.100 \times 0.6065\\ &= 0.0607 \end{align*}

4. When $$\ce{[A]}$$ is reduced by 90%, we have

$$0.01 = 0.1\, e^{(\large{-0.100 \times t})}$$

$$0.100 \times t = \ln (10)$$

$$\mathrm{t = \dfrac{2.303}{0.100} = 23.03\: min}$$

Check:

\begin{align*} \mathrm{[A]} &= 0.1\, e^{(\large{-0.1 \times 23.03})}\\ &= 0.010 \end{align*}

Example 2

The dimerization reaction of butadiene is second order process:

$\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}} \nonumber$

The rate constant at some temperature is 0.100 /min with an initial concentration of butadiene ([B]) of 1 M. Calculate the concentration of butadiene at 1, 2, 5, 10, 20, 30, and 70 minutes.

Solution

Many of the following values can be evaluated without using a calculator.

 t / min [B]/[M] 0 1 2 5 10 20 30 70 1 0.909 0.833 0.667 0.5 0.33 0.25 0.125

Discussion:

This is an applicaiton of this equation:

$\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t$

Exercise

From the data and results in Example 2:

1. Calculate k. (What is the rate constant?)
2. Calculate the half life for [B] = 1.0. (What is the half life for an initial concentration of 1.0?)
3. Calculate [B] when t = 40 min.
4. Calculate t when [B] = 0.100. (Estimate the time required for 90% of B to polymerize.)
5. How does the total pressure change with time?

## Questions

1. For the polymerization of butadiene,

$$\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}}$$

the rate of decreasing of partial pressure of butadiene is 2 kPa/min. What is the rate of increasing of partial pressure of (octadiene)?
Hint: 1 kPa/min

Discussion -
Consider the relationship

$$-\ce{\dfrac{d[C4H6]}{dt}} = \ce{2 \dfrac{d[C8H12]}{dt}}$$

2. 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 $$\ce{C8H12}$$ (octadiene) formed?

$$\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}}$$

Hint: 6 kPa

Discussion -
The drop in pressure is 101-95 = 6 kPa. Consider the reaction:

\begin{alignat}{3} \ce{2 &C4H_{6\large{(g)}} \rightarrow\: &&C8H_{12\large{(g)}}}\\ \ce{&12\: kPa &&\:\:6\: kPa} \end{alignat}

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?

3. 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?
Hint: 0.000133 kPa/min

Discussion -
The reaction is:

$$\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}}$$

Since the total pressure drops from 101 to 95 kPa after 10 min, the partial pressure of $$\ce{C4H6}$$ goes from 101 to [101 - 2*(101-95)] = 89 kPa. Using the integrated rate law:

$$\dfrac{1}{89} - \dfrac{1}{101} = k \times 10$$

$$k =\mathrm{1.33\,e^{-4}}$$

What is the total pressure when the reaction is completed?

4. Radioactive decay always follows 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?
Hint: 0.0341 /m

Skill -
Evaluate half-life according to conditions. Apply the equation,

$$k \times t_{\frac{1}{2}} = \ln 2$$

in this case.