(A) For reaction $-I$: The rate law is $-\frac{d[A]}{dt} = k_1[A]^0$. Since the order is $0$,the unit of $k_1$ is $\text{concentration} \times \text{t

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(A) For reaction $-I$: The rate law is $-\frac{d[A]}{dt} = k_1[A]^0$. Since the order is $0$,the unit of $k_1$ is $\text{concentration} \times \text{time}^{-1} = M \ sec^{-1}$.For reaction $-II$: The rate law is $-\frac{d[B]}{dt} = k_2[B]^1$. Since the order is $1$,the unit of $k_2$ is $\text{time}^{-1} = sec^{-1}$.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

Medium

Consider the following two reactions:
$A \to \text{Product}; -\frac{d[A]}{dt} = k_1[A]^0$
$B \to \text{Product}; -\frac{d[B]}{dt} = k_2[B]$
The units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as:

A
$M \ sec^{-1}, sec^{-1}$
B
$M \ sec^{-1}, M \ sec^{-1}$
C
$sec^{-1}, M^{-1} \ sec^{-1}$
D
$M \ sec^{-1}, M \ sec^{-1}$

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Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

The possible mechanism for the reaction $2NO + Br_2 \to 2NOBr$ is:
$NO + Br_2 \rightleftharpoons NOBr_2$ (Fast)
$NOBr_2 + NO \to 2NOBr$ (Slow)
The rate law expression is:

Medium

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Why can’t molecularity of any reaction be equal to zero?

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For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively. The concentration of $N_2O_5$ in $mol \ L^{-1}$ will be:

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$A$ bacterial infection in an internal wound grows as $N(t) = N_0 \exp(t)$,where the time $t$ is in hours. $A$ dose of antibiotic,taken orally,needs $1 \ hour$ to reach the wound. Once it reaches there,the bacterial population decreases as $\frac{dN}{dt} = -5N^2$. What will be the plot of $\frac{N_0}{N}$ vs. $t$ after $1 \ hour$?

DifficultJEE MAIN 2019

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For a reaction,$A + B \longrightarrow$ Product; the rate law is given by $r = k[A]^2[B]^{1/2}$. What is the order of the reaction?

Medium

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Consider the following two reactions:$A \to \text{Product}; -\frac{d[A]}{dt} = k_1[A]^0$$B \to \text{Product}; -\frac{d[B]}{dt} = k_2[B]$The units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as:

Similar Questions

The possible mechanism for the reaction $2NO + Br_2 \to 2NOBr$ is:$NO + Br_2 \rightleftharpoons NOBr_2$ (Fast)$NOBr_2 + NO \to 2NOBr$ (Slow)The rate law expression is:

Why can’t molecularity of any reaction be equal to zero?

For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively. The concentration of $N_2O_5$ in $mol \ L^{-1}$ will be:

For a reaction,$A + B \longrightarrow$ Product; the rate law is given by $r = k[A]^2[B]^{1/2}$. What is the order of the reaction?

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Consider the following two reactions:
$A \to \text{Product}; -\frac{d[A]}{dt} = k_1[A]^0$
$B \to \text{Product}; -\frac{d[B]}{dt} = k_2[B]$
The units of $k_1$ and $k_2$ are expressed in terms of molarity $(M)$ and time $(sec^{-1})$ as:

The possible mechanism for the reaction $2NO + Br_2 \to 2NOBr$ is:
$NO + Br_2 \rightleftharpoons NOBr_2$ (Fast)
$NOBr_2 + NO \to 2NOBr$ (Slow)
The rate law expression is: