2 n 2 g 3h 2 g 1 3 n2 g h 2 g 2 2 2 nh 3

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Unformatted text preview: d on Law of Mass Action. 2 N 2 ( g ) + 3H 2 ( g ) 1 3 N2 (g ) + H 2 (g ) 2 2 2 NH 3 ( g ) [p NH K′ = [p N NH 3 ( g ) N 2 ( g ) + 3H 2 ( g ) 1 K =K = K ′′′ c ' c "2 c 3 3 p ] [p H 2 [p NH K″ = 2 NH 3 ( g ) p] [p N 3 2 p ] [p H 2 2 p] 3 1 p] p] 2 2 3 K ′′′ = c [p N 2 p ] [p H p] 2 2 [p NH 3 p] Equilibrium expression depends on the stoichiometry Homogeneous Equilibrium How to express the equilibrium constants for the following reactions based on Law of Mass Action. 2 2 P ( g ) + 3Cl2 ( g ) PCl3 ( g ) + Cl2 ( g ) 2 P ( g ) + 5Cl2 ( g ) 2 PCl3 ( g ) PCl5 ( g ) 2 PCl5 ( g ) [p PCl 3 K1 = 3 2 [p P p ] [pCl2 p ] K2 = (1) + 2 × (2) = (3) K 3 = K1 × ( K 2 ) p] p PCl5 p p PCl3 p p Cl2 p 2 2 K3 = [p PCl 5 2 p] 5 [p P p ] [p Cl2 p ] Equilibrium expression depends on the stoichiometry Heterogeneous equilibrium Equilibrium Constant expression: include all gaseous and aqueous reactants and products - Their concentrations are variable. Ignore all pure solids and pure liquids - their concentrations remain the same; and their activities equal 1. CaCO3 (...
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This document was uploaded on 02/26/2014.

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