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A reaction is first order in `A` and second order in `B`. How is rate affected when concentration of `B` is tripled? |
Answer» Correct Answer - 9 Rate `r_(1)=K[A]^(1)[B]^(2) …(1)` Now if concentration of `B` is triple So, `r_(2)=K[A]^(1)[3B]^(2) …(2)` Compare the eqs. (1) and (2), `r_(1)/r_(2)=1/9 :. R_(2)=9 r_(1)` `therefore` Rate becomes `9` times. |
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