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Chemical Kinetics

Question
CBSEENCH12006285

The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

exp.

 

[A]/
mol L–1

[B]/M

Initial rate of formation
of D/mol L–1 min–1

I

0.1

0.1


6.0 x 10–3

II

0.3

0.2

7.2 x 10–2

III

0.3

0.4

2.88 x 10–1

IV

0.4

0.1

2.40 x 10–2


Determine the rate law and the rate constant for the reaction.

Solution
Let rate law of reaction be,

rate = kAx By

where x and y are order of reaction w.r.t. A and B respectively.

From experiments I and IV, we can write
6.0 x 10–3 = k[0.1][0.1]y ...(i)
2.4 x 10–2 = k[0.4][0.1]y ...(ii)

Dividing (ii) by (i), we can write

               2.4×10-26.0×10-3 = 0.40.1x = 4x

or                 4=4x      or      x = 1.

   From experiments I and III, we can write
                   7.2×10-2 = k0.3x 0.2y              ...(iii)2.88×10-1 = k0.3x 0.4y            ...(iv)

Dividing (iv) by (iii), we can write

                2.88×10-17.2×10-1 = 0.40.2y = 2y
or                        4=2y
or                        y=2.
               Rate = kA B2

Order of reaction w.r.t. A = 1
Order of reaction w.r.t. B = 2
Overall order of reaction = 1 + 2 = 3.
Substituting the values of initial rate of formation of experiment (I) (equ. V) we can write 6.0 x 103 mol L–1 min–1.
= k[0.1 mol L–1] [0.1 mol L–1]2

or      
                 k=6.0×10-3M min-11×10-3M3   = 6 M-2 s-1.