CHEM 113 1st Edition Lecture 7 Outline of Last Lecture I. Reaction Mechanisms a. Elementary stepsII. Correlating mechanisms with the rate lawIII. The rate- determining step of a reactionOutline of Current Lecture II. Practicing example problems of reaction mechanismsCurrent LectureIn class we discussed and solved the following problem.The overall reaction 2NO (g) + O2 2NO2 (g) has an experimental rate law Rate= k[NO]^2[O2]. A proposed mechanism is1). NO(g) + O2(g) NO3 [fast]2). NO3(g) + NO(g) 2NO2(g) [slow; rate determining]- Let’s go through all three criteria to see if this reaction is valido 1). The elementary steps must add up to the overall balanced equation.NO + O2 + NO3 + NO NO3 + 2NO2NO + O2 + NO3 + NO NO3 + 2NO22NO+O2 2NO2 These workings show that the first criterion is met!o 2). The elementary steps must be reasonable.These 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. Both steps are bimolecular, so this criterion is met!o 3). The mechanism must correlate with the observed rate law To see that this is met, we first write rate laws for the elementary steps- *Note that the first reaction occurs both forwards and in reverse, therefore there are two rates.Rate 1 (forward)= k1 [NO][O2]Rate 1(reverse)= k-1 [NO3]*k-1 is the rate constant and NO3 is the reactant for the reverse reaction.Rate 2= k2[NO3][NO] Next we must show that the rate law for the rate-determining step (step 2 in this case) gives the overall rate law. How we have it written does not because it contains the intermediate NO3, where an overall rate law only includes the reactants and products. - We eliminate NO3 from the rate law for step 2 by expressing it in terms of reactants.Rate 1(forward)= Rate 1 (reverse) or k1[NO][O2]= k-1[NO3]To express [NO3} in terms of reactants, we isolate it algebraically[NO3] = (k1/k-1)[NO][O2]- We then substitute this into rate 2Rate 2 = k2{(k1/k-1)[NO][O2]} * [NO]Rate 2= (k2k1/k-1) [NO]^2[O2]- This is identical to the overall rate law! o To summarize, we asses the validity of a mechanism with a fast initial step by 1). Writing rate laws for the fats step (both directions) and for the slow step. 2). Expressing [intermediate] in terms of [reactant] by setting the forward rate law of the reversible step equal to the reverse rate law, and solving for [intermediate]. 3). Substituting the expression for [intermediate] into the rate law for the slop step to obtain the overall rate law.* Note that for any mechanism, only reactants involved up to and including the slow (rate-determining) step appear in the overall rate
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