(D) For the given reaction,the rate law expression is given by: $\text{Rate} = k [N_2O_5]$.Given that the rate constant $k = 3 \times 10^{-5} \, s^{-1

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(D) For the given reaction,the rate law expression is given by: $\text{Rate} = k [N_2O_5]$.Given that the rate constant $k = 3 \times 10^{-5} \, s^{-1}$ and the rate of reaction is $2.40 \times 10^{-5} \, mol \, L^{-1} \, s^{-1}$.Substituting these values into the rate law equation:$2.40 \times 10^{-5} = (3 \times 10^{-5}) \times [N_2O_5]$Solving for $[N_2O_5]$:$[N_2O_5] = \frac{2.40 \times 10^{-5}}{3 \times 10^{-5}} = 0.8 \, mol \, L^{-1}$.

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

Medium

The rate constant for the reaction,$2N_2O_5 \to 4NO_2 + O_2$ is $3 \times 10^{-5} \, s^{-1}$. If the rate is $2.40 \times 10^{-5} \, mol \, L^{-1} \, s^{-1}$,then the concentration of $N_2O_5$ (in $mol \, L^{-1}$) is

IIT 2000 All IIT

A
$1.4$
B
$1.2$
C
$0.04$
D
$0.8$

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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 the reaction,$2N_2O_5 \to 4NO_2 + O_2$ is $3 \times 10^{-5} \, s^{-1}$. If the rate is $2.40 \times 10^{-5} \, mol \, L^{-1} \, s^{-1}$,then the concentration of $N_2O_5$ (in $mol \, L^{-1}$) is

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If the relation between half-life time $(t_{1/2})$ and initial concentration $[R]_0$ is given by $t_{1/2} \propto \frac{1}{[R]_0^{n-1}}$,what is the order of the reaction?

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