If the half-life of a reaction is halved when the initial concentration of the reactant is doubled,what is the order of the reaction?

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $0$

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The following results have been obtained during the kinetic studies of the reaction: $2 \ NO + 2 \ H_2 \longrightarrow N_2 + 2 \ H_2O$
Expt$\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$$[NO] \ (mol \ L^{-1})$$[H_2] \ (mol \ L^{-1})$
$1$$4.8 \times 10^{-5}$$1 \times 10^{-2}$$1 \times 10^{-3}$
$2$$43.2 \times 10^{-5}$$3 \times 10^{-2}$$1 \times 10^{-3}$
$3$$86.4 \times 10^{-5}$$3 \times 10^{-2}$$2 \times 10^{-3}$

The rate of a reaction depends upon the

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

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The rate law for the reaction $A + B \rightarrow P$ is given by $\text{rate} = k[A]^2[B]$. The rate constant of the reaction at $300 \text{ K}$ is $6.0 \text{ M}^{-2} \text{s}^{-1}$. Calculate the rate of the reaction when $[A] = 1 \text{ M}$ and $[B] = 0.2 \text{ M}$. (in $\text{ M s}^{-1}$)

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 $2O_3 \to 3O_2$.

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