Question

For a gas phase reaction ${\mathrm{Cl}}_2+{\mathrm{CHCl}}_3\to \mathrm{HCl}$ $+{\mathrm{CCl}}_4$, the rate law is given as $\mathrm{r}=\mathrm{K}{\left[{\mathrm{Cl}}_2\right]}^{1/2}$ $\left[{\mathrm{CHCl}}_3\right]$. Explain how the rate of reaction varies when the concentration of chlorine is doubled. Give units of rate constant.

Answer 1

1. For the given reaction, $$\begin{gathered} {\mathrm{Cl}}_2\left(\mathrm{g}\right)+{\mathrm{CHCl}}_3\left(\mathrm{aq}\right)\to \mathrm{HCl}\left(\mathrm{aq}\right)+{\mathrm{CCl}}_4\left(\mathrm{aq}\right) \\ \mathrm{Chlorine}\mathrm{trichloromethane}\mathrm{hydrochlroic}\mathrm{acid}\mathrm{carbon}{\mathrm{tetrachloride}}^{} \end{gathered}$$
   Rate of this reaction is : $r=\mathrm{K}{\left[{\mathrm{Cl}}_2\right]}^{1/2}\left[{\mathrm{CHCl}}_3\right]$
2. If the concentration of chlorine is double,
   $r_1=\mathrm{K}{\left[2{\mathrm{Cl}}_2\right]}^{1/2}\left[{\mathrm{CHCl}}_3\right]=\sqrt{2}\mathrm{K}{\left[{\mathrm{Cl}}_2\right]}^{1/2}\left[{\mathrm{CHCl}}_3\right]$
   $r_1=\sqrt{2}\times r$
3. Hence, the rate of the reaction $\left(r_1\right)$ becomes$\sqrt{2}$ times of the initial rate(r).
4. Now, the order of this reaction is 1.5.
5. Therefore, the unit of rate constant is ${\mathrm{mol}}^{0.5}.{\mathrm{L}}^{-0.5}.{\mathrm{s}}^{-1}$.