What is the concentration (in mol L^{-1}) of | vedclass

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(C) For a zero-order reaction,$A \longrightarrow 3B$,the rate law is $[A] = [A]_0 - kt$.From the table,at $t = 15 \ s$,$[A] = 0.05 \ mol \ L^{-1}$ and

Vedclass · Accepted Answer

$1.98 	imes 10^{-1}$

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(C) For a zero-order reaction,$A \longrightarrow 3B$,the rate law is $[A] = [A]_0 - kt$.From the table,at $t = 15 \ s$,$[A] = 0.05 \ mol \ L^{-1}$ and $[A]_0 = 0.1 \ mol \ L^{-1}$.Substituting these values: $0.05 = 0.1 - k(15) \implies 15k = 0.05 \implies k = \frac{0.05}{15} = \frac{1}{300} \ mol \ L^{-1} \ s^{-1}$.Now,at $t = 20 \ s$,the concentration of $A$ is $[A] = [A]_0 - kt = 0.1 - (\frac{1}{300}) 	imes 20 = 0.1 - \frac{20}{300} = 0.1 - 0.0667 = 0.0333 \ mol \ L^{-1}$.The amount of $A$ reacted is $\Delta[A] = [A]_0 - [A] = 0.1 - 0.0333 = 0.0667 \ mol \ L^{-1}$.According to the stoichiometry $A \longrightarrow 3B$,the concentration of product $B$ formed is $3 	imes \Delta[A] = 3 	imes 0.0667 = 0.2001 \ mol \ L^{-1}$.Rounding to the nearest option,we get $0.198 \ mol \ L^{-1}$ or $1.98 	imes 10^{-1} \ mol \ L^{-1}$.

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