Tang 03 equilibrium law - keq 2

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Tang 03 equilibrium law - keq 2

  1. 1. EQUILIBRIUM LAW - K eq
  2. 2. EQUILIBRIUM LAW - K eq A simple mathematical relationship defines each reaction at chemical equilibrium. H 2(g) + I 2(g) <===> 2 HI (g) K eq = [HI (g) ] 2 [H 2(g) ][I 2(g) ]
  3. 3. EQUILIBRIUM LAW - K eq <ul><li>can always be developed from the balanced chemical equation </li></ul><ul><li>K eq will always have a specific value at specific environmental conditions </li></ul><ul><ul><li>if the conditions change, the K eq will also change </li></ul></ul><ul><li>units for K eq will never be used </li></ul>
  4. 4. EQUILIBRIUM LAW - K eq <ul><li>N 2(g) + 3 H 2(g) <===> 2 NH 3(g) </li></ul><ul><li>2 H 2(g) + O 2(g) <===> 2 H 2 O (g) </li></ul><ul><li>2 NO 2(g) <===> N 2 O 4(g) </li></ul>k eq = [NH 3(g) ] 2 [N 2(g) ][H 2(g) ] 3 k eq = [H 2 O (g) ] 2 [H 2(g) ] 2 [O 2(g) ] k eq = [N 2 O 4(g) ] [NO 2(g) ] 2 Example #1
  5. 5. EQUILIBRIUM LAW - K eq <ul><li>How are the equilibrium laws of the following equations related? </li></ul><ul><li>N 2(g) + 3 H 2(g) <===> 2 NH 3(g) </li></ul><ul><li>2NH 3(g) <===> N 2(g) + 3 H 2(g) </li></ul>ii) k eq = [N 2(g) ][H 2(g) ] 3 [NH 3(g) ] 2 i) k eq = [NH 3(g) ] 2 [N 2(g) ][H 2(g) ] 3
  6. 6. EQUILIBRIUM LAW - K eq If the direction of an equation is reversed, the equilibrium constant is the reciprocal of the original equilibrium constant. K eq forward = 1 / K eq reverse
  7. 7. EQUILIBRIUM LAW - K eq <ul><li>How are the equilibrium laws of the following equations related? </li></ul><ul><li>PCl 3(g) + Cl 2(g) <===> PCl 5(g) </li></ul><ul><li>2 PCl 3(g) + 2 Cl 2(g) <===> 2 PCl 5(g) </li></ul>i) k eq = [PCl 5(g) ] [PCl 3(g) ][Cl 2(g) ] ii) k eq = [PCl 5(g) ] 2 [PCl 3(g) ] 2 [Cl 2(g) ] 2
  8. 8. EQUILIBRIUM LAW - K eq When the coefficients of a balanced equation are multiplied by a factor, the equilibrium constant is raised to the exponent of the same factor. K eq rxn ii) = [K eq rxn i) ] x
  9. 9. EQUILIBRIUM LAW - K eq <ul><li>What is the equilibrium law of the sum of the following reactions? </li></ul><ul><li>2 N 2(g) + O 2(g) <=> 2 N 2 O (g) </li></ul><ul><li>2 N 2 O (g) + 3 O 2(g) <=> 4 NO 2(g) </li></ul>k eq = [N 2 O (g) ] 2 [N 2(g) ] 2 [O 2(g) ] k eq = [NO 2(g) ] 4 [N 2 O (g) ] 2 [O 2(g) ] 3 k eq = [NO 2(g) ] 4 [N 2(g) ] 2 [O 2(g) ] 4 K eq final = [N 2 O (g) ] 2 [N 2(g) ] 2 [O 2(g) ] [NO 2(g) ] 4 [N 2 O (g) ] 2 [O 2(g) ] 3 X
  10. 10. EQUILIBRIUM LAW - K eq When chemical equilibria are added together, the equilibrium constants are multiplied together. K eq final rxn = K eq rxn 1 x K eq rxn 2
  11. 11. EQUILIBRIUM LAW - K eq Example #2 At 25°C, K eq = 7.0 x 10 25 for: 2 SO 2(g) + O 2(g) <===> 2 SO 3(g) What is the value of K eq for: SO 3(g) <===> SO 2(g) + ½ O 2(g) K eq = 7.0 x 10 25 inversed, to the power of 0.5 = 1.195 x 10 -13 = 1.2 x 10 -13
  12. 12. EQUILIBRIUM LAW - K eq MAGNITUDE OF K eq The value of K eq (large or small) can provide a hint to the ratio of reactants to products at equilibrium.
  13. 13. EQUILIBRIUM LAW - K eq MAGNITUDE OF K eq <ul><li>K eq is very large (K eq > 1) </li></ul><ul><ul><li>[products] > [reactants] </li></ul></ul><ul><li>K eq ≈ 1 </li></ul><ul><ul><li>[products] ≈ [reactants] </li></ul></ul><ul><li>K eq is very small (K eq < 1) </li></ul><ul><ul><li>[products] < [reactants] </li></ul></ul>
  14. 14. EQUILIBRIUM LAW - K eq MAGNITUDE OF K eq Example #3 <ul><li>Which of the following reactions will tend to proceed farthest toward completion? </li></ul><ul><li>H 2(g) + Br 2(g) <===> 2 HBr (g) </li></ul><ul><ul><li>K eq = 1.4 x 10 -21 </li></ul></ul><ul><li>2 NO (g) <===> N 2(g) + O 2(g) </li></ul><ul><ul><li>K eq = 2.1 x 10 30 </li></ul></ul><ul><li>2 BrCl (g) <===> Br 2(g) + Cl 2(g) </li></ul><ul><ul><li>K eq = 0.195 </li></ul></ul>
  15. 15. EQUILIBRIUM LAW - K eq EQUILIBRIUM INVOLVING PURE SOLIDS AND LIQUIDS
  16. 16. EQUILIBRIUM LAW - K eq EQUILIBRIUM – SOLIDS AND LIQUIDS What is the concentration of a solid or liquid? (i.e. H 2 O) Does the concentration of these pure compounds change? ex. 1 mol NaHCO 3 occupies 38.9 cm 3 2 mol NaHCO 3 occupies 77.8 cm 3 Molar concentration remains the same. Solids and liquids are unaffected by concentration
  17. 17. EQUILIBRIUM LAW - K eq EQUILIBRIUM – SOLIDS AND LIQUIDS In the equilibrium law, solids and liquids do not need to be included as it becomes part of the equilibrium constant. NH 3(aq) + H 2 O (l) <===> NH 4 + (aq) + OH - (aq) k eq = [NH 4 + (aq) ][OH - (aq) ] [NH 3(aq) ]
  18. 18. EQUILIBRIUM LAW - K eq EQUILIBRIUM – SOLIDS AND LIQUIDS Example #4 <ul><li>Write the equilibrium law for the following reactions: </li></ul><ul><li>CaCO 3(s) <===> CaO (s) + CO 2(g) </li></ul><ul><li>2 NaHCO 3(s) <===> Na 2 CO 3(s) + H 2 O (g) + CO 2(g) </li></ul><ul><li>CaO (s) + SO 2(g) <===> CaSO 3(s) </li></ul>k eq = [CO 2(g) ] k eq = [H 2 O (g) ][CO 2(g) ] k eq = ___ 1__ [SO 2(g) ]
  19. 19. EQUILIBRIUM LAW - K eq EQUILIBRIUM – SOLIDS AND LIQUIDS Example #4 <ul><li>d) 2 Hg (l) + Cl 2(g) <===> Hg 2 Cl 2(s) </li></ul><ul><li>NH 3(g) + HCl (g) <===> NH 4 Cl (s) </li></ul>k eq = __ 1__ [Cl 2(g) ] k eq = _____ 1______ [NH 3(g) ][HCl (g) ]
  20. 20. EQUILIBRIUM LAW - K eq EFFECT OF TEMPERATURE ON K eq
  21. 21. EQUILIBRIUM LAW - K eq EQUILIBRIUM – TEMPERATURE Will temperature change the value of K eq ? Why or why not? Reactions are endothermic or exothermic and therefore will be affected by the addition or removal of heat.
  22. 22. EQUILIBRIUM LAW - K eq EQUILIBRIUM – TEMPERATURE <ul><li>3 H 2(g) + N 2(g) <===> 2 NH 3(g) + heat </li></ul><ul><li>What is the equilibrium law? </li></ul><ul><li>Which way does equilibrium shift when temperature increases? How will K eq change? </li></ul><ul><li>When temperature decreases? </li></ul>k eq = ___ [NH 3(g) ] 2 ___ [H 2(aq) ] 3 [N 2(g) ] Shifts to the left. Since the denominator increases, K eq decreases (is smaller in value) Shifts to the right. Since the numerator increases, K eq increases (is larger in value)
  23. 23. EQUILIBRIUM LAW - K eq EQUILIBRIUM – TEMPERATURE <ul><li>heat + PCl 5(g) <===> PCl 3(g) + Cl 2(g) </li></ul><ul><li>What is the equilibrium law? </li></ul><ul><li>Which way does equilibrium shift when temperature increases? How will K eq change? </li></ul><ul><li>When temperature decreases? </li></ul>k eq = [PCl 3(g) ][Cl 2(g) ] [PCl 5(g) ] Shifts to the right. Since the numerator increases, K eq increases (is larger in value) Shifts to the left. Since the denominator increases, K eq decreases (is smaller in value)

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