The rate of reaction between two reactants $A$ and $B$ decreases by a factor of $4$ if the concentration of reactant $B$ is doubled. The order of this reaction with respect to reactant $B$ is

  • A
    $-1$
  • B
    $-2$
  • C
    $1$
  • D
    $2$

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Similar Questions

The decomposition of ozone in the upper atmosphere is catalyzed by nitric oxide. The mechanism of the reaction is as follows:
$2NO \rightleftharpoons N_2O + [O]$
$O_3 + [O] \to 2O_2$ (slow)
Determine the order of the reaction.

For the reaction $2 NO_{(g)} + Cl_{2_{(g)}} \rightarrow 2NOCl_{(g)}$,when the concentration of $Cl_2$ is doubled,the rate of the reaction becomes twice the original rate. When the concentration of $NO$ is doubled,the rate becomes four times the original rate. What is the overall order of the reaction?

Ozone decomposes into oxygen as follows:
$O_3 \rightleftharpoons O_2 + [O]$
$O_3 + [O] \to 2O_2$ (slow)
Determine the order of the reaction $2O_3 \to 3O_2$.

The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:
Experiment $[A] / mol \, L^{-1}$ $[B] / mol \, L^{-1}$ Initial rate / $mol \, L^{-1} \, min^{-1}$
$I$ $0.1$ $0.1$ $2.0 \times 10^{-2}$
$II$ $-$ $0.2$ $4.0 \times 10^{-2}$
$III$ $0.4$ $0.4$ $-$
$IV$ $-$ $0.2$ $2.0 \times 10^{-2}$

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For a reaction $A \longrightarrow P$,a plot between $[A]_0$ $Vs$ $\frac{1}{t_{1/2}}$ is a straight line having a positive slope. When the initial concentration is $1 \times 10^{-2} \ M$,its half-life period is found to be $20 \ min$. When the concentration of $A$ is $2 \times 10^{-2} \ M$,then the half-life will be (in $min$)?

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