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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?

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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?

Numerical
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Solution

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.

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