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  • A first order reaction has a specific reaction rate of . H

A first order reaction has a specific reaction rate of 10^{-3} \mathrm{~s}^{-1}. How much time will it take for 40g of the reactant to reduce to 10g ?

Option: 1

2386 second


Option: 2

1386 second


Option: 3

3465 second


Option: 4

693 second


Answers (1)

For a first orer reaction, half-time \mathrm{t_{1 / 2}} is given as
\mathrm{t_{3 / 2}=0.693 / K} where K is rate constant
\mathrm{\Rightarrow t_{3 / 2}=0.693 / 10^{-3} \mathrm{~s} \\}

\mathrm{\Rightarrow t_{3 / 2}=693 \mathrm{~s}}

For reduction of reactant concentration from 40g to 10g, it shall take 2 half-times.
So time taken =2 x 693 s =1386 second.

Posted by

Kshitij

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