The rate law of a chemical reaction : `2NO+O_(2) to 2NO_(2)`, is given as rate `=k[NO]^(2)[O_(2)]`. How will the rate of reaction change if the volume of reaction vessel is reduced to `1//4`th of its original value ?

Explanation

For, `2NO+O_(2)to2NO_(2)`
Rate `=k[NO]^(2)[O_(2)]`
Let a moles of `NO` and `b` moles of `O_(2)` be taken to start a reaction at any time in a vessel of V litre
`r_(1)=k[(a)/(V)]^(2)[(b)/(V)]`…………`(i)`
If volume of vessel is reduced to `V//4`, then for same moles of `NO` and `O_(2)`
`r_(2)=k[(a)/(V//4)]^(2)[(b)/(V//4)]=64k[(a)/(V)]^(2)[(b)/(V)]`................`(ii)`
By Eqs. `(i)` and `(ii)` , `(r_(2))/(r_(1))=64`
`r_(2)` is `64` times of `r_(1)`