Question

The following data were obtained at 300 K for the reaction 2A + B → C + D: 

Experiment number	[A] (mol L−1)	[B] (mol L−1)	Rate of formation of D (mol L−1 min−1)
1	0.1	0.1	6.0 × 10−3
2	0.3	0.2	7.4 × 10−2
3	0.3	0.4	2.88 × 10−1
4	0.4	0.1	2.4 × 10−2

Calculate the rate of formation of D when [A] = 0.5 mol L⁻¹ and [B] = 0.2 mol L⁻¹.

Answer 1

Given:

2A + B → C + D

And experimental data for different concentrations and rates of formation of D.

Assume the rate law is 

Rate = k[A]^m [B]ⁿ

We’ll use the experimental data to find m and n.

From experiments 1 & 4:

Exp 1: [A] = 0.1, Rate = 6.0 × 10⁻³

Exp 4: [A] = 0.4, Rate = 2.4 × 10⁻²

[B] is constant = 0.1

`"Rate"_4/"Rate"_1 = ([A]_4/([A]_1))^m`

⇒ `(2.4 xx 10^-2)/(6.0 xx 10^-3)`

= `(0.4/0.1)^m`

⇒ 4

= 4^m

= m = 1

From experiments 2 & 3:

[A] = 0.3

Exp 2: [B] = 0.2, Rate = 7.4 × 10⁻²

Exp 3: [B] = 0.4, Rate = 2.88 × 10⁻¹

`(2.88 xx 10^-1)/(7.4 xx 10^-1) = (0.4/0.2)^n`

⇒ 3.89 = 2ⁿ

⇒ n = 2

By using experiment 1

Rate = 6.0 × 10⁻³, [A] = 0.1, [B] = 0.1

Rate = k[A]¹ [B]²

= k(0.1)(0.1)2

= k × 0.001

`k = (6.0 xx 10^-3)/0.001`

k = 6.0

Calculation of rate when [A] = 0.5, [B] = 0.2

Rate = 6.0 × (0.5)¹ × (0.2)²

= 6.0 × 0.5 × 0.04

= 6.0 × 0.02

= 1.2 × 10⁻¹ mol L⁻¹ min⁻¹