(A) Let $R_1, R_2$,and $R_3$ be the rates of three reactions of first,second,and third-order,respectively,and $k$ be the rate constant for all three r

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(A) Let $R_1, R_2$,and $R_3$ be the rates of three reactions of first,second,and third-order,respectively,and $k$ be the rate constant for all three reactions.The rate laws are:$R_1 = k[A]^1$$R_2 = k[A]^2$$R_3 = k[A]^3$where $[A]$ is the concentration of reactant $A$ in moles per litre.Given that $[A] = 1$,we have:$R_1 = k(1)^1 = k$$R_2 = k(1)^2 = k$$R_3 = k(1)^3 = k$Therefore,$R_1 = R_2 = R_3$.

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

Medium

The rate constant is the same for $3$ reactions of order $I, II$,and $III$,respectively,with the unit of concentration being in moles per litre. If the concentration of the reactant is unity,the rates of reaction $R_1, R_2, R_3$ will be:

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A
$R_1 = R_2 = R_3$
B
$R_1 < R_2 < R_3$
C
$R_1 > R_2 > R_3$
D
$R_1 = R_2 \neq R_3$

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Class 12

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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For a reaction: $3X \to 4Y$,calculate the rate of reaction based on the following graph when the concentration of $X$ is $0.1 \ M$.

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For a reaction between substances $A$ and $B$,the rate law is given by: $\text{Rate} = K[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what will be the ratio of the new rate to the original rate?

Medium

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Calculate the order of the reaction with respect to $A$ and $B$ based on the following data:

$[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$

$0.05$ $0.05$ $1.2 \times 10^{-3}$

$0.10$ $0.05$ $2.4 \times 10^{-3}$

$0.05$ $0.10$ $1.2 \times 10^{-3}$

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The rate constant of esterification is given by $k = k^{\prime} [H_2O]$. If the rate constant $k = 2.0 \times 10^{-3} \ min^{-1}$,calculate $k^{\prime}$. (Assume $[H_2O] = 55.5 \ mol \ L^{-1}$)

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The rate of reaction between $A$ and $B$ increases by a factor of $100$ when the concentration of $A$ is increased $10$ folds. The order of reaction with respect to $A$ is

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$[A] \ (mol/L)$	$[B] \ (mol/L)$	Rate $(mol/L \cdot s)$
$0.05$	$0.05$	$1.2 \times 10^{-3}$
$0.10$	$0.05$	$2.4 \times 10^{-3}$
$0.05$	$0.10$	$1.2 \times 10^{-3}$

The rate constant is the same for $3$ reactions of order $I, II$,and $III$,respectively,with the unit of concentration being in moles per litre. If the concentration of the reactant is unity,the rates of reaction $R_1, R_2, R_3$ will be:

Similar Questions

For a reaction: $3X \to 4Y$,calculate the rate of reaction based on the following graph when the concentration of $X$ is $0.1 \ M$.

For a reaction between substances $A$ and $B$,the rate law is given by: $\text{Rate} = K[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what will be the ratio of the new rate to the original rate?

Calculate the order of the reaction with respect to $A$ and $B$ based on the following data: $[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$ $0.05$ $0.05$ $1.2 \times 10^{-3}$ $0.10$ $0.05$ $2.4 \times 10^{-3}$ $0.05$ $0.10$ $1.2 \times 10^{-3}$

The rate constant of esterification is given by $k = k^{\prime} [H_2O]$. If the rate constant $k = 2.0 \times 10^{-3} \ min^{-1}$,calculate $k^{\prime}$. (Assume $[H_2O] = 55.5 \ mol \ L^{-1}$)

The rate of reaction between $A$ and $B$ increases by a factor of $100$ when the concentration of $A$ is increased $10$ folds. The order of reaction with respect to $A$ is

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Calculate the order of the reaction with respect to $A$ and $B$ based on the following data:

$[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$

$0.05$ $0.05$ $1.2 \times 10^{-3}$

$0.10$ $0.05$ $2.4 \times 10^{-3}$

$0.05$ $0.10$ $1.2 \times 10^{-3}$