Chemical Equilibrium - General College Chemistry I - Lecture Notes | CHEM 112, Study notes of Chemistry

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CHEMICAL EQUILIBRIUM Chapter 7 “When a system is in chemical equilibrium, a change in one of the parameters of the equilibrium produces a shift in such a direction that, were no other factors involved in this shift, it would lead to a change of opposite sign in the parameter involved.” Henri Louis Le Chatelicr, 1888 Nature of Equilibrium equilibria \ | | static dynamic (mechanical) | thermal phase | chemical Arrows of Chemistry reaction: Ag’ (aq) + Cl'(aqg) ~> AgCl(s) resonance: O=S8-0 <> O-S=0 equilibrium: H,0() <=> H,O0(g) N2O4(g) <=> 2 NO.(g) Phase Equilibrium HOW) <=> H,O(g) CO(s) <=> COr(g) attributes of equilibrium isolated from outside interference macroscopic properties constant spontaneously reach equilibrium state forward rate = reverse rate (dynamically balanced) reached from products or reactants equilibrium constant, K Kp ~ pressure K,.- concentration K, - ionization of woak acid Kj, - ionization of weak base K,, - dissolution of slightly soluble salt FIG I. Equilibrium vapor pressure of water ‘Vapor pressure (atm) Oo "10" 20-300 «4050 GO ‘Temperatura (°C) * -2- Form of Equilibrium Constant Expressions Law of Mass Action . for aA+bB <=> cC+dD the equilibrium constants K, and K, are _ Poli Podto x, ~ lkelP lio P Pavol adeo “ [AlfglBlzg where pure solids and pure liquids are represented by the number 1 EX 1. Write the equilibrium constant expression for each of the following reactions. CaCO;(s) <=> Ca0(s) + CO,(g) H,0@) <=> Ha(g) + Ww 0,(g) Fe,03(s) + 3 Ho{g) <=> Fo(s) + 3 H,O(g) Ch(g) + 2HgO(s) + H,O() <=> HgO * HgCl(s) + 2 HOCKag) Determining Equilibrium Constants (table of changes on pp. 70-71 - ICE) EX 2. At 1000 K the equilibrium gas mixture contains 0,562 atm ‘SOp, 0.101 atm O,, and 0.332 atm SO3. What is K? 4 S0.(8) + O2(g) <=> 2 SO3(8) EQ , EX 3. What is the value of K,, if an equilibrium mixture contains 1.0 mol Fe, 1.0.x 10 mol Oy, and 2.0 mol Fe,03(s) in a 2.0-L container? 4 Fe(s) + 3 O2(g) <=> 2 Fe, 03{s) EQ 5- Equilibrium Constant and Reaction Quotient Want to tell how far a reaction has gone (reactions seldom go to completion) * reactions eventually reach a state of equilibrium * want a number which defines this equilibrium situation * no net change => some fixed relationship between reactants and products * equilibrium constant gives the relationship, some examples at 25°C) [H,0] ast 1/20, => H,O = 2g) + n(g) <=> H20(g) °* Tal = 5.6x10°° P2 ClLO(g) + H,O(g) <=> 2HOCI(g) K,y= PoP = 0.0900 ci,0fH,0 PNo, 31 Na(g) + On(g) <=> 2 NO2(g) Ky => C= 4.7 10 1) K > 1 => product-favored; K >> 1 => reaction essentially complete 2) K <1 => reactant-favored; K << 1 => essentially no reaction * reaction quotient, Q, tells how reaction approaches equilibrium: _ (ch DF _ PEA C*iaepre 2 = paps 1) Q< K = reaction proceeds left to right 2) Q = K => reaction is at equilibrium 3) Q > K => reaction proceeds right to left Example: N,O4(g) <=> 2.NO,(g) initial equilibrium exp [N2O4]_ [NO,] — [N,0,] — [NOz] 1 0.25 0 0.10 0.30 2 0 0.50 0.10 0.30 3 0.25 0.50. 0.26 0.48 Concentration Concentration _— i i { i z ] bea Concentration “Time ‘Time Equilibrium Calculations 1. Using K to Determine Equilibrium Amounts of Reactants and Products Given K and all equilibrium partial pressures but one, find the missing pressure. EX 8. At 425°C K, = 55.6 for the following reaction. If Py, = 2. 12 atm and Py =0.12 atm at 425°C what is the equilibrium partial pressure of HI? Hols) + L(g) <=> 2HICg) EQ Given K find the equilibrium partial pressures of the gaseous products from decomposition of a solid. EX 9. IfK,=2.9x 107 at 25°C what are the equilibrium partial pressures of NH3 and CO,? NH,OCONH,(s) <> 2. NHs(g) + CO,(g) EQ Given K find the equilibrium concentration of the ions from a sparingly soluble salt, (Chapter 9 - Dissolution and Precipitation Equilibria) EX 10. What are the equilibrium concentration of the ions at 25°C if Ky, = 3.2.x 10°59 AuCl3(s) <=> Au**(aq) + 3 CI(aq) EQ -7- 2. Using Initial and Equilibrium Amounts EX 11. The equilibrium concentration of gaseous chlorine is 0.030 M when 0.075 and 0.033 M hydrogen chloride and oxygen gas, respectively, are initially placed in a reaction vessel, How much hydrogen chloride and oxygen gas are left unreacted at equilibrium? 4HCl(g) + 0,(g) <=> 2 Cha(g) + 2H,0(g) I Cc E 3. Using K and Initial Amounts EX 12. If 2.00 mol of HBr were introduced into a 1.00 L vessel at 1495 K what would be the equilibrium concentration of all species if K, = 2.86 X 105? 2 HBr(g) <=> H,(g) + Br,(g) I Cc E 4, Equilibrium Calculations Involving the Quadratic Equation —b + Vb? —4ac 2a for ax?+bx+c = 0 the solutionis x = . -10- I. temperature - only stress that can change value of K 1) phase equilibria H,0@) <=> H,O(g), K=Puog, Pog) 2) exothermic and endothermic reactions - sign of AH® EX 17. For each of the following reactions determine whether the value of K would be larger or smaller at a higher temperature. a) At 500K K = 90 for the following exothermic reaction. No(g) + 3 Ha(g) <=> 2 NEL(g) b) At 25°C K = 107! for the following endothermic reaction. N2(g) + O2(g) <=> 2 NO(g) Il. total pressure 1) pressure induced phase transition H,O(s) <=> =H, 0(2) d=0.917 d=0.999 2) increase of Pror by decreasing volume No(g) + 3 Ap{g) <=> 2.NH3(g) EX 18. For the above chemical reaction what would be the effect on the equilibrium if the volume were changed so that the total pressure were increased 10-fold? yt x II. concentration EX 19. For the following reaction if 0.500 M of each reactant were initially present what would be the equilibrium concentrations if K, = 24? PCL (g) + Ch(g) <=> PCI5(g) I 0.500 M 0.500 M OM c E : [PCl5]o a) add PCI; to double molarity; Q= = ° . PCI [Chale I Cc E [PCls], b) then decrease V by 25%; = , Y 2 PahiChh I Cc