Q.

A reaction $A \to B$ is of the order 2.5 with respect to A. The half-life of the reaction will be given

Moderate

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By Expert Faculty of Sri Chaitanya

$0.693 / k$

$1 / ([A]_{0} k)$

$k / [A]_{0}$

$[1 / (1.5 k [A]_{0}^{1.5})] (2^{1.5} - 1)$

answer is D.

(Detailed Solution Below)

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Detailed Solution

$- \frac{d [A]}{d t} = k [A]^{2.5} .$

Hence, $- \frac{d [A]}{[A]^{2.5}} = k d t .$

Thus

$\frac{1}{[A]^{1.5}} - \frac{1}{[A]_{0}^{1.5}} = (1.5) k t$

For $t_{0.5}, [A] = [A]_{0} / 2$

Hence

$\frac{1}{{([A]_{0} / 2)}^{1.5}} - \frac{1}{[A]_{0}^{1.5}} = 1.5 k t_{0.5} \Rightarrow t_{0.5} = \frac{1}{1.5 k [A]_{0}^{1.5}} (2^{1.5} - 1)$

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