JEE Main · 2025 · Shift-IhardCK-140

A person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

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 Ist^{\text{st}} order kinetics.]

Options
  1. a

    image

  2. b

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  3. c

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  4. d

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Correct Answerb

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Detailed Solution

Step 1: Analyse bacterial growth (Before medicine)

Bacterial growth follows first order kinetics:

dAdt=k[A]\frac{dA}{dt} = k[A]

AA0=ekt\frac{A}{A_0} = e^{kt}

This gives an exponential growth curve (N/N₀ vs t is exponential, concave up).

Step 2: Analyse bacterial decay (After medicine)

Rate of decay r=dAdt[A]2r = -\tfrac{dA}{dt} \propto [A]^2 (second order in bacteria)

r=k[A]2r = k[A]^2

So rr vs NN is a parabola (y=kx2y = kx^2).

Step 3: Match with graphs

  • Before: N/N0N/N_0 vs tt → exponential growth (starts slow, then rapid) ✓
  • After: rr vs NN → parabola (second order) ✓

Graph (b) shows: Before = exponential growth curve, After = parabolic r vs N relationship.

Answer: (b)

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