A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial

Question

A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine?
[Given : $N =$ No. of bacteria, $t =$ time, bacterial growth follows $I^{st}$ order kinetics.]

Options:

Answer:

B

Solution:

*Before applying medicine
$\frac{dA}{dt} = K [A]$ (First order growth) (Rate law)
$\frac{A}{A _{0}} = \frac{N}{N _{0}} = e^{Kt}$
*After applying medicine

$r = - \frac{dA}{dt} = K [A]^{2}$ (Rate law)
$y = Kx^{2}$ Parabola

TestHub · Accepted Answer

*Before applying medicine dt dA ​ = K [ A ] (First order growth) (Rate law) A 0 ​ A ​ = N 0 ​ N ​ = e Kt *After applying medicine r = − dt dA ​ = K [ A ] 2 (Rate law) y = Kx 2 Parabola

Chemistry - Chemical Kinetics Question with Solution | TestHub

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