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

Question:

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

2A + B → C + D

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

Solution:

Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

Rate $=k[\mathrm{~A}]^{x}[\mathrm{~B}]^{y}$

According to the question,

$6.0 \times 10^{-3}=k[0.1]^{x}[0.1]^{y}$         ...(i)

$7.2 \times 10^{-2}=k[0.3]^{x}[0.2]^{y}$            ...(ii)

$2.88 \times 10^{-1}=k[0.3]^{x}[0.4]^{y}$          ...(iii)

$2.40 \times 10^{-2}=k[0.4]^{x}[0.1]^{y}$        ...(iv)

Dividing equation (iv) by (i), we obtain

$\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}}=\frac{k[0.4]^{x}[0.1]^{y}}{k[0.1]^{x}[0.1]^{y}}$

$\Rightarrow 4=\frac{[0.4]^{x}}{[0.1]^{x}}$

$\Rightarrow 4=\left(\frac{0.4}{0.1}\right)^{x}$

$\Rightarrow(4)^{1}=4^{x}$

$\Rightarrow x=1$

Dividing equation (iii) by (ii), we obtain

$\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}}=\frac{k[0.3]^{x}[0.4]^{y}}{k[0.3]^{x}[0.2]^{y}}$

$\Rightarrow 4=\left(\frac{0.4}{0.2}\right)^{y}$

$\Rightarrow 4=2^{y}$

$\Rightarrow 2^{2}=2^{y}$

$\Rightarrow 4=2^{y}$

$\Rightarrow 2^{2}=2^{y}$

$\Rightarrow y=2$

Therefore, the rate law is

Rate = [A] [B]2

$\Rightarrow k=\frac{\text { Rate }}{[\mathrm{A}][\mathrm{B}]^{2}}$

From experiment I, we obtain

$k=\frac{6.0 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

= 6.0 L2 mol−2 min−1

From experiment II, we obtain

$k=\frac{7.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.2 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

= 6.0 L2 mol−2 min−1

From experiment III, we obtain

$k=\frac{2.88 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.3 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

= 6.0 L2 mol−2 min−1

From experiment IV, we obtain

$k=\frac{2.40 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}}{\left(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\right)\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^{2}}$

= 6.0 L2 mol−2 min−1

Therefore, rate constant, k = 6.0 L2 mol−2 min−1