Reaction rate between two substance $$A$$ and $$B$$ is expressed as following: $$\text{rate} = k[A]^n[B]^m$$. If the concentration of $$A$$ is doubled and concentration of $$B$$ is made half of initial concentration, the ratio of the new rate to the earlier rate will be:

The given rate law is
$$\text{rate}_1 = k[A]^n[B]^m$$

Apply the stated changes in concentrations:
  • $$[A]$$ is doubled  →  $$[A]' = 2[A]$$
  • $$[B]$$ is halved  →  $$[B]' = \tfrac12[B]$$

The new rate becomes
$$\text{rate}_2 = k([A]')^n([B]')^m = k\,(2[A])^n\!\left(\tfrac12[B]\right)^{\!m}$$

Simplify the powers of 2:
$$\text{rate}_2 = k\,2^{\,n}[A]^n \cdot 2^{-\,m}[B]^m = 2^{\,n-m}\,k[A]^n[B]^m = 2^{\,n-m}\,\text{rate}_1$$

Hence the ratio of the new rate to the original rate is
$$\frac{\text{rate}_2}{\text{rate}_1} = 2^{\,n-m}$$

Therefore, the correct choice is:
Option D which is: $$2^{(n-m)}$$