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In formation of compound [B] follows the first order of kinetics and after 90 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x s^{-1}. Find the value of x.

Option: 1

128 \times 10^{-6} \mathrm{sec}


Option: 2

134 \times 10^{-4} \mathrm{sec}


Option: 3

119 \times 10^{-5} \mathrm{sec}


Option: 4

230 \times 10^{-6} \mathrm{sec}


Answers (1)

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                                 [A]  \mathrm{ \rightarrow }  [B]
since the follows the first order kinetics and after 90 minutes the concentration of [A] was found to be half of its initial concentration. That means 90 min is the half life time for the reaction. And we know that for a first order reaction half life time is given as
\mathrm{ \begin{aligned} t_{\frac{1}{2}} & =\frac{0.693}{k} \\ k & =\frac{0.693}{t \frac{1}{2}}=\frac{0.693}{90 \times 60 \mathrm{sec}} \\ k & =\frac{6930}{9 \times 6} \times 10^{-6} \\ k & =128 \times 10^{-6} \mathrm{sec} \end{aligned} }
 

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vishal kumar

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