The kinetic study of a reaction like vA right | vedclass

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(D) The rate law for the reaction is given by $r = k[A]^n$.Taking the square root on both sides, we get $\sqrt{r} = \sqrt{k} 	imes [A]^{n/2}$.The giv

Vedclass · Accepted Answer

$n=2, k=16.0 \ dm^3 \ mol^{-1} \ min^{-1}$

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(D) The rate law for the reaction is given by $r = k[A]^n$.Taking the square root on both sides, we get $\sqrt{r} = \sqrt{k} 	imes [A]^{n/2}$.The given graph is a plot of $\sqrt{r_0}$ versus $[A]_0$, which is a straight line passing through the origin.Comparing this with the equation $y = mx$, we have $y = \sqrt{r_0}$, $x = [A]_0$, and slope $m = \sqrt{k}$.From the graph, the slope is $4.0$.Therefore, $\sqrt{k} = 4.0$, which implies $k = (4.0)^2 = 16.0$.Also, comparing the powers of $[A]$, we have $n/2 = 1$, which gives $n = 2$.Thus, the order of the reaction is $n = 2$ and the rate constant is $k = 16.0 \ dm^3 \ mol^{-1} \ min^{-1}$.Hence, option $(d)$ is the correct answer.

The kinetic study of a reaction like $vA \rightarrow P$ at $300 \ K$ provides the following curve, where concentration is taken in $mol \ dm^{-3}$ and time in $min$. Identify the correct order $(n)$ and rate constant $(k)$.

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