JEE Chemistry 2025 Question with Solution

Given: Bacterial growth before treatment follows first-order kinetics. After treatment, the rate of bacterial decay is proportional to the square of the number of bacteria.

Find: The correct set of before and after graphs.

For bacterial growth before treatment,

So,

This represents an exponential increase of $N$ with $t$. Therefore, the before graph must be an exponentially rising curve.

After treatment, decay follows

So the decay rate magnitude is proportional to

Hence, the graph of $r$ versus $N$ must increase non-linearly as $N$ increases. This gives an upward-curving graph, not a straight line or a decreasing curve.

Thus, the correct set is the one with an exponential rise before treatment and a curved increasing $r$ versus $N$ graph after treatment.

Therefore, the correct option is B.

Given:

• Before treatment, bacterial growth follows first-order kinetics.
• After treatment, bacterial decay rate is proportional to the square of the existing bacteria.

Find: Which graphical option matches both situations.

Before treatment, first-order growth means the rate is directly proportional to the number present:

Integrating gives

So $\frac{N}{N_0}$ versus $t$ shows exponential growth.

After treatment, the statement says the decay rate is proportional to the square of bacteria count:

Therefore, when plotted against $N$, the rate $r$ must rise like a parabola-type curve from the origin.

Checking the options:

• Option A has a decreasing after curve, so it does not represent $r \propto N^2$.
• Option B has exponential increase before treatment and a curved increasing after graph.
• Option C has a linear after graph, which is inconsistent with square dependence.
• Option D also shows a linear after graph.

Therefore, the correct option is B.