1.

Areaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half ?

Answer»

SOLUTION :For a second order reaction, `R=k[A]^(2)` where A is a reactant.
(i) If the concentration of A is doubled then,
`R_(1)=k[2A]^(2)=4K[A]^(2)`
`THEREFORE (R_(1))/(R)=(4k[A]^(2))/(k[A]^(2))=4`
`therefore ` RATE will increase 4 time original rate.
(ii) When the concentration of A is reduced to half,
`R_(2)=k[(A)/(2)]^(2)=(1)/(4)k[A]^(2)`
`therefore (R_(2))/(R)=(1//4xxk[a]^(2))/(k[A]^(2))=(1)/(4)`
`therefore ` Rate will reduce to `1//4` th initial value.


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