Question

Chlorine is commonly used to disinfect drinking water, and the inactivation of pathogens by chlorine follows first-order kinetics. The following data show E. coli inactivation:

Contact time (min)	Percent (%) inactivation
0.00	0.0
0.50	68.3
1.00	90.0
1.50	96.8
2.00	99.0
2.50	99.7
3.00	99.9

(a) Determine the first-order inactivation constant, k. [Hint: % inactivation

$\mathbf{=}\mathbf{100}\mathbf{\times }\left(1-\frac{{\left[A\right]}_t}{{\left[A\right]}_0}\right)$

(b) How much contact time is required for 95% inactivation?

Short answer

1. Rate constant is$\left(K=-2.29s^{-1}\right)$
2. Contact time is$\left(t=1.30s\right)$

Step-by-step solution

First-order kinetics

The reaction is said to be in first order when the reaction rate is linearly dependent upon reactant concentration. As the concentration doubles, the rate of the reaction also doubles.

First order inactivation constant

Given as the percentage of inactivation is:

% inactivation$=100\times \left(1-\frac{{\left[A\right]}_t}{{\left[A\right]}_0}\right)$

and,

role="math" localid="1663363821482" $\%inactivation=68.3$

So,

$$\begin{gathered} 68.3=100\times \left(1-\frac{{\left[A\right]}_t}{{\left[A\right]}_0}\right) \\ 0.683=1-\frac{{\left[A\right]}_t}{{\left[A\right]}_0} \\ \frac{{\left[A\right]}_t}{{\left[A\right]}_0}=1-0.683 \\ \frac{{\left[A\right]}_t}{{\left[A\right]}_0}=0.317 \end{gathered}$$

For first-order kinetics rate constant at t=0.50;

$$\begin{gathered} ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0 \\ 2.303log\frac{{\left[A\right]}_t}{{\left[A\right]}_0}=Kt \\ 2.303log0.317=K\left(0.50\right) \\ 2.303(-0.49)=K(0.50) \\ -1.14=K(0.50) \\ K=-2.29s^{-1} \end{gathered}$$

Time required for 95% of inactivation

Given as the percentage of inactivation is:

% inactivation$=100\times \left(1-\frac{{\left[A\right]}_t}{{\left[A\right]}_0}\right)$

% inactivation=95

So,

$$\begin{gathered} 95=100\left(1-\frac{{\left[A\right]}_t}{{\left[A\right]}_0}\right) \\ 0.95=\left(1-\frac{{\left[A\right]}_t}{{\left[A\right]}_0}\right) \\ \frac{{\left[A\right]}_t}{{\left[A\right]}_0}=1-0.95 \\ \frac{{\left[A\right]}_t}{{\left[A\right]}_0}=0.05 \end{gathered}$$

For first-order kinetics rate constant is -2.29;

$$\begin{gathered} ln{\left[A\right]}_t=Kt+ln{\left[A\right]}_0 \\ 2.303log\frac{{\left[A\right]}_t}{{\left[A\right]}_0}=Kt \\ 2.303log0.05=-2.29\left(t\right) \\ 2.303(-1.3)=-2.29\left(t\right) \\ -2.99=-2.29\left(t\right) \\ t=1.30s \end{gathered}$$