Now the reaction (2NO+2H2== 2H20+N2) is believed to take place in three steps, the first of which is the fast reversible dimerization of NO to form N2O2, and the last of which (again fast) is the reaction N20 + H2 -- N2 + H2O. So, what is the slow (second) step? Then show, using the steady-state approximation, that the mechanism is consistent with the observed rate law. Why is it only approximately true that the reaction is first-order in H2?

Question

Now the reaction (2NO+2H2==> 2H20+N2) is believed to take place in three steps, the first of which is the fast reversible dimerization of NO to form N2O2, and the last of which (again fast) is the reaction N20 + H2 --> N2 + H2O. So, what is the slow (second) step? Then show, using the steady-state approximation, that the mechanism is consistent with the observed rate law. Why is it only approximately true that the reaction is first-order in H2?