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Chapter 13: Equilibrium

Chapter 13: Equilibrium. AP Chemistry. Label the Forward Rate and Reverse Rate on the graph below. CO + H 2 O CO 2 + H 2. Equilibrium. The state in which there are no observable changes as time goes by Rate of fwd rxn = Rate of rev rxn [R] and [P] are constant

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Chapter 13: Equilibrium

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  1. Chapter 13: Equilibrium AP Chemistry

  2. Label the Forward Rate and Reverse Rate on the graph below

  3. CO + H2O CO2 + H2

  4. Equilibrium • The state in which there are no observable changes as time goes by • Rate of fwd rxn = Rate of rev rxn • [R] and [P] are constant • Examples: acetone, NaCl(s), NO2 tubes

  5. N2O4 2NO2 K E K E K E Label the region in which the study of kinetics is concerned Label the region that indicates the region has reached equilibrium

  6. 4 Conditions Necc for Equilibrium • Both Reactants and Products are Present • Rate of Forward Rxn = Rate of Reverse Rxn • Closed System • No macroscopic changes (Pressure, Temp, color, etc.)

  7. Which of these are at equilibrium? • A lit Bunsen burner • An Alka Seltzer tablet in water • Vapor pressure in a sealed flask • A saturated solution of PbCl2 • Student population of OHS over a year • Salt dissolving in water

  8. N2O4 2 NO2

  9. N2O4 2NO2 Law of Mass Action aA + bB cC + dD No units!!

  10. If K = 1 If K >> 1 If K << 1 Concentration of products = concentration of Reactants Concentration of products > concentration of Reactants “Equilibrium lies far to the right” Concentration of products < concentration of Reactants “Equilibrium lies far to the left”

  11. Homogeneous Equilibrium All species are in the same phase (i.e. all gases) N2O4 (g) 2 NO2(g) But…with gases Kc ≠ Kp

  12. CO +2Cl COCl2 Sample Problem: The equilibrium concentrations for the rxn between carbon monoxide & molecular chlorine to form COCl2 (g) at 74 oC are [CO] = 0.012 M, [Cl]=0.054 M, and [COCl2] = 0.14 M. Calculate Kc and Kp.

  13. CaCO3 (s) CaO (s) + CO2 (g) Heterogeneous Equilibria Reactants and products are in different phases How are the 2 pictures different from each other? How are they the same? dPCO2

  14. CaCO3 (s) CaO (s) + CO2 (g) Kc = [CO2] Kp = PCO2 Solids and liquids do not affect K; they have a constant concentration (i.e. density)

  15. Water Softeners Problem: “Hard Water” contains lots of Ca2+ Solution: “Soften” water by replacing Ca2+ with Na+

  16. N2O4(g) 2NO2(g) 2NO2(g) N2O4(g) ‘ K = K = = 4.63 x 10-3 [NO2]2 [N2O4] [N2O4] [NO2]2 1 K = Modifying the Balanced Eqn Original rxn Reverse rxn = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

  17. N2O4(g) 2NO2(g) ½N2O4(g) NO2(g) 2N2O4(g) 4NO2(g) K = K = K = = 4.63 x 10-3 = K1/2 = K2 [NO2]2 [NO2] [NO2]4 [N2O4]1/2 [N2O4] [N2O4]2 Modifying the Rxn con’t Original rxn New rxn = (4.63 X 10-3)1/2 New rxn = (4.63 X 10-3)2

  18. aA + bB cD + dD Reaction Quotient (Q) initial concentrations Substitute of the reactants and products into the equilibrium constant (Kc) expression.

  19. Reactants Products • IF • Qc > Kc • system proceeds from right to left to reach equilibrium • (reverse reaction is favored) • Qc = Kc • system is at equilibrium • Qc < Kc • system proceeds from left to right to reach equilibrium • (forward reaction is favored)

  20. [Br]2 [Br2] Kc = Br2 (g) 2 Br (g) = 1.1 x 10-2 Kc = (0.0120 + 2x)2 0.163 - x At a given temp, (Kc) = 1.10 x 10-2 for the reaction Br2(g)2 Br (g) If initially [Br2] i = 0.163 M and [Br] i = 0.0120 M, calculate the concentrations of these species at equilibrium. Are we at equilibrium????? ICE Let x be the change in concentration of Br2 Initial (M) 0.163 0.0120 Change (M) -x +2x Equilibrium (M) 0.163 - x 0.0120 + 2x Solve for x

  21. -b ± b2 – 4ac x = 2a Br2 (g) 2Br (g) Initial (M) 0.163 0.0120 Change (M) -x +2x Equilibrium (M) 0.163 - x 0.0120 + 2x = 1.10 x 10-2 Kc = (0.0120 + 2x)2 0.163 - x ax2 + bx + c =0 4x2 + 0.0590x - 0.00165 = 0 Recall the quadratic equation gives 2 solutions! We must determine which is correct to find the equilibrium concentrations x = 0.0142 x = -0.0290 [Br2] eq = 0.148 M [Br] eq= 0.0404 M

  22. [CO]2[O2] 2CO2 (g) 2 CO (g) + O2 (g) [CO2]2 Kc = (2x)2 (x) = 2.0 x 10-6 Kc = (0.4 - 2x)2 At a particular temperature, K = 2.0 X 10-6 for the reaction: 2CO2 (g) 2CO(g) + O2 (g) If 2.0 mol of CO2 is initially placed into a 5.0 L vessel, calculate the equilibrium concentrations of all species Are we at equilibrium????? 0 0 0.40 Initial (M) +2x +x -2x Change (M) x Equilibrium (M) 0.4 - 2x 2x Solve for x

  23. 2CO2 (g) 2 CO (g) + O2 (g) (2x)2 (x) 4x3 (4x3) 4x3 = 2.0 x 10-6 = 2.0 x 10-6 = 2.0 x 10-6 = 2.0 x 10-6 Kc = Kc = Kc = Kc = (0.4 - 2x)2 (0.4)2 (0.4 - 2x)2 (0.4 - 2x)2 LOOK! K IS REALLY SMALL YUCK!!! 0 0 0.40 Initial (M) +2x +x -2x Change (M) x Equilibrium (M) 0.4 - 2x 2x ≈0.4 x = 4.3 x 10-3M = [O2]eq [CO2]eq= 0.40 M [CO] eq= 8.6 X 10-3M

  24. Calculator MATH “0: Solver” 0= AX2 + BX + C ENTER After X = , type 0 “Solve” “QUIT” “X,t,theta,n” button ENTER

  25. 2CO2 (g) 2 CO (g) + O2 (g) Check Assumption 0 0 0.40 Initial (M) +2x +x -2x Change (M) x Equilibrium (M) 0.4 - 2x 2x ≈0.4 0.40 – 2x = 0.40 – 2(4.3 X 10-3) = 0.3914

  26. When can we use the assumption? K is very small (K < 10-3) K is very big (K > 103)

  27. Henri Louis Le Chatlier(1850-1936) Inventor of acetylene torch Professor of Industrial Chemistry and Metallurgy Instrumental in the development of cement and Plaster of Paris

  28. LeChatlier’s Principle When a stress is applied to a system at equilibrium, the system will respond to partially undo the stress Add Reactant, Add Product, Remove Reactant, Remove Product, Add Heat, Increase Pressure,…

  29. Predicting adjustments Haber process N2 + 3 H2 2 NH3 + energy produced used produced used Use energy Produce NH3

  30. Synthesis of Ammonia N2 (g) + 3H2 (g) <===> 2NH3 (g) + heat Problem: The forward Rxn is slow at Room Temp, BUT also exothermic. Adding heat favors the REVERSE Rxn! Le Chatelier’s Solution: Use an iron catalyst, at apressure of 200 ATM! Accidental contamination of the Rxn chamber with air results in an huge explosion blowing fragments of steel through the lab floor and ceiling. Le Chatelier abandons the project, and less than five years later, Haber and Claude were successful in producing ammonia on a commercial scale using a similar process. "I let the discovery of the ammonia synthesis slip through my hands. It was the greatest blunder of my scientific career."

  31. produced used produced used 2 H+ + 2 CrO42-Cr2O72- + H2O [Cr2O72-] [CrO42-] [H2O] Use H+ ORANGE (Add H+) [H+] [CrO42-] [Cr2O72-] [H2O] (Use H+) Produce H+ YELLOW Reaches equilibrium faster---no change! H+ H+ H+ Na+

  32. 2 H+ + 2 CrO42-Cr2O72- + H2O Add H+ X= CrO4-2 O= Cr2O7-2 x x xx x x x x xox xx x x x xx xxxx o x x x x x x x x ox o oo o o o ooo o o o x o oo oo oo o ooo oo Add OH-

  33. produced used produced used 2 NO2 N2O4 + energy [NO2] [N2O4] DARKER Use Heat Produce Heat LIGHTER Darker; then lighter Decrease Pr. Lighter; then Darker Increase Pr.

  34. Le Chatelier Explains Tooth Decay! Ca5 (PO4)3OH (s) 5 Ca+2(aq) + 3 PO4–3 (aq) + OH- (aq)

  35. N2(g) + 3H2(g) 2NH3(g) Equilibrium shifts left to offset stress Add NH3 Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. • Changes in Concentration 14.5

  36. aA + bB cC + dD Le Châtelier’s Principle • Changes in Concentration continued Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left 14.5

  37. A (g) + B (g) C (g) Le Châtelier’s Principle • Changes in Volume and Pressure Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas 14.5

  38. Le Châtelier’s Principle • Changes in Temperature Change Exothermic Rx Endothermic Rx Increase temperature K decreases K increases Decrease temperature K increases K decreases • Adding a Catalyst • does not change K • does not shift the position of an equilibrium system • system will reach equilibrium sooner 14.5

  39. Change Equilibrium Constant Le Châtelier’s Principle Change Shift Equilibrium Concentration yes no Pressure yes no Volume yes no Temperature yes yes Catalyst no no 14.5

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