(A) For the reaction,$A + 2B \rightarrow$ Products.Since the half-life $(t_{1/2})$ remains constant when the concentration of $B$ is increased,the rea

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(A) For the reaction,$A + 2B \rightarrow$ Products.Since the half-life $(t_{1/2})$ remains constant when the concentration of $B$ is increased,the reaction is of $1^{st}$ order with respect to $B$.Since the rate remains constant when the concentration of $A$ is doubled,the reaction is of $0^{th}$ order with respect to $A$.The rate law expression is: $\text{Rate} = k[A]^0[B]^1$.The overall order of the reaction is $0 + 1 = 1$.The unit of the rate constant $(k)$ for a $1^{st}$ order reaction is $s^{-1}$.

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

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For a reaction,$A + 2B \rightarrow$ Products,when the concentration of $B$ alone is increased,the half-life remains the same. If the concentration of $A$ alone is doubled,the rate remains the same. The unit of the rate constant for the reaction is:

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A
$s^{-1}$
B
$L \ mol^{-1} \ s^{-1}$
C
$mol \ L^{-1} \ s^{-1}$
D
$atm^{-1}$

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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

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The rate law for the reaction $A^{+} + B^{+} + C \longrightarrow \text{Product}$ is expressed as $\text{Rate} = k[A]^{2}[B]^{1}[C]^{0}$. What is the overall order of the reaction?

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The data for the reaction $A + B \to C$ is given below. The rate law corresponding to the above data is:

$Exp.$ $[A]_0$ $[B]_0$ Initial rate

$(1)$ $0.012$ $0.035$ $0.10$

$(2)$ $0.024$ $0.070$ $0.80$

$(3)$ $0.024$ $0.035$ $0.10$

$(4)$ $0.012$ $0.070$ $0.80$

$Exp.$	$[A]_0$	$[B]_0$	Initial rate
$(1)$	$0.012$	$0.035$	$0.10$
$(2)$	$0.024$	$0.070$	$0.80$
$(3)$	$0.024$	$0.035$	$0.10$
$(4)$	$0.012$	$0.070$	$0.80$

MediumAIPMT 1994

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Why is the hydrolysis of an ester slow at the beginning and fast after some time?

Medium

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The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?

Experiment $[S]$ $[Nu]$ Rate

$1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$

Experiment	$[S]$	$[Nu]$	Rate
$1$	$0.1$	$0.1$	$2.2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4.4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$4.4 \times 10^{-3}$

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The rate of reaction $A + 2B \to 3C$ becomes $72$ times when the concentration of $A$ is tripled and the concentration of $B$ is doubled. The order of reaction with respect to $A$ and $B$ respectively is:

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For a reaction,$A + 2B \rightarrow$ Products,when the concentration of $B$ alone is increased,the half-life remains the same. If the concentration of $A$ alone is doubled,the rate remains the same. The unit of the rate constant for the reaction is:

Similar Questions

The rate law for the reaction $A^{+} + B^{+} + C \longrightarrow \text{Product}$ is expressed as $\text{Rate} = k[A]^{2}[B]^{1}[C]^{0}$. What is the overall order of the reaction?

The data for the reaction $A + B \to C$ is given below. The rate law corresponding to the above data is: $Exp.$ $[A]_0$ $[B]_0$ Initial rate $(1)$ $0.012$ $0.035$ $0.10$ $(2)$ $0.024$ $0.070$ $0.80$ $(3)$ $0.024$ $0.035$ $0.10$ $(4)$ $0.012$ $0.070$ $0.80$

Why is the hydrolysis of an ester slow at the beginning and fast after some time?

The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur? Experiment $[S]$ $[Nu]$ Rate $1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$ $2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$ $3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$

The rate of reaction $A + 2B \to 3C$ becomes $72$ times when the concentration of $A$ is tripled and the concentration of $B$ is doubled. The order of reaction with respect to $A$ and $B$ respectively is:

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The data for the reaction $A + B \to C$ is given below. The rate law corresponding to the above data is:

$Exp.$ $[A]_0$ $[B]_0$ Initial rate

$(1)$ $0.012$ $0.035$ $0.10$

$(2)$ $0.024$ $0.070$ $0.80$

$(3)$ $0.024$ $0.035$ $0.10$

$(4)$ $0.012$ $0.070$ $0.80$

The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?

Experiment $[S]$ $[Nu]$ Rate

$1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$