DownloadChapters 16 and 17 Chemical Equilibrium and Solubility and Precipitation






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Goals for Chapters16 and 17. Understand the concept of equilibriumWrite equilibrium expressions
Chapters 16 and 17 Chemical Equilibrium and Solubility and...

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1. Chapters 16 and 17 ? Chemical Equilibrium and Solubility and Precipitation Chapters 16 and 17

2. Goals for Chapters16 and 17 Understand the concept of equilibrium Write equilibrium expressions ? solve for Kc and Kp Solve problems for equilibrium concentrations Use Qc to predict where the equilibrium will shift Understand Le Ch?telier?s Principle Understand catalysts Write solubility products (Ksp) Basic calculations involving Ksp Understand the Common Ion Effect

3. Equilibrium ? the extent of reaction Until now we have discussed reactions that go to completion ? theoretical yield of 100%. The Silver Lab went to 100% completion. However, many reactions do not actually proceed to 100% completion. Why not? Equilibrium looks at the extent of a chemical reaction.

4. The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. We cannot tell if a reaction is happening or not ? but it is. Molecules are colliding with each other causing reactions - forming reactants from products or products from reactants.

5. What is happening? As the substance warms it begins to decompose: N2O4(g) ? 2NO2(g) When enough NO2 is formed, it can react to form N2O4: 2NO2(g) ? N2O4(g) At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4 The double arrow implies the process is dynamic.

6. It does not matter what you start with!

7. Experimental Data

8. A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. This does not mean the concentrations are the same.

9. How do we show a reaction is at equilibrium? We know that both the forward and reverse reactions are being carried out ? so we can just use a double-headed arrows.

10. The Equilibrium Constant To generalize this expression, consider the reaction:

11. Derivation of Kp Relationship between concentration and pressure obtained from the ideal gas law. Recall PV = nRT or Substitute for P in equilibrium expression. Consider the reaction: aA + bB ? cC + dD Use this relationship to relate KP and Kc

12. Write these reactions and also write the equilibrium expression (E) Ozone going to molecular oxygen Nitrogen monoxide and chlorine gas yielding NOCl gas. Hydrogen gas plus gaseous iodine producing hydrogen iodide gas.

13. Pretty easy thus far - how about some numbers? (M) The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate Kc.

14. What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium.

15. Homogeneous versus heterogeneous equilibrium Thus far all reactants and products have been in the same phase ? however, sometimes reactions take place in different phases.

16. Picture of what is happening

17. Review

18. Equilibrium Calculations ? a different kind of ICE than you are used to. A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448?C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448?C for the reaction taking place, which is

19. ICE tables are important

20. So, the equilibrium constant is calculated as follows

21. You try one Sulfur trioxide gas decomposes at high temperature in a sealed container to sulfur dioxide gas and oxygen gas. The vessel is at 1000K with a sulfur trioxide concentration of 6.09 x 10-3 M. At equilibrium the SO3 concentration is 2.44 x 10-3 M. Calculate Kc.

22. What if all you know is the Kc and the starting concentrations? (H) At 12800C the equilibrium constant for the reaction below is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

23. How do we solve for x?

25. You try one! Phosphorus pentachloride gas decomposes into phosphorus trichloride gas and chlorine gas. The Kp (just use atm) is 0.497 at 500K. What are the equilibrium pressures of all three chemicals at this temperature? Starting pressure of phosphorus pentachloride is 1.66 atm. What is the Kc?

26. Approximating ? makes life simpler We know that when Keq is small the reaction will not proceed very far to the right (reactant favored). This means that the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants. e.g. 0.20 ? x is just about 0.20 (x is small) If the difference between Keq and initial concentrations is around 3 orders of magnitude or greater the you can ignore x if it is subtracted or added in a term. Otherwise, you have to use the quadratic.

27. Example of approximating

28. Can we predict the direction of a reaction? Yes ? it is called a Q calculation. (E) To calculate Q, one substitutes the initial concentrations into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

29. What affects Equilibrium? Le Ch?telier?s Principle ? if a system at equilibrium is disturbed by a change in the concentration, T, V, or P of one of the components, the system will shift its equilibrium position so as to counteract the affect of the disturbance.

30. Change in V or P

31. Changes in T Think about the chapters you just finished ? reactions give of or take up heat. How do you think this plays into Le Ch?telier?s Principle

32. The Effect of Changes in Temperature

33. How about adding a catalyst? Adding a Catalyst does not change K, and is not consumed in the reaction does not shift the position of an equilibrium system system will reach equilibrium sooner

35. Solubility Equilibria Remember your little chart that tells you if a compound is soluble? Remember when I said that compound is insoluble ?as far as you know??

36. The usual Rules regarding Solubility Salts are generally more soluble in HOT water (Gases are more soluble in COLD water) Alkali Metal salts are very soluble in water. NaCl, KOH, Li3PO4, Na2SO4 etc... Ammonium salts are very soluble in water. NH4Br, (NH4)2CO3 etc? Salts containing the nitrate ion, NO3-, are very soluble in water. Most salts of Cl-, Br- and I- are very soluble in water - exceptions are salts containing Ag+ and Pb2+. soluble salts: FeCl2, AlBr3, MgI2 etc... ?insoluble? salts: AgCl, PbBr2 etc...

37. Dissolving a salt ? what happens? A salt is an ionic compound - usually a metal cation bonded to a non-metal anion. The dissolving of a salt is an example of equilibrium. The cations and anions are attracted to each other in the salt. They are also attracted to the water molecules. The water molecules will start to pull out some of the ions from the salt crystal.

38. At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions. However, soon the ions floating in the water begin to collide with the salt crystal and are ?pulled back in? to the salt. (precipitation) Eventually the rate of dissociation is equal to the rate of precipitation. The solution is now saturated. It has reached equilibrium.

39. Solubility Equilibrium: Dissociation = Precipitation

40. Dissolving silver sulfate, Ag2SO4, in water When silver sulfate dissolves it dissociates into ions. When the solution is saturated, the following equilibrium exists: Ag2SO4 (s) ? 2 Ag+ (aq) + SO42- (aq) Since this is an equilibrium, we can write an equilibrium expression for the reaction: Ksp = [Ag+]2[SO42-] Notice that the Ag2SO4 is left out of the expression! Why? Since K is always calculated by just multiplying concentrations, it is called a ?solubility product? constant - Ksp.

41. Calculating Ksp of Silver Chromate A saturated solution of silver chromate, Ag2CrO4, has [Ag+] = 1.3 x 10-4 M. What is the Ksp for Ag2CrO4? Ag2CrO4 (s) ? 2 Ag+ (aq) + CrO42- (aq) No need for a table here ? if the silver concentration is 1.3 x 10-4M, what must then be the chromate anion concentration? Ksp = [Ag+]2[CrO42-] Ksp = (1.3 x 10-4 )2 (6.5 x 10-5) = 1.1 x 10-12

42. Calculating the Ksp of silver sulfate The solubility of silver sulfate is 0.0144 mol/L. This means that 0.0144 mol of Ag2SO4 will dissolve to make 1.0 L of saturated solution. Calculate the value of the equilibrium constant, Ksp for this salt. Ag2SO4 (s) ? 2 Ag+ (aq) + SO42- (aq) 0 0 + 2s + s 2s s Ksp = [Ag+]2[SO42-] = (2s)2(s) = (4s2)(s) = 4s3 We know: s = 0.0144 mol/L Ksp = 4(0.0144)3 = 1.2 x 10-5

43. Calculating solubility given Ksp The Ksp of NiCO3 is 1.4 x 10-7 at 25?C. Calculate its molar solubility. NiCO3 (s) ? Ni2+ (aq) + CO32- (aq) --- --- + s + s s s Ksp = [Ni2+][CO32-] 1.4 x 10-7 = s2 s = = 3.7 x 10-4 M

44. Calculate the solubility of MgF2 in water. What mass will dissolve in 2.0 L of water? Ksp for MgF2 is 7.4 x 10-11 MgF2 (s) ? Mg2+ (aq) + 2 F- (aq) ---- ---- + s + 2s s 2s Ksp = [Mg2+][F-]2 = (s)(2s)2 = 4s3 Ksp = 7.4 x 10-11 = 4s3 s = 2.6 x 10-4 mol/L

45. The solubility of MgF2 in pure water is 2.6 x 10-4 mol/L. What happens to the solubility if we dissolve the MgF2 in a solution of NaF (which is quite soluble), instead of pure water? The Common Ion Effect on Solubility

46. Explaining the Common Ion Effect The presence of a common ion in a solution will lower the solubility of a salt. Le Ch?telier?s Principle: The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower!

47. Ksp and Solubility Generally, we can state that salts with very small Ksp values are only sparingly soluble in water. It is OK to compare two salts for solubility if they have the same number of ions. CuI has Ksp = 5.0 x 10-12 and CaSO4 has Ksp = 6.1 x 10-5. Since the Ksp for calcium sulfate is larger than that for the copper (I) iodide, we can state that calcium sulfate is more soluble in water.

48. But!

49. Application: Mixing Solutions - Will a Precipitate Form?

50. Pb(NO3)2 (aq) + K2CrO4 (aq) ? PbCrO4 (s) + 2 KNO3 (aq) Step 1: Is a sparingly soluble salt formed? We can see that a double replacement reaction can occur and produce PbCrO4. Since this salt has a very small Ksp, it may precipitate from the mixture. The solubility equilibrium is: PbCrO4 (s) ? Pb2+ (aq) + CrO42- (aq) Ksp = 2 x 10-16 = [Pb2+][CrO42-] If a precipitate forms, it means the solubility equilibrium has shifted BACKWARDS. This will happen only if Qsp > Ksp.

51. Step 2: Find the concentrations of the ions that form the sparingly soluble salt. Since we are mixing two solutions in this example, the concentrations of the Pb2+ and CrO42- will be diluted. We have to perform a dilution calculation. Dilution: M1V1 = M2V2 [Pb2+] = [CrO42-] =

52. Step 3: Calculate Qsp for the mixture. Qsp = [Pb2+][CrO42-] = (0.0080 M)(0.020 M) Qsp = 1.6 x 10-4 Step 4: Compare Qsp to Ksp. Since Qsp >> Ksp, a precipitate will form when the two solutions are mixed! Note: If Qsp = Ksp, the mixture is saturated If Qsp < Ksp, the solution is unsaturated Either way, no ppte will form!

53. What?s Next? Chapters 18 and 19 ? Acids, Bases and Salts & Reactions of Acids and Bases Please R&O Chapters 18 and 19 ? print off all materials listed for this chapter.


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