The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular. Furthermore, the rate law can be determined directly from the number of each type of molecule in an elementary step. For example, the rate law for step 1 is RATE= k[NO2]^2 The exponent "2" is used because the reaction involves two molecules. The rate law for step 2 is RATE= k[N2O4]^1 = k[N2O4][CO] because the reaction involves only one molecule of each reactant the exponents are ommitted. Consider the following elementary steps that make up the mechanism of a certain reaction: 1. 2A----> B + C 2. B + D----> E + C PART A: WHAT IS THE OVERALL REACTION? (EXPRESS YOUR ANSWER AS A CHEMICAL EQUATION) PART B: WHICH SPECIES IS A REACTION INTERMEDIATE? 1. A 2. B 3. C 4. D 5. E PART C: What is the rate law for step 1 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is type k*[A]*[C]^3. PART D: What is the rate law for step 2 of this reaction? Express your answer in standard MasteringChemistry notation. For example, if the rate law is type k*[A]*[C]^3.