The mechanism of the reaction,
$2NO(g) + 2{H_2}(g) \to {N_2}(g) + 2{H_2}O(g)$ is :
Step $1$ : $2NO(g) + {H_2}(g)\xrightarrow{{slow}}{N_2} + {H_2}{O_2}$
Step : $2$ ${H_2}{O_2} + {H_2}\xrightarrow{{fast}}2{H_2}O$
Then the correct statement is
Rate $= k{\left[ {NO} \right]^2}\,{\left[ {{H_2}} \right]^2}$
Rate $= k \left[ {{H_2}{O_2}} \right]\,\left[ {{H_2}} \right]$
On doubling the concentration of $H_2$ , keeping the concentration, of $'NO'$ constant, the rate will become double
If the initial concentration of $H_2$ and $NO$ is $C_0$ and after time $'t'$ the concentration of $N_2$ is $x$ , then Rate $= K{\left( {{C_0} - 2x} \right)^2}$
The rate of the reaction becomes twice when the concentration of reactant becomes $8$ times then the order of the reaction is
Write differential rate expression of following reaction and give its order of reaction :
$5 B r^{-}+B r O_{3}^-+6 H^{+} \rightarrow 3 B r_{2}+3 H_{2} O$
Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
take 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
If doubling the initial concentration of reactant doubles $t_{1/2}$ of reaction, the order of reaction is
For the reaction $C{H_3}COOC{H_3} + {H_2}O\xrightarrow{{{H^ + }}}$ $C{H_3}COOH + C{H_3}OH$ The progress of the process of reaction is followed by