The kinetic study of a reaction like vA right | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 give

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $(k)$.

Similar Questions

For a zero order reaction,will the molecularity be equal to zero? Explain.

Assertion : The kinetics of the reaction $mA + nB + pC \to m'X + n'Y + p'Z$ obey the rate expression as $\frac{dX}{dt} = k[A]^m[B]^n$.
Reason : The rate of the reaction does not depend upon the concentration of $C$.

Which of the following determines the rate of a multistep reaction?

For a reaction $r = k[A][B]^2$,if the concentration of $A$ is doubled,the rate of reaction:

Vedclass Test Series

Exam Paper Generator

Online Exam Module