What is the effect on rate of following elementary reaction when volume of the reaction vessel is doubled?

2NO + O₂ → 2NO₂ 

\(\frac{1}{8}\) of the initial rate

For elementary reactions, Rate law is written according to law of mass action.

For the reaction, 2NO + O₂ → 2NO₂ 

r = k[NO]²[O₂]¹ 

Molar concentration ∝ \(\frac{1}{volume}\)

as volume is doubled so concentration gets halved

r' = k[\(\frac{NO}{2}\)]²[\(\frac{O_2}{2}\)]¹

r' = \(\frac{1}{8}\)k[NO]²[O₂]¹ = \(\frac{1}{8}\)r