For the non-stoichiometry reaction, $$2A + B \to C + D$$, the following kinetic data were obtained in three separate experiments, all at 298 K.
Initial Concentration   Initial Concentration  Initial rate of formation of C
(A)                                   (B)                                (mol L$$^{-1}$$ S$$^{-1}$$)
0.1 M                               0.1 M                          $$1.2 \times 10^{-3}$$
0.1 M                               0.2 M                          $$1.2 \times 10^{-3}$$
0.2 M                               0.1 M                          $$2.4 \times 10^{-3}$$

The rate law for the formation of C is:

We have the reaction $$2A + B \rightarrow C + D$$ and we want to obtain a mathematical expression for the initial rate of formation of $$C$$. In chemical kinetics, the general differential rate law for such a reaction is written as a product of the concentrations of the reactants raised to unknown powers (orders of reaction). So we write

$$\text{rate}= \frac{dc}{dt}=k[A]^m[B]^n$$

where $$k$$ is the rate constant at the given temperature, and $$m$$ and $$n$$ are the partial reaction orders with respect to $$A$$ and $$B$$ respectively. Our task is to determine $$m$$ and $$n$$ from the experimental data.

The table of data is:

Experiment 1: $$[A]_0 = 0.1\ \text{M},\quad [B]_0 = 0.1\ \text{M},\quad \text{rate}_1 = 1.2\times10^{-3}$$

Experiment 2: $$[A]_0 = 0.1\ \text{M},\quad [B]_0 = 0.2\ \text{M},\quad \text{rate}_2 = 1.2\times10^{-3}$$

Experiment 3: $$[A]_0 = 0.2\ \text{M},\quad [B]_0 = 0.1\ \text{M},\quad \text{rate}_3 = 2.4\times10^{-3}$$

First we compare Experiments 1 and 2. The concentration of $$A$$ is kept the same, $$[A]=0.1\ \text{M}$$, while $$[B]$$ is doubled from $$0.1\ \text{M}$$ to $$0.2\ \text{M}$$. Mathematically, we write

$$\frac{\text{rate}_2}{\text{rate}_1}= \frac{k[A]^m[B]_2^n}{k[A]^m[B]_1^n}=\left(\frac{[B]_2}{[B]_1}\right)^n.$$

Substituting the numerical values, we have

$$\frac{1.2\times10^{-3}}{1.2\times10^{-3}} = \left(\frac{0.2}{0.1}\right)^n.$$

The left side is $$1$$, and the right side simplifies to $$2^n$$. Hence

$$1 = 2^n \quad\Rightarrow\quad 2^n = 1 \quad\Rightarrow\quad n = 0.$$

So the reaction is zero-order with respect to $$B$$.

Next we compare Experiments 1 and 3. This time $$[B]$$ is kept constant at $$0.1\ \text{M}$$, while $$[A]$$ is doubled from $$0.1\ \text{M}$$ to $$0.2\ \text{M}$$. Using the same procedure,

$$\frac{\text{rate}_3}{\text{rate}_1}= \frac{k[A]_3^m[B]^n}{k[A]_1^m[B]^n}= \left(\frac{[A]_3}{[A]_1}\right)^m.$$

Substituting the values,

$$\frac{2.4\times10^{-3}}{1.2\times10^{-3}} = \left(\frac{0.2}{0.1}\right)^m.$$

The left side is $$2$$, and the right side is $$2^m$$. Therefore

$$2 = 2^m \quad\Rightarrow\quad 2^m = 2 \quad\Rightarrow\quad m = 1.$$

Thus the reaction is first-order with respect to $$A$$.

Combining the two results $$m=1$$ and $$n=0$$, the rate law becomes

$$\text{rate}= \frac{dc}{dt}= k[A]^1[B]^0 = k[A].$$

This matches the expression given in option D.

Hence, the correct answer is Option D.