The half-life of a substance in a certain enz | vedclass

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(C) Enzyme-catalysed reactions follow first-order kinetics.The concentration of the substance decreases from $1.28 \; mg \; L^{-1}$ to $0.04 \; mg \;

Vedclass · Accepted Answer

$690$

Vedclass · Answer

(C) Enzyme-catalysed reactions follow first-order kinetics.The concentration of the substance decreases from $1.28 \; mg \; L^{-1}$ to $0.04 \; mg \; L^{-1}$.We can calculate the number of half-lives $(n)$ as follows:$1.28$ $\xrightarrow{t_{1/2}} 0.64$ $\xrightarrow{t_{1/2}} 0.32$ $\xrightarrow{t_{1/2}} 0.16$ $\xrightarrow{t_{1/2}} 0.08$ $\xrightarrow{t_{1/2}} 0.04$This sequence shows that the concentration reduces by half $5$ times,so $n = 5$.The total time required is $t = n 	imes t_{1/2}$.$t = 5 	imes 138 \; s = 690 \; s$.

The half-life of a substance in a certain enzyme-catalysed reaction is $138 \; s$. The time required for the concentration of the substance to fall from $1.28 \; mg \; L^{-1}$ to $0.04 \; mg \; L^{-1}$ is ....... $s$.

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