The reaction Cl 2 (g) + CHCl 3 (g) --> CCl 4 (g) + HCl(g) proceeds through the f

Question

The reaction Cl2(g) + CHCl3(g) --> CCl4(g) + HCl(g) proceeds through the following mechanism:

Cl2(g) --> 2 Cl(g)
Cl(g) + CHCl3(g) --> HCl(g) + CCl3(g)
Cl(g) + CCl3(g) --> CCl4(g)

(a) The second step of this mechanism is rate-determining (slow). What is the rate law for this reaction?

Rate = k [Cl2] [CHCl3]Rate = k [Cl2]2 [CHCl3]    Rate = k [Cl2] [CHCl3]2Rate = k [Cl2]1/2 [CHCl3]Rate = k [Cl2] [CHCl3]1/2Rate = k [Cl2]2Rate = k [Cl2]2 [CHCl3]1/2

(b) What would the rate law be if the first step of this mechanism were rate-determining?

Rate = k [Cl2] [CHCl3]Rate = k [Cl2]2 [CHCl3]    Rate = k [Cl2] [CHCl3]2Rate = k [Cl2]1/2 [CHCl3]Rate = k [Cl2] [CHCl3]1/2Rate = k [Cl2]2Rate = k [Cl2]2 [CHCl3]1/2Rate = k [Cl2]

Explanation / Answer

SLOW IS ALWAYS IS CALLED AS RATE DETERMINING STEP

a) RATE = K[Cl(g)]^1[CHCl3(g)]^1

               0r

Rate = k [Cl2]1/2 [CHCl3]

b) Rate = k [Cl2]

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