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Chemical Equilibrium

Chemical Equilibrium. Chapter 15. The Concept of Chemical Equilibrium. Chemical equilibrium occurs when opposing reactions are proceeding at equal rates. Like a bridge. Equilibrium is reached when you have a mixture of products and reactants whose concentrations no longer change with time.

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Chemical Equilibrium

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  1. Chemical Equilibrium Chapter 15

  2. The Concept of Chemical Equilibrium • Chemical equilibrium occurs when opposing reactions are proceeding at equal rates. • Like a bridge. • Equilibrium is reached when you have a mixture of products and reactants whose concentrations no longer change with time. • N2O4 2NO2

  3. At equilibrium, the concentrations of reactants and products are no longer changing with time. • For equilibrium to occur, neither reactants nor products can escape from the system. • At equilibrium a particular ratio of the concentration terms equals a constant.

  4. The Equilibrium Constant • aA + bBcC + dD • Kc = [C]c[D]d/[A]a[B]b • The equilibrium constant depends only on the stoichiometry of the reaction, not on its mechanism.

  5. Write the equilibrium expression for the following rations: • 2O3 3O2 • 2NO + Cl2 2NOCl • Ag+ + 2NH3 Ag(NH3)2+

  6. The value of Keq is independent of the starting concentrations.

  7. Equilibrium Constant and Pressure • When the reactants and products of a reaction are gases we can express the equilibrium constant expression in terms of partial pressure. • aA(g) + bB(g) cC(g) + dD(g) • N2O4(g)  2 NO2(g) • Note Kp is usually numerically different from Kc

  8. In the synthesis of ammonia from nitrogen and hydrogen Kc = 9.6 at 300 oC. Calculate Kp for this reaction.

  9. Working with Equilibrium Constants • CO(g) + Cl2(g)  COCl2(g) • Kc = 4.56 x 109 • We would say that this equilibrium lies to the right. Meaning that there is a larger concentration of products than reactants.

  10. The Direction of Chemical Equilibrium and K • Equilibrium can be reached from either direction. • N2O4 2NO2 • Kc = 0.212 • 2NO2 N2O4 • Kc = 4.72

  11. Relating Chemical Equations and Equilibrium Constants • If we multiply a chemical equation by a certain factor it changes the equilibrium constant. • 2 N2O4 4 NO2 • Kc = 0.0449

  12. Sometimes we have overall reaction equations that are the result of two or more steps. • 2 NOBr 2 NO + Br2Kc = • Br2 + Cl2 2 BrClKc = • Net process: • 2 NOBr + Cl2 2 NO + 2 BrCl

  13. Summary • The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium constant of the forward reaction. • The equilibrium constant of the reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number. • The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

  14. Heterogeneous Equilibrium • Many equilibria involve substances that are all in the same physical state. • These are called homogeneous equilibria • An equilibrium system in which the substances are in different physical states is called a heterogeneous equilibrium. • Example: • PbCl2(s)  Pb2+(aq) + 2 Cl-(aq) • Whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium constant expression.

  15. Write the equilibrium constant expressions for the following equations: • CO2(g) + H2(g) CO(g) + H2O(l) • SnO2(s) + 2 CO Sn(s) + 2 CO2 • CaCO3(s) CaO(s) + CO2(g)

  16. Calculating Equilibrium Constants • A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472 oC. The equilibrium mixture of the gases was analyzed and found to contain 7.38 atm H2, 2.46 atm N2 and 0.166 atm NH3. From this data calculate the equilibrium constant. • Kp = 2.79 x 10-5

  17. Sometimes we will not know the equilibrium concentrations for all of the substances. • In this case we will use an I.C.E. chart. • Problem: • A closed system initially containing 1.0 x 10-3 M H2 and 2.0 x 10-3 M I2 at 448 oC 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 for the reaction.

  18. Applications of Equilibrium Constants • Remember that if K is very large the equilibrium mixture will contain mostly products. • If K is small the mixture will contain mostly reactants. • The equilibrium constant allows us to: • Predict the direction in which a reaction mixture will proceed to achieve equilibrium. • Calculate the concentrations of reactants and products when equilibrium has been reached.

  19. A 1.0 L flask is filled with 1.0 mol of H2 and 2.0 mol of I2 at 448 oC. The value of the equilibrium constant Kc for the reaction: H2(g) + I2(g)  2 HI(g) What are the equilibrium concentrations of H2, I2, and HI?

  20. Le Chatelier’s Principle • This principle states that: • If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of on of the components, the system will shift its equilibrium position to counter act the effect of the disturbance. • If the temperature of the equilibrium remains constant so must Kc and Kp

  21. Acids and Bases Chapter 16

  22. Acids and Bases • An acid is a substance that, when dissolved in water, increases the concentration of H+ ions in solution. • HCl(aq) H+(aq) + Cl-(aq) • A base is a substance that, when dissolved in water, increases the concentration of OH- ions in solution. • NaOH(aq) Na+(aq) + OH-(aq)

  23. Bronsted-Lowry Acids and Bases • The arrhenius model of acids and bases is restricted to aqueous solutions. • The Bronsted-Lowry model is based on the transfer of protons (H+) • Acids are substances that donate protons to other substances. • Bases are substances that accept protons.

  24. Conjugate Acid Base Pairs • Any acid and base that only differ in the presence or absence of a proton are referred to as a conjugate acid-base pair. • The stronger the acid, the weaker its conjugate base. • The stronger the base, the weaker its conjugate acid.

  25. Strengths of Acids and Bases • A strong acid completely transfers its protons to water, leaving no undissociated molecules in solution. • A weak acid only partially dissociates in aqueous solution. They exist as a mixture of protonated and the constituent ions. • The conjugate base of a weak acid is a weak base. • A substance with negligible acidity, such as CH4, contains hydrogen but does not demonstrate acidic behavior in water. Its conjugate base is a strong base.

  26. Water (Is totally sweet) • Water can act as either an acid or a base. • Autoionization:

  27. The pH scale • pH related to the concentration of H+ ions in solution. • pH = - log[H+] • Solutions were [H+] > [OH-] will have a pH < 7 • Solutions where [H+] < [OH-] will have a pH > 7 • Calculate the [H+] of a solution with a pH of 3.76

  28. calculate the pH for solutions with the following [H+]: • [H+] = 1 x 10-12 M • 12.00 • [H+] = 2 x 10-6 M • 2.3 • [H+] = 3.8 x 10-4 M • 3.42 What is the [H+] of a solution with a pH of 8.28 • 5.3 x 10-9 M

  29. Strong Acids • HNO3(aq) + H2O(l)  H3O+(aq) + NO3-(aq) • Complete ionization • What is the pH of a solution of 0.2 M HNO3? • 0.7 • What is the pH of a 0.04M solution of HClO4? • 1.40 • An aqueous solution of HNO3 has a pH of 2.34. What is the concentration of the acid? • 0.0046 M

  30. Strong Bases • NaOHNa+(aq) + OH-(aq) • Completely ionized • What is the pH of a 0.028 M solution of NaOH? • 12.45 • What is the [H+] in a solution of Ca(OH)2 with a pOH of 2.66? • 4.55 x 10-12 M

  31. Weak acids • HA(aq) + H2O(l)  H3O+(aq) + A-(aq) • The larger the value of Ka the stronger the acid.

  32. Calculating Ka • A student prepared a 0.10 M solution of formic acid HCOOH and measure its pH. At 25oC the pH was found to be 2.38. Calculate Ka for this acid. • Niacin, on of the B vitamins has the formula NC5H4COOH. Calculate the Ka of this acid if a 0.02 M solution has a pH of 3.62.

  33. Percent Ionization • Percent ionization = • Calculate the percent ionization of a 0.035 M solution of HNO2 wit a pH of 2.43. • 11%

  34. Using Ka to calculate pH • Calculate the pH of a 0.3 M solution of CH3COOH (Ka = 1.8 x 10-5)

  35. Polyprotic Acids • H2SO3(aq) H+(aq) + HSO3-(aq) Ka1 = 1.7 x 10-2 • HSO3-(aq) H+(aq) + SO3-(aq) Ka2 = 6.4 x 10-8 • In general its always easier to remove the first hydrogen than the second hydrogen. • In some cases Ka1 >>> Ka2 • These acids can be treated as monoprotic strong acids.

  36. What is the pH of a 0.0037 M solution of H2CO3? • Since Ka1>>>Ka2 we can consider the equilibrium as:

  37. Weak Bases • B(aq) + H2O(l) HB+(aq) + OH-(aq) • Kb = • Calculate [OH-] in a 0.15 M solution of NH3

  38. Relationship Between Ka and Kb • In general stronger acids have weaker conjugate bases. • NH4+ and NH3 • NH4+(aq)  NH3(aq) + H+(aq) • NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

  39. Acid-Base Behavior and Chemical Structure • A molecule containing H will only transfer a proton if the H-X bond is polarized: • The strength of the bond also determines the strength of an acid. • Very strong bonds, like in HF, do not dissociate as easily. • The H-X bond strength tends to decrease as X increases in size. • HCl is stronger than HF

  40. The Common Ion Effect • If we examine a solution with a weak acid and a salt of that weak acid we would see that the components of the solution contain a similar ion. • The presences of extra CH3COO- causes the acid to ionize less than it normally would.

  41. What is the pH of a solution made by adding 0.3 mol of acetic acid and 0.3 mol of sodium acetate in a 1.0 L aqueous solution.

  42. Buffer Solutions • Solutions that contain weak conjugate acid-base pairs act as buffers • Buffers resist drastic change in pH upon the addition of small amounts of strong acid or strong base. • HX(aq) H+(aq) + X-(aq) • Ka =

  43. The pH of a buffer solution • pH = pKa + Log(Base/Acid) • What is the pH of a buffer that is 0.12M in lactic acid (HC3H5O3) and 0.1M in sodium lactate (NaC3H5O3)?

  44. How many moles of NH4Cl must be added to 2.0 L of 0.1 M NH3 to form a buffer whose pH is 9.00?

  45. Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M sodium lactate. (b) Calculate the pH of a buffer formed by mixing85 mL of 0.13 M lactic acid with 95 mL of 0.15 M sodium lactate

  46. Calculate the pH of a buffer that is 0.105 M in NaHCO3 and 0.125 M in

  47. Buffer Capacity and pH Range • Buffer capacity refers to the amount of acid or base the buffer can neutralize before the pH begins to change. • The buffer capacity is directly related to the amount of acid and base in the buffer. • The pH range of any buffer is the pH range over which the buffer acts effectively.

  48. Addition of strong acids and bases to buffers • Consider a buffer that contains a weak acid, HX and its conjugate base X-. • When a strong acid is added the H+ that is produced is consumed by X- to produce HX • When a strong base is added the OH- that is produced is consumed by HX to produce X-. • To calculate how a buffer responds to the addition of a strong acid or base: • Condister the acid-base neutralization reaction, and determine its effect on [HX] and [X-]. • Use Ka and the new concentrations of HX and X- to calculate [H+].

  49. A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol of CH3OONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. • Calculate the pH of this solution after 0.020 Mol of NaOH is added. • For comparison calculate the pH that would result if 0.020 mol of NaOH were added to 1.00 L of pure water. (neglect any volume changes)

  50. Acid Base Titrations • Titration is a process used to determine the concentration of an unknown solution. • Acid base titrations are monitored by using the pH of the solution as either the acid or base is added.

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