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(a) For a reaction A+BrarrP, the rate is...

(a) For a reaction `A+BrarrP,` the rate is given by Rate `=K[A][B]^(2)`
(i) How is the rate of reaction affected if the concentration of B is doubled ?
(ii) What is the overall order of reaction if A is present in large excess ?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.

Text Solution

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(a) (i) `A+BrarrP" Rate "=k[A] [B]^(2)`
Differential rate equation of reaction is
`(dn)/(dt)=k[A]'[B]^(2)=k[A][B]^(2)`
When concentration of B is doubled it means concentration of B becomes `[2xxB]" "therefore` New rate of reaction,
`(dn)/(dt)=k[A] [2B]^(2)=4k[A] [B]^(2)=4[(dn)/(dt)]`
i.e., rate of reaction will become 4 times.
(ii) A useful protocol for determining the order of reaction with respect to a particular components is to measure the concentration dependence of rate when all other reactants are in great excess.
`v=K[A]_("excess")^(a)[B]^(2)`
`v=K'[B]^(2)["Over all order "=2]`
`k'=K[A]_("excess")^(a)`
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