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Equilibrium Law

Equilibrium Law. 1) Equilibrium Constant, Keq:. If, aA + bB  cC + dD. then. unitless. temperature dependent!. eg. For A 2 + B 2  2AB. then. Note: all “eq”. 2) Significance of Keq magnitude:. If Keq is very large then.

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Equilibrium Law

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  1. Equilibrium Law 1) Equilibrium Constant, Keq: If, aA + bB  cC + dD then unitless • temperature dependent!

  2. eg. For A2 + B2 2AB then Note: all “eq”

  3. 2) Significance of Keq magnitude: • If Keq is very large then is large and the reaction nears completion. • If Keq is very small then there is no reaction.

  4. 3) Keq and the Balanced Chemical Eq’n • any action performed on the chemical rxn, the Keq expression is raised to that action. eg. H2 + I22HI Flip the rxn or x by -1: eg. 2HI  H2 + I2

  5.  by 2 or x 1/2: eg. 1/2 H2 + 1/2 I2HI

  6. 4) Keq and Reaction Kinetics: If A + B  C + D and both the forward and reverse reactions are elementary steps, then Ratef = kf [A]1[B]1 and Rater = kr [C]1[D]1

  7. At equilibrium: Ratef = Rater and kf [A]1[B]1 = kr [C]1[D]1 then

  8. 5) Keq and the Effect of Temperature: • usually Eaf ≠ Ear. • an  in T won’t affect the forward and reverse reactions equally, so, Ratef ≠ Rater and the equilibrium will change

  9. If exothermic: At higher temperature, Rater creates more reactants than at lower T, and If endothermic then the opposite occurs:

  10. 6) Heterogeneous Equilibria: • reactants and products in different physical states (s, l ,g, aq) • Pure liquids (not aq) and solids have constant densities • as a result, pure solids and liquids are not written in the Keq expression.

  11. eg. 2H2O(l) 2H2 (g) + O2 (g) then eg. Zn(s) + Cu2+(aq) Cu (s) + Zn2+(aq)

  12. Equilibrium and Spontaneity ∆Gº = -RT ln Keq R = 8.314 J/mol•K Calculate ∆Gº from Keq eg.PCl3 + Cl2 PCl5 ; Keq = 0.18 ∆Gº = -RT ln Keq = (-8.314 J/ mol•K x 298K) x ln (0.18) = 4.2 kJ/mol

  13. Calculate Keq from ∆Gº ∆Gº = - RT ln Keq ln Keq = - (∆Gº) / (RT) Keq = e-(∆Gº)/(RT) * Note : ∆Gº in J/mol eq. 2CO(g) + O2(g) 2CO2(g) and ∆Gº = -514.5 kJ Keq = e-(∆Gº)/(RT) = e-(-514.5 x 103 J)/(8.314 J/mol•K x 298 K) = 1.54 x 1090 the products are highly favoured!

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