For the reaction $2A + 2B \rightarrow 2C + D$,the rate law is expressed as $\text{rate} = k[A]^2[B]$. Calculate the rate constant if the rate of reaction is $0.24 \ mol \ dm^{-3} \ s^{-1}$ where $[A] = 0.5 \ M$ and $[B] = 0.2 \ M$.

In a multistep reaction,the overall rate of reaction is equal to the

$A_2 + B_2 \to 2AB$; $R.O.R = k[A_2]^a[B_2]^b$

Initial $[A_2]$	Initial $[B_2]$	$R.O.R. (r) \ M s^{-1}$
$0.2$	$0.2$	$0.04$
$0.1$	$0.4$	$0.04$
$0.2$	$0.4$	$0.08$

Order of reaction with respect to $A_2$ and $B_2$ are respectively:

$Zn + 2H^{+} \to Zn^{2+} + H_2$
The half-life period is independent of the concentration of zinc at constant $pH$. For the constant concentration of $Zn$,the rate becomes $100$ times when $pH$ is decreased from $3$ to $2$. Identify the correct statements $(pH = -\log [H^{+}])$:
$(A)$ $\frac{dx}{dt} = k[Zn]^0[H^{+}]^2$
$(B)$ $\frac{dx}{dt} = k[Zn][H^{+}]^2$
$(C)$ Rate is not affected if the concentration of zinc is made four times and that of $H^{+}$ ion is halved.
$(D)$ Rate becomes four times if the concentration of $H^{+}$ ion is doubled at constant $Zn$ concentration.

For the reaction $2 \ NOBr_{(g)} \rightarrow 2 \ NO_{(g)} + Br_{2_{(g)}}$,the rate law is $r = k[NOBr]^{2}$. If the rate constant is $1.62 \ M^{-1} \ s^{-1}$ and the concentration of $NOBr$ is $2.00 \times 10^{-3} \ M$,what is the rate of reaction?

The order of a reaction for an esterification process is . . . . . . .