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 $(\frac{r _{2}}{r _{1}})$ is

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r 1 ​ = k [ A ] n [ B ] m Now A is doubled \& B is halved in concentration ⇒ r 2 ​ = k 2 n [ A ] n ⋅ 2 m [ B ] m ​ Now r 1 ​ r 2 ​ ​ = 2 ( n − m )

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