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] are 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 = \[\ce{\frac{Rate}{[A][B]^2}}\]

= \[\ce{\frac{mol L^{-1}min^{-1}}{mol L^{-1}(mol L^{-1})^2}}\]

= mol⁻² L² min⁻¹

k_I = \[\ce{\frac{6.0 \times 10^{-3}}{0.1 \times (0.1)^2}}\] = 6.0

k_II = \[\ce{\frac{7.2 \times 10^{-2}}{0.3 \times (0.2)^2}}\] = 6.0

k_III = \[\ce{\frac{2.88 \times 10^{-1}}{0.3 \times (0.4)^2}}\] = 6.0

k_IV = \[\ce{\frac{2.4 \times 10^{-2}}{0.4 \times (0.1)^2}}\] = 6.0

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

Answer 2

Suppose the reaction is of order p with respect to A and of order q with respect to B. Therefore,

Rate = k[A]^p [B]^q

For set I, 6.0 × 10⁻³ = k (0.1)^p (0.1)^q    ...(i)

For set II, 7.2 × 10⁻² = k (0.3)^p (0.2)^q    ...(ii)

For set III, 2.88 × 10⁻¹ = k (0.3)^p (0.4)^q    ...(iii)

For set IV, 2.40 × 10⁻² = k (0.4)^p (0.1)^q    ...(iv)

Dividing eq. (ii) by eq. (iii), we get

`(7.2 xx 10^-2)/(2.88 xx 10^-1) = (0.2/0.4)^q`

or `(1/2)^2 = (1/2)^q`

∴ q = 2

Dividing eq. (i) by eq. (iv), we get

`(6.0 xx 10^-3)/(2.40 xx 10^-2) = (0.1/0.4)^q`

or `1/4 = (1/4)^p`

∴ p = 1

Hence, the rate law is Rate = k [A][B]²

From set I, `(6.0 xx 10^-3)/((0.1)^1 xx (0.1)^2)` = 6.0 mol⁻² L² min⁻¹

From set I, `(7.2 xx 10^-2)/((0.3)^1 xx (0.2)^2)` = 6.0 mol⁻² L² min⁻¹

From set I, `(2.88 xx 10^-1)/((0.3)^1 xx (0.4)^2)` = 6.0 mol⁻² L² min⁻¹ 

From set I, `(2.40 xx 10^-2)/((0.4)^1 xx (0.1)^2)` = 6.0 mol⁻² L² min⁻¹

Hence, k = 6.0 mol⁻² L² min⁻¹