Following reaction takes place in one step: `2NO +O_(2) + 2NO_(2)` How will the rate of above reaction change if the volume of reaction vessel is reduced to 1/3 rd of original volume? Will there be any change in the order of reaction?

Explanation

Rate `=k[NO]^(2)[O_(2)]`
Suppose initially, moles of NO = a, moles of `O_(2)=b`, volume of the vessel = V L. Then
`[NO]=(a)/(V)M," "[O_(2)]=(b)/(V)M:."Rate"(r_(1))=k((a)/(V))^(2)((b)/(V))=k(a^(2)b)/(V^(3))" "...(i)`
New volume = `V//3`.
`:." New concentrations ":[NO]=(a)/(V//3)=(3a)/(V),[O_(2)]=(b)/(V//3)=(3b)/(V)`
`:." New rate"(r_(2))=k((3a)/(V))^(2)((3b)/(V))=(27ka^(2)b)/(V^(3))" "...(ii)`
`:.(r_(2))/(r_(1))=27" or "r_(2)=27r_(1)," i.e., rate becomes 27 times"`
There is no effect on the order of reaction.