The reaction 2 NO 2 (g) + F 2 (g) --> 2 NO 2 F(g) proceeds through the following
ID: 1023759 • Letter: T
Question
The reaction 2 NO2(g) + F2(g) --> 2 NO2F(g) proceeds through the following mechanism:
NO2(g) + F2(g) --> NO2F(g) + F(g)
F(g) + NO2(g) -> NO2F(g)
(a) The first step of this mechanism is rate-determining (slow). What is the rate law for this reaction
Rate = k [NO2] [F2]
Rate = k [NO2]2 [F2]
Rate = k [NO2] [F2]2
Rate = k [NO2]1/2
[F2] Rate = k [NO2] [F2]1/2
Rate = k [NO2]2
Rate = k [NO2]2 [F2]1/2
(b) What would the rate law be if the second step of this mechanism were rate-determining?
Rate = k [NO2] [F2]
Rate = k [NO2]2 [F2]
Rate = k [NO2] [F2]2
Rate = k [NO2]1/2 [F2]
Rate = k [NO2] [F2]1/2
Rate = k [NO2]2
Rate = k [NO2]2 [F2]1/2
Rate = k [NO2]
Explanation / Answer
mechanism
NO2(g) + F2(g) --> NO2F(g) + F(g) slow step
F(g) + NO2(g) -> NO2F(g)
Rate law
Rate = K[NO2][F2]
Rate = k [NO2] [F2]
b.Rate law of second step
1/2F2(g) + NO2(g) -> NO2F(g)
Rate = k [NO2] [F2]1/2
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