The order of a reaction whose half life changes from 150 mins to 75 m…

The correct answer is Option (1) → \(\frac{3}{2}\)

For a reaction of order n (n ≠ 1) the half-life depends on the initial concentration as:

    \( \displaystyle t_{1/2} \propto \frac{1}{[A]_0^{\,n-1}} \)

For zero order (n = 0), t_1/2 = [A]₀/2k is proportional to [A]₀; for first order (n = 1) it does not depend on [A]₀. Taking the ratio of the two experiments:

    \( \displaystyle \frac{t_1}{t_2} = \left(\frac{[A]_2}{[A]_1}\right)^{n-1} \)

    \( \displaystyle \frac{150}{75} = \left(\frac{1.6}{0.4}\right)^{n-1} \)

    \( \displaystyle 2 = 4^{\,n-1} = 2^{\,2(n-1)} \)

    \( \displaystyle 2(n-1) = 1 \;\Rightarrow\; n = \frac{3}{2} \)

So the order of the reaction is 3/2.

Note: The question is from beyond the NCERT textbook.