# 10 - 17.41 Since all equilibrium concentrations are given...

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

17.39 When x mol of CH4 reacts, 2x mol of H2O also reacts to form x mol of CO2 and 4x mol of H2. The final (equilibrium) concentration of each reactant is the initial concentration minus the amount that reacts. The final (equilibrium) concentration of each product is the initial concentration plus the amount that forms. 17.40 a) The approximation applies when the change in concentration from initial to equilibrium is so small that it is insignificant. This occurs when K is small and initial concentration is large. b) This approximation will not work when the change in concentration is greater than 5%. This can occur when [reactant]initial is very small, or when [reactant]change is relatively large due to a large K.
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 17.41 Since all equilibrium concentrations are given in molarities and the reaction is balanced, construct an equilibrium expression and substitute the equilibrium concentrations to find Kc 17.43 The reaction table requires that the initial [PCl5] be calculated: [PCl5] = 0.15 mol/2.0 L = 0.075 M x = [PCl5] reacting (-x), and the amount of PCl3 and of Cl2 forming (+x). Concentration (M) PCl5(g) ' PCl3(g) + Cl2(g) Initial 0.075 0 0 Change -x +x +x Equilibrium 0.075 - x x x 17.44 The reaction table requires that the initial [H2] and [F2] be calculated: [H2] = 0.10 mol / 0.50 L = 0.20 M; [F2] = 0.050 mol / 0.50 L = 0.10 M x = [H2] reacting Concentration (M) H2(g) + F2(g) ' 2HF(g) Initial 0.20 0.10 0...
View Full Document

## This note was uploaded on 03/14/2012 for the course CHEN 654 taught by Professor Nimos during the Winter '12 term at Beaufort County Community College.

Ask a homework question - tutors are online