Following reaction takes place in one step : 2" NO "(g)+O_(2)(g)iff2" NO"_(2)(g) How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one-third of its original volume ? Will there be any change in the order of the reaction with reduced volume ?

Following reaction takes place in one step : 2" NO "(g)+O_(2)(g)iff2"

1.	Following reaction takes place in one step : 2" NO "(g)+O_(2)(g)iff2" NO"_(2)(g) How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one-third of its original volume ? Will there be any change in the order of the reaction with reduced volume ?
Answer» SOLUTION :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.