Question

If units of rate constant is mol- L sec- then what is the order 'n' ?​

Answer 1

By rate law, the rate constant is given by,

$\tt{\longrightarrow r=k[R]^n}$

where $\tt{r}$ is rate of the reaction, $\tt{k}$ is rate constant, $\tt{[R]}$ is concentration of reactant, and $\tt{n}$ is order of the reaction, so this equation represents the rate of an $\tt{n^{th}}$ order reaction.

• Unit of $\tt{r}$ is $\tt{mol\,L^{-1}\,s^{-1}.}$

• Unit of $\tt{k}$ is given as $\tt{mol^{-1}\,L\,s^{-1}.}$

• Unit of $\tt{[R]}$ is $\tt{mol\,L^{-1}.}$

Taking units of each term in the equation,

$\tt{\longrightarrow mol\,L^{-1}\,s^{-1}=mol^{-1}\,L\,s^{-1}\left(mol\,L^{-1}\right)^n}$

$\tt{\longrightarrow mol\,L^{-1}\,s^{-1}=mol^{n-1}\,L^{1-n}\,s^{-1}}$

Equating power of $\tt{mol,}$

$\tt{\longrightarrow n-1=1}$

$\tt{\longrightarrow\underline{\underline{n=2}}}$

Hence the order of the reaction is 2.

Answer 2

By rate law, the rate constant is given by,

\tt{\longrightarrow r=k[R]^n}⟶r=k[R]

n

where \tt{r}r is rate of the reaction, \tt{k}k is rate constant, \tt{[R]}[R] is concentration of reactant, and \tt{n}n is order of the reaction, so this equation represents the rate of an \tt{n^{th}}n

th

order reaction.

Unit of \tt{r}r is \tt{mol\,L^{-1}\,s^{-1}.}molL

−1

s

−1

.

Unit of \tt{k}k is given as \tt{mol^{-1}\,L\,s^{-1}.}mol

−1

Ls

−1

.

Unit of \tt{[R]}[R] is \tt{mol\,L^{-1}.}molL

−1

.

Taking units of each term in the equation,

\tt{\longrightarrow mol\,L^{-1}\,s^{-1}=mol^{-1}\,L\,s^{-1}\left(mol\,L^{-1}\right)^n}⟶molL

−1

s

−1

=mol

−1

Ls

−1

(molL

−1

)

n

\tt{\longrightarrow mol\,L^{-1}\,s^{-1}=mol^{n-1}\,L^{1-n}\,s^{-1}}⟶molL

−1

s

−1

=mol

n−1

L

1−n

s

−1

Equating power of \tt{mol,}mol,

\tt{\longrightarrow n-1=1}⟶n−1=1

\tt{\longrightarrow\underline{\underline{n=2}}}⟶

n=2

Hence the order of the reaction is 2.