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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 |
Answer» Correct Answer - 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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