For a reaction of order $\mathrm{n}$, the unit of the rate constant is :

  • [JEE MAIN 2021]
  • A

    $\mathrm{mol}^{1-\mathrm{n}} \mathrm{L}^{1-\mathrm{n}} \mathrm{s}^{-1}$

  • B

    $\mathrm{mol}^{1-\mathrm{n}} \mathrm{L}^{\mathrm{n}-1} \mathrm{~s}^{-1}$

  • C

    $\mathrm{mol}^{1-\mathrm{n}} \mathrm{L}^{1-\mathrm{n}} \mathrm{s}$

  • D

    $\mathrm{mol}^{1-\mathrm{n}} \mathrm{L}^{2 \mathrm{n}} \mathrm{s}^{-1}$

Similar Questions

The rate constant for the reaction $2N_2O_5 \to  4NO_2 + O_2$ is $3.0\times10^{-5}\, sec^{-1}$. If rate is $2.40\times10^{-5}\, M\, sec^{-1}$, then the concentration of $N_2O_5$ (in $M$) is ?

A study of chemical kinetics of the reaction $A + B \to$  Products, gave the following data at $25\,^oC$.

Exp. No. [A] [B] Rate
$1.$ $1.0$ $0.15$ $4.2 × 10^{-6}$
$2.$ $2.0$ $0.15$ $8.4 × 10^{-6}$
$3.$ $1.0$ $0.20$ $5.6 × 10^{-6}$

Find out rate law

For the reaction $A + B \to $ products, what will be the order of reaction with respect to $A$ and $B$ ?

   Exp.    $[A]\,(mol\,L^{-1})$   $[B]\,(mol\,L^{-1})$   Initial rate    $(mol\,L^{-1}\,s^{-1})$ 
   $1.$  $2.5\times 10^{-4}$  $3\times 10^{-5}$  $5\times 10^{-4}$
   $2.$   $5\times 10^{-4}$  $6\times 10^{-5}$  $4\times 10^{-3}$
   $3.$   $1\times 10^{-3}$  $6\times 10^{-5}$  $1.6\times 10^{-2}$

For a reaction $\mathrm{A} \xrightarrow{\mathrm{K}_4} \mathrm{~B} \xrightarrow{\mathrm{K}_2} \mathrm{C}$

If the rate of formation of $B$ is set to be zero then the concentration of $B$ is given by :

  • [JEE MAIN 2024]

Why can’t molecularity of any reaction be equal to zero ?