CHEM 1415 1st EditionLecture 16Unit: Chemical Equilibrium ConcentrationOutline of Current Lecture I. Chemical EquilibriumII. The Equilibrium ExpressionIII. Equilibrium Concentrationsa. Example 1Current Lecture Chemical Equilibriumo Dynamic Equilibrium Rates of evaporation or condensation become equal and do not stop at equilibrium, and do not go to zero Individual molecules continue to move from the liquid to the vapor phaseand back againo The rate of the forward reaction equals the rate of the reverse reactiono Arrows indicate emphasize the dynamic character of the processR ←→Po Reverse reaction Raterev=krev[ P]o Two rates equal at equilibrium Ratefor=Raterevokfor[R]=krev[P]∧kforkrev=[P]eq[R ]eq The Equilibrium Expressiono Using a general chemical reaction aA+bB ←→cC +dD o The ratio of concentration is Q=[C ]c[D]d[ A ]a[B]bo The expression for the equilibrium constant is always based directly on the reaction stoichiometryo Homogeneous equilibria – reactants and products are in the same phase, either gaseous or aqueouso Heterogeneous equilibrium – multiphase system Equilibrium Concentrationso Strategies to solve equilibrium concentration problems Write a balanced chemical equation for the relevant equilibrium or equilibriaThese 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. Write the corresponding equilibrium expression Create a table of concentrations for all of the reacting specieso Example 1 Hydrogen reacts with Iodine gas at elevated temperatures, k = 59.3, T = 400℃, 0.050 M of hydrogen and 0.050 M of Iodine. What are the equilibrium concentrations? -H2(g )+I2 (g)←→2 H I(g)-K=[HI ]2[H2][I2]=59.3H2I2HIInitial Concentration0.050 M 0.050 M 0 MChange in Concentration-x -x +2xFinal Concentration0.050 - x 0.050 - x 2x-K=[HI]2[H2] [I2]=59.3 = (2 x)2(0.050−x)2=59.3-2 x0.050−x=7.70-9.70 x=0.39- X = 0.040- Therefore[H2]=[I2]=0.050−0.040=0.010 M ∧[HI]=2 x=0.080
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