(C) For a second order reaction,the rate law is given by $R = k[A]^2$.Given:$k = 1.62 \ M^{-1} \ s^{-1}$$[A] = 2 \times 10^{-3} \ M$Substituting the v

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(C) For a second order reaction,the rate law is given by $R = k[A]^2$.Given:$k = 1.62 \ M^{-1} \ s^{-1}$$[A] = 2 \times 10^{-3} \ M$Substituting the values into the rate law:$R = 1.62 \times (2 \times 10^{-3})^2$$R = 1.62 \times 4 \times 10^{-6}$$R = 6.48 \times 10^{-6} \ M \ s^{-1}$

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

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The rate constant for a second order reaction,$A \rightarrow \text{Product}$ is $1.62 \ M^{-1} \ s^{-1}$. What will be the rate of reaction when concentration of reactant is $2 \times 10^{-3} \ M$ (in $M \ s^{-1}$)?

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A
$3.24 \times 10^{-3}$
B
$3.24 \times 10^{-6}$
C
$6.48 \times 10^{-6}$
D
$2 \times 10^{-3}$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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The rate constant for a second order reaction,$A \rightarrow \text{Product}$ is $1.62 \ M^{-1} \ s^{-1}$. What will be the rate of reaction when concentration of reactant is $2 \times 10^{-3} \ M$ (in $M \ s^{-1}$)?

Similar Questions

Consider the plots for the types of reaction $nA \to B + C$. These plots respectively correspond to the reaction orders:

For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate equation can be expressed in two ways $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$ and $+\frac{d[NO_2]}{dt} = k'[N_2O_5]$. $k$ and $k'$ are related as:

For a reaction $A + B \rightarrow \text{Product}$,the rate of reaction doubles when the concentration of $A$ is doubled. The rate becomes four times when the concentrations of both $A$ and $B$ are doubled. What is the order of the reaction with respect to $A$ and $B$ respectively?

For the reaction $2N_2O_5 \to 4NO_2 + O_2$,the rate of reaction 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 $N_2O_5$ at that time will be:

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