The reaction 2 NO2(g) + F2(g) --> 2 NO2F(g) proceeds through the following mecha
ID: 1023746 • 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
We know the elementary steps can be used to write rate law correctly.
If the first step is rate determining step then the rate is directly proportional to the reactants concentrations in elementary steps.
NO2(g) + F2(g) --> NO2F(g) + F(g) (k1)
F(g) + NO2(g) -> NO2F(g) (k2)
Therefore
1)
Rate = k1 [NO2][F2]
So rate = k [NO2][F2]
Answer is option first is correct
2)
When the rate determining step is second then,
We write
Rate = k2 [F][NO2]
But we know since [F] depends on the concentration of NO2 and F2 therefore,
[F] = k1 [NO2][F2]
Lets plug this value of F into above rate equation
Rate = k1 k2 [NO2]2[F2]
K1*k2 = K (constant)
Therefore
Rate of this step would be
Rate = K [NO2]2[F2]
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