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

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

Dividing eqn (ii) by (i) we get, Again, Dividing iii by ii we gt, 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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