Question

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

Answer 1

The given reaction is:-

$X⟶Y$

Since it follows second order kinetics so,

$(Rate{)}_1=K[X{]}^2$ $-\left(i\right)$

where $K$= rate constant

$\left[X\right]$= concentration of $X$

If the concentration of $\left[X\right]$ is made to increase $3$ times then rate will increase by $9$ times as given by:-

$(Rate{)}_2$=$K[3X{]}^2=9K[X{]}^2$ $-\left(ii\right)$

$(Rate{)}_2=9\times (Rate{)}_1$

So, the rate of formation of $Y$ will increase by $9$ times