• # question_answer For the reaction $A+2B\to$ products(started with concentrations taken in stoichiometric proportion), the experimentally determined rate law is $-\frac{d[B]}{dt}=k\sqrt{[A]}\sqrt{[B]}$ The half time of the reaction would be A) $\frac{\sqrt{2}\times 0.693}{k}$             B) $\frac{0.693}{1/k}$ C) $\frac{0.693}{\sqrt{2}k}$                      D) not defined

Solution :

 A + 2B$\xrightarrow{{}}$Product a 2a 0 a - x 2a - 2x x
$\therefore$      $r=k\,\,\sqrt{A}\,\,\,\sqrt{B}$ $=\sqrt{(a-x)}\sqrt{2a-2x}$ $=\sqrt{2}\,k(a-x)$ Solve it for ${{t}_{1/2}}$

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