Rate law for a reaction between A and B is given by $$R = k[A]^n[B]^m$$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $$\left(\dfrac{r_2}{r_1}\right)$$ is

The given rate law is $$r = k[A]^n[B]^m$$, where $$r$$ is the rate, $$k$$ is the rate constant, and $$n,m$$ are the orders with respect to A and B respectively.

Initial rate $$r_1$$ is therefore
$$r_1 = k[A]^n[B]^m \quad -(1)$$

For the new experiment, the concentration changes are:
A is doubled  →  $$[A]_{\text{new}} = 2[A]$$
B is halved  →  $$[B]_{\text{new}} = \frac{1}{2}[B]$$

Substituting these new concentrations into the rate law gives the new rate $$r_2$$:
$$\begin{aligned} r_2 &= k\,(2[A])^n\!\left(\frac{1}{2}[B]\right)^m \\ &= k\,2^n[A]^n \; 2^{-m}[B]^m \\ &= k\,2^{\,n-m}[A]^n[B]^m \quad -(2) \end{aligned}$$

Divide equation $$(2)$$ by equation $$(1)$$ to obtain the required ratio:
$$\frac{r_2}{r_1} = \frac{k\,2^{\,n-m}[A]^n[B]^m}{k[A]^n[B]^m} = 2^{\,n-m}$$

Thus, $$\dfrac{r_2}{r_1} = 2^{\,n-m}$$.

The correct choice is Option A.