If the initiation step is most endothermic and therefore RDS, does that mean H*
ID: 884354 • Letter: I
Question
If the initiation step is most endothermic and therefore RDS, does that mean H* the SS approximation is not likely to apply?
2H2 + O2 2H2O A proposed mochanism for this reaction involves the following steps: Inititation: Branching: H. +02OH. +O. Propagation: OH. +H2 >H. + H20 Termination: H. + wall H--wall a) Using the following bond energies (0-0 : 5.1 eV: H-H : 4.5 eV: O-H : 4.4 eV) determine which of the steps is most endothermic. b) Assuming this is the slowest step, state for which (one) of the intermediate species the steady- state approximation is not likely to apply. c) Apply the steady-state approximation to the rate equations for the other intermediates, and derive a rate law for the overall reaction.Explanation / Answer
a) branching steps are more endothermic
b) initiation step is a slowest step.