The order of a reactions where half life changes from 150 mins to 75 min when the concentration of reactant is changed from 0.4 M to 1.6 M respectively is.

$\frac{3}{2}$

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

Given:

When the concentration of the reactant changes from 0.4 M to 1.6 M, the half-life changes from 150 minutes to 75 minutes.

Relationship between half-life and order of reaction:

The general relation between half-life \( t_{1/2} \) and concentration \( [A] \) for a reaction of order \( n \) is given as:

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

Where:

\( n \) is the order of the reaction.

\( t_{1/2} \) is the half-life.

\( [A] \) is the concentration of the reactant.

Initial concentration: \( [A_1] = 0.4 \, \text{M} \), \( (t_{1/2})_1 = 150 \, \text{minutes} \)

Final concentration: \( [A_2] = 1.6 \, \text{M} \), \( (t_{1/2})_2 = 75 \, \text{minutes} \)

The concentration has increased by a factor of 4, and the half-life has decreased by a factor of 2.

Using the relationship \( t_{1/2} \propto [A]^{1-n} \), we can write:

\(\frac{(t_{1/2})_2)}{(t_{1/2})_1)} = \left(\frac{[A_2]}{[A_1]}\right)^{1-n}\)

Substitute the known values:

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

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

Take the logarithm of both sides:

\(1-n = \frac{1}{2}\)

\(n = \frac{3}{2}\)

Conclusion

The order of the reaction is \( \frac{3}{2} \), so the correct answer is Option (1).