166. The rate law for the gas phase reaction of chlorform with chlorine.
CHCl₃(g) + Cl₂(g) → CCl ₄(g) + HCl(g)
is given by rate = k[CHCl₃] [Cl₂]^1/2. How would the rate of reaction vary when (a) the concentration of CHCl₃ is doubled (b) the concentration of Cl₂ is doubled. What is the effect of each of these two changes on the rate constant.

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The rate law for the gas phase reaction of chlorform with chlorine.
CHCl₃(g) + Cl₂(g) → CCl ₄(g) + HCl(g)
 the rate is given by 

 rate₁ =k[CHCl₃] [Cl₂]^1/2 .....1

if the concentration of CHCl₃ is doubled
then 

rate₂ =k[CHCl₃]² [Cl₂]^1/2 .....2
divide 1by 2 we get

$$\begin{gathered} \frac{rat{e}_1 =\mathrm{k}\left[CHC{l}_3\right] [C{l}_2{]}^{1/2} .}{rat{e}_2 =\mathrm{k}[CHC{l}_3{]}^2 [C{l}_2{]}^{1/2} } \\ rate become double \\ \frac{rat{e}_1}{rat{e}_2}= \frac{1}{2} \\ 2rat{e}_1= rat{e}_2 \\ thus rate 1 become double \end{gathered}$$

if the concentration of Cl₂  is doubled
then .
rate₃ =k[CHCl₃] [Cl₂]^2/2........3

divide 1 by 3 
we get 

$$\begin{gathered} \frac{rat{e}_1 =\mathrm{k}\left[CHC{l}_3\right] [C{l}_2{]}^{1/2} .}{rat{e}_3 =\mathrm{k}\left[CHC{l}_3\right] \left[C{l}_2\right]} \\ rate become double \\ \frac{rat{e}_1}{rate3}= \frac{1}{2} \\ 2rat{e}_1= rat{e}_2 \\ thus rate 1 become double \end{gathered}$$

thus the rate reaction is double in both case.