In a chemical reaction $$A$$ is converted into $$B$$. The rates of reaction, starting with initial concentrations of $$A$$ as $$2\times 10^{-3}$$ M and $$1\times 10^{-3}$$ M, are equal to $$2.40\times 10^{-4}$$ Ms$$^{-1}$$ and $$0.60\times 10^{-4}$$ Ms$$^{-1}$$ respectively. The order of reaction with respect to reactant $$A$$ will be

For any reaction whose rate depends only on the concentration of reactant $$A$$, the rate law can be written as
$$r = k [A]^{\,n}$$
where $$k$$ is the rate constant and $$n$$ is the order of the reaction with respect to $$A$$.

Two experiments are provided:

Experiment 1: $$[A]_1 = 2\times10^{-3}\,\text{M}$$, $$r_1 = 2.40\times10^{-4}\,\text{M s}^{-1}$$

Experiment 2: $$[A]_2 = 1\times10^{-3}\,\text{M}$$, $$r_2 = 0.60\times10^{-4}\,\text{M s}^{-1}$$

Divide the rate expressions for the two experiments to eliminate $$k$$:

$$\frac{r_1}{r_2} \;=\; \frac{k [A]_1^{\,n}}{k [A]_2^{\,n}} \;=\; \left(\frac{[A]_1}{[A]_2}\right)^{\!n}$$

Insert the numerical values:

$$\frac{2.40\times10^{-4}}{0.60\times10^{-4}} \;=\; \left(\frac{2\times10^{-3}}{1\times10^{-3}}\right)^{\!n}$$

$$4 = 2^{\,n}$$

Since $$2^{\,2}=4$$, we obtain

$$n = 2$$

Therefore, the reaction is second‐order with respect to $$A$$.

Option D which is: $$2$$