Section 1 Review What constitutes a reversible reaction? Describe what it means when a reaction is in a state of chemical equilibrium. How do the rates of the forward and reverse reactions change throughout a reaction?

Chapter 16: Chemical Equilibrium 16-2: The Law of Chemical Equilibrium Chapter 16: Chemical Equilibrium

The Law of Chemical Equilibrium Development of the concept of equilibrium was empirical and depended on the chemist being able to measure the concentration or reactants and products, since rate of reaction depends upon concentration of reactants and products

The Equilibrium Constant Guldberg and Waage proposed the law of mass action Relates the concentrations of reactants and products at equilibrium by using the equilibrium constant (Keq) Equation for a reversible reaction: aA + bB cC + dD a, b, c, and d are coefficients for substances A,B,C, and D The equilibrium expression for this reaction is Keq =([C]c[D]d )/([A]a[B]b) The square brackets denote equilibrium concentrations (usually in M) Concentrations are raised to the power of its coefficient Products of forward reaction in numerator Reactants of forward reaction in denominator

Law of Chemical Equilibrium Every reversible reaction (at a given temperature) proceeds to an equilibrium state that has a specific ratio of the concentrations of reactants and products expressed by Keq Regardless of initial concentrations each reaction will reach equilibrium that can be expressed by the equilibrium constant Equilibrium position- each set of different equilibrium concentrations resulting from different initial concentrations. There are an infinite number of equilibrium positions but only one equilibrium constant (fig. 16-7 p. 542)

Practice

Equilibrium Constant Measure of the extent to which a reaction proceeds to completion. It does not tell you the amount of time needed to reach equilibrium Keq>>1, is read equilibrium constant is significantly greater than 1 so the numerator must be larger than the denominator (concentration of products must be much greater than the concentration of the reactants) Keq<<1, denominator must be larger than the numerator (concentration of reactants must be much greater than the concentration of products) Keq~ 1, there are considerable concentrations of both reactants and products present at equilibrium

Homogeneous and Heterogenous Equilibria Every example so far has been with gases Homogeneous Equilibria- equilibrium conditions for reactions where all reactants and products are in the same state Heterogeneous Equilibria- equilibrium conditions for reactions with substances in more than one state NH4Cl(s) NH3(g) + HCl(g) We have been expressing concentrations of gases in molarity. For pure solids or liquids we can leave the concentration out, because it doesn’t change. So equation for this type of reaction is Keq=[NH3][HCl]

Practice

The Reaction Quotient Q is used to determine if a reaction is at equilibrium Calculated like Keq except the concentrations used are the concentrations taken at the time of measurement, not the equilibrium concentrations Example: N2(g) +3H2(g) 2NH3(g) where Keq=.105 When you measure the concentrations of each compound they are: [NH3]=.15M, [N2]=.002M, and [H2]=.10 M Then plug them in: Keq= [NH3]2/[N2][H2]3 =(.15)2/(.002)(.10)3 and Q= 1.1x104 Since Q = Keq the reaction is not at equilibrium

How will the reaction proceed? If Q=Keq then the reaction is at equilibrium If Q<Keq then the reaction will consume reactants and form products to reach equilibrium If Q > Keq then the reaction will consume products and form reactants to reach equilibrium

Practice Read through Sample Problem 3 Phosphorus pentachloride decomposes partially to phosphorus trichloride and chlorine at elevated temperatures. All the substances are gases. At a given temperature the concentration of PCL5 is .80 mol/L, the concentration of PCl3 is .20 mol/L, and the concentration of Cl2 is 2.20 mol/L. The equilibrium constant for the reaction is .055. Is the reaction at equilibrium? If not which direction will it proceed?

Practice At 740 degrees C, Keq=.0060 for the decomposition of calcium carbonate, which is described by the equation CaCO3 CaO + CO2. Find Q and predict how the reaction will proceed if [CO2]=.004M

Practice For the reaction CO + H2O H2 +CO2, where Keq=5.10 at 527 degrees C. If [CO]=.15M, [H2]=.25M and {CO2]=.37M, calculate Q and determine how the reaction will proceed.

Video Opposing Reactions in Equilibrium System