Difficult
For the following rate law,determine the unit of the rate constant: Rate $= -\frac{d[R]}{dt} = k[A]^{\frac{1}{2}}[B]^{2}$
The total order of reaction $n = \frac{1}{2} + 2 = \frac{5}{2} = 2.5$.
Rate $= k[A]^{\frac{1}{2}}[B]^{2} = k[Concentration]^{\frac{5}{2}}$.
$\therefore k = \frac{\text{Rate}}{[Concentration]^{\frac{5}{2}}}$.
$\therefore$ Unit of $k = \frac{\text{mol } L^{-1} s^{-1}}{(\text{mol } L^{-1})^{\frac{5}{2}}} = (\text{mol } L^{-1})^{1 - 2.5} s^{-1} = (\text{mol } L^{-1})^{-1.5} s^{-1}$.
$= \text{mol}^{-1.5} L^{1.5} s^{-1}$ or $L^{1.5} \text{mol}^{-1.5} s^{-1}$.
Rate $= k[A]^{\frac{1}{2}}[B]^{2} = k[Concentration]^{\frac{5}{2}}$.
$\therefore k = \frac{\text{Rate}}{[Concentration]^{\frac{5}{2}}}$.
$\therefore$ Unit of $k = \frac{\text{mol } L^{-1} s^{-1}}{(\text{mol } L^{-1})^{\frac{5}{2}}} = (\text{mol } L^{-1})^{1 - 2.5} s^{-1} = (\text{mol } L^{-1})^{-1.5} s^{-1}$.
$= \text{mol}^{-1.5} L^{1.5} s^{-1}$ or $L^{1.5} \text{mol}^{-1.5} s^{-1}$.
Explore More
Similar Questions
The rate constant value for a reaction is $1.75 \times 10^2 \ L^2 \ mol^{-2} \ sec^{-1}$. The half-life period $t_{1/2} \propto$ . . . . . . .
Consider the following reaction $A + B \rightarrow C$.
The time taken for $A$ to become $1/4$ of its initial concentration is twice the time taken to become $1/2$ of the same. Also,when the change of concentration of $B$ is plotted against time,the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is . . . . .
The time taken for $A$ to become $1/4$ of its initial concentration is twice the time taken to become $1/2$ of the same. Also,when the change of concentration of $B$ is plotted against time,the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is . . . . .
The reaction $2NO + O_2 \rightarrow 2NO_2$ is an example of which order of reaction?
Easy
View SolutionFor a chemical reaction $A \to B$,it is found that the rate of reaction doubles when the concentration of $A$ is increased four times. The order of the reaction with respect to $A$ is:
MediumAIIMS 1997
View SolutionWhich function of $[X]$ plotted against time will give a straight line for a second order reaction? $X \to \text{Product}$
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo