Reasontheconcentrationofapuresolidorpure

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Unformatted text preview: 3x10 K3 = ??? 4 [CH 4 ] [H 2 O] [CS2 ] [H 2 ] K 2 = [CO] [H 2 ]3 [CH 4 ] [H 2 S]2 K1K 2 = [ CH 4 ] [H 2 O] [CS2 ] [H ] 4 2 x = K 3 3 [CO] [ H ] [ CH ] [H 2 S]2 2 4 · The equilibrium constant for an overall equat ion is equal to the product of the equilibrium constants of the individual equat ions. K3 = K1 x K2 · Because an equilibrium can be approached from either direct ion, the direct ion in which we write the chemical equat ion is arbitrary. For example, for the reaction shown below: ¾¾ ® N2O4 (g) ¬¾ 2 NO2 (g) ¾ K c = [NO2 ]2 = 0.212 (at100 o C) [N 2 O 4 ] · When considering the equilibrium in the reverse direction: ¾¾ ® 2 NO2 (g) ¬¾ N2O4 (g) ¾ K c = [N 2 O 4 ] 1 = = 4.72 (at 100 o C) 2 [NO 2 ] 0.212 Kfwd = 1/Krev 10 Chemistry 102 Chapter 14 CLASSIFICATION OF CHEMICAL EQUILIBRIA · Chemical Equilibria can be classified according to the physical state of the reactants and products present: I. Homogeneous Equilibrium An equilibrium that invo lves reactants and products in a single phase. Example: Catalyt ic methanat ion: CO(g) + 3 H2 (g) ¾¾ ® ¬¾ ¾ CH4 (g) + H2O (g) II. Heterogeneous Equilibrium An equilibrium invo lving reactants and products...
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This document was uploaded on 03/18/2014 for the course CHEM 102 at Los Angeles Mission College.

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