The following data is given for reaction between $A$ and $B$

$S.NO.$  $[A]$    $mol.L^{-1}$  $[B]$    $mol.L^{-1}$ $Rate$    $mol.L^{-1}\,sec^{-1}$
$I$ $1 \times 10^{-2}$ $2 \times 10^{-2}$ $2 \times 10^{-4}$
$II$ $2 \times 10^{-2}$ $2 \times 10^{-2}$ $4 \times 10^{-4}$
$III$ $2 \times 10^{-2}$ $4 \times 10^{-2}$ $8 \times 10^{-4}$ 

  Which of the following are correct statements -

$(a)$  Rate constant of the reaction $10^{-4}$

$(b)$  Rate law of the reaction is $k[A][B]$

$(c)$  Rate of reaction increases four times on doubling the concentration of both the reactant

  • A

    $a,\, b$ and $c$

  • B

    $a$ and $b$

  • C

    $b$ and $c$

  • D

    $c$ alone

Similar Questions

In the given reaction, ; $K_3 > K_2 > K_1$

then rate determining step will be

The following results have been obtained during the kinetic studies of the reaction:

$2 A+B \rightarrow C+D$

Experiment  $[ A ] / mol L ^{-1}$ $[ B ] / mol L ^{-1}$ Initial rate of formation of $D / mol \,L ^{-1} \,min ^{-1}$
$I$ $0.1$ $0.1$ $6.0 \times 10^{-3}$
$II$ $0.3$ $0.2$ $7.2 \times 10^{-2}$
$III$ $0.3$ $0.4$ $2.88 \times 10^{-1}$
$IV$ $0.4$ $0.1$ $2.40 \times 10^{-2}$

Determine the rate law and the rate constant for the reaction.

The rate of a gaseous reaction is given by the expression $K\,[A]\,[B]$. If the volume of the reaction vessel is suddenly reduced to $1/4^{th} $ of the initial volume, the reaction rate relating to original rate will be

Write differential rate expression of following reaction and give its order of reaction :

$2 N _{2} O _{5} \rightarrow 4 NO _{2}( g )+ O _{2}$

$C _{4} H _{9} Cl + OH ^{-} \rightarrow C _{4} H _{9} OH + Cl ^{-}$

For the reaction

$2H_2 + 2NO \to N_2 + 2H_2O$

the following mechanism has been proposed

$(i)$  $2NO \rightleftharpoons N_2O_2\,$   (fast)

$(ii)$ $N_2O_2 + H_2 \xrightarrow{{{k_2}}} N_2O + H_2O\,$    (slow)

$(iii)$  $N_2O + H_2 \to N_2 + H_2O\,$   (fast)

then what will be the rate law of this reaction ?