Question

The following mechanism has been proposed for the gas-phase reaction of chloroform (CHCl3) and chlorine:
Step 1: Cl2(g)⇌2Cl(g)(fast)
Step 2: Cl(g)+CHCl3(g)→HCl(g)+CCl3(g)(slow)
Step 3: Cl(g)+CCl3(g)→CCl4(fast)
what is the overall rate constant?

Answer 1

The slow reaction is the one that is going to limit the speed of the reaction. Because they are elementary steps the rate of the reaction is going to follow the kinetics of this slow step. But if you express the rate law in function of the reactants of the slow reaction you will be including intermediate and not the reactants of the overall reaction. You have to replace the intermediate concentration in function of the reactant of the overall reaction. Since the first step reaches equilibrium the rate of the forward is equal to the rate of the reverse reaction. \[R = k_{1} [Cl _{2}] = k _{2}[Cl ]^{2}\] then we can express the intermedia concentration of [Cl] as \[[Cl] = \sqrt{\frac{ k _{1} }{ k _{2} } [Cl _{2}]} = \left( \frac{ k _{1} }{ k _{2} } \right)^{\frac{ 1 }{ 2 }}[Cl _{2}]^{\frac{ 1 }{ 2 }}\]
if we replace this in the rate law for the Step 2 and we combine the k1, k2, and k in one new k, we can express the rate law for the overall reaction as
\[R = k [Cl _{2}]^{\frac{ 1 }{ 2 }} [CHCl _{3}]\]

Step 1: Cl2(g)⇌2Cl(g)(fast)
Step 2: Cl(g)+CHCl3(g)→HCl(g)+CCl3(g)(slow) R= k [Cl] [CHCl3]
Step 3: Cl(g)+CCl3(g)→CCl4(fast)

Answer 2

Isn't the Cl-Cl bond in Cl2 meant to be broken by UV light to produce two Cl free radicals? And then one of the Cl free-radicals reacts with the carbon trichloride to produce HCl and a CCl3 free-radical? And in then the other Cl free-radical reacts with the CCl3 free-radical to produce carbon tetrachloride?