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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?
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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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