1 / 38

CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM. Properties of an Equilibrium. Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction. Pink to blue Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O. Blue to pink

bluma
Download Presentation

CHEMICAL EQUILIBRIUM

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CHEMICAL EQUILIBRIUM

  2. Properties of an Equilibrium Equilibrium systems are • DYNAMIC (in constant motion) • REVERSIBLE • can be approached from either direction Pink to blue Co(H2O)6Cl2 ---> Co(H2O)4Cl2 + 2 H2O Blue to pink Co(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2

  3. Fe(SCN)(H2O)53+ + H2O Fe(H2O)63+ + SCN- Chemical EquilibriumFe3+ + SCN- FeSCN2+ e  + 

  4. Chemical EquilibriumFe3+ + SCN- FeSCN2+ e • After a period of time, the concentrations of reactants and products are constant. • The forward and reverse reactions continue after equilibrium is attained.

  5. Examples of Chemical Equilibria Phase changes such as H2O(s) H2O(liq) 

  6. Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq) 

  7. Chemical Equilibria CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq) At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT.

  8. Equilibrium achieved Screen 16.4 & Fig. 16.2 Reaction Quotient & Equilibrium Constant Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium

  9. Reaction Quotient & Equilibrium Constant At any point in the reaction H2 + I2 ↔2 HI

  10. Equilibrium achieved Reaction Quotient & Equilibrium Constant In the equilibrium region

  11. The Reacton Quotient, Q In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium.

  12. THE EQUILIBRIUM CONSTANT Hi Mr. Rierson! For any type of chemical equilibrium of the type a A + b Bc C + d D the following is a CONSTANT (at a given T) If K is known, then we can predict concs. of products or reactants.

  13. Determining K 2 NOCl(g) ↔2 NO(g) + Cl2(g) Place 2.00 mol of NOCl in a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of an “ICE” table of concentrations [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change Equilibrium 0.66

  14. Determining K 2 NOCl(g) ↔ 2 NO(g) + Cl2(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of a table of concentrations [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33

  15. Determining K 2 NOCl(g) ↔ 2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33

  16. Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O2(g)  SO2(g)

  17. Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)

  18. Messing with K • If you reverse the reaction . . . • K (forward) = K-1(reverse) • If you multiply the coefficients by a factor x . . • If you add two reactions together . . . • Multiply the K’s together • If you switch from measuring concentration to measuring pressure . . . • This is related through Ideal Gas Law PV = nRT or P = (n/V) RT or P = C RT… This is primarily impacted by the change in the number of moles (n) of gas from reactants to products • Kp = Kc(RT)change in n

  19. The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. For N2(g) + 3 H2(g) ↔ 2 NH3(g) Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

  20. The Meaning of K For AgCl(s) ↔ Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 1.8 x 10-5 Conc. of products is much less than that of reactants at equilibrium. The reaction is strongly reactant-favored. Ag+(aq) + Cl-(aq) e AgCl(s) is product-favored.

  21. Product- or Reactant Favored Product-favored Reactant-favored

  22. The Meaning of K K comes from thermodynamics. See Chapter 19, page 812-813 ∆G˚ < 0: reaction is product favored ∆G˚ > 0: reaction is reactant-favored If K > 1, then ∆G˚ is negative If K < 1, then ∆G˚ is positive

  23. The Meaning of K 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.

  24. The Meaning of K If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? If not, which way does the reaction “shift” to approach equilibrium?

  25. The Meaning of K All reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. Q (2.33) < K (2.5) Reaction is NOT at equilibrium, so [iso] must become ________ and [n] must ____________.

  26. Typical Calculations PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calc. equilibrium concentrations. H2(g) + I2(g) ↔ 2 HI(g)

  27. H2(g) + I2(g) ↔ 2 HI(g)Kc = 55.3 Step 1. Set up ICE table to define EQUILIBRIUM concentrations. [H2] [I2] [HI] Initial 1.00 1.00 0 Change Equilib

  28. H2(g) + I2(g) ↔ 2 HI(g)Kc = 55.3 Step 1. Set up ICE table to define EQUILIBRIUM concentrations. [H2] [I2] [HI] Initial 1.00 1.00 0 Change -x -x +2x Equilib 1.00-x 1.00-x 2x where x is defined as am’t of H2 and I2 consumed on approaching equilibrium.

  29. H2(g) + I2(g) ↔ 2 HI(g)Kc = 55.3 Step 2. Put equilibrium concentrations into Kc expression.

  30. H2(g) + I2(g) ↔ 2 HI(g)Kc = 55.3 Step 3. Solve Kc expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H2] = [I2] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M

  31. Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g) e

  32. Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g) If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N2O4] [NO2] Initial 0.50 0 Change Equilib

  33. Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g) If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N2O4] [NO2] Initial 0.50 0 Change -x +2x Equilib 0.50 - x 2x

  34. Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g) Step 2. Substitute into Kc expression and solve. Rearrange: 0.0059 (0.50 - x) = 4x2 0.0029 - 0.0059x = 4x2 4x2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax2 + bx + c = 0 a = 4 b = 0.0059c = -0.0029

  35. Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g) Solve the quadratic equation for x. ax2 + bx + c = 0 a = 4 b = 0.0059c = -0.0029 x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027

  36. Nitrogen Dioxide EquilibriumN2O4(g) ↔ 2 NO2(g) x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion: x = 0.026 M [N2O4] = 0.50 - x = 0.47 M [NO2] = 2x = 0.052 M x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027

  37. Solving Quadratic Equations • Recommend you solve the equation exactly on a calculator (do you have a quadratic program? Get one…) or use the “method of successive approximations” • See Appendix A.

  38. Your turn: • Now: some group-work problems for you to work in class and if time, put solutions on board • Webassign tonight as HW there will be two assignments. • First one is practice (due Thursday) • Second one is graded (due Thursday) • Read 16.4, 16.5 • Prepare for 1 question quiz tomorrow

More Related