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What is the effect on rate of following elementary reaction when volume of the reaction vessel is doubled? 2NO + O2 → 2NO2 |
8 times \(\frac{1}{8}\) of the initial rate 4 times \(\frac{1}{4}\) of the initial rate |
\(\frac{1}{8}\) 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 |
