Medium
The following are the rate constants of two different reactions. What is the overall order of each reaction?
$(a)$ $2.1 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$
$(b)$ $4.5 \times 10^{-3} \ min^{-1}$
$(a)$ $2.1 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$
$(b)$ $4.5 \times 10^{-3} \ min^{-1}$
(A) The order of a reaction can be determined from the units of the rate constant $(k)$.
The general unit for the rate constant is $(mol \ L^{-1})^{1-n} \ s^{-1}$,where $n$ is the order of the reaction.
For $(a)$: The unit is $mol \ L^{-1} \ s^{-1}$.
Comparing this with $(mol \ L^{-1})^{1-n} \ s^{-1}$,we get $1-n = 1$,which implies $n = 0$. Thus,it is a $0^{th}$ order reaction.
For $(b)$: The unit is $min^{-1}$ (or $s^{-1}$).
Comparing this with $(mol \ L^{-1})^{1-n} \ s^{-1}$,we get $1-n = 0$,which implies $n = 1$. Thus,it is a $1^{st}$ order reaction.
The general unit for the rate constant is $(mol \ L^{-1})^{1-n} \ s^{-1}$,where $n$ is the order of the reaction.
For $(a)$: The unit is $mol \ L^{-1} \ s^{-1}$.
Comparing this with $(mol \ L^{-1})^{1-n} \ s^{-1}$,we get $1-n = 1$,which implies $n = 0$. Thus,it is a $0^{th}$ order reaction.
For $(b)$: The unit is $min^{-1}$ (or $s^{-1}$).
Comparing this with $(mol \ L^{-1})^{1-n} \ s^{-1}$,we get $1-n = 0$,which implies $n = 1$. Thus,it is a $1^{st}$ order reaction.
Explore More
Similar Questions
If the reaction between $A$ and $B$ to give $C$ shows first-order kinetics in $A$ and second-order in $B$,the rate equation can be written as:
Easy
View Solution$2 \ NO_{(g)} + Cl_{2(g)} \rightleftharpoons 2 \ NOCl_{(g)}$
This reaction was studied at $-10^{\circ} C$ and the following data was obtained:
$Run$ $[NO]_{0}$ $[Cl_{2}]_{0}$ $r_{0}$ $1$ $0.10$ $0.10$ $0.18$ $2$ $0.10$ $0.20$ $0.35$ $3$ $0.20$ $0.20$ $1.40$
$[NO]_{0}$ and $[Cl_{2}]_{0}$ are the initial concentrations and $r_{0}$ is the initial reaction rate.
The overall order of the reaction is ..........
(Round off to the Nearest Integer).
This reaction was studied at $-10^{\circ} C$ and the following data was obtained:
| $Run$ | $[NO]_{0}$ | $[Cl_{2}]_{0}$ | $r_{0}$ |
| $1$ | $0.10$ | $0.10$ | $0.18$ |
| $2$ | $0.10$ | $0.20$ | $0.35$ |
| $3$ | $0.20$ | $0.20$ | $1.40$ |
$[NO]_{0}$ and $[Cl_{2}]_{0}$ are the initial concentrations and $r_{0}$ is the initial reaction rate.
The overall order of the reaction is ..........
(Round off to the Nearest Integer).
The rate constant for the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$ is $3.0 \times 10^{-5} \text{ s}^{-1}$. If the rate of reaction is $2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$, then the concentration of $N_2O_5$ in $\text{mol L}^{-1}$ is:
DifficultMHT CET 2026
View SolutionFor the inversion of cane sugar,the half-life is $500 \ min$ at all concentrations of sugar when $pH = 5$. However,when $pH = 6$,the half-life becomes $50 \ min$. Determine the rate law for the inversion of cane sugar.
Difficult
View SolutionIn the reaction,$P + Q \longrightarrow R + S$,the time taken for $75 \%$ reaction of $P$ is twice the time taken for $50 \%$ reaction of $P$. The concentration of $Q$ varies with reaction time as shown in the figure. The overall order of the reaction is
MediumIIT 2013
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