Equilibrium Basic ConceptsThe Equilibrium ConstantSlide 3Basic ConceptsSlide 5Slide 6Slide 7Slide 8Slide 9Variation of Kc with the Form of the Balanced EquationPartial Pressures and the Equilibrium ConstantSlide 12Relationship Between Kp and KcSlide 14Slide 15Slide 16Slide 17Heterogeneous EqulibriaSlide 19Slide 20Slide 21Slide 22Uses of the Equilibrium Constant, KcSlide 24The Reaction QuotientSlide 26Disturbing a System at Equlibrium: PredictionsDisturbing a System at Equlibrium: PredictionsSlide 29Slide 30Slide 31Slide 32Slide 33Slide 34Disturbing a System at Equilibrium: CalculationsSlide 36The Haber Process: An Application of Equilibrium∆G, ∆G˚, and KeqSlide 39∆G, ∆G˚, and KeqSlide 41Slide 42Thermodynamics and KeqRelationship Between Gorxn and the Equilibrium ConstantEquilibrium Basic Concepts•Reversible reactions do not go to completion.–They can occur in either direction       ggggD d + C cB b + A a•Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate.–A chemical equilibrium is a reversible reaction that the forward reaction rate is equal to the reverse reaction rate.•Chemical equilibria are dynamic equilibria.–Molecules are continually reacting, even though the overall composition of the reaction mixture does not change.The Equilibrium Constant•Kc is the equilibrium constant .•Kc is defined for a reversible reaction at a given temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation. •Equilibrium constants are dimensionless because they actually involve a thermodynamic quantity called activity.–Activities are directly related to molarityThe Equilibrium ConstantBasic Concepts•One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction. (g)(g)(g)(g)D C B A•The rates of the forward and reverse reactions can be represented as:    rate. reverse therepresents which DCkRaterate. forward therepresents which BAkRaterrff•When system is at equilibrium:Ratef = RaterThe Equilibrium Constant•Write equilibrium constant expressions for the following reactions at 500oC. All reactants and products are gases at 500oC.The Equilibrium ConstantThe Equilibrium Constant•One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate Kc for the reaction.•Equil []’s 0.028 M 0.172 M 0.086 M 235ClPClPClThe Equilibrium Constant•At a given temperature 0.80 mole of N2 and 0.90 mole of H2 were introduced in an evacuated 1.00-liter container after equilibrium was established the equilibrium concentrations were determined top be 0.20 mole of NH3, 0.70 moles N2, and 0.06 moles H2. Calculate Kc for the reaction.Product-favoredProduct-favoredK > 1K > 1Reactant-favoredReactant-favoredK < 1K < 1The Equilibrium ConstantProducts favored103 > K > 10-3 Reactants favoredPCl3 + Cl2 PCl5Variation of Kc with the Form of the Balanced Equation•The value of Kc depends upon how the balanced equation is written. From the prior examples we have:PCl5 PCl3 + Cl2Kc = [PCl3][Cl2]/[PCl5] = 0.53Equil. []’s 0.172 M 0.086 M 0.028 M Equil []’s 0.028 M 0.172 M 0.086 MPCl3 + Cl2 PCl5Kc’ = [PCl5]/[PCl3][Cl2] = 1.89Equil. []’s 0.172 M 0.086 M 0.028 M 2PCl5 2PCl3 + 2Cl2Kc “ = [PCl5]2/[PCl3]2[Cl2]2 = 3.56Kc ’’’ = [PCl3] 2[Cl2] 2/[PCl5] 2 = 0.28Equil []’s 0.028 M 0.172 M 0.086 MPartial Pressures and the Equilibrium Constant•For gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations.•For gases, the pressure is proportional to the concentration.•We can see this by looking at the ideal gas law.–PV = nRT–P = nRT/V –n/V = M–P= MRT and M = P/RTPartial Pressures and the Equilibrium Constant•Consider this system at equilibrium at 5000C.2SO2(g) + O2(g) 2SO3(g)Relationship Between Kp and Kc•Kc is 49 for the following reaction at 450oC. If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach equilibrium in a 3.0-liter vessel,(a) How many moles of I2 remain unreacted at equilibrium?     gg2g2HI 2I H(b) What are the equilibrium partial pressures of H2, I2 and HI?(c) What is the total pressure in the reaction vessel?(a) How many moles of I2 remain unreacted at equilibrium?     gg2g2HI 2I H     gg2g2HI 2I H(b) What are the equilibrium partial pressures of H2, I2 and HI?     gg2g2HI 2I H(c) What is the total pressure in the reaction vessel?Relationship Between Kp and Kc•Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of Kp?     g2ggBr + NO 2 NOBr 2Heterogeneous Equlibria•Heterogeneous equilibria have more than one phase present.–For example, a gas and a solid or a liquid and a gas.•How does the equilibrium constant differ for heterogeneous equilibria?–Pure solids and liquids have activities of unity.–Solvents in very dilute solutions have activities that are essentially unity.–The Kc and Kp for the reaction shown above are:2COp2cP=K ][CO=K     C500at CO CaO CaCOog2ss3 Heterogeneous EqulibriaHeterogeneous Equlibria•What are Kc and Kp for this reaction?     C)25at (F 2CaCaFo-1aq2aqs2 Heterogeneous Equlibria•What are Kc and Kp for this reaction?       C)500at (H 4OFe OH 4Fe 3og2s43g2s An aside: Similar to Hess’ Law we can add several reactions together to get to an overall reaction not listed in a table.To determine K for the sum of the reactions simply multiply all the values for each intermediate step to get the K for the overall reaction.Uses of the Equilibrium Constant, Kc•The equilibrium constant, Kc, is 3.00 for the following reaction