Rate law for a reaction is $r=k[A]^2[B]$. If rate constant is $6.25 \ mol^{-2} \ dm^6 \ s^{-1}$,what is the rate of reaction when $[A]=1 \ mol \ dm^{-3}$ and $[B]=0.2 \ mol \ dm^{-3}$?

The rate of a certain reaction is given by, $Rate = k[H^{+}]^n$. The rate increases $100$ times when the $pH$ changes from $3$ to $1$. The order $(n)$ of the reaction is:

If the rate law of a reaction $n A \rightarrow B$ is expressed as Rate $= -\frac{1}{n} \frac{d[A]}{dt} = \frac{d[B]}{dt} = k[A]^{x}$,the unit of rate constant $k$ will be:

During the kinetic study of the reaction $2A + B \rightarrow C + D$,the following results were obtained:

Experiment	$[A] \ (M), [B] \ (M)$ and Initial rate of formation of $D$
$i. \ [A]=0.1, [B]=0.1$	$6.0 \times 10^{-3} \ M \ s^{-1}$
$ii. \ [A]=0.3, [B]=0.2$	$7.2 \times 10^{-2} \ M \ s^{-1}$
$iii. \ [A]=0.3, [B]=0.4$	$2.88 \times 10^{-1} \ M \ s^{-1}$
$iv. \ [A]=0.4, [B]=0.1$	$2.40 \times 10^{-2} \ M \ s^{-1}$

Based on the above data,the overall order of the reaction is:

The decomposition of dimethyl ether leads to the formation of $CH_4$,$H_2$ and $CO$ and the reaction rate is given by
Rate $= k [CH_3OCH_3]^{3/2}$
The rate of reaction is followed by increase in pressure in a closed vessel,so the rate can also be expressed in terms of the partial pressure of dimethyl ether,i.e.
Rate $= k (p_{CH_3OCH_3})^{3/2}$
If the pressure is measured in $bar$ and time in $minutes$,then what are the units of rate and rate constants?

$A \rightarrow$ products ($1^{st}$ order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions. Run $1 \Rightarrow 100 \ mL$ of $10 \ M$ solution of reactant $A$. Run $2 \Rightarrow 200 \ mL$ of $10 \ M$ solution of reactant $A$. Run $3 \Rightarrow 100 \ mL$ of $10 \ M$ solution of reactant $A + 100 \ mL$ of $H_2O$ added. The correct variation of rate of reaction is: