Determination of order of reaction by initial concern method



For the reaction,

            2NO + Cl2            →       2NOCl

Following datas were obtained

Expt no.

Initial concern (mol litre-1)

Initial rate (mol l-1S-1)

0

1

2

3

0.010                         cl2

0.010                        0.010

0.010                        0.020

0.020                        0.020

1.2 × 10-4

2.4 × 10-4

9.6 × 10-4

 

Determine the order of the reaction with respect to NO, Cl2 and overall order. Write the rate law expression and determine the value of rate constant.

 

Soln:

Let the order w.r.t. NO and Cl2 be in and n. The rate low expression will be,

Rate = k [NO]m [Cl]n

From the given experimental data,

1.2 × 10-4 =      k [0.01]m [0.01]n          ———-(i)

2.4 × 10-4  =     k [0.010]m [0.02]n        ———-(ii)

9.6 × 10-4 =      k [0.02]m [0.02]n          ———-(iiii)

 

Dividing eqn (ii) by (i) we get,

order of reaction

Dividing eqn (iii) by (ii), we get,

order of reaction

order w.r.t. NO = 2

order w.r.t. Cl2 = 1

 

Overall order of r × n = 2 + 1 = 3

 

Then, the rate law expression,

Rate = k [NO]2 [Cl2]

 Putting the value of m and n in equation (i), we get,

order of reaction

 

The experimental data for the reaction

 2A  + B2          →      2AB is as follows,

 

 Exp no, [A]                 [B]                   Rate

            Mole l-1            moll-1               Mole l-1 s-1

 1.         0.50                 0.50                 1.6 × 10-4

2.         0.50                 1.00                 3.2 × 10-4

3.         1.00                 1.00                 3.2 × 10-4

 

Determine the order of the r × n wrt A and B and overall r × n. Write the rate law expression and determine the value of rate of constant.

 Solution:

Let the order w.r.t. A and B be m and n. The rate law expression will be,

Rate =  k  [A]m  [B]n

From the given experimental data,

1.6 × 10-4 =      k [0.50]m [0.50]n…………………(i)

3.2 × 10-4  =     k [0.50]m [1]n……………………(ii)

3.2 × 10-4  =     k [1]m [1]n……………………..…(iii)

 Dividing eqn (ii) by (i) we get,

order of reaction

Again, Dividing iii by ii we gt,

order of reaction

Order w.r.t. A = 0

Order w.r.t. B = 1

Over all Order of the r × n = 1 + 0 = 1

 

Then the rate law expression.

Rate = [ A ]0 [ B]1

Again,

Putting the value of m and n in eqn (i) we get

Or, 1.6 × 10-4 = k [0.50]0 [0.50]1

Or, 1.6 × 10-4 = k [0.50]1

 k = 3.2 × 10-4 Sec-1

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