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Topic Extension

Topic Extension. Solubility Equilibrium Complex-Ions Equilibrium Qualitative Analysis. Acid-Base Equilibrium. Equilibrium. Chemical Reactions. Two Types. A. COMPLETE Reactants  Products B. EQUILIBRIUM Reactants  Products. time. time. Chapter 17: Equilibrium

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Topic Extension

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  1. Topic Extension Solubility Equilibrium Complex-Ions Equilibrium Qualitative Analysis Acid-Base Equilibrium Equilibrium

  2. Chemical Reactions Two Types A. COMPLETE ReactantsProducts B. EQUILIBRIUM Reactants  Products time time

  3. Chapter 17: Equilibrium The Extent of Chemical Reactions 17.1 The Dynamic Nature of the Equilibrium State 17.2 The Reaction Quotient and the Equilibrium Constant 17.3 Expressing Equilibria with Pressure Units: Relation Between Kc and Kp 17.4 Reaction Direction: Comparing Q and K 17.5 How to Solve Equilibrium Problems 17.6 Reaction Conditions and the Equilibrium State: Le Châtelier’s Principle

  4. Reaching Equilibrium on the Macroscopic and Molecular Level 2 NO2 (g) N2O4 (g) Fig. 17.1

  5. Law of Mass Action • At a given T, a chemical system reaches a state in which the ratio of reactants and products has a constant value. • Ratio calledREACTION QUOTIENT, Q [final] [Products]p Q = = [initial] [ Reactants]R At equilibrium Q = K

  6. [C]c [D]d Kc [A]a[B]b Writing the Mass-Action Expression for Equilibrium For the general reaction: aA + bB  cC + dD Rate forward = Rate reverse When not at equilibrium use Q [ ] = Molarity = Example: The Haber process for ammonia production: N2 (g) + 3 H2 (g) 2 NH3 (g) K =

  7. Example: Write the balanced chemical reaction Write the equilibrium constant, Kc 3.0 - M o l e s Moles H2 2.0 - Equilibrium amounts 1.0 - Moles CO Moles CH4 = moles H2O 0 time

  8. Properties of Kc A. Can be based on: Molarity = [ ] identified as KC Partial pressure = atm identified as Kp details next slide B. Constant for given temperature C. Kc = details on next slide D. Homogeneous or Heterogeneous Included in K expression: (aq) and (g) species only No solids, liquids in expression - M is constant. 1 Kc reverse

  9. Expressing K with Pressure Units n V For gases, PV=nRT can be rearranged to give: P = RT nP VRT n V or: = Since = Molarity, and R is a constant if we keep the temperature constant then the molar concentration is directly proportional to the pressure. Therefore for an equilibrium between gaseous compounds we can express the reaction quotient in terms of partial pressures. For: 2 NO(g) + O2 (g) 2 NO2 (g) If there is no change in the number of moles of reactants and products then n = 0 then Kc = Kp , or if there is a change in the number of moles of reactants or products then: P 2NO2 Qp = P 2NO x PO2 Kp = Kc(RT)ngas

  10. [SO2]2[O2] 1 Qc(rev) = = [SO3]2 Qc(fwd) The Form of K for a Forward and Reverse Reaction The production of sulfuric acid depends upon the conversion of sulfur dioxide to sulfuric trioxide before the sulfur trioxide is reacted with water to make the sulfuric acid. 2 SO2 (g) + O2 (g) 2 SO3 (g) [SO3]2 Qc(fwd) = [SO2]2[O2] For the reverse reaction: 2 SO3 (g) 2 SO2 (g) + O2 (g) at 1000K Kc(fwd) = 261 1 1 and: Kc(fwd) = = = 3.83 x 10 -3 Kc(rev) 261

  11. ApplicationsOf Equilibrium Qualitative Value of KC large number favors ____________ small number favors ____________ Examples: Favors Kc = 5.79 x 10-7 ___________________ Kc = 0.00469 _____________ Kc = 2.37 x 1012 ______________

  12. Quantitative Example Problem 1: Calculating Kc from concentrations data. Class Practice Problem #2 Example Problem 2: Find equilibrium concentration of one species given Kc and the other concentrations. Class Practice Problem #3

  13. Example Problem 3: Predict extent of reaction Use reaction quotient - Qc Used when we do not know if the system is at equilibrium. Given: Kc and [ ] step 1: calculate Qc using concentrations step 2: compare Qc with Kc [see below] If Qc Kc reaction need to go toward products If Qc Kc reaction need to go toward reactants If Qc= Kc reaction is at equilibrium Problem solving strategy

  14. Example Problem 4:I.C.E. • Problem: Given the that the reaction to form HF from molecular hydrogenand fluorine has a reaction quotient of 115 at a certain temperature. If 3.000 mol of each component is added to a 1.500 L flask, calculate the equilibrium concentrations of each species. • Write the balanced reaction and Kc expression Plan: 1. Calculate the concentrations of each component. 2. Figure out the changes and express in terms of x 3. Express the final or equil. Conc. In terms of x 4. Solve the equilibrium equation for x 5. Determine the actual concentration. Concentration (M) H2 + F2 2 HF Initial Change Equilibrium

  15. Qualitative Le Chatelier’s Principle Quantitative 1. Calculate actual concentrations after shifts. 2. Calculate new Kc after temperature change. ln = - Changes K2 - H 1 1 K1 R T2 T1 R = 8.31 J/mol-K

  16. CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) Add NH3 CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) The Effect of a Change in Concentration Given an equilibrium equation such as : Forces equilibrium to produce more product. Remove NH3 Forces the reaction equilibrium to go back to the left and produce more of the reactants.

  17. Summary:Effect of Various Disturbances on an Equilibrium System Disturbance Net Direction of Reaction Effect on Value of K Concentration Increase [reactant] Toward formation of product None Decrease [reactant] Toward formation of reactant None Pressure (volume) Increase P Toward formation of lower amount (mol) of gas None Decrease P Toward formation of higher amount (mol) of gas None

  18. Disturbance Net Direction of Reaction Effect on Value of K Temperature Increase T Toward absorption of heat K Increases if H0rxn> 0 K Decreases if H0rxn< 0 Decrease T Toward release of heat K Increases if H0rxn< 0 K Decreases if H0rxn> 0 Catalyst addedNone; rates of forward and reverse reactions increase equally None

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