Question

For zero order and second order reactions units of rate constant k are respectively​

Answer 1

$\underline{ \large{ \bf{Units \: of \: Rate \: Constant \ (k)}}}$

$\boxed{ \bf{k = {(concentration)}^{(1 - n)} {time}^{ - 1} }}$

n → Order of reaction

SI units:

Concentration → mol/L

Time → s

For zero order reaction: (n = 0)

$\rm \leadsto k = {(mol \: {L}^{ - 1} )}^{(1 - 0)} {s}^{ - 1} \\ \\ \rm \leadsto k = mol \: {L}^{ - 1} \: {s}^{ - 1}$

For second order reaction: (n = 2)

$\rm \leadsto k = {(mol \: {L}^{ - 1} )}^{(1 - 2)} {s}^{ - 1} \\ \\ \rm \leadsto k = {(mol \: {L}^{ - 1} )}^{ - 1} {s}^{ - 1} \\ \\ \rm \leadsto k = {mol}^{ - 1} \: L \: {s}^{ - 1}$

Additional Information:

For first order reaction: (n = 1)

$\rm k = s^{-1}$

For third order reaction: (n = 3)

$\rm k = mol^{-2} \ L^2 \ s^{-1}$

Answer 2

Explanation:

AnswEr :

• Consider the following rate equation for an nth order reaction.

$\sf \: r \propto k [A]^n$

Expressing the equation in form of their SI units :

$\implies \sf mol.L {}^{ - 1} {s}^{ - 1} = k \times (mol \: {L}^{ - 1}) {}^{n} \\ \\ \implies \boxed{ \boxed{ \sf \: k_n = (mol {L}^{ - 1} ) {}^{1 - n} {s}^{ - 1} }}$

For a zero order reaction,

$\longrightarrow \sf \: k = (mol.L {}^{ - 1} ) {}^{1 - 0} {s}^{ - 1} \\ \\ \longrightarrow \underline{ \boxed{\sf \: k_0 = mol.L {}^{ - 1} {s}^{ - 1} }}$

For a zero order reaction,the SI unit of the Rate Constant is equal to that of Rate of Reaction.

For a second order reaction,

$\longrightarrow \sf \: k = (mol.L {}^{ - 1} ) {}^{1 - 2} {s}^{ - 1} \\ \\ \longrightarrow \sf k = (mol.L^{-1})^{-1} s^{-1} \\ \\ \longrightarrow \underline{ \boxed{\sf \: k_2 =L mol{}^{ - 1} {s}^{ - 1} }}$

Time taken for the completion of a reaction can be expressed in $\sf min^{-1}$ and $\sf hour^{-1}$ also.