[kirushanth]

Consider the simple reaction A+2B ---> Products. If we double the volume of the vessel where this reaction takes place, how will the rate of disappearance of B get affected?

[curiouscat]

1/8 assuming they are gases.

[juanrga]

The ordinary rate is
[tex]R = k [A] [ B ]^2[/tex]
or, equivalently,
[tex]R = k \frac{n_A}{V} \left( \frac{n_B}{V} \right)^2 \propto \frac{1}{V^3}[/tex]
Therefore for reactions in different volumes
[tex]\frac{R´}{R} = \left( \frac{V}{V´} \right)^3 [/tex]
if [itex]V' = 2V[/itex] we obtain the 1/8 mentioned by curiouscat

[adianadiadi]

Unless we know the orders with respect to each term in the rate equation, it is not possible to tell what will happen exactly.

[curiouscat]

Unless we know the orders with respect to each term in the rate equation, it is not possible to tell what will happen exactly.

With only the given info. fair to assume its an elementary reaction....

[juanrga]

The original poster says that it is a "simple reaction" and this is sometimes used as synonym of elementary reaction {*} in whose case the orders are known.

{*} http://www.chem.arizona.edu/~salzmanr/480a/480ants/reacmech/reacmech.html