Question

The rate law for the reaction ${\mathbf{Cl}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{CHCl}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{HCl}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{CCl}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$is$\mathbf{Rate}\mathbf{=}\mathbf{k}{\left[{\mathrm{Cl}}_2\right]}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\left[{\text{CHCl}}_3\right]$. What are the units for k assuming time in seconds?

Short answer

The unit of k for the reaction is ${\mathrm{L}}^{1/2}{\mathrm{mol}}^{-1/2}{\mathrm{s}}^{-1}$.

Step-by-step solution

Define the rate laws.

We have two types of rate laws:

The rate of reaction a reaction depend on concentrations is the differential rate law. While the concentrations of species in the reaction depend on time is the integrated rate law.

Determine the unit for k.

The order of the reaction is determined by adding the order of reactants i.e.,

$\begin{array}{rcl}\mathrm{n}& =& {\mathrm{n}}_{{\mathrm{Cl}}_2}+{\mathrm{n}}_{{\mathrm{CHCl}}_3}\\ & =& 1+\frac{1}{2}\\ & =& \frac{3}{2}\end{array}$

Now the unit of the reaction:

$\begin{array}{c}\left[\mathrm{k}\right]={\left(\frac{\mathrm{mol}}{\mathrm{L}}\right)}^{1-\frac{3}{2}}{\mathrm{s}}^{-1}\\ ={\mathrm{L}}^{1/2}{\mathrm{mol}}^{-1/2}{\mathrm{s}}^{-1}\end{array}$