CHEM 1442 1st Edition Lecture 10Outline of Last Lecture I. ExamplesOutline of Current Lecture II. Le Chatelier’s principleIII. pHCurrent LectureLe Chatelier’s principle- The system will try to “reduce/ oppose” the effect of the disturbance as the system wil reach a new state of equilibriumExample:Consider the equilibriumPCl5(g)  PCl3(g) + Cl2(g) ∆Ho= 87.9 kJIn which direction ill the reaction shift when the temperature is increased?What is the disturbance? Temperature increaseHow do you increase temperature? Heat is added, which is the real disturbanceWhich direction will “oppose” the disturbance? RightExample:Consider the following reactionN2(g) + O2(g) 2 NO(g) Kc = 4.10 x 10-4 (at 2000oC)0.200 moles of N2 and 0.150 moles of O2 are placed in a 1.00 L flaskThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Set up the ICE table so you can determine the concentrations of each chemical species at equilibrium. N2(g) + O2(g)  2NO(g) Initial 0.200 0.150 0Change -x -x +2xEquilibrium 0.200-x 0.150-x 2xKc = [NO]2 / [N2][O2] = (2x)2/ (0.200-x)(0.150-x) = 4.10 x 10-4Approximation: X will be small relative to 0.200 and 0.150.x = 0.00175357Verify the approximation by the 5% rule:0.00175357/ 0.150 x 100% = 1.17% < 5%Accept the answer (0.00175357).Now plug in x-value and find answer.pHThe p-scalepX = -log XThe definition:pH = -log[H3O+]Example: Calculate the pH of a solution whose concentration of H3O+ is 8.3 x 10-12 MpH = 11.08Example: The concentration of H3O+is increased to 1.9 x 10-4 M. Calculate the pH of the solution.pH = 3.72As the concentration of H3O+increases, the pH value decreases.What is the pH of “pure” water?[H3O+] = 1.0 x 10-7 MpH = 7.00The Auto-Ionization of Water (Self-Ionization/ Auto-Protolysis)Example: Write the equilibrium constant expression forH2O(l) + H2O(l) OH- (aq) + H3O+(aq)Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25oC) = The ion-product constant for water[H3O+] = 1.0 x 10-7 M [OH-] = 1.0 x 10-7 MIn any aqueous solution:Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25oC)pKw = -(log [H3O+] + log [OH-]) = 14.00pKw = pH + pOH = 14.00 (at 25oC)In an aqueous solution:H3O+(aq) = H+(aq)[H3O+] = [H+]Bronsted-Lowry (BL) Definition of Acids and Bases- An acid is a proton (H+) donor- A base is a proton (H+)