Chemistry - Chemical Kinetics Question with Solution | TestHub

The rate of the reaction is given by $r=k[A{]}^2[B]$.

Concentration is inversely proportional to the volume of the vessel ($C=\frac{n}{V}$).

When the volume is reduced to half, the concentration of each reactant is doubled.

Let the new concentrations be $[A{]}^{\prime }=2[A]$ and $[B{]}^{\prime }=2[B]$.

The new rate of reaction is $r^{\prime }=k[A{]}^{\prime 2}[B{]}^{\prime }=k(2[A]{)}^2(2[B])=8k[A{]}^2[B]$.

$r^{\prime }=8r$

The rate of reaction is increased by eight times.

Answer: increased by eight times.