JEE Main 2025 Apr 7 Shift 1, Chemistry Q17: Reactions of Various Orders

JEE Main2025Apr 7, Shift 1Chemistry

Q.

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.]

A
B
C
D

Solution

Before treatment (growth, first order in N):

$\frac{d N}{d t} = k N \Rightarrow \frac{N}{N _{0}} = e^{k t}$

So $N / N_{0}$ versus $t$ is an exponential rise starting from 1 at $t = 0$ .

After treatment (decay, second order in N):

$r = - \frac{d N}{d t} = k N^{2}$

So $r$ versus $N$ is a parabola of the form $y = k x^{2}$ , passing through the origin and opening upward.

The graph set that shows an exponential rise for 'Before' and a rising parabola for 'After' is Option (2).

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