In: Civil Engineering
6. A certain contaminant was studied in the laboratory by measuring over time the concentration in a sample placed in a closed container. The values of the observed concentrations are given below. Determine the reaction order and the decay rate for this contaminant.
Time (hrs) |
0 |
1 |
3.5 |
6.5 |
10 |
Conc (mg/L) |
100 |
80 |
50 |
26 |
10 |
As both the reaction order and decay rate is unknown let us go for trial and error method .
STEP 1 : Assume it is a zero order reaction
Zero order reaction : graph of A versus time is straight line plot
Thus Af = -Kt + Ai
Here the slope is negative because concentration decreases as time increases
where Af = final concentration
K = decay rate
t = time taken to decrease from initial to final concentration
Ai = initial concentration
let us find K from the above table (values taken from first and last columns )
10 = -K(10) + 100
-90 = -K(10)
K = 9mg/L hrs-1
Check for K (values taken from first and second columns)
80 = -K(1) + 100
-20 = -K
K = 20 mg/L hrs-1
As the K value is different in both the cases, our assumption is wrong.
STEP 2 : Assume it as a first order reaction
First order reaction : graph of ln [A] versus time is a straight line plot
Thus ln Af = -Kt + ln Ai
ln [Af / Ai ] = -Kt
Here the slope is negative because ln of concentration decreases as time increases
where Af = final concentration
K = decay rate
t = time taken to decrease from initial to final concentration
Ai = initial concentration
let us find K from the above table (values taken from first and last columns )
ln [ 10 / 100 ] = -K(10)
K = 0.23 hrs-1 = 0.2 hrs-1
check for K (values taken from first and second columns)
ln [ 80/ 100 ] = - K (1)
K= 0.223hrs-1 = 0.2 hrs-1
Double check the value of K (values taken from first and third column)
ln [ 50 / 100] = -K (3.5)
K = 0.198 hrs-1 = 0.2 hrs-1
Thus , as the value of K is constant it can be concluded that it is a first order reaction and the decay rate is 0.2 hrs-1