Question
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.
Solution
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
if the concentration of Cl2 is doubled
then .
rate3 =k[CHCl3] [Cl2]2/2........3
divide 1 by 3
we get
thus the rate reaction is double in both case.
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
if the concentration of Cl2 is doubled
then .
rate3 =k[CHCl3] [Cl2]2/2........3
divide 1 by 3
we get
thus the rate reaction is double in both case.
