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Chapter 15

Chapter 15. AP Chemistry. Chemical Equilibrium. Occurs when opposing reactions are proceeding at equal rates Reaction still proceeds, so then [ ] of reactant and product stay the same [r] ≠ [p], but whatever the eqlm concentration value was, remains unchanged K f [A] = K r [B] A B.

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Chapter 15

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  1. Chapter 15 AP Chemistry

  2. Chemical Equilibrium • Occurs when opposing reactions are proceeding at equal rates • Reaction still proceeds, so then [ ] of reactant and product stay the same • [r] ≠ [p], but whatever the eqlm concentration value was, remains unchanged Kf [A] = Kr [B] A B Kf [A] rate Kr[B] eqlm time

  3. aA + bB cC + dD • Equilibrium expression • Keq or Kc = [C]c [D] d • [A]a [B]b • Kc = eqlm constant • Unit less number • Constant value regardless of conc. • Will vary with temp

  4. Direction of Eqlm and Kc • Eqlm can be established from either reactant side or product side • Kc for one direction = 1/Kc for reverse • Ex: N2(g) + 3H2(g) 2NH3(g) Habor Process • Kp = 4.34 x 10 -3 • Would one expect large amounts of product? • What is the Kp for the reverse reaction? Kpforward = 1/Kprev rev = 4.34 x 10-3 = 2.30 x102

  5. Heterogeneous Eqlm • Eqlm with solid or pure liquids • Conc. of solids/pure liquids remain constant • Not part of eqlm expression, but are necessary for eqlm to be established • Ex: 4H2O(g) + 3Fe(s) Fe3O4(s) + 4H2(g) Kc = ? Kc = [H2]4 [H2O]4

  6. 2NO2Cl(g) 2NO2(g) + Cl2(g) • eqlm [ ] are [NO2] = 0.0108 M • [Cl2] = 0.00538 M • [NO2Cl] = 0.00106 M • What is Kc? • Kc = [NO2]2 [Cl2] • [NO2Cl]2 • = 0.558 • Direct substitution for Kc is used when eqlm concentrations are known for all species • What do you do when you do not know eqlm conc.?

  7. 2SO3(g) 2SO2(g) + O2(g) • initial [SO3] = 6.09 x 10-3M • eqlm [SO3] = 2.44 x 10-3M • Kc = ? • 2SO3 2SO2 + O2 • Δ = 6.09 x 10 -3M • = 2.44 x 10-3M • Initial 6.09 x 10-3 M 0 0 • ∆ -3.65 x 10-3M + 3.65 x 10-3 + 1.83 x 10-3 • 2.44 x 10-3M3.65 x 10-3 1.83 x 10-3 • Kc = [SO2]2 [O2] = [3.65 x 10-3M] [1.83 x 10-3M] • [SO3]2 [2.44 x 10-3M]2

  8. Direction of Eqlm • Using reaction quotient, Q, the direction of the eqlm can be determined • Calculate by substituting [rxn] into eqlm expression • Q>K product react, rxn moves left • Q<K reactants react, rxn moves right • Q=K eqlm conc.

  9. Example: 2SO3 2SO2 + O2 Kc = 4.08 x 10-3 With initial conc. of [SO3] = 2 x 10-3 M [SO2] = 5 x 10-3 M [O2] = 3 x 10-2 M Calculate Q and direction Q = [SO2]2 [O2] = [5 x 10-3]2 [3 x 10-2] [SO3]2 [2 x 10-3]2 Q > K, reaction goes from right to left, forming SO3

  10. *calculating eqlm conc. when one is unknown* • Example: PCl5(g) PCl3(g) + Cl2(g) Kp = 0.497 At eqlm PPCl5(g) = .560 atm, PPCl3 = .350 atom. What is PCl2? PCl5 PCl3 + Cl2 .860 atm .350 atm x KP = [PCl3] [Cl2] substituting [Cl2] = Kp [PCl5] [PCl5] [PCl3] [Cl2] = 1.22 atm

  11. *calculating eqlm conc. from initial conc.* PCl5(g) PCl3(g) + Cl2(g) Kp =.497 What are eqlm conc. when PCl5 initially is PPCl5= 1.66 atm? Kp = [PCl3] [Cl2] [PCl5] .497 = x2 1.66 atm – x .825 atm - .497x = x2 x2 + .497x - .825 atm = 0 x = .693 atm PPCl5 = 1.66 - .693 = 0.97 atm PPCl3 = .693 atm PCl2 = .693 atm

  12. Le Chatelier’s Principle • If a reaction at eqlm is disturbed by changes in temperature, pressure, or concentration, then the eqlm will shift to counteract the disturbance

  13. Changes in [ ] • Addition of reactant or product will shift the eqlm temporarily away from the addition until the eqlm re-established • Removal of reactant or product will shift eqlm towards the removal

  14. Changes in Volume • Decrease volume will increase pressure • Eqlm will shift in the direction of fewest gas molecules • Equal molecules in reactant and product show no shift with regards to pressure changes

  15. Changes in Temperature • [ ] and volume shift eqlm, but do not change K • Temp. will shift and change K • Increase temp., eqlm will shift to the side that absorbs heat  endo shifts right, K inc exo shifts left, K dec • Decrease temp., eqlm shifts to side that produces heat • End shifts left, K dec • Exo shift right, K inc

  16. Ex: PCl5(g) PCl3(g) ΔH° = 87.9 KJ • Add Cl2 shift left • Inc. temp shift right • Dec. volume shift left • Add PCl5 shift right • Ex: 2POCl3(g) 2PCl3(g) + O2(g) • determine ΔH reaction (Appendix C) ΔHrxn= ΔHprod - Δhreact ΔHrxn= 508 KJ B.Keq will do what with an increase in temperature? Keq will increase because eqlm will shift to the right

  17. Effect of Catalysts • Catalysts increase rate of forward and reverse reactions • So eqlm is reaches faster, but eqlm concentration and also Keq are unchanged

  18. Entropy • The amount of randomness or disorder in a system • Increases with increase in KE • Pure crystalline solids have entropy values of 0 • Entropy of a reaction can be calculated similar to ΔHrxn ΔS° = ΣS°product – ΣS°reactant

  19. Gibbs Free Energy • Used to measure spontaneity of a reaction • ΔH and ΔS are approximations for spontaneity • Ex: ΔH large and negative • Ex: ΔS large and positive • Free energy gives definite criteria for spontaneous reactions • Portion of energy is a spontaneous reaction available to do work ΔG = Δ H - T Δ S Δ G is negative, spontaneous forward reaction Δ G = 0, equilibrium Δ G is positive, nonspontaneous forward reaction or Δ G° = ΣG°f (product) – ΣGf (reactants) Indicate spontaneous reactions Free energy

  20. Free Energy and Keq Δ G = Δ G° + RT ln Q ** at eqlm ΔG =0, Q = K, so ΔG° = -RT ln k ΔG° (-) K>1 ΔG° =0 K=1 ΔG° (+) K<1 *relate spontaneity and eqlm position* Free energy at non std. conditions Reaction quotient

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