122 Chap 14 - Chapter 14 Chemical Equilibrium...

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Unformatted text preview: Chapter 14 Chemical Equilibrium End-of-Chapter Problems: see next page Online assignment #3 (Chapters 13&14) Online assignment #4 (Review - Not Graded) Due before 11:30 pm Thursday, March 11 End-of-Chapter Problems - PP 612 to 622 6 7 10 12 13 14 15 16 18 21 27 33 35 49 55 56 63 71 73 74 75 81 83 87 89 91 I. Introduction Many chemical reactions are reversible; after they run awhile, the products react to produce original reactants. Equilibrium: when the forward and reverse reactions proceed at equal rates - then the concentrations of reactants and products remain constant ( equilibrium concentrations ). Assume the time it takes to reach equilibrium is fast unless told otherwise. Example: See next page for data on the following reaction. 1CO + 3H 2 1CH 4 + 1H 2 O 1 CO + 3 H 2 1 CH 4 + 1 H 2 O Moles vs Time Rate vs Time I. Introduction Typical Problem: Mix 1.0 mol CO & 3.0 moles H 2 in a 1.0 L container. At equilibrium an analysis reveals that you have 0.40 mol H 2 O. What is M of each at equilibrium? Let X = M of CO that reacts at equilibria. 1 CO + 3H 2 1CH 4 + 1H 2 O Starting M : 1.0 3.0 0.0 0.0 at Equilibria 1.0-X 3.0-3X X X We know that X = 0.40; therefore, at equilibria: M CO = 1.0-X = 1.0-0.40 = 0.60 M H 2 = 3.0-3X = 3.0 - 1.2 = 1.8 M H 2 O = X = 0.40 II. Equilibrium Expression The equilibrium expression can always be obtained directly from the balanced chemical equation. Note that [ ] = M concentration AT EQUILIBRIUM . You are given the following equilibrium reaction : a A + b B c C + d D The following is the equilibrium expression where K is the equilibrium constant & a,b,c&d = balancing coefficients II. Equilibrium Expression [C] c [D] d K = for a A + b B c C + d D II. Equilibrium ExpressionII....
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122 Chap 14 - Chapter 14 Chemical Equilibrium...

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