Question

The data given below is for the reaction between [NO] and [Cl₂] to form NOCl at 25°C.

S. No.	Conc. of [NO] mol L−1	Conc. of [Cl2] mol L−1	Rate: mol L−1 sec−1
1.	2.0	2.0	2.0 × 10−3
2.	2.0	6.0	6.0 × 10−3
3.	6.0	2.0	1.8 × 10−2

Answer the following questions.

1. What is the order of reaction with regard to NO and Cl₂?
2. Calculate the overall order of the reaction.
3. Find the value of rate constant (k).

Answer 1

i. Given the reaction:

\[\ce{NO + Cl2 -> NOCl}\]

The rate law is generally expressed as:

Rate = k[NO]^m[Cl₂]ⁿ

Where m and n are the orders of reaction with respect to NO and Cl₂, respectively.

Now, we will determine the order of reaction with respect to NO and Cl₂.

Compare experiments 1 and 2 where [NO] is constant.

`"Rate"_2/"Rate"_1 = (([Cl_2]_2)/([Cl_2]_1])^n`

`(6.0 xx 10^-3)/(2.0 xx 10^-3) = (6.0/2.0)^n`

3 = 3ⁿ

Thus, n = 1

Finding m (order with respect to NO):

Compare experiments 1 and 3, where [Cl₂] is constant.

`"Rate"_3/"Rate"_1 = (([NO]_3)/([NO]_1])^m`

`(1.8 xx 10^-2)/(2.0 xx 10^-3) = (6.0/2.0)^m`

9 = 3^m

Thus, m = 2

ii. The overall order of the reaction is:

m + n = 2 + 1 = 3

iii. Using the rate equation:

Rate = k[NO]²[Cl₂]¹

Substituting values from experiment 1:

2.0 × 10⁻³ = k(2.0)²(2.0)¹

2.0 × 10⁻³ = k(4.0 × 2.0)

2.0 × 10⁻³ = k(8.0)

k = `(2.0 xx 10^-3)/8.0`

k = 2.5 × 10⁻⁴ mol⁻² L² s⁻¹