The Proposed mechanism for the reaction A(g) + B(g) ---> D(g) + E(g) is as follows:
Step 1. B(g) ---> 2C(g) (fast)
Step 2. A(g) + C(g) ---> E(g)+ F(g) (slow)
Step 3. C(g) + F(g) ---> D(g) (fast)
The overall rate law for the reaction should be
A. Rate = k[B]
B. Rate = k[A][C]
C. Rate = k[A][B][C]
D. Rate = k[A][B]1/2
E. Rate = k[A]1/2[B]
The correct answer is D but I'm not sure how exactly to get to it. I'm lost in that the fast reactions aren't in equilibrium and I'm not sure what to do.
A(g) + B(g) ---> D(g) + E(g)
The rate law be r = k[A]m[B]n ----(1)
Where m , n are the order of the reactions with respect to A & B respectively.
The mechanism is
Step 1. B(g) ---> 2C(g) (fast)
Step 2. A(g) + C(g) ---> E(g)+ F(g) (slow)
Step 3. C(g) + F(g) ---> D(g) (fast)
The slowest step is the rate determining step, so step 2 is the rate determining step.
It involves 1 mole of A(g) so the order of the reaction with respect to A is 1 ---> m = 1
In the slowest step there is no involvement of B , but B can produces 2 moles of C in the step(1).
Then in the slowest step(2) , there involves only one mole of C , so the order with respect to B is 1/2.
(ie) n = 1/2
Plug the values of m & n in (i) we get r = k[A]1[B]1/2
Therefore option (D) is correct
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