387mo lh2o calculatetheequilibriumconstantfor

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: g) ¾ Kc = Example 2: The Equilibrium Constant for a reaction is: K c = [NH 3 ]4 [O 2 ]5 4 [NO] [H 2 O]6 What is the Equilibrium Constant expressio n when the equation for this reaction is halved and then reversed? Original Equat ion: ¾¾ ® 4 NO + 6 H2O ¬¾ 4 NH3 ¾ + 5 O2 Halved and Reversed: Kc = 6 Chemistry 102 Chapter 14 CALCULATING KC FOR REACTIONS · Molar concentrations (mo larit ies) of products and reactants at equilibrium must be subst ituted in the expressio n of Kc Example 1: CO (g) + ¾¾ ® 3 H2 (g) ¬¾ ¾ CH4 (g) + H2O (g) When 1.00 mo l CO and 3.00 mol H2 are placed in a 10.0 L vessel at 927 °C and allowed to come to equilibrium, the mixture is found to contain 0.387 mo l H2O. Calculate the equilibrium constant for this react ion. ¾¾ ® CO (g) + 3 H2 (g) ¬¾ CH4 (g) + H2O (g) ¾ Init ial 1.00 mo l 3.00 mo l 0 0 D – 0.387 – 1.16 + 0.387 + 0.387 Equilibrium 0.613 mol 1.84 mol 0.387 mol 0.387 mol CO: H2 : CH4: H2O: Equilibrium amounts: 0.613 mo l 1.84 mo l 0.387 mo l 0.387 mo l K c = Equilibrium Concentrations(M) 0.613 mo l/10.00...
View Full Document

This document was uploaded on 03/18/2014 for the course CHEM 102 at Los Angeles Mission College.

Ask a homework question - tutors are online