Question

For the reaction: \[\ce{2H2 + 2NO <=> 2H2O + N2}\], the following rate data was obtained: 

S.No.	[NO] mol L−1	[H2] mol L−1	Rate: mol L−1 sec−1
1	0.40	0.40	4.6 × 10−3
2	0.80	0.40	18.4 × 10−3
3	0.40	0.80	9.2 × 10−3

Calculate the following:

1. The overall order of a reaction.
2. The rate law.
3. The value of rate constant (k).

Answer 1

The initial rate expression can be written as

Rate = k [NO]^a [H₂]^b

Substituting the values of concentration of NO and H₂ from experimental data 1 and 2

(Rate)₁ = k (0.40)^a (0.40)^b    ...(i)

(Rate)₂ = k (0.80)^a (0.40)^b    ...(ii)

Dividing equation (ii) by equation (i)

`("Rate")_2/("Rate")_1 = (k (0.80)^a (0.40)^b)/(k (0.40)^a (0.40)^b)`

= `(18.4 xx 10^-3)/(4.6 xx 10^-3)`

⇒ (2)^a = 4

a = `4/2`

a = 2

Thus, the order of reaction with respect to NO is 2. On comparing experiment data (1) and (3)

(Rate)₃ = k (0.40)^a (0.80)^b    ...(iii)

(Rate)₁ = k (0.40)^a (0.40)^b    ...(iv)

Dividing equation (iii) by equation (iv)

`("Rate")_3/("Rate")_1 = (k (0.40)^a (0.80)^b)/(k (0.40)^a (0.40)^b)`

= `(9.2 xx 10^-3)/(4.6 xx 10^-3)`

⇒ (2)^b = 2

b = `2/2`

b = 1

Thus, the order of reaction with respect to H₂ is 1.

1. Overall order of the reaction = 2 + 1 = 3

2. The rate law for the reaction

Rate = k [NO]² [H₂]

3. Rate constant can be calculated by substituting the values of rate of [NO] and [H₂] for any experiment

k = `"Rate"/(["NO"]^2 ["H"_2])`

= `(4.6 xx 10^-3)/((0.40)^2 (0.40))`

= `(4.6 xx 10^-3)/0.064`

= 71.875 × 10⁻³

Thus, the value of rate constant

k = 71.875 × 10⁻³ mol⁻² L² s⁻¹