Question

For the reaction, $A+B⟶$ Products, rate is given as, rate $=k\left[A\right]{\left[B\right]}^2$.
How will the rate change if volume of the reaction mixture is reduced to $1/3$ of its original volume?

Answer 1

When volume is reduced to $1/3$rd, then concentration will increase three times.
$\therefore r_1=k\left[A\right]{\left[B\right]}^2$ ...(i)
$r_2=k\left[3A\right]{\left[3B\right]}^2$ ...(ii)
From equations (i) and (ii),
$\frac{r_1}{r_2}=\frac{1}{27}$
$r_2=27r_1$
$\therefore$ Rate will increase $27$ times.