For the reaction $A \to B$ , the rate law expression is $- \frac{d [A]}{d t} = k {[A]}^{1 / 2}$ . If initial concentration of A is $A_{0}$ . Which of the following statement is/are correct regarding this reaction?

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There is a linear decrease of rate of reaction with time

The integrated form of rate expression is $\sqrt{A} = \frac{- K t}{2} + \sqrt{A_{0}}$

The half life period is $t_{1 / 2} = \frac{\sqrt{2} (\sqrt{2} - 1)}{K} \sqrt{A_{0}}$

The plot of $\sqrt{A}$ against ‘t’ will be non-linear.

answer is B, C, D.

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Detailed Solution

For $n^{t h}$ order
$K = \frac{1}{(n - 1) t} [\frac{1}{{[A]}^{n - 1}} - \frac{1}{{[A_{0}]}^{n - 1}}]$
$n = 1 / 2$
$K = \frac{2}{t} [\sqrt{[A_{0}]} - \sqrt{[A]}]$
$\sqrt{[A]} = - \frac{K t}{2} + \sqrt{[A_{0}]}$
$∴ \sqrt{[A]} vs t$ is linear with negative slope