Question

5) You are tasked with determining the photolytic behavior of a new pesticide in rice field water. To this end, you collect field water, add the pesticide to the water in a lab, and put samples in a photochamber, which exposes them to UV light. You then analyze your samples over time and record the average concentration of the pesticide at each time you took a sample. The values are listed below: (20) Time (hours) Average Concentration (mg/L) 50.0 4.2 45.6 8.5 38.3 17 26.7 28 19.4 42 l 1.2 72 5.90 120 1.80 a) Using this data, calculate the photolysis rate constant, k and provide the equation you obtained kphoto from. You are encouraged to use Microsoft Excel or a similar program for this question. Please remember your units and significant figures b) Please calculate the photolysis half-life, tin. Please remember your units and significant figures! c) How much time is required for 98% dissipation of this pesticide due to photolysis? d) After pesticide application, California rice field water is required to be contained in the field for 30 days before it can be released into the Sacramento River. Will there be a significant issue regarding this pesticide's potential release into the Sacramento River? Why or why not?

Calculating the photolysis rate constant, photolysis half-life, >98% dissipation.

Answer 1

(a)

Photolysis is first order reaction.So rate equation is:

-d[C]/dt=k[C]

Where C=concentration and t= time

So integrating

lnC -lnC₀=-kt

So lnC=lnC₀-kt

So plot ln C vs t and intercept is lnC_{0 and slope is -k.}

Time (hr)	C(mg/lit)	Time (s)	ln c
0	50	0	3.912023
4.2	45.6	15120	3.819908
8.5	38.3	30600	3.64545
17	26.7	61200	3.284664
28	19.4	100800	2.965273
42	11.2	151200	2.415914
72	5.9	259200	1.774952
120	1.8	432000	0.587787

4.5 In C vs t 3.5 2.5 In C In C vs t 1.5 Linear (In C vs t) 0.5 0 100000 200000 300000 400000 500000 time (s)

So from graph k=8*10^-6 sec^-1

(b) photolysis half-life

for first order reaction

t_1/2=0.693/k=0.693/8*10^-6=86625 sec=24.0625 hr

(c) for >98% dissipation means

C=C₀-0.98*C₀=50.0-0.98*50=1 mg/lit

So lnC-lnC₀=-kt

ln50-ln1=8*10^-6*t

t=489002.8 sec=135.834 hr