(B) The rate law for the reaction is given by: $\text{Rate} = k[HI]^n$.Using the data from the table:For experiment $1$: $8.80 \times 10^{-10} = k(0.0

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(B) The rate law for the reaction is given by: $\text{Rate} = k[HI]^n$.Using the data from the table:For experiment $1$: $8.80 \times 10^{-10} = k(0.0500)^n$ (Equation $1$)For experiment $2$: $3.52 \times 10^{-9} = k(0.1000)^n$ (Equation $2$)Dividing Equation $2$ by Equation $1$:$\frac{3.52 \times 10^{-9}}{8.80 \times 10^{-10}} = \left(\frac{0.1000}{0.0500}\right)^n$$4 = (2)^n$Since $2^2 = 4$,we find that $n = 2$.Thus,the order of the reaction with respect to $HI$ is $2.0$.

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

Medium

Consider the following gas-phase reaction.
$2HI_{(g)} \longrightarrow H_{2(g)} + I_{2(g)}$
and the following experimental data obtained at $555 \ K$. What is the order of the reaction with respect to $HI_{(g)}$?

$[HI] \ (M)$ Rate $(M \ s^{-1})$

$0.0500$ $8.80 \times 10^{-10}$

$0.1000$ $3.52 \times 10^{-9}$

$0.1500$ $7.92 \times 10^{-9}$

A
$1.0$
B
$2.0$
C
$2.5$
D
$3.0$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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The possible mechanism for the reaction $2NO + Br_2 \to 2NOBr$ is:
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$[HI] \ (M)$	Rate $(M \ s^{-1})$
$0.0500$	$8.80 \times 10^{-10}$
$0.1000$	$3.52 \times 10^{-9}$
$0.1500$	$7.92 \times 10^{-9}$

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$A$ reaction between two different reactants cannot be a .....

The rate of reaction,$A + B + C \longrightarrow P$ is given by$r = K[A]^{1/2} [B]^{1/2} [C]^{1/4}$The order of reaction is

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