For the reaction,

2NO + Cl_{2} → 2NOCl

Following datas were obtained

Expt no. |
Initial concern (mol litre |
Initial rate (mol l |

0 1 2 3 |
0.010 cl 0.010 0.010 0.010 0.020 0.020 0.020 |
– 1.2 × 10 2.4 × 10 9.6 × 10 |

Determine the order of the reaction with respect to NO, Cl_{2} and overall order. Write the rate law expression and determine the value of rate constant.

Soln:

Let the order w.r.t. NO and Cl_{2} 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. Cl_{2} = 1

Overall order of r × n = 2 + 1 = 3

Then, the rate law expression,

Rate = k [NO]^{2} [Cl_{2}]

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

**The experimental data for the reaction**

2A + B_{2} → 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,

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}