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Unformatted text preview: 21309•A === B + C•Small K K< 106•K = [B][C]/A• Ao= 3 Bo= Co= 0•[A]eq= 31x [B]eq=[C]eq= +1x•3 x ≈ 3 if K small•1B + 1C• Bo+ x x..•K = 3x108 = x2/3x ≈ x2/3•X2 = 9x108•X = 3x104•Note 3x = 3  .0003 ≈ 3.•K = 3x108 = x2/3x ≈ x2/3•X2 = 9x108•X = 3x104•Note 3x = 3  .0003 ≈ 3Bonus•A) k1[A][B]/{k2+ k1}• B) k1[A]/{k2+ k1}• C) k1[A][B]/{k1[A] + k2}•D) k1[A]/{k1[A] + k2}•E) k1[A][B]/{k1[B] + k2}LeChatelier.•Temperature•If A → B exothermic ΔH negativeT increases A increases, B decreasesTo relieve “stress” absorb extra heat A → B + H Heat “product” in exothermic“K” = H[B]/[A] H increases with T, “K” is constant so [B][A] goes down as T goes up .•Endothermic•H + A → B•“K” = [B]/H[A] H and T up, [B]/[A] up•This means new [B]/[A] = K.•Arrhenius ln(k) = ln(A) – Ea/RT• Eaalways positive gives downslope•Ln(k) vs. 1/T graph•Endothermic reaction gives downslope for ln(K) vs !/T plot• Slope = ΔHreaction/RExothermic reaction...
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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.
 Spring '09
 DR.BRENNAN

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