Chemistry - Chemical Kinetics Question with Solution | TestHub

The initial rate of reaction is given by:

$r_1=k[A{]}^a[B{]}^b$

When the concentration of $A$ is doubled and $B$ is halved, the new concentrations become $[A{]}^{\prime }=2[A]$ and $[B{]}^{\prime }=\frac{[B]}{2}$.

The new rate of reaction is:

$r_2=k[A{]}^{\prime a}[B{]}^{\prime b}$

Substituting the new concentrations, we get:

$r_2=k(2[A]{)}^a{\left(\frac{[B]}{2}\right)}^b$

$r_2=k\cdot 2^a[A{]}^a\cdot 2^{-b}[B{]}^b$

$r_2=2^{a-b}\cdot k[A{]}^a[B{]}^b$

Since $r_1=k[A{]}^a[B{]}^b$, we can write:

$r_2=2^{a-b}\cdot r_1$

Therefore, the ratio is:

$\frac{r_2}{r_1}=2^{a-b}$

Answer: $2^{(a-b)}$