Chemical Equilibrium, Study notes of Law

Download Chemical Equilibrium and more Study notes Law in PDF only on Docsity! 1 Chemical Equilibrium • When some types of chemical reactions occur in the gas or solution phases, these reaction attain “chemical equilibrium”, i.e., the reaction does not go to completion, but the reaction vessel will contain both reactant species and product species mixed together. Chemical Equilibrium • This occurs when the concentrations of the reactants stop decreasing, and the concentrations of the products stop increasing. 2 NO2 N2O4 (I will use to indicate an equilibrium process in my lecture notes) 2 Chemical Equilibrium • NO2 is a brown gas while N2O4 is colorless 2 NO2 N2O4 • At any given time in a container of NO2, some fraction of the gas will be in the form of NO2, and some fraction will be in the form of N2O4. Chemical Equilibrium • Chemical equilibrium is a dynamic process—an individual molecule will repeatedly move from the NO2 form to the N2O4 form, the overall concentrations of NO2 and N2O4 do not change at a given temperature 2 NO2 N2O4 5 Equilibrium Constants Example (con’t.) H2CO3(aq) + H2O(l) HCO3 -(aq) + H3O +(aq) HCO3 -(aq) + H2O(l) CO3 2-(aq) + H3O +(aq) Net: H2CO3(aq) + 2 H2O(l) CO3 2-(aq) + 2 H3O +(aq) Equilibrium Constants Example (con’t.) For the first reaction: H2CO3(aq) + H2O(l) HCO3 -(aq) + H3O +(aq) KC1 = [HCO3 - (aq)][H3O + (aq)] [H2CO3(aq)] = 4.2 x 10 -7 6 Equilibrium Constants Example (con’t.) For the second reaction: HCO3 -(aq) + H2O(l) CO3 2-(aq) + H3O +(aq) KC2 = [CO3 2- (aq)][H3O + (aq)] [HCO3 (aq)] = 4.8 x 10 -11 Equilibrium Constants Example (con’t.) For the net reaction: H2CO3(aq) + 2 H2O(l) CO3 2-(aq) + 2 H3O +(aq) KC = [CO3 2- (aq)][H3O + (aq)] 2 [H2CO3(aq)] = KC1 KC2 = 2.0 x 10 -17 7 Equilibrium Constant and Pressure • How does the expression for the equilibrium constant change if pressure is used as the variable instead of concentration? • Using the Ideal Gas Law: P A = nRT V = [A]RT [A] = P A RT Equilibrium Constant and Pressure • For the generic reaction aA + bB cC + dD we can write the equilibrium constant in terms of pressure K P = P C c P D d P A a P B b 10 Using Equilibrium Constants Example: Determine [SO4 2-] when a solution of 1.00 M H2SO4(aq) is prepared: • Step 4—solve equilibrium constant expression for unknowns [SO4 2- (aq)] [H3O + (aq)] [HSO4 - (aq)] = x(1.00 + x) 1.00 - x = 1.2 x 10 -2 x 2 + 1.012x - 0.012 = 0 Using the quadratic equation to solve for x, we get x = 0.012 M = [SO4 2-] Using Equilibrium Constants Example: H2SO3 is formed in an equilibrium reaction between SO2 and H2O SO2(g) + H2O(g) H2SO3(aq) SO2 has an average concentration of .006 ppm, and H2O has a vapor pressure of ~20 Torr Determine the amount of H2SO3 in the troposphere Step 1—write an expression for the equilibrium constant KP = PH2SO3 PSO2 PH2O = KC (RT) n = 8.47 x 10 3 (.0821) (298) = 3.46 x 10 2 11 Using Equilibrium Constants Example: Determine the amount of H2SO3 in the troposphere SO2(g) + H2O(g) H2SO3(aq) Step 2—determine pressures of reactants (pressures must be given in units of atm because the R used has atm units H2O: 20 Torr = .026 atm SO2: (.006 ppm)(1 atm) = 6 x 10-9 atm Step 3—solve expression for PH2SO3 PH2SO3 = (3.46 x 10 2 ) PSO2 PH2O = (3.46 x 10 2 )(6 x 10 -9 )(.026) = 5.4 x 10 -8 atm Interpreting Equilibrium Constants • If KC >> 1, then the reaction is strongly product-favored, i.e., the mixture will contain more products than reactants • If KC << 1, then the reaction is strongly reactant-favored, i.e., the mixture will contain more reactants than products • If KC 1, the mixture will contain approximately equal amount of reactant and products 12 Interpreting Equilibrium Constants Example: Acetic acid, CH3COOH, has a KC of 1.8 x 10-5 Determine the relative concentrations of CH3COOH, CH3COO-, and H+ in an aqueous solution balanced equation: CH3COOH CH3COO- + H+ K C = [CH 3 COO - ][H + ] [CH 3 COOH] = 1.8 x 10 -5 Interpreting Equilibrium Constants Example (con’t.): build concentration table [CH3COOH] [CH3COO-] [H+] initially 1.00 0.00 0.00 equilibrium 1.00 – x x x solve equilibrium constant expression for unknown concentrations x x 1.00 - x = 1.8 x 10 -5 15 Haber-Bosch Process Increase in Concentration or Partial Pressure For the process: N2(g) + 3 H2(g) 2 NH3(g) an increase in N2 and/or H2 concentration or pressure, will cause the equilibrium to shift towards the production of NH3 16 Decrease in Concentration or Partial Pressure For the process: N2(g) + 3 H2(g) 2 NH3(g) likewise, a decrease in NH3 concentration or pressure will cause more NH3 to be produced Changes in Temperature For the process: N2(g) + 3 H2(g) 2 NH3(g) + heat an increase in temperature will cause the reaction to shift back towards reactants because the reaction is exothermic 17 Increase in Volume For the process: N2(g) + 3 H2(g) 2 NH3(g) + heat an increase in volume, causes the equilibrium to shift to the left where there are more gaseous molecules Decrease in Volume For the process: N2(g) + 3 H2(g) 2 NH3(g) + heat a decrease in volume, causes the equilibrium to shift to the right where there are fewer gaseous molecules