(A) The rate law for the reaction is given by $r = k[A]^1[B]^1$.For the first experiment: $0.10 = k(20)(0.5)$ $\Rightarrow 0.10 = 10k$ $\Rightarrow k

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(A) The rate law for the reaction is given by $r = k[A]^1[B]^1$.For the first experiment: $0.10 = k(20)(0.5)$ $\Rightarrow 0.10 = 10k$ $\Rightarrow k = 0.01 \ L \ mol^{-1} \ s^{-1}$.For the second experiment: $0.40 = k(x)(0.5)$. Substituting $k = 0.01$: $0.40 = 0.01(x)(0.5)$ $\Rightarrow 0.40 = 0.005x$ $\Rightarrow x = 80 \ mol \ L^{-1}$.For the third experiment: $0.80 = k(40)(y)$. Substituting $k = 0.01$: $0.80 = 0.01(40)(y)$ $\Rightarrow 0.80 = 0.4y$ $\Rightarrow y = 2 \ mol \ L^{-1}$.Therefore,the values are $x = 80$ and $y = 2$.

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

Medium

For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.
Rate $(mol \ L^{-1} \ s^{-1})$ $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$
$0.10$ $20$ $0.5$
$0.40$ $x$ $0.5$
$0.80$ $40$ $y$

What is the value of $x$ and $y$?

Rate $(mol \ L^{-1} \ s^{-1})$	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$
$0.10$	$20$	$0.5$
$0.40$	$x$	$0.5$
$0.80$	$40$	$y$

JEE MAIN 2023 All JEE MAIN

A
$80$ and $2$
B
$40$ and $4$
C
$160$ and $4$
D
$80$ and $4$

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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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Consider the following data for the given reaction $2 HI_{(g)} \rightarrow H_{2_{(g)}} + I_{2_{(g)}}$. The order of the reaction is:

Experiment $1$ $2$ $3$

$[HI] \ (mol \ L^{-1})$ $0.005$ $0.01$ $0.02$

Rate $(mol \ L^{-1} \ s^{-1})$ $7.5 \times 10^{-4}$ $3.0 \times 10^{-3}$ $1.2 \times 10^{-2}$

Experiment	$1$	$2$	$3$
$[HI] \ (mol \ L^{-1})$	$0.005$	$0.01$	$0.02$
Rate $(mol \ L^{-1} \ s^{-1})$	$7.5 \times 10^{-4}$	$3.0 \times 10^{-3}$	$1.2 \times 10^{-2}$

DifficultJEE MAIN 2024

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The alkaline hydrolysis of an ester is a ........ reaction.

Difficult

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The rate of reaction is determined by the slow step of the reaction. This step is called:

Easy

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The following data are for the decomposition of ammonium nitrate in aqueous solution. The order of the reaction is:
$N_2$ volume in $cc$ $(V_t)$ $6.25$ $9.50$ $11.42$ $13.65$ $35.05$ $(V_{\infty})$
Time $(t)$ in minutes $10$ $15$ $20$ $25$ $\infty$

Medium

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For the decomposition of a compound $AB$ at $600 \ K$,the following data were obtained:
$[AB] \ (mol \ dm^{-3})$ Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$
$0.20$ $2.75 \times 10^{-8}$
$0.40$ $11.0 \times 10^{-8}$
$0.60$ $24.75 \times 10^{-8}$

The order for the decomposition of $AB$ is:

DifficultKCET 2009

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$N_2$ volume in $cc$ $(V_t)$	$6.25$	$9.50$	$11.42$	$13.65$	$35.05$ $(V_{\infty})$
Time $(t)$ in minutes	$10$	$15$	$20$	$25$	$\infty$

$[AB] \ (mol \ dm^{-3})$	Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$
$0.20$	$2.75 \times 10^{-8}$
$0.40$	$11.0 \times 10^{-8}$
$0.60$	$24.75 \times 10^{-8}$

For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.Rate $(mol \ L^{-1} \ s^{-1})$$[A]$ $(mol \ L^{-1})$$[B]$ $(mol \ L^{-1})$$0.10$$20$$0.5$$0.40$$x$$0.5$$0.80$$40$$y$What is the value of $x$ and $y$?

Similar Questions

Consider the following data for the given reaction $2 HI_{(g)} \rightarrow H_{2_{(g)}} + I_{2_{(g)}}$. The order of the reaction is: Experiment $1$ $2$ $3$ $[HI] \ (mol \ L^{-1})$ $0.005$ $0.01$ $0.02$ Rate $(mol \ L^{-1} \ s^{-1})$ $7.5 \times 10^{-4}$ $3.0 \times 10^{-3}$ $1.2 \times 10^{-2}$

The alkaline hydrolysis of an ester is a ........ reaction.

The rate of reaction is determined by the slow step of the reaction. This step is called:

The following data are for the decomposition of ammonium nitrate in aqueous solution. The order of the reaction is:$N_2$ volume in $cc$ $(V_t)$$6.25$$9.50$$11.42$$13.65$$35.05$ $(V_{\infty})$Time $(t)$ in minutes$10$$15$$20$$25$$\infty$

For the decomposition of a compound $AB$ at $600 \ K$,the following data were obtained: $[AB] \ (mol \ dm^{-3})$Rate of decomposition of $AB \ (mol \ dm^{-3} \ s^{-1})$$0.20$$2.75 \times 10^{-8}$$0.40$$11.0 \times 10^{-8}$$0.60$$24.75 \times 10^{-8}$ The order for the decomposition of $AB$ is:

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For a chemical reaction $A + B \rightarrow \text{Product}$,the order is $1$ with respect to $A$ and $B$.
Rate $(mol \ L^{-1} \ s^{-1})$ $[A]$ $(mol \ L^{-1})$ $[B]$ $(mol \ L^{-1})$
$0.10$ $20$ $0.5$
$0.40$ $x$ $0.5$
$0.80$ $40$ $y$

What is the value of $x$ and $y$?

Consider the following data for the given reaction $2 HI_{(g)} \rightarrow H_{2_{(g)}} + I_{2_{(g)}}$. The order of the reaction is:

Experiment $1$ $2$ $3$

$[HI] \ (mol \ L^{-1})$ $0.005$ $0.01$ $0.02$

Rate $(mol \ L^{-1} \ s^{-1})$ $7.5 \times 10^{-4}$ $3.0 \times 10^{-3}$ $1.2 \times 10^{-2}$

The following data are for the decomposition of ammonium nitrate in aqueous solution. The order of the reaction is:
$N_2$ volume in $cc$ $(V_t)$ $6.25$ $9.50$ $11.42$ $13.65$ $35.05$ $(V_{\infty})$
Time $(t)$ in minutes $10$ $15$ $20$ $25$ $\infty$