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Chemical Equilibrium Chapter 15 Homogeneous equilibrium: When all reacting species are in the same phase. Heterogeneous equilibrium: When all reacting species are NOT in the same phase. Heterogeneous Equilibrium: When reacting species are in DIFFERENT phases. SOLID and LIQUID phases are EXCLUDED from the equilibrium expression because they do not have “concentrations” Chemical Equilibrium A state achieved when the RATES of the forward and reverse reactions are EQUAL & the concentrations of the reactants and products remain constant. Equilibrium Constant The equilibrium constant (Keq) is independent of concentration changes, but dependent on the temperature. Writing Equilibrium Constant Expressions For the general reaction α A + β B ↔ c C + d D α β Forward Rate = kF [A] [B] c d Reverse Rate = kR [C] [D] At Equilibrium Forward Rate = Reverse Rate Forward Rate = Reverse Rate Writing Equilibrium Constant Expressions For the general reaction α A + β B ↔ c C + d D At Equilibrium α β c d kF [A] [B] = kR [C] [D] Put CONSTANTS on one side of equation And Concentrations on other side of equation Equilibrium Constant • The equilibrium expression compares reactant & product concentrations. p [Products] K = r [Reactants] Size of Equilibrium Constant 3 If Kc > 10 , products predominate –3 If Kc < 10 , reactants predominate –3 3 If Kc is in the range 10 to 10 , appreciable concentrations of both reactants and products are present. Writing Equilibrium Constant Expressions α A + β B ↔ c C + d D At Equilibrium α β c d kF [A] [B] = kR [C] [D] » ÿ c d k F … C ⁄Ÿ [D ] K eq = = β » k R α B … A [ ] TThhee CCoonncceepptt ooff EEqquuiilliibbrriiuumm The point at which the rate of decomposition N2O4(g) → 2NO2(g) equals the rate of dimerization 2NO2(g) → N2O4(g). is a dynamic equilibrium. The equilibrium is dynamic because the reaction has not stopped: the opposing rates are equal. N 2 O 4 ( g ) 2 N O 2 ( g ) At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4 The double arrow implies the process is dynamic Forward reaction: Rate = kf[N2O4] Reverse reaction: B → A Rate = kr[NO2] At equilibrium kf[N2O4] = kr[NO2] The Direction of the Chemical Equation and Keq An equilibrium can be approached from any direction. For Example: N2O4(g) 2NO2(g) has P2 K = NO2 = 6.46 eq P N2O4 In the reverse direction: 2NO2(g) N2O4(g) PN O 1 K = 2 4 = 0.155 = eq P2 6.46 NO2 Other Ways to Manipulate Chemical Equations and Keq Values The reaction 2N2O4(g) 4NO2(g) P4 NO2 Has Keq = P2 N2O4 • which is the square of the equilibrium constant for N2O4(g) 2NO2(g) Equilibrium Concentration Amounts of components are given as molarity { moles solute / liters of solution or partial pressure of a gas 2 2 [NO ] P = 2 NO2 Kc [ ] Kp = N2O4 PN2O4 Equilibrium Constant Write the Kp expressions for: 2 N2O5(g) ↔ 4 NO2(g) + O2(g) 4 2 Kp = (P NO2 ) (P O2) / (P N2O5) Equilibrium Constant Write the Kc expressions for: 2 N2O5(g) ↔ 4 NO2(g) + O2(g) 4 2 KC = [NO2 ] [ O2] / [N2O5] 2 N2O5(g) ↔ 4 NO2(g) + O2(g) The Kp and Kc expressions : 4 2 Kp = (P NO2 ) (P O2) / (P N2O5) 4 2 KC = [NO2 ] [ O2] / [N2O5] Is Kp = KC ? No ! Relation between Gas Pressure and Concentration • P V = n R T Ideal Gas Equation • Concentration (M) = moles / Liter • M = n / V • Rearranging P V = n R T n P M = = V RT When are KP and KC EQUAL ? For which of the following is KP = KC (a) CO2(g) + C(s) ↔ 2 CO(g) 2+ 2+ (b) Hg(l) + Hg (aq) ↔ Hg2 (aq) (c)2Fe(s) + 3H2O(g) ↔ Fe2O3(s) + 3H2(g) (d) 2 H2O(l) ↔ 2 H2(g) + O2(g) Homogeneous And Heterogeneous Equilibria • Homogeneous Equilibrium When all reacting species are in the same phase. • Heterogeneous Equilibrium: When all reacting species are NOT in the same phase. Heterogeneous Equilibrium: When reacting species are in DIFFERENT phases. SOLID and LIQUID phases are EXCLUDED from the equilibrium expression because they do not have “concentrations” Equilibrium Constant The equilibrium concentrations at 1000 K for the reaction CO(g) + Cl2(g) ↔ COCl2(g) are –2 [CO] = 1.2x10 M, [Cl2]=0.054 M, [COCl2] = 0.14 M Calculate Kc and Kp. CO(g) + Cl2(g) ↔ COCl2(g) –2 [CO] = 1.2x10 M, [Cl2]=0.054 M, [COCl2] = 0.14 M Kc = [COCl2] / [CO][Cl2] –2 Kc = 0.14 /(1.2x10 )(0.054 ) = 21578.2984 and KC = KP (RT) Why ? Kp = KC / RT = KC / (0.0821)(1000) = Sulfur dioxide + Oxygen = Sulfur trioxide SO2 (g) + O2 (g) = SO3 (g) 2 SO2 (g) + O2 (g) = 2 SO3 (g) 2 [SO3 ] KC = 2 [SO2 ] [O2 ] The Initial Change Equilibrium method Is used to calculate equilibrium constant 2 SO2 (g) + O2 (g) = 2 SO3 (g) Initial Moles # # # Change in Moles –2x -x + 2x Equilibrium # - 2x # - x # + 2x Finally Concentrations at Equilibrium The Initial Change Equilibrium method Is used to calculate equilibrium constant 1.00 mole of SO2 and 1.00 mole of O2 are put into a 1.00 liter flask. At equilibrium, 0.0925 mole of SO3 is formed. Calculate Kc at 1000 K for the reaction 2 SO2 (g) + O2 (g) = 2 SO3 (g) The Initial Change Equilibrium method Is used to calculate equilibrium constant 2 SO2 (g) + O2 (g) = 2 SO3 (g) Initial Moles 1.00 1.00 0 Change in Moles -0.925 -0.925/2 +0.925 Equilibrium 0.075 0.537 0.925 Equil. Conc 0.075 M 0.537 M 0.925 M The Initial Change Equilibrium method Is used to calculate equilibrium constant 2 [SO3 ] KC = 2 [SO2 ] [O2 ] 2 0.925 K = = 2.8x102 C 2 0.075 0.537 Methane (CH4) reacts with hydrogen sulfide to yield H2 and carbon disulfide What is the value of Kp if the partial pressures in an equilibrium mixture are 0.20 atm of CH4, 0.25 atm of H2S, 0.52 atm of CS2, and 0.10 atm of H2? How do you solve the problem? • 1st Write Reaction • 2nd Balance Reaction • 3rd Write Equilibrium Constant Equation • 4th Put Equilibrium Pressures in Equation • Do the Arithmetic Using Equilibrium Constants The equilibrium constant (K c ) for the formation of nitrosyl chloride, 2 NO(g) + Cl2(g) ↔ 2 NOCl(g) from nitric oxide & chlorine gas is 6.5 x 10 4 at 35 °C. 4 2NO(g) + Cl2(g) ↔ 2NOCl(g) Kc= 6.5x10 2.0 x 10–2 moles of NO, –3 8.3 x 10 moles of Cl2, & 6.8 moles of NOCl are mixed in a 2.0–L flask. Is the system at equilibrium? If not, what will happen? 4 2NO(g) + Cl2(g) ↔ 2NOCl(g) Kc= 6.5x10 2 4 [ NOCl ] K = 6 .5 x10 = 2 [ NO ] [ Cl 2 ] 2.0 x 10–2 moles of NO, 8.3 x 10–3 moles of Cl2, 6.8 moles of NOCl in a 2.0–L flask. [3.4]2 = 2.8x107 [1.0x10−2 ]2[4.15x10−3 ] The reaction quotient (Qc) Predicts reaction direction. Qc > Kc System proceeds to form reactants. Qc = Kc System is at equilibrium. Qc < Kc System proceeds to form products. Using Equilibrium Constants •A mixture of 0.500mol H2 and 0.500mol I2 was placed in a 1.00–L stainless steel flask at 700°C. The equilibrium constant Kc is 57 H2(g) + I2(g) ↔ 2HI(g) •Calculate the equilibrium concentrations. Le Chatelier’s principle: If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset. Le Chatelier’s principle: Stress may be changes in • concentration, • pressure, volume, or • temperature that removes the system from equilibrium. Concentration Changes: The concentration stress of an added reactant or product is relieved by reaction in the direction that consumes the added substance. Concentration Changes: The concentration stress of a removed reactant or product is relieved by reaction in the direction that replenishes the removed substance. Use the synthesis of ammonia as an example N2(g) + 3H2(g) ↔ 2NH3(g) At equilibrium [N2 ] = 0.50M [H2 ] = 3.00M and [NH3 ] = 1.98 M at 700K N2(g) + 3H2(g) ↔ 2NH3(g) At equilibrium [N2 ] = 0.50M [H2 ] = 3.00M and [NH3 ] = 1.98 M at 700K what happens when the concentration of N2 is increased to 1.50M? N2(g) + 3H2(g) ↔ 2NH3(g) when the concentration of N2 is increased Le Châtelier’s principle says the reaction will relieve the stress by converting the N2 to NH3 N2(g) + 3H2(g) ↔ 2NH3(g) Kc = 0.291 at 700K If H2 is increased Le Châtelier’s principle tells us the reaction will relieve the stress by converting the extra H2 to NH3 N2(g) + 3H2(g) ↔ 2NH3(g) Kc = 0.291 at 700K If NH3 is removed Le Châtelier’s principle tells us the reaction will relieve the stress by producing more NH3 Volume and Pressure Changes P V = n R T Pressure is inversely proportional to volume Increasing pressure = Decreasing volume From P V = n R T P = M (RT) increasing pressure or decreasing volume increases concentration. N2(g) + 3H2(g) ↔ 2NH3(g) Use the synthesis of ammonia as an example There are Four (4) units of gas reactants but only Two (2) units of gas products Therefore, an increase in pressure would shift the equilibrium to the right Le Châtelier’s Principle Volume and Pressure Changes Only reactions containing gases are affected by changes in volume and pressure.
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