Question

For the reaction \[\ce{2NO_{(g)} + C2_{(g)} -> 2NOCl_{(g)}}\], following data were obtained. 

Experiment	Initial conc. of NO (mol L−1)	Initial conc. of Cl2 (mol L−1)	Initial rate (mol L−1 s−1)
1	0.010	0.020	2.40 × 10−4
2	0.030	0.020	2.16 × 10−3
3	0.030	0.040	4.32 × 10−3

Determine the orders with respect to NO and Cl₂, the rate law and the rate constant.

Answer 1

The given reaction is \[\ce{2NO_{(g)} + C2_{(g)} -> 2NOCl_{(g)}}\] 

And experimental data for different initial concentrations and reaction rates.

Let the rate law be 

Rate = k[NO]^m[Cl₂]ⁿ

We’ll find m, n, and then calculate k.

Compare Experiments 1 & 2:

[NO] changes from 0.010 to 0.030 (× 3)

[Cl₂] is constant at 0.020

Rate increases from 2.40 × 10⁻⁴ to 2.16 × 10⁻³ (× 9)

`(2.16 xx 10^-3)/(2.40 xx 10^-4)`

= `(0.030/0.010)^m`

⇒ 9 = 3m

⇒ m = 2

So, order with respect to NO = 2

Compare experiments 2 & 3:

[Cl₂] changes from 0.020 to 0.040 (× 2)

[NO] is constant at 0.030

Rate changes from 2.16 × 10⁻³ to 4.32 × 10⁻³ (× 2)

`(4.32 xx 10^-3)/(2.16 xx 10^-3) = (0.040/0.020)^n`

⇒ 2 = 2ⁿ

n = 1

So order with respect to Cl₂ = 1

Rate = k[NO]²[Cl₂]

Use experiment 1:

Rate = 2.40 × 10⁻⁴

[NO] = 0.010, [Cl₂] = 0.020

2.40 × 10⁻⁴ = k(0.010)²(0.020)

= k × 2.0 × 10⁻⁶

`k = (2.40 xx 10^-4)/(2.0 xx 10^-6)`

= 120