Chapter Fourteen Chemical Equilibrium

Chemical equilibrium is achieved when: Chapter Fourteen/ Chemical Equilibrium The Concept of Equilibrium and the Equilibrium Constant Few chemical reactions proceed in only one direction, most are reversible. At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process begins to take place and reactant molecules are formed from product molecules. Chemical equilibrium is achieved when: The rates of the forward and reverse reactions are equal and The concentrations of the reactants and products remain constant Physical equilibrium H2O (l) H2O (g) Chemical equilibrium N2O4 (g) 2NO2 (g)

Start with NO2 Start with N2O4 Start with NO2 & N2O4 Chapter Fourteen/ Chemical Equilibrium The Concept of Equilibrium and the Equilibrium Constant N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2 Start with N2O4 Start with NO2 & N2O4

Where K is equilibrium constant. Chapter Fourteen/ Chemical Equilibrium The Concept of Equilibrium and the Equilibrium Constant Where a, b, c, & d: are the stoichiometric coefficients for A, B, C , & D. Where K is equilibrium constant. The equilibrium constant: for a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentrations has a constant value, K, called the equilibrium constant. Always The concentration of solids and pure liquids and solvent are not included in the expression for the equilibrium constant. K dos not have a unit Law of Mass Action

Chapter Fourteen/ Chemical Equilibrium The Concept of Equilibrium and the Equilibrium Constant The magnitude of the equilibrium constant tells us whether an equilibrium reaction favors the products or reactants. If K is much greater than 1 (that is, K >> 1), the equilibrium will lie to the right and favors the products. if the equilibrium constant is much smaller than 1 (that is, K << 1), the equilibrium will lie to the left and favor the reactants. In this context, any number greater than 10 is considered to be much greater than 1, and any number less than 0.1 is much less than 1.

Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria homogeneous equilibrium applies to reactions in which all reacting species are in the same phase. Example K can be given as Note that the subscript in Kc indicates that the concentrations of the reacting species are expressed in molarity or moles per liter.

Relationship between Kc and Kp: Kc ≠ Kp Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria Relationship between Kc and Kp: Kc ≠ Kp ∆n = moles of gaseous products – moles of gaseous reactants Kc = Kp, when ∆n = 0

Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria Example Write expressions for Kc , and Kp if applicable, for the following reversible reactions at equilibrium?

Because there are no gases then there is no Kp Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria Always do not write solvent in the expression of equilibrium constant. Normally water is solvent Because there are no gases then there is no Kp

Writing Equilibrium Constant Expressions Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria

Because there are no gases then there is no Kp Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria Always do not write solvent in the expression of equilibrium constant. Normally water is solvent Because there are no gases then there is no Kp

The following equilibrium process has been studied at 230 oC: Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria Example The following equilibrium process has been studied at 230 oC: In one experiment, the concentrations of the reacting species at equilibrium are found to be [NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M. Calculate the equilibrium constant (Kc) of this reaction at this temperature. Stopped here

The equilibrium constant Kp for the decomposition of Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria Example The equilibrium constant Kp for the decomposition of Is found to be 1.05 at 250 oC . If the equilibrium partial pressures of PCl5, and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250 oC.

∆n = moles of products – moles of reactants Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Homogeneous Equilibria Example The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. ∆n = moles of products – moles of reactants CO (g) + Cl2 (g) COCl2 (g) [COCl2] [CO][Cl2] Dn = 1 – 2 = -1 R = 0.0821 Kc = T = 273 + 74 = 347 K = 0.14 0.012 x 0.054 = 220 Kp = 220 x (0.0821 x 347)-1 = 7.7

Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Heterogeneous Equilibria heterogeneous equilibrium applies to reactions in which all reacting species are in different phases. Example All rules applied for homogeneous equilibria also applies for heterogeneous. CaCO3 (s) CaO (s) + CO2 (g)

Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Example Write the equilibrium constant expression Kc, and Kp if applicable, for each of the following heterogeneous systems: Heterogeneous Equilibria

X Writing Equilibrium Constant Expressions Heterogeneous Equilibria Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Heterogeneous Equilibria X

AgCl is solid therefore it dose not go to equilibrium constant. Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions AgCl is solid therefore it dose not go to equilibrium constant. There are no gases thus there are no Kp,. Heterogeneous Equilibria

Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions P4 is solid and PCl3 is liquid therefore they do not go to constant of equilibrium. Heterogeneous Equilibria

Consider the following heterogeneous equilibrium: Chapter Fourteen/ Chemical Equilibrium Writing Equilibrium Constant Expressions Example Consider the following heterogeneous equilibrium: At 800 oC, the pressure of CO2 is 0.236 atm. Calculate (a) Kp and (b) Kc for the reaction at this temperature. CaCO3 and CaO are solid therefore they do not go to equilibrium of constant. (b) Heterogeneous Equilibria R = 0.0821 T = 273 + 800 = 1073 K ∆n = moles of products – moles of reactants Dn = 1 – 0 = 1

Thank you for listening