Question

The relative rates of the reaction A + B ? AB in vessels a

Answer 1

A + B    AB

Let the rate law of the equation is -

r = k[A]^a[B]^b[C]^c

Relative rates in vessel a-d are 1:2:1:2

(a). Let the rates invessel a-d are x, 2x, x, 2x

For vessel a -

x = k[3]^a[3]^b[1]^c ..............................(1)

For vessel b -

2x = k[6]^a[3]^b[1]^c   ............................(2)

For vessel c -

x = k[3]^a[6]^b[1]^c   .............................(3)

For vessel c -

2x = k[3]^a[3]^b[2]^c   ................................(4)

Dividing equation 1 by 2 -

1/2 = [3]^a/[6]^a

1/2 = [1/2]^a

a = 1

Dividing equation 1 by 3 -

1 = [3]^b/[6]^b

1 = [1/2]^b

b = 0

Dividing equation 1 by 4 -

1/2 = [1]^c/[2]^c

1/2 = [1/2]^c

c = 1

Therefore, order of reaction in -

A = 1

B = 0

C = 1

(b). rate law is given as -

r = k[A]¹[B]⁰[C]¹

r = k[A][C]

(c). Mechanism that agrees with rate law :

A + B    C fast step

A + C AB slow step

As the slow step is the rate determining step,

rate of reaction r = k[A][C]

(d). C doesn't appear in the reaction because C is a reaction intermediate, it doesn't appear in the reaction but the formation of product depends on its concentration.