Chemical EquilibriumThe Equilibrium ConstantKp vs KcKc for the Sum of ReactionsHeterogeneous EquilibriumUsing the Equilibrium ConstantLeChatelier's PrincipleExam Review TopicsChemical Equilibrium-A Dynamic EquilibriumWhen compounds react, they eventually form a mixture of products and (unreacted) reactants, in a dynamicequilibrium.Much like water in a U-shape tube, the water constantly mixes back and forth through the lower portion of thetube, as if the forward and reverse "reactions" were occurring at the same rate. This makes the system appear tobe "static" (or, stationary), when in reality, it is "dynamic" (in constant motion). For example, The Haber process for producing ammonia from nitrogen and hydrogen gas does not go tocompletion, but instead reaches an equilibrium state where all three participants are presentN2(g) + 3 H2(g) 2 NH3(g)Chemical Equilibrium is the state reached by a reaction mixture when the rates of the forward and reversereactions have become equal.Example of Applying Stoichiometry to an Equilibrium MixtureYou place 1.000 mol N2 and 3.000 mol H2 into a reaction vessel at 450oC. and 10.0 atm. The reaction is: N2(g) + 3 H2(g) 2 NH3(g)What is the composition of the equilibrium mixture if you obtain 0.080 mol of NH3 from it?SolutionUsing the information given in the problem, you set up the following table.Amount (mol) N2(g) + 3 H2(g) 2 NH3(g)Starting 1.000 3.000 0Change − x − 3x +2xEquilibrium 1.000 −x 3.000−3x 2x = 0.080 (or x = 0.040)The problem statement gives the equilibrium amount of NH3. This tells you that 2x is 0.080 mol (x = 0.040). Youcalculate equilibrium amounts for other substances from the expressions given in the table, using this value of x.Equilibrium amount N2 = 1.000 − x = 1.000 − 0.040 = 0.960 mol N2Equilibrium amount H2 = 3.000 − 3x = 3.000 − (3 x 0.040) = 2.880 mol H2Equilibrium amount NH3 = 2x = 0.080 mol NH3 Page 1 Once the water seeks the proper level, that is,reaches equilibrium it continues to migrate backand forth across the "arrow" but at the same ratemaking it appear to be static.Reactants ProductsThe Equilibrium ConstantEvery equilibrium reaction has its own "balance" point under any given set of conditions. That is, the ratio ofproducts produced to unreacted reactants remains constant under constant conditions of pressure andtemperature.Consider the reaction a A + b B c C + d D where A,B,C, and D denote the reactants andproducts and a,b,c and d are the coefficients in the balanced chemical equation.The equilibrium-constant expression for a reaction is an expression obtained by multiplying theconcentrations of products, dividing by the concentrations of reactants, and raising each concentration to apower equal to the coefficient in the chemical equation.The equilibrium constant, Kc, is the value obtained for the equlibrium-constant expression when equilibriumconcentrations are substituted.For the general reaction above, the equilibrium-constant expression would beThe Law of Mass Action is a relation that states that the values of the equilibrium-constant expression Kc areconstant for a particular reaction at a given temperature, whatever equilibrium concentrations aresubstituted.For example, the equilibrium-constant expression for the equation CO(g) + 3 H2(g) CH4(g) + H2O(g) isEquilibrium....from a Kinetics StandpointConsider the reaction N2(g) + 3 H2(g) 2 NH3(g)If we were to consider the Rate Laws outlined in the Kinetics chapter for the forward and reverse reactions, wewould get:Rate(forward) = kf[N2][H2]3andRate(reverse) = kr[NH3]2At equilibrium, the rate of the forward and reverse reactions would be equal, therefore:kf[N2][H2]3 = kr[NH3]2If we rearrange the equations to get both constants on one side of the equal sign, we get:Therefore, we can identify the Equilibrium Constant, Kc, as Page 2 Kc=[C]c[D]d[A]a[B]bKc=[CH4][H2O][CO][H2]3kfkr=[NH3]2[N2][H2]3kfkrObtaining the Equilibrium Constant for a ReactionEquilibrium concentrations for a reaction must be determined experimentally and then substituted into theequilibrium-constant expression in order to calculate Kc.Consider the reaction CO(g) + 3 H2(g) CH4(g) + H2O(g)Suppose we started with initial concentrations of CO and H2 of 0.100 M and 0.300 M respectively. When thereaction finally settled into equilibrium we determined the equilibrium concentrations to be as follows: Reactants Products[CO] = 0.0613 M [CH4] = 0.0387 M[H2] = 0.1839 M [H2O] = 0.0387 MThe equilibrium constant expression for this reaction is:If we substitute the equilibrium concentrations, we getNote that regardless of what initial concentrations you begin with whether they be reactants or products, the Lawof Mass Action dictates that the reaction will always settle into an equilibrium where the equilibrium-constantexpression will equal Kc.For example, if we repeat the experiment on the previous page, only this time, we'll start with initialconcentrations of products; [CH4]init = 0.1000 M and [H2O]init = 0.1000 M We find that with these initial conditions, as the reaction settles into equilibrium, the equilibrium concentrationsare as follows. Reactants Products[CO] = 0.0613 M [CH4] = 0.0387 M[H2] = 0.1839 M [H2O] = 0.0387 MSubstituting these into the equilibrium-constant expression, we obtain the same result.The notable thing here is that whether we start with reactants initially or products initially, the reactionwill settle into the same equilibrium with the value of Kc remaining constant. Page 3 Kc=[CH4][H2O][CO][H2]3Kc=(0.0387)(0.0387)(0.0613)(0.1839)3=3.93Kc=(0.0387)(0.0387)(0.0613)(0.1839)3=3.93The Equilibrium Constant, KpIn gas-phase equilibria, it is usually more convenient to express the equilibrium concentrations in terms of partialpressures rather than Molarities.It should be noted that the partial pressure of a gas