(A) The given reaction is $2 \ NOBr_{(g)} \rightarrow 2 \ NO_{(g)} + Br_{2_{(g)}}$.Given: $k = 1.62 \ M^{-1} \ s^{-1}$ and $[NOBr] = 2.00 \times 10^{-

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$6.48 \times 10^{-6} \ M \ s^{-1}$

(A) The given reaction is $2 \ NOBr_{(g)} \rightarrow 2 \ NO_{(g)} + Br_{2_{(g)}}$.Given: $k = 1.62 \ M^{-1} \ s^{-1}$ and $[NOBr] = 2.00 \times 10^{-3} \ M$.The rate law is $r = k[NOBr]^{2}$.Substituting the values:$r = 1.62 \ M^{-1} \ s^{-1} \times (2.00 \times 10^{-3} \ M)^{2}$$r = 1.62 \times 4.00 \times 10^{-6} \ M \ s^{-1}$$r = 6.48 \times 10^{-6} \ M \ s^{-1}$.

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

Medium

For the reaction $2 \ NOBr_{(g)} \rightarrow 2 \ NO_{(g)} + Br_{2_{(g)}}$,the rate law is $r = k[NOBr]^{2}$. If the rate constant is $1.62 \ M^{-1} \ s^{-1}$ and the concentration of $NOBr$ is $2.00 \times 10^{-3} \ M$,what is the rate of reaction?

MHT CET 2020 All MHT CET

A
$6.48 \times 10^{-6} \ M \ s^{-1}$
B
$4.05 \times 10^{-5} \ M \ s^{-1}$
C
$2.46 \times 10^{-6} \ M \ s^{-1}$
D
$5.24 \times 10^{-6} \ M \ s^{-1}$

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Rate law , Rate constant , Order of Reaction and Molecularity

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For the reaction $2 \ NOBr_{(g)} \rightarrow 2 \ NO_{(g)} + Br_{2_{(g)}}$,the rate law is $r = k[NOBr]^{2}$. If the rate constant is $1.62 \ M^{-1} \ s^{-1}$ and the concentration of $NOBr$ is $2.00 \times 10^{-3} \ M$,what is the rate of reaction?

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For a reaction $A \to B$,the rate of reaction quadrupled when the concentration of $A$ is doubled. The rate expression of the reaction is $r = K[A]^n$. The value of $n$ is

The mechanism for the reaction is given below:$2P + Q \to S + T$$P + Q \to R + S$ (slow)$P + R \to T$ (fast)The rate law expression for the reaction is:

For the reaction $A + B \rightarrow \text{Product}$,if the concentration of $A$ is doubled,the reaction rate doubles. When the concentration of $B$ is doubled,the reaction rate remains unchanged. What is the overall order of the reaction?

The rate of the reaction: $2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$\frac{d[NO_2]}{dt} = k'[N_2O_5]$ ; $\frac{d[O_2]}{dt} = k''[N_2O_5]$The relationship between $k$ and $k'$ and between $k$ and $k''$ are:

The rate law for the reaction $A + B \rightarrow \text{product}$ is $\text{rate} = k[A][B]$. When will the rate of reaction increase by a factor of $2$?

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The mechanism for the reaction is given below:
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$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$ ; $\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are: