The rate law of a chemical reaction given below:
`2NO+O_(2)rarr 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//4th` of its original valur?

Explanation

When volume is reduced to 1/3rd , then concentration will increase three times.
`r_(1)=k[A][B]^(2)" ".....(i)`
`r_(2)=k[3A][3B]^(2)" ".....(ii)`
From eqs. (i) and (ii),
`r_(1)/(r_(2))=(1)/(27)`
`r_(2)=27r_(1)`
`:.` Rate will increases 27 times.