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(C) For a zero-order reaction $A \longrightarrow 3 B$,the rate law is $[A] = [A_0] - kt$.From the table,at $t = 0$,$[A_0] = 0.1 \ mol \ L^{-1}$.At $t

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$1.98 	imes 10^{-1}$

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(C) For a zero-order reaction $A \longrightarrow 3 B$,the rate law is $[A] = [A_0] - kt$.From the table,at $t = 0$,$[A_0] = 0.1 \ mol \ L^{-1}$.At $t = 15 \ s$,$[A] = 0.05 \ mol \ L^{-1}$.Substituting these values: $0.05 = 0.1 - k(15)$,which gives $k = 0.05 / 15 = 1/300 \ mol \ L^{-1} \ s^{-1}$.At $t = 20 \ s$,the concentration of $A$ remaining is $[A] = 0.1 - (1/300) 	imes 20 = 0.1 - 0.0667 = 0.0333 \ mol \ L^{-1}$.The amount of $A$ reacted is $0.1 - 0.0333 = 0.0667 \ mol \ L^{-1}$.According to the stoichiometry,$1 \ mol$ of $A$ produces $3 \ mol$ of $B$.Therefore,concentration of $B = 3 	imes 0.0667 = 0.2001 \ mol \ L^{-1} \approx 1.98 	imes 10^{-1} \ mol \ L^{-1}$.

Time $(s)$	Concentration of the reactant $(mol \ L^{-1})$
$0$	$0.1$
$15$	$0.05$
$20$	$0.1 - x$

Time $(sec)$	Rate $(mole \ litre^{-1} \ sec^{-1})$
$0$	$2.8 \times 10^{-2}$
$10$	$2.78 \times 10^{-2}$
$20$	$2.81 \times 10^{-2}$
$30$	$2.79 \times 10^{-2}$

For the reaction $A \longrightarrow 3 B$,the rate is given by $rate = k[A]^0$. Based on the data provided in the table,what is the concentration of the product $B$ (in $mol \ L^{-1}$) after $20 \ s$?
Time $(s)$ Concentration of the reactant $(mol \ L^{-1})$
$0$ $0.1$
$15$ $0.05$
$20$ $0.1 - x$

Similar Questions

The rates of a certain reaction $(dc/dt)$ at different times are as follows:
Time $(sec)$ Rate $(mole \ litre^{-1} \ sec^{-1})$
$0$ $2.8 \times 10^{-2}$
$10$ $2.78 \times 10^{-2}$
$20$ $2.81 \times 10^{-2}$
$30$ $2.79 \times 10^{-2}$

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$II. \ t_{1/2} = \frac{[R]_0}{4k}$

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For the reaction $A \longrightarrow 3 B$,the rate is given by $rate = k[A]^0$. Based on the data provided in the table,what is the concentration of the product $B$ (in $mol \ L^{-1}$) after $20 \ s$?Time $(s)$Concentration of the reactant $(mol \ L^{-1})$$0$$0.1$$15$$0.05$$20$$0.1 - x$