Question 15.7: Reaction Mechanisms II The gas-phase reaction of chlorine wi...

Two questions must be answered. First, does the mechanism give the correct overall stoichiometry? Adding the three steps does yield the correct balanced equation:

  Cl_{2}  (g)\xrightleftharpoons[]{} 2  Cl  (g)

Cl (g)  +  CHCl_{3} (g)  →  HCL (g) +  CCl_{3} (g)

\underline{CCl_{3} (g)  + Cl  (g)  →  CCl_{4} (g)}    

Cl_{2} (g)   +  \cancel{Cl  }(g) +  CHCl_{3} (g)   + \cancel{ CCl_{3}} (g)  +   \cancel{Cl  }(g)   →   \cancel{2  Cl }(g)  + HCL  (g) +  \cancel{CCl_{3}} (g)   +  CCl_{4} (g)

Overall reaction:                          Cl_{2}   (g)  +  CHCl_{3} (g)    →  HCL  (g)  +  CCl_{4} (g)

Second, does the mechanism agree with the observed rate law? Since the overall reaction rate is determined by the rate of the slowest step,

Overall rate = rate of second step  =   k_{2} [Cl] [CHCl_{3}]

Since the chlorine atom is an intermediate, we must find a way to eliminate [Cl] in the rate law. This can be done by recognizing that since the first step is at equilibrium, its forward and reverse rates are equal:

k_{1} [Cl_{2}]  =  k_{- 1} [Cl]²

Solving for  [Cl]² gives

[Cl]²  =  \frac{k_{1} [Cl_{2}] }{ k_{- 1}}

Taking the square root of both sides yields

[Cl] =   (\frac{k_{1}}{k_{- 1}})^{1 / 2} [Cl_{2}] ^{1 / 2}

and

Rate = k_{2}[Cl] [CHCl_{3}] =  k_{2} (\frac{k_{1}}{k_{- 1}})^{1 / 2} [Cl_{2}] ^{1 / 2} [CHCl_{3}]  =  k  [Cl_{2}] ^{1 / 2} [CHCl_{3}]

where                     k =     k_{2} (\frac{k_{1}}{k_{- 1}}  )^{1 / 2}

The rate law derived from the mechanism agrees with the experimentally observed rate law. This mechanism satisfies the two requirements and thus is an acceptable mechanism.