Chem 0120 1st Edition Lecture 6Outline of Last Lecture I. Expressing Rates of ReactionsII. Integrated Rate LawsIII. Example 13.5IV. Half-LifeV. Elementary Reactions Outline of Current Lecture I. Elementary ReactionII. Exercise 13.8III. Mechanism with initial fast step generates intermediateIV. Rate Discovery Step (R.D.S.)V. Elementary Reaction VI. EquilibriumCurrent LectureI. Elementary ReactionNO2(g) + CO(g)  NO(g) CO2(g) Step 1: NO2 + NO2  NO3 + NOStep 2: NO3 + CO  NO2 + CO2 * NO3 is a reaction intermediate (formed in one step & consumed in another)II. Exercise 13.8 Rate Law and Mechanism2H2O2 2H2O + O2Step 1: H2O2+ I- H2O+ IO-Step 2: H2O2 + IO- H2O+ O2 + I-Catalyst: I-Reaction Intermediate: IO-III. Mechanism with initial fast step generates intermediateThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.2N2O5 (g) 4NO2 (g) +O2 (g)Step 1:N2O5 NO2+ NO3Step 2: NO2+ NO3Step 3: IV. Rate Discovery Step (R.D.S.)Rate = k [NO2][NO3]*rate forward = rate reversed at equilibrium R.D.S. in fast step 1k 1[N2O5]=K −1[NO3][ NO2]K 1K −1[ N 2 O 5][NO 2]=[NO3]R.D.S. in slow step 2 Rate=K 2[NO2] [NO3]Rate=(K 2 K 1K −1)=[N 2O 5][ NO 2][ NO 2]=K [ N2O5]V. Elementary Reaction OH- + CH3Cl CH3OH + Cl-2nd OrderBiomolecularNucleophobic Substitution (SN)Rate = k [CH3Cl] [OH-]VI. Equilibrium (Chapter 14) *Due to formatting a two way arrow indicating the reaction occurring in both directions will be written as such:  please do not write like this on exams or homework because it indicates something different which will be taught in later sections*Rate of Forward Reaction = Rate of Reverse Reaction aA + bBcC + dDKc=[C ]c[ D ]d[ A ]a[B]bKc is obtained from rate constant at every step of