CHEM 108 1st Edition Lecture 16Chapter 15: Chemical equilibrium continued Catalysts and Equilibrium Systems:- Systems reach equilibrium faster, when the reaction is catalyzed - No change in K or position of equilibrium, because the concentrations of the reactants and products don’t change.- Calculating Equilibrium Concentrations:o When the value of Kc and the initial concentrations are known. When the Kc expression is a perfect square: solving a linear equation. When the Kc expression is not a perfect square: solving a quadratic equation.I.C.E. Table: - Simple way of solving Equilibrium constant problems- I: initial conditions (concentrations, partial pressures). - C: changes as system moves toward equilibrium. - E: equilibrium values.Example of the I.C.E table:- Reaction: N2 (g) + O2 (g) ⇌ 2 NO(g)o Initial N2 = 0.79 atm; Initial O2 = 0.21 atm. o Initial NO(g) = 0.0 atm (always assume that the product is zero because the reaction hasn’t started yet and there are no products)o Kp = 1.0 × 10−5 o What are the partial pressures of reactants and products when the system achievesequilibrium?P N2(atm) P O2 (atm) P NO(atm)Initial conditions 0.79 0.21 0Changes -x -x +2xEquilibrium values 0.79 - x 0.21-x 2x- Kp= (PNO)2/( PO2)( PN2) 1.0 × 10−5(atm) = (2x)2/(0.21-x)(0.79 - x)- Now you just solve for x using K and the quadratic equationoo For the equation: o {0.0163, -0.0163}= the values of
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