Which of the following are the correct repres | vedclass

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(B) For a zero order reaction: $1$. The integrated rate law is $[A] = [A]_0 - kt$. Thus,a plot of $[A]$ versus $t$ is a straight line with a negative

Vedclass · Accepted Answer

$(i), (ii), (iv)$

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(B) For a zero order reaction: $1$. The integrated rate law is $[A] = [A]_0 - kt$. Thus,a plot of $[A]$ versus $t$ is a straight line with a negative slope $(-k)$,which matches graph $(i)$. $2$. The half-life is $t_{1/2} = \frac{[A]_0}{2k}$. Thus,a plot of $t_{1/2}$ versus $[A]_0$ is a straight line passing through the origin with a slope of $\frac{1}{2k}$,which matches graph $(ii)$. $3$. The rate law is $	ext{Rate} = k[A]^0 = k$. The rate is independent of the reactant concentration. Thus,a plot of $	ext{Rate}$ versus $[A]$ is a horizontal line,which matches graph $(iv)$. Graph $(iii)$ shows $	ext{Rate} \propto [A]$,which is characteristic of a first order reaction. Therefore,the correct representations are $(i), (ii),$ and $(iv)$.

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

Which of the following are the correct representations of a zero order reaction,where $[A]$ represents the reactant concentration and $[A]_0$ represents the initial concentration?

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

The reaction is:

$A \rightarrow P$ is a zero order reaction. At $298 \ K$ the rate constant of the reaction is $1 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$. Initial concentration of '$A$' is $0.1 \ mol \ L^{-1}$. What is the concentration of '$A$' after $10 \ s$?

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