A person's wound was exposed to some bacteria and then bacterial 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 1ˢᵗ order kinetics.]

Solution

Chemistry ma’am · 3 steps

Turn each sentence into a rate law first, then sketch it: growth is eᵏᵗ, decay is a parabola in N.

1. Before the medicine. First-order growth means dN/dt = kN. Integrating gives N = N₀ eᵏᵗ, so N/N₀ = eᵏᵗ. It starts at exactly 1 when t = 0 and rises exponentially, gently at first and then more and more steeply.
2. After the medicine. The decay rate is proportional to the square of the number of bacteria. That is a second-order rate law.

   r = k' N²

   Option	Before: N/N₀ vs t	After: r vs N	Verdict
   (a)	starts at 1 and rises	straight line, r ∝ N	after graph is first order
   (b)	starts at 0 and has a kink	upward parabola	N/N₀ must be 1 at t = 0
   (c)	starts at 1, exponential rise	upward parabola from origin	both correct
   (d)	starts at 1, exponential rise	r falls as N grows	rate cannot drop as N rises

3. So. r = k'N² is an upward-opening parabola through the origin. Only (c) pairs an exponential rise from 1 with that parabola, so the answer is (c).

The idea to keep. For any graph question in kinetics, write the rate law first. Rate vs concentration is a straight line for first order and a parabola through the origin for second order.

Answer from the NTA final answer key. Solution by PrepWiser.