The rate law of the reaction $A + 2B \to \text{Product}$ is given by $\frac{d[B]}{dt} = k[B^2]$. If $A$ is taken in excess,the order of the reaction will be

$[A]_0 / \text{mol } L^{-1}$	$t_{1/2} / \text{min}$
$0.100$	$200$
$0.025$	$100$

For a given reaction $R \rightarrow P$,$t_{1/2}$ is related to $[A]_0$ as given in the table:
Given: $\log 2 = 0.30$
Which of the following is true?
$A.$ The order of the reaction is $1/2$.
$B.$ If $[A]_0$ is $1 \text{ M}$,then $t_{1/2}$ is $200 \sqrt{10} \text{ min}$.
$C.$ The order of the reaction changes to $1$ if the concentration of reactant changes from $0.100 \text{ M}$ to $0.500 \text{ M}$.
$D.$ $t_{1/2}$ is $800 \text{ min}$ for $[A]_0 = 1.6 \text{ M}$.
Choose the correct answer from the options given below:

Write the unit of the rate constant for the following reactions:
$1.$ $5/2$ order
$2.$ $n$ order

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.

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 $1/4$ of the initial volume,the reaction rate relating to the original rate will be

For the reaction $2NO + Cl_2 \rightarrow 2NOCl$,the following mechanism is proposed. What will be the rate law for the reaction? $NO + Cl_2 \rightleftharpoons NOCl_2$ (fast); $NOCl_2 + NO \rightarrow 2NOCl$ (slow)