Question

The conversion of molecules A to B follow second order kinetics. If concentration of A is increased to three times, how will it affect the rate of formation of B?

Answer 1

For the reaction A → B,

Since it follows second-order kinetics so,

Rate of reaction (r) = k[A]²  ...(1)

If the concentration of reactant increased to three times.

Rate of reaction (r') = k[3A]²  ...(2)

Thus, on dividing equations (1) and (2)

`r/r^' = (k[A]^2)/(k[3A]^2)`

= `1/9`

Therefore, the rate of formation of B increases to 9 times.