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For the reaction $A + 2 B ⟶$ products (started with concentrations taken in stoichiometric proportion), the experimentally determined rate law is :
$- \frac{d [A]}{d t} = k \sqrt{[A]} \sqrt{[B]}$
The half time of the reaction would be:

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$\frac{0.698}{k}$

$\frac{0.693}{\sqrt{2 k}}$

not defined

$\frac{0.693}{1 / k}$

answer is C.

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Detailed Solution

Given,

$A + 2 B \to$ Products

$\begin{array}{cc} a & 2 a \\ a - x & 2 a - 2 x \end{array}$

Here,

$\frac{d x}{d t} = k \sqrt{2} (α - x) t_{1 / 2} = \frac{\ln 2}{\sqrt{2} k} t_{1 / 2} = \frac{0.693}{1 / k}$

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