The Proposed mechanism for the reaction A (g) + B (g) ---> D (g) + E (g) is as f

Question

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.

Explanation / Answer

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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