Question

Consider the following reaction. 4 A + B + 2 C →→ 4 D + E...

Consider the following reaction.

4 A + B + 2 C →→ 4 D + E

The following data was collected on the reaction.

Trial [A]0 [B]0 [C]0 Rate (M/s)
1 0.22 0.75 0.22 9.506
2 0.44 0.75 0.22 19.012
3 0.22 1.5 0.22 19.012
4 0.22 0.75 0.44 9.506

Two mechanisms are proposed:

Mechanism 1:

A + B →→ D + Y
Y + C →→ X
X + A + C →→ D + Z
Z + A →→ E

Mechanism 2:

A + C →→ D + Y
Y + B + C →→ X + D
X + A + C →→ E

Which mechanism best fits the data? Select an answer ( Mechanism 1, Mechanism 2 )

Which is most likely the rate-determining step? Select an answer( Step 1 ,Step 2, Step 3, Step 4 )

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Answer #1

rate law is

rate = K[A]x[B]y[C]z

from trial 1

rate = 9.506 = K *(0.22)x (0.75)y (0.22)z (1)

from trial 2

rate = 19.012 = K* (0.44)x *(0.75)x(0.22)z (2)

trial 3

rate = 19.012 = K *(0.22)x ( 1.5)y (0.22)z (3)

trial 4

rate = 9.506 = K*(0.22)x(0.75)y(0.44)z (4)

now, trial 1/ trial 4

(0.22)/(0.44)z = (9.506/9.506) = 1

or, Z = 0.

trial 1/trial 2

(0.22)x/(0.44)x = (9.506/19.012) = (1/ 2)

or, X = 1

trial 1/ trial 3

(0.75/0.15)y = (9.506/19.012) = (1/2)

or, Y = 1

then rate law is

rate = K*[ A] [B]

overall order = (1+1) = 2

so, rate of the reaction is independent on concentration of C.

In mechanism 2, in all the steps C is present. But in mechanism 1 , C is not present in all the steps.

Hence, mechanism 1 , best fits the data.

rate of the reaction depends on the slowest step.

By rate law the rate of the reaction depends only on the concentration of A and B, so rate-determining step is step 1.

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