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The conversion of A rarr B follows secon...

The conversion of `A rarr B` follows second-order kinetics. Doubling the concentration of `A` will increase the rate of formation of `B` by a factor

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The correct Answer is:
4
`Ararr B` follows `II` order
`:.` Rate `r_(1)=K[A]^(2) …(1)`
Now if concentration of A is doubled
So rate, `r_(2)=K[2A]^(2) …(2)`
Compare the eqs. `(1)` and `(2)`
`r_(1)/r_(2)=1/4` or `r_(2) =4r_(1)`
`:.` Rate become `4` times
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