Writing the rate law implied by a simple mechanism Suppose the first step is muc
ID: 717958 • Letter: W
Question
Writing the rate law implied by a simple mechanism
Suppose the first step is much faster than the second
Step Elementary reaction rate constant 1 NO2(g)+F2(g)=NO2F(g)+F(g) k1 2 F(g)+NO2(g)=NO2F(g) k2 Write the balanced chemical equation for the overall chemical reaction: Write the experimentally- observable rate law for the overall chemical reaction. Note: your answer should not contain the concentrations of any intermediates Express the rate constant k for the overall chemical reaction in terms of k, k2, and (if necessary) the rate constants k-1 and k-2 for the reverse of the two elementary reactions in the mechanism.Explanation / Answer
The given reactions are
NO2(g) + F2(g) = NO2F(g) + F(g)
F(g) + NO2(g) = NO2F(g)
Part a
Overall balanced reaction by adding the above equations
2NO2(g) + F2(g) = 2NO2F(g)
Part b
Second reaction is slow reaction
Rate law is determined by the slowest step.
Rate law = k [F] [NO2]
But F is an intermediate.
d[F] /dt = 0 = k1[NO2] [F2] - k-1[NO2F][F] - k2[NO2][F] + K-2 [NO2F]
k1[NO2] [F2] - k-1[NO2F][F] - k2[NO2][F] + K-2 [NO2F] = 0
- k-1[NO2F][F] - k2[NO2][F] = - K-2 [NO2F] - k1[NO2] [F2]
k-1[NO2F][F] + k2[NO2][F] = K-2 [NO2F] + k1[NO2] [F2]
[F] (k-1[NO2F] + k2[NO2] ) = K-2 [NO2F] + k1[NO2] [F2]
[F] = ( K-2 [NO2F] + k1[NO2] [F2] ) / (k-1[NO2F] + k2[NO2] )
Rate law = k [F] [NO2]
Rate law = k [NO2] ( K-2 [NO2F] + k1[NO2] [F2] ) / (k-1[NO2F] + k2[NO2] )
Part c
Rate law = kk1/k2 [NO2] ( K-2/k1 [NO2F] + [NO2] [F2] ) / (k-1 /k2[NO2F] + [NO2] )
= k' [NO2] ( k'' [NO2F] + [NO2] [F2] ) / (k''' [NO2F] + [NO2] )
= k'k" [NO2] [NO2F] + k' [NO2]2 [F2] ) / (k''' [NO2F] + [NO2] )
= [ ( k'k" + k') / k"'] ( [NO2] [NO2F] + [NO2]2 [F2] ) / ([NO2F] + [NO2] )
Rate constant K = [(kk1/K2) (k-2/k1) + (kk1/k2)] / (k-1/k2)
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