For the reaction `2NO(g)+O_(2)(g)rarr2NO_(2)(g)` , the rate law is `r=k[NO]^(2)[O_(2)]` . The reaction is carried out in a V L closed container . If the container had a volume equal to one-fourth of V L , then what would the reaction - rate be ?

Explanation

Rate `(r)=k[NO]^(2)[O_(2)]`
Let , the initial number of moles of NO and `O_(2)` are a and b respectively . So initial concentration of NO `((a)/(V)) " mol.L"^(-1) "and that of " O_(2)=((b)/(V))"mol.L"^(-1)` .
So , initial reaction -rate , `r_(1)=k[NO]^(2)[O_(2)]=k((a)/(V))^(2)((b)/(V))" " ....[1]`
If the volume of the container were `(V)/(4)L`,
then the concentration of `NO=(a)/(V/(4))=((4a)/(V))"mol .L"^(-1)`
and that of `O_(2)=(b)/(V/(4))=((4b)/(V))"mol .L"^(-1)`
So , the reaction -rate would be ,
`r_(2)=k[NO]^(2)[O_(2)]=k((4a)/(V))^(2)((4b)/(V))=64xxk((a)/(V))^(2)((b)/(V))" "..[2]`
Comparing equations [1] and [2] we get , `r_(2)=64r_(1)`
Therefore , the reaction -rate in the container of volume `(V)/(4)L` would be 64 times higher than that in the container of volume VL.