The reaction $3ClO^{-} \rightarrow ClO_{3}^{-} + 2Cl^{-}$ occurs in the following two steps:
$(i)$ $ClO^{-} + ClO^{-} \xrightarrow{K_{1}} ClO_{2}^{-} + Cl^{-}$ (Slow step)
$(ii)$ $ClO_{2}^{-} + ClO^{-} \xrightarrow{K_{2}} ClO_{3}^{-} + Cl^{-}$ (Fast step)
Then the rate of the given reaction is equal to . . . . . . .

For a hypothetical reaction $A + B \rightarrow C$,the following data is provided from three different experiments:
$1$. $[A] = 0.01 \ M$,$[B] = 0.01 \ M$ - Rate of reaction $= 1.0 \times 10^{-4} \ M \ s^{-1}$.
$2$. $[A] = 0.01 \ M$,$[B] = 0.03 \ M$ - Rate of reaction $= 9.0 \times 10^{-4} \ M \ s^{-1}$.
$3$. $[A] = 0.03 \ M$,$[B] = 0.03 \ M$ - Rate of reaction $= 2.70 \times 10^{-3} \ M \ s^{-1}$.
Determine the rate law.

If in a certain reaction,two different reactants take part,then:

For the reaction $A + B \longrightarrow \text{product}$,the rate law equation is $\text{rate} = k[A]^2[B]$. If the rate of reaction is $0.22 \ mol \ L^{-1} \ s^{-1}$,calculate the rate constant $k$. Given: $[A] = 1 \ mol \ L^{-1}, [B] = 0.25 \ mol \ L^{-1}$.

The following graph shows the relationship between $(a-x)^{-1}$ and time $t$ for a second-order reaction. If $\theta = \tan^{-1}(1/2)$ and $OA = 2 \ L \ mol^{-1}$,then the rate at the start of the reaction will be (in $mol \ L^{-1} \ min^{-1}$):

For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively. The concentration of $N_2O_5$ in $mol \ L^{-1}$ will be: