The reaction,X to product follows first order | vedclass

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Question

(C) For a first order reaction,the concentration decreases as $[A]_t = [A]_0 	imes (1/2)^n$,where $n$ is the number of half-lives.Given $[A]_0 = 0.1

Vedclass · Accepted Answer

$3.47 	imes 10^{-4} \, M \, min^{-1}$

Vedclass · Answer

(C) For a first order reaction,the concentration decreases as $[A]_t = [A]_0 	imes (1/2)^n$,where $n$ is the number of half-lives.Given $[A]_0 = 0.1 \, M$ and $[A]_t = 0.025 \, M$,we have $0.025 = 0.1 	imes (1/2)^n$,which implies $(1/2)^n = 0.25 = (1/2)^2$.Thus,$n = 2$ half-lives.Since $2 	imes T_{1/2} = 40 \, min$,the half-life $T_{1/2} = 20 \, min$.The rate constant $k = \frac{0.693}{T_{1/2}} = \frac{0.693}{20} \, min^{-1} = 0.03465 \, min^{-1}$.The rate of reaction is given by $Rate = k 	imes [X]$.At $[X] = 0.01 \, M$,$Rate = 0.03465 	imes 0.01 = 3.465 	imes 10^{-4} \, M \, min^{-1} \approx 3.47 	imes 10^{-4} \, M \, min^{-1}$.

Time $(s)$	$0$	$30$	$60$	$90$
Concentration $(mol \ L^{-1})$	$0.551$	$0.312$	$0.173$	$0.085$

The reaction,$X \to$ product follows first order kinetics. In $40 \, min$ the concentration of $X$ changes from $0.1 \, M$ to $0.025 \, M$. Then the rate of reaction when concentration of $X$ is $0.01 \, M$ is:

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