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
$a,\, b$ and $c$
$a$ and $b$
$b$ and $c$
$c$ alone
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 ?