1 / 22

CHAPTER 13

CHAPTER 13. Chemical Equilibrium. Equilibrium. Some reactions go to completion. All the reactants are converted into products. There are many reactions that do not go to completion: Example: 2 NO 2 (g)  N 2 O 4 (g) Dark Brown Colorless

meara
Download Presentation

CHAPTER 13

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. CHAPTER 13 Chemical Equilibrium

  2. Equilibrium • Some reactions go to completion. • All the reactants are converted into products. • There are many reactions that do not go to completion: • Example: 2 NO2 (g)  N2O4 (g) Dark Brown Colorless • If the reaction went to completion the container would become colorless – it doesn’t

  3. EXAMPLE • Should: • 2 NO2 (g)  N2O4 (g) Dark Brown Colorless • Actual: • 2 NO2 (g)  N2O4 (g) Light Tan

  4. CHEMICAL EQUILIBRIUM • Some Reactions go both way. • Sometimes very few products are created. 2 CaO (s)  2 Ca (s) + O2 (g) • Very Little appears to happen. • Sometimes a lot of products are created. Cu (s) + 2 AgNO3 (aq)  2 Ag (s) + Cu(NO3)2 (aq) • Very Little of the reactant remains.

  5. CHEMICAL EQUILIBRIUM • CuCl2 (aq)  Cu+2 (aq) + 2 Cl- (aq) + Heat Green Blue Colorless • Observe what happens when the following occurs: • Add Heat • Remove Heat • Add NaCl • Add AgNO3

  6. CHEMICAL EQUILIBRIUM • Reaction Graph – Concentration of CuCl2 nevers reaches zero.

  7. EQUILIBRIUM • Is not static. • It is Highly Dynamic. • At the Macro Level everything appears to have stopped. • At the Micro Level, reactions are continuing. • The reaction is travels both directions. • As much Reactant is being created as Products. • The Reaction Rates in both directions are in Equilibrium. • Not the concentrations of Reactants and Products.

  8. CHEMICAL EQUILIBRIUM EQUATION • jA + kB  mC + nD j is coef for Reactant A k is coef for Reactant B m is coef for Product C n is coef for Product D Keq = [C]m[D]n [A]j[B]k

  9. EXAMPLE PROBLEM The following reaction is allowed to go to equilibrium. CuCl2 (aq)  Cu+2 (aq) + 2 Cl- (aq) + Heat The final concentrations of all the reactants and products CuCl2 = [0.0250] Cu+2 = [1.25] Cl- = [0.625] What is the Equilibrium constant?

  10. SOLUTION Keq = [C]m[D]n [A]j[B]k Keq = [1.25][0.625]2 = 19.5 [0.0250]

  11. What does it mean? • If the Keq < 1, then Reactants are favored • If the Keq = 1, then Products and Reactants are equal • If the Keq > 1 then Products are favored • Since the Keq is 19.5 and 19.5 is greater then 1, more Products will be present then reactants.

  12. Another Example • The Keq for the following reaction is 130, If the concentration of Nitrogen is 0.100 M and the concentration of Hydrogen is 0.200 M, what is the concentration of Ammonia? N2 + 3 H2 2 NH3

  13. CONTINUED • 130 = [NH3]2 [N2][H2]3 • 130 = [NH3]2 [0.100 ][3 0.200]3 • 130 = [NH3]2 [0.0008] • 130 * 0.0008 = NH32 = 0.322 M/2 = 0.161 M

  14. HETEROGENOUS EQUILIBRIUM • Reactants or Products that are solids, and/or water are not included in the expression. H2SO4 (aq) + 2 NaOH (aq)  Na2SO4(aq) + 2 H2O (l) • H2SO4 = 0.100 M • NaOH = 0.200 M • Na2SO4 = 0.150 M • What is the Keq?

  15. Continued Keq = [0.150] = 37.5 [0.100][0.200]2 In this reaction, are the products or reactants favored? How do you know?

  16. A BIT HARDER 3 NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3 H2O (l) Na3PO4 = 0.200 M The Keq is 130 What is the concentration of H3PO4 and NaOH? We have two unknowns so we must have two equations: First Equation is: Keq = [0.200] = 130 [NaOH]3[H3PO4] What is the second equation?

  17. A BIT HARDER CONTINUED 3 NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3 H2O (l) Let’s replace NaOH with X and H3PO4 with Y Let’s look at the equation now: Keq = [0.200] = 130 [X]3[Y] If we look at the coefficents, we can see that: 3 NaOH = 1 H3PO4 We replace NaOH with X and H3PO4 with Y we get: 3 X = Y Let’s look at the equation now: Keq = [0.200] = 130 [X]3[3X]

  18. A BIT HARDER CONTINUED #2 Keq = [0.200] = [X]3[3X] 130 0.00154 = 3X4 0.000513 = X4 0.1505 = X = [NaOH] 0.1505 * 3 = [.4515] = [H3PO4]

  19. Ksp vs Keq • PbCl2, AgCl, and HgCl2 are considered insoluble in water. That mean they do not dissolve in water, right? • Actually, a very, very, very tiny amount will dissolve in water. • We use the Ksp (Constant for Solid Products) to determine the amount that will dissolve.

  20. Ksp vs Keq • jAB (s)  mA (aq) + nB (aq) Ksp = [A]m[B]n • Notice the Ksp equation uses only the concentration of the PRODUCTS, reactants are not included.

  21. EXAMPLE • Lead (II) Chloride is considered insoluble in water, but experiments show that a very tiny amount will dissolve in water, if the Ksp for PbCl2 is 3.40 x 10-15 what is the concentration of Lead and Chlorine ions? • Ksp = [A]m[B]n • PbCl2 (s)  Pb2+ (aq) + 2 Cl- (aq) • 3.40 x 10-15 = [Pb2+] [Cl-]2

  22. Ksp Continued • Pb2+ = 2 Cl- • X = 2Y • 3.40 x 10-15 = XY2 = 2Y*Y2 • 3.40 x 10-15 = 2Y3 • 3.40 x 10-15 = 2Y3 2 2 • 1.70 x 10-15 = [Y]3 • 1.19 x 10-5 = [Y] = [Cl-] • 2 (1.19 x 10-5) = [2.38 x 10-5] = [Pb2+]

More Related