Chem 104 1st Edition Lecture 31Outline of Last Lecture I. Worked examples of chapter 14Outline of Current LectureI. Calculating the equilibrium concentrationsCurrent LectureI. Use the change in concentration of a material you know to determine the change in other chemicals in the reaction. One way to solve for the concentrations is to use theice method. The ice method stands for initial molarity, change in concentration, and equilibrium molarity.Example: 2A + B ↔ 4C. The intial concentractions A=1.00 M, B=1.00 M, C=0. Measure the equilbirum concentration of C = 0.50 M:This is how the ice method starts off:[A] [B] [C]Initial molarity 1.00 1.00 0Change inconcentrationEquilibrium molarity 0.50C ¿4¿A ¿2[B]¿¿K =¿Since [C] would have had to increase by .50, it would have used ¼ of [B] according to stoichiometry:.50 C(1 B4 C)=.125 mol B.50 C(2 A4 C)=.25 mol ASince [A] and [B] are reactants they must be subtracted from the initial concentration. Thus the final ice method table would look like this:[A] [B] [C]I 1.00 1.00 0C -0.25 -0.125 +0.50E 0.75 0.875 0.50II. Example 2: Find Kc for 2CH4 ↔ C2H2 + 3H2 at 1700 °C if the intial concentration [CH4]=0.115 M and the equilibrium molarity of [C2H2] = 0.035 M?[A] [B] [C]I 0.115 0 0CE 0.0350.035 M(3 H21 C2H2)=.105 M H20.035 M(2 C H41 C2H2)=0.07 M C H4Kc=(0.035) (0.105)(0.045)=0.082So…[A] [B] [C]I 0.115 0 0C -0.07 +0.035 +0.105E 0.045 0.035 0.105III. Example 3: Finish the ICE table for 2NO2 ↔ N2O4:NO2N2O4I 0.0200 0CE 0.01720.0172−0.0200=−0.00280.0028(1 N2O42 N O2)=0.0014So…NO2N2O4I 0.0200 0C -0.0028 +0.0014E 0.0172
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