Question

The following results have been obtained during the kinetic studies of the reaction:

\[\ce{2A + B -> C + D}\]

Experiment	[A]/mol L−1	[B]/mol L−1	Initial rate of formation of D/mol L−1 min−1
I	0.1	0.1	6.0 × 10−3
II	0.3	0.2	7.2 × 10−2
III	0.3	0.4	2.88 × 10−1
IV	0.4	0.1	2.40 × 10−2

Determine the rate law and the rate constant for the reaction.

Answer 1

In experiments I and IV [B] is the same but [A] has increased four times and the velocity of the reaction has also increased four times.

∴ Rate relative to A ∝ [A]   ...(i)

In experiments II and III [A] is the same but [B] has doubled and the rate of reaction I has also quadrupled.

Rate relative to B ∝ [B]²   ...(ii)

On combining equations (i) and (ii), we get the reaction \[\ce{2A + B -> C + D}\] the rate law is obtained.

Rate =  k [A] [B]²

The overall order of reaction = 1 + 2 + 3

Calculating the rate constant:

k = `"Rate "/(["A"]["B"]^2)`

= `("mol L"^-1  "min"^-1)/(("mol L"^-1)("mol L"^-1)^2)`

= mol⁻² L² min⁻¹

k_I = `(6.0 xx 10^-3)/(0.1 xx (0.1)^2)` = 6.0

k_II = `(7.2 xx 10^-2)/(0.3 xx (0.2)^2)` = 6.0

k_III = `(2.88 xx 10^-1)/(0.3 xx (0.4)^2)` = 6.0

k_IV = `(2.4 xx 10^-2)/(0.4 xx (0.1)^2)` = 6.0

Hence, the rate constant = 6.0 mol⁻² L² min⁻¹