February 11, 2009

# February 11, 2009 - 1x •[C eq = C o 3x • K c =[C o 3x...

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February 11, 2009

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. Clicker question Kinetics Use the steady state assumption to find the concentration of A* in the kinetic mechanism and the rate of formation of product A + B A* + B K 1 bimolecular A* + B → A + B k -1 bimolecular A* → P (Product) k 2 unimolecular
. Equilibrium constant ( P or c) aA + bB ==== cC • K p = p C c / [p A a p B b ] ? Always stoichiometric

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. K c = [C] c /[A] a [B] b Units A === B K c = [B]/[A] If [A] eq = 5M, [B] eq = 1M K c = 1/5 = 0.2 if [B] = 2 [A] eq = [B] eq /K c = 2/0.2 = 10
. What if K = 3 and [A] = 3, [B] = 5 Check 5/3 = 1.67 not 3 non equilibrium concentrations [A] eq = 3- 1x x change in A If A decreases, B must increase to eqm [B] eq = 5 + 1x

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. 3 = (5 + x)/ (3-x) 9-3x = 5+ x 4 = 4x x=1 [A] eq = 3 –x = 3 – 1 = 2 [B] eq = 5+ x = 6 6/2 = 3 = K
. 2A + 1B = 3C K = [C] 3 /[A] 2 [B] 1 [A] eq = A o -2x • [B] eq = B o

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Unformatted text preview: - 1x • [C] eq = C o + 3x • K c = [C o + 3x] 3 /{[A o -2x] 2 [B o – 1x] 1 . • Problem permutations • 1) one non eqm concentration zero – changes by x • 2) value of x given like lab • 3) comes to eqm value find x, then other eqm concentrations . • Special Soluble cases • H 2 + I 2 = 2HI • K p = p HI 2 /p I2 p H2 K p = 4 p H2 = p I2 = 3atm p HI = 0 3-x 2x . • K = 4 = (2x) 2 /(3-x) 2 • 2 = 2x/(3-x) • 6-2x = 2x • 4x = 6 • X = 3/2 = 1.5 • P HI = 3atm p H2 = p I2 = 3-1.5 = 1.5atm . • Small K K< 10-6 • A === B + C • K = [B][C]/A • A o = 2 B o = C o = 0 • [A] eq = 2-1x [B] eq =[C] eq = +1x • 2- x ≈ 2 if K small . • K = 3x10-8 = x 2 /3-x ≈ x 2 /3 • X 2 = 9x10-8 • X = 3x10-4 • Note 3-x = 3 - .0003 ≈ 3...
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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.

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February 11, 2009 - 1x •[C eq = C o 3x • K c =[C o 3x...

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