CHEM1909
2005-N-14
November 2005
 It has been proposed that the reaction
Cl2(g) + CHCl3(g)  HCl(g) + CCl4(g)
proceeds by the following mechanism:
Cl2(g)
k1
k–1
Cl(g) + CHCl3(g)
CCl3(g) + Cl(g)
2Cl(g)
k2
k3
(fast equilibrium)
HCl(g) + CCl3(g)
CCl4(g)
(slow)
(fast)
Derive the rate expression for this mechanism.
As the second reaction is slow, it is rate determining. From the mechanism, the
rate of this step is given by:
rate = k2[Cl(g)][CHCl3(g)]
As Cl is a highly reactive intermediate, its concentration cannot be included in
the rate equation which is to be experimental tested. As the first step is fast, the
equilibrium between Cl2(g) and Cl(g) will be set up rapidly and maintained for
most of the reaction. For an equilibrium,
rate forward reaction = rate backward reaction
k1[Cl2(g)] = k-1[Cl(g)]2
or [Cl(g)]2 =
k1
[Cl 2 (g)]
k-1
Hence,
rate = k2[Cl(g)][CHCl3(g)] = k2 ×
= k2
k1
[Cl 2 (g)] × [CHCl3(g)
k1
k1
[CHCl 3 (g)][Cl(g)]1 2 = k[CHCl 3 (g)][Cl(g)]1 2
k-1
k1
where k = k2
k1
Answer: rate = k[CHCl 3 (g)][Cl(g)]1 2
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.
Marks
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