Question

The conversion of molecules $X$ to $Y$ follows second order kinetics. If concentration of $X$ is increased to three times, how will it affect the rate of formation of $Y$?

Answer 1

The given reaction is: $X\to Y$
As the reaction follows second order
kinetics, the rate law equation for this
reaction can be given as -
Rate $\left(r\right)=k[X{]}^2$

Let Initial concentration $=a\text{mol}{L}^{-1}$
$\therefore$ Rate, $r_1=k(a{)}^2=k{a}^2$

Now, if concentration is increased to three
times, new concentration will be $3a$
mol $L^{-1}$.
$\therefore$ Rate, $r_2=k(3a{)}^2=9k{a}^2$
Hence, $r_2=9r_1$

Final answer:

The rate will become $9$ times the initial
rate.