Chapter 11, Problem 11.49PAE

The rate of photodecomposition of the herbicide piclo- ram in aqueous systems was determined by exposure to sunlight for a number of days. One such experiment produced the following results. (Data from R.T. Hedlun and C.R. Youngson, “The Rates of Photodecomposition of Picloram in Aqueous Systems," Fate of Organic Pesticides in tbe Aquatic Environment, Advances in Chemistry Series, #111, American Chemical Society (1972), 159—172.)

Exposure Time, t (days)

[Pidoram]

(mol L_1) 0 4.14 X 10-6 7 3.70 X 10-6 14 3.31 X 10-6 21 2.94 X 10~6 28 2.61 X 10~6 35 2.30 X 10-6 42 2.05 X 10-6 49 1.82 X 10"6 56 1.65 X 10-6

Determine the order of reaction, the rate constant, and the half-life for the photodecomposition of picloram.

Interpretation Introduction

Interpretation:

To interpret the data collected and to determine the order of reaction, rate constant and the half-life for the photochemical decomposition of Picloram

Concept Introduction:

For a first order reaction: $\text{k = }\frac{2.303}{\text{t}}\text{log}\frac{{\text{N}}_{\text{o}}}{{\text{N}}_{\text{t}}}$

And half-life is given by: ${\text{t}}_{1/2}=\frac{0.693}{\text{k}}$

Where,

k = decay constant

t = time taken

${\text{N}}_{\text{o}}$ = initial concentration

${\text{N}}_{\text{t}}$ = concentration at time t

${\text{t}}_{1/2}$ = half life

Answer to Problem 11.49PAE

Solution:

The order of reaction:

First

The value of rate constant:

$\text{0}{\text{.016 day}}^{-1}$

Half life of reaction:

$43.31{\text{ day}}^{-1}$

Explanation of Solution

Given information:

Data collected for decomposition of Picloram is:

Time, t (days)	[Picloram] (mol L-1)
0	4.14 $\times {10}^{-6}$
7	3.70 $\times {10}^{-6}$
14	3.31 $\times {10}^{-6}$
21	2.94 $\times {10}^{-6}$
28	2.61 $\times {10}^{-6}$
35	2.30 $\times {10}^{-6}$
42	2.05 $\times {10}^{-6}$
49	1.82 $\times {10}^{-6}$
56	1.65 $\times {10}^{-6}$

From the given data we will first calculate if the reaction is first order,

For the first order reaction, we will use the formula $\text{k = }\frac{2.303}{\text{t}}\text{log}\frac{{\text{N}}_{\text{o}}}{{\text{N}}_{\text{t}}}$

Case 1)

When time (t) is 7 days

${\text{N}}_{\text{o}}$ = 4.14 $\times {10}^{-6}$ mol L-1

${\text{N}}_{\text{t}}$ = 3.70 $\times {10}^{-6}$ mol L-1

Substituting the values,$\begin{array}{l}{\text{k}}_1\text{ = }\frac{2.303}{7}\text{log}\frac{4.14\times {10}^{-6}}{3.70\times {10}^{-6}}\\ {\text{k}}_1\text{ = 0}\text{.329 }\times \text{ 0}\text{.049}\\ {\text{k}}_1\text{ = 0}{\text{.016 day}}^{-1}\end{array}$

Case 2)

When time (t) is 14 days

${\text{N}}_{\text{o}}$ = 4.14 $\times {10}^{-6}$ mol L-1

${\text{N}}_{\text{t}}$ = 3.31 $\times {10}^{-6}$ mol L-1

Substituting the values,$\begin{array}{l}{\text{k}}_2\text{ = }\frac{2.303}{14}\text{log}\frac{4.14\times {10}^{-6}}{3.31\times {10}^{-6}}\\ {\text{k}}_2\text{ = 0}\text{.1645 }\times \text{ 0}\text{.097}\\ {\text{k}}_2\text{ = 0}{\text{.016 day}}^{-1}\end{array}$

The value of ${\text{k}}_1$ and ${\text{k}}_2$ are equal. Therefore, the reaction is first order reaction which is independent of the initial concentration with rate constant $\text{0}{\text{.016 day}}^{-1}$

The half-life of the reaction can be calculated with the help of formula ${\text{t}}_{1/2}=\frac{0.693}{\text{k}}$

Substituting the value of k ${\text{t}}_{1/2}=\frac{0.693}{0.016}=43.31{\text{ day}}^{-1}$

Hence, the half-life is $43.31{\text{ day}}^{-1}$

Conclusion

The decomposition of Picloram is a first order reaction with rate constant $\text{0}{\text{.016 day}}^{-1}$ and half-life of $43.31{\text{ day}}^{-1}$