# Jan_30_09 - Chemistry 108 Friday, Jan 30 Making Rate...

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Friday, Jan 30 Chemistry 108

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Making Rate Equations k 1 A B k -1 A: d[A]/dt = -k 1 [A] + k -1 [B] B d[B]/dt = + k 1 [A] – k -1 [B] Which one? Degree of Advancement (DOA) Rules: If rate involves product (right side), divide by stoichiometric coefficient and multiply by +1 If rate involves reactant (left side) divide by stoichiometric coeff and multiply by -1
2 DOA = +1/1 d[B]/dt = -1/1 d[A]/dt • = + k 1 [A] – k -1 [B] For A ===== 2B DOA = -1/1 d[A]/dt = +1/2 d[B]/dt Problem: Given 2A → 3B and DOA = 0.15M/s, find d[B]/dt DOA = 0.15 = +1/3 d[B]/dt; d[B]/dt = 3x0.15M/s = 0.45M/s

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Other Rate Equations A → B → C k 1 k 2 d[A]/dt = -k 1 [A] d[B]/dt = +k 1 [A] – k 2 [B] • d[C]/dt = k 2 [B]
Second Order/Bimolecular A + B → Product • d[A]/dt = d[B]/dt = -k 1 [A][B] 1 st order in A, B second order overall Bimolecular DOA = +k 1 [A][B] Messy if A o ≠ B o , also if reversible

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## This note was uploaded on 02/26/2012 for the course CHEM 108 taught by Professor Dr.brennan during the Spring '09 term at Binghamton University.

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Jan_30_09 - Chemistry 108 Friday, Jan 30 Making Rate...

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