Question

The following rate data were obtained at 300 K for the reaction, \[\ce{2A + B -> C + D}\]

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	4.5 × 10−3
2	0.3	0.2	5.4 × 10−2
3	0.3	0.4	2.16 × 10−1
4	0.4	0.1	1.8 × 10−2

Calculate the rate of formation of D when [A] = 0.5 mol L⁻¹ and [B] = 0.3 mol L⁻¹.

Answer 1

The given reaction is \[\ce{2A + B -> C + D}\]

And the rate law takes the form Rate = k[A]^m[B]ⁿ

We are given a data table with concentrations and rates, and we need to:

1. Determine the reaction orders m and n

2. Determine the rate constant k

3. Use the rate law to find the rate when [A] = 0.5 mol L⁻¹ and [B] = 0.3 mol L⁻¹

From exp. 2: [A] = 0.3, [B] = 0.2, Rate = 5.4 × 10⁻²
From exp. 4: [A] = 0.4, [B] = 0.1, Rate = 1.8 × 10⁻²

`(5.4 xx 10^-2)/(1.8 xx 10^-2)`

= `([0.3]^m [0.2]^n)/([0.4]^m [0.1]^n)`

`3 = (3/4)^m * 2^n`

Now, try to solve by guessing values of mmm and nnn. Try m = 2, n = 2:

`(3/4)^2 * 2^2 = 9/16 * 4`

= `36/16`

= 2.25

Too low.

Try m = 1, n = 2:

`3/4 * 4 = 3`

m = 1, n = 2

Use experiment 1:
[A] = 0.1 mol L⁻¹, [B] = 0.1 mol L⁻¹, Rate = 4.5 × 10⁻³

Rate = k[A]¹[B]²

⇒ 4.5 × 10⁻³

= k × (0.1) × (0.1)²

⇒ 4.5 × 10⁻³ = k × (0.1) × (0.01)

= k × 0.001

`k = (4.5 xx 10^-3)/0.001`

= 4.5 mol⁻² L² min⁻¹

By using rate law to calculate rate when [A] = 0.5 mol L⁻¹, [B] = 0.3 mol L⁻¹

Rate = 4.5 × [0.5]¹ × [0.3]²

= 4.5 × 0.5 × 0.09

= 4.5 × 0.045

= 0.2025 mol L⁻¹ min⁻¹