chap-15-2011W-a

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 (...
View Full Document

## This document was uploaded on 02/26/2014.

Ask a homework question - tutors are online