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
In a reaction,if the concentration of reactant $A$ is tripled,the rate of reaction becomes twenty-seven times. What is the order of the reaction?
Easy
View SolutionThe rate of a reaction is given by $r = K[x][y] / [OH^-]$. If the concentration of $[OH^-]$ is increased,the order of the reaction will be ........
Easy
View SolutionTotal order of reaction $X + Y \rightarrow XY$ is $3$. The order of reaction with respect to $X$ is $2$. State the differential rate equation for the reaction.
For the reaction $2A + B \rightarrow C + D$,select the correct rate law based on the following data:
$1$. $[A] = 0.1, [B] = 0.1, \text{Initial Rate} = 7.5 \times 10^{-3}$
$2$. $[A] = 0.3, [B] = 0.2, \text{Initial Rate} = 9.0 \times 10^{-2}$
$3$. $[A] = 0.3, [B] = 0.4, \text{Initial Rate} = 3.6 \times 10^{-1}$
$4$. $[A] = 0.4, [B] = 0.1, \text{Initial Rate} = 3.0 \times 10^{-2}$
$1$. $[A] = 0.1, [B] = 0.1, \text{Initial Rate} = 7.5 \times 10^{-3}$
$2$. $[A] = 0.3, [B] = 0.2, \text{Initial Rate} = 9.0 \times 10^{-2}$
$3$. $[A] = 0.3, [B] = 0.4, \text{Initial Rate} = 3.6 \times 10^{-1}$
$4$. $[A] = 0.4, [B] = 0.1, \text{Initial Rate} = 3.0 \times 10^{-2}$
Easy
View SolutionThe rate of the reaction $CCl_3CHO + NO \to CHCl_3 + NO + CO$ is given by $Rate = K [CCl_3CHO] [NO]$. If concentration is expressed in $mol \ L^{-1}$,the units of $K$ are:
Easy
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