What is the effect on rate of following elementary reaction when volume of the reaction vessel is doubled?
2NO + O2 → 2NO2
Answer & explanation
Correct answer: option 2
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