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.
Answer & explanation
Correct answer: option 1
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).