Rate law for a reaction between A and B is given by
$\mathrm{R}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$
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 $\left(\frac{r_2}{r_1}\right)$ is

Question

Rate law for a reaction between A and B is given by

\mathrm{R}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{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(\frac{r_2}{r_1}\right)

is

Quantrex Academy · Accepted Answer

Correct answer: A. 2^ (n-m)

r_1=k[~A]^ n [~B]^ m Now A is doubled \& B is halved in concentration r_2=k 2^ n [~A]^ n · [~B]^ m 2^ m Now (r_2)/(r_1)=2^ (n-m)

Rate law for a reaction between A and B is given by R=k[~A]^ n...