- #1
annnoyyying
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Homework Statement
Overall reaction: A + F = I + J (just take = as the equilibrium sign)
It is first order in both A and B.
The following mechanism has been proposed:
1. A + B = C + D
2. C + D = E + B (B is the catalyst)
3. E + F = G + H
4. G + H = I + J
These are all elementary steps and are all reversible.
Derive the mechanistic rate equation for the reaction.
Homework Equations
none, a matter of playing around with the equations above and making assumptions...
The Attempt at a Solution
The rate law (not mechanistic) should be r = k[A][F], according to the overall chemical equation.
C, D, E, G, H are all intermediates and their concentrations cannot be measured, so should not appear in the final mechanistic rate law. Steady state approximation can be applied to eliminate these when deriving the rate law. The concentration of B should not really change because it is consumed in 1 and regenerated in 2. 1 and 2 are in fast equilibrium.
The other thing I know is that F affects the rate of the reaction.
Is it possible to assume that the 3rd reaction as a rate determining, slow, forward step?
Thanks in advance