1.

For gaseous substances having rate of reaction is k[A][BN].If volume of container become (1)/(4) than rate becomes …….times of initial rate.

Answer»

<P>`(4)/(1)`
`(1)/(8)`
`(1)/(16)`
`(16)/(1)`

Solution :Initial rate `r_(1)=k(p_(A))(p_(B))`
If rate BECOMES `(1)/(4)` ,then pressure will be 4 TIME
`THEREFORE` New pressure will be be `4p_(A)` and `4p_(B)`.
So,rate `r_(2)=k(4p_(A))(4p_(B))=16k(p_(A))(p_(B))`
So, `(r_(2))/(r_(1))=(16k(p_(A))(p_(B)))/(k(p_(A))(p_(B)))=16`


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