T_{50} of a first-order reaction is 10 min. | vedclass

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(B) For a first-order reaction,the half-life $t_{1/2} = 10 \ min$.Initial concentration $[A]_0 = 10 \ mol \ L^{-1}$.After $20 \ min$ (which is $2 	im

Vedclass · Accepted Answer

$0.0693 	imes 2.5 \ mol \ L^{-1} \ min^{-1}$

Vedclass · Answer

(B) For a first-order reaction,the half-life $t_{1/2} = 10 \ min$.Initial concentration $[A]_0 = 10 \ mol \ L^{-1}$.After $20 \ min$ (which is $2 	imes t_{1/2}$),the concentration $[A]$ will be:$[A] = [A]_0 	imes (1/2)^n = 10 	imes (1/2)^2 = 10 / 4 = 2.5 \ mol \ L^{-1}$.The rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \ min^{-1}$.The rate of reaction is given by $Rate = k 	imes [A]$.Substituting the values: $Rate = 0.0693 	imes 2.5 \ mol \ L^{-1} \ min^{-1}$.

$T_{50}$ of a first-order reaction is $10 \ min$. Starting with $10 \ mol \ L^{-1}$,the rate after $20 \ min$ is:

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