(C) Let the rate law be $r = k[A]^x[B]^y$.From experiment $1$ and $2$,$[B]$ is constant. Comparing the rates:$\frac{8.4 \times 10^{-6}}{4.2 \times 10^

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(C) Let the rate law be $r = k[A]^x[B]^y$.From experiment $1$ and $2$,$[B]$ is constant. Comparing the rates:$\frac{8.4 \times 10^{-6}}{4.2 \times 10^{-6}} = (\frac{2.0}{1.0})^x \implies 2 = 2^x \implies x = 1$.From experiment $1$ and $3$,$[A]$ is constant. Comparing the rates:$\frac{5.6 \times 10^{-6}}{4.2 \times 10^{-6}} = (\frac{0.20}{0.15})^y \implies \frac{4}{3} = (\frac{4}{3})^y \implies y = 1$.Thus,the rate law is $r = k[A]^1[B]^1$ or $r = k[A][B]$.

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

Medium

$A$ study of chemical kinetics of the reaction $A + B \to$ Products,gave the following data at $25 \ ^oC$.

$Exp. \ No.$ $[A]$ $[B]$ $Rate$

$1.$ $1.0$ $0.15$ $4.2 \times 10^{-6}$

$2.$ $2.0$ $0.15$ $8.4 \times 10^{-6}$

$3.$ $1.0$ $0.20$ $5.6 \times 10^{-6}$

Find out the rate law.

A
$r = k[A]$
B
$r = k[B]$
C
$r = k[A][B]$
D
$r = k[A][B]^2$

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Class 12

Chemistry Questions

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is

Difficult

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The correct difference between $1^{st}$-order and $2^{nd}$-order reaction is that:

NEET 2018

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For the following graphs,choose the correct option regarding the order of reaction:

DifficultJEE MAIN 2021

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For the reaction $2A + B \rightarrow \text{Product}$,the rate constant $(K)$ is $2.5 \times 10^{-5} \text{ L mol}^{-1} \text{ s}^{-1}$ after $15 \text{ s}$,$2.60 \times 10^{-5} \text{ L mol}^{-1} \text{ s}^{-1}$ after $30 \text{ s}$,and $2.55 \times 10^{-5} \text{ L mol}^{-1} \text{ s}^{-1}$ after $50 \text{ s}$. What is the order of the reaction?

Medium

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If the concentration of reactant $A$ is increased by $10$ times,the rate of reaction increases $100$ times. What is the order of reaction if the rate law is $r = k[A]^x$?

MediumMHT CET 2025

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$Exp. \ No.$	$[A]$	$[B]$	$Rate$
$1.$	$1.0$	$0.15$	$4.2 \times 10^{-6}$
$2.$	$2.0$	$0.15$	$8.4 \times 10^{-6}$
$3.$	$1.0$	$0.20$	$5.6 \times 10^{-6}$

$A$ study of chemical kinetics of the reaction $A + B \to$ Products,gave the following data at $25 \ ^oC$. $Exp. \ No.$ $[A]$ $[B]$ $Rate$ $1.$ $1.0$ $0.15$ $4.2 \times 10^{-6}$ $2.$ $2.0$ $0.15$ $8.4 \times 10^{-6}$ $3.$ $1.0$ $0.20$ $5.6 \times 10^{-6}$ Find out the rate law.

Similar Questions

Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$takes place in two steps :$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$The order of the reaction is

The correct difference between $1^{st}$-order and $2^{nd}$-order reaction is that:

For the following graphs,choose the correct option regarding the order of reaction:

If the concentration of reactant $A$ is increased by $10$ times,the rate of reaction increases $100$ times. What is the order of reaction if the rate law is $r = k[A]^x$?

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$A$ study of chemical kinetics of the reaction $A + B \to$ Products,gave the following data at $25 \ ^oC$.

$Exp. \ No.$ $[A]$ $[B]$ $Rate$

$1.$ $1.0$ $0.15$ $4.2 \times 10^{-6}$

$2.$ $2.0$ $0.15$ $8.4 \times 10^{-6}$

$3.$ $1.0$ $0.20$ $5.6 \times 10^{-6}$

Find out the rate law.

Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is