Question

Following reaction takes places in one step:

\[\ce{2NO_{(g)} + O2_{(g)} -> 2NO2_{(g)}}\]

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 the reduced volume?

Answer 1

The given reaction is \[\ce{2NO_{(g)} + O2_{(g)} -> 2NO2_{(g)}}\]

Assume the rate law follows the stoichiometry (since it’s a one-step reaction):

Rate = k[NO]²[O₂]

When volume is reduced to 1/3, the concentration of each gas increases by 3 times (since concentration ∝ 1/volume for gases).

New concentrations are

\[\ce{NO -> 3 NO}\]

\[\ce{O2 -> 3O2}\]

New Rate = k(3[NO])²(3[O₂]) = k × 9[NO]² × 3[O₂] = 27 × Original Rate

∴ The rate increases by 27 times.

There is no change in the order of the reaction.

The order depends only on the exponents in the rate law, not on concentration or volume.