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Today…

Today…. Turn in: Get out POGIL (Day 1) Our Plan: Test Results/Calendar POGIL discussion Intro to Equilibrium Notes Homework (Write in Planner): Begin Homework Problems Gizmos Due next class. Test Results. Chapter Fourteen. Chemical Equilibrium. What is Equilibrium?.

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Today…

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  1. Today… • Turn in: • Get out POGIL (Day 1) • Our Plan: • Test Results/Calendar • POGIL discussion • Intro to Equilibrium Notes • Homework (Write in Planner): • Begin Homework Problems • Gizmos Due next class

  2. Test Results

  3. Chapter Fourteen Chemical Equilibrium

  4. What is Equilibrium? • Reactions don’t always have a beginning and an end. They find a balance and have to work through stress to find it. • Many reaction never EVER finish. • They are reversible • Can go forward or backward without extra energy used

  5. What is Equilibrium? • Equilibrium state is reached when a reaction’s forward progress is perfectly balanced with the reverse process.

  6. Dynamic Nature of Equilibrium When a system reaches equilibrium, the forward and reverse reactions continue to occur … but at equal rates. We are usually concerned with the situation after equilibrium is reached. After equilibrium the concentrations of reactants and products remain constant.

  7. Concentration vs. Time Beginning with 1 M H2 and 1 M I2, the [HI] increases and both [H2] and [I2] decrease. If we begin with only 1 M HI, the [HI] decreases and both [H2] and [I2] increase. Beginning with 1 M each of H2, I2, and HI, the [HI] increases and both [H2] and [I2] decrease.

  8. Regardless of the starting concentrations; once equilibrium is reached … … the expression with products in numerator, reactants in denominator, where each concentration is raised to the power of its coefficient, appears to give a constant.

  9. [G]g[H]h Kc = [A]a[B]b The Equilibrium Constant Expression For the general reaction: aA + bBgG + hH The equilibrium expression is: Each concentration is simply raised to the power of its coefficient Products in numerator. Reactants in denominator.

  10. The Equilibrium Constant • The equilibrium constant is constant regardless of the initial concentrations of reactants and products. • This constant is denoted by the symbol Kcand is called the concentration equilibrium constant. • The constant is only good at a given temperature. If temperature is changed, so is the constant. • Concentrations of the products appear in the numerator and concentrations of the reactants appear in the denominator. • The exponents of the concentrations are identical to the stoichiometric coefficients in the chemical equation.

  11. Quick Check • Write the equilibrium constant expression for:

  12. Example 14.1 If the equilibrium concentrations of COCl2 and Cl2 are the same at 395 °C, find the equilibrium concentration of CO in the reaction: CO(g) + Cl2(g) COCl2(g) Kc = 1.2 x 103 at 395 °C

  13. Consider the reaction: 2 NO(g) + O2(g) 2 NO2(g) Now consider the reaction: 2 NO2(g) 2 NO(g) + O2(g) Modifying the Chemical Equation [NO2]2 Kc = ––––––––– = 4.67 x 1013 (at 298 K) [NO]2 [O2] What will be the equilibrium constant K'cfor the new reaction? [NO]2 [O2] 1 K'c = ––––––––– = ––––––––––– = [NO2]2 [NO2]2 ––––––––– [NO]2 [O2] 1 1 –– = ––––––––– = 2.14 x 10–14 Kc 4.67 x 1013

  14. Consider the reaction: 2 NO(g) + O2(g) 2 NO2(g) Now consider the reaction: NO2(g) NO(g) + ½ O2(g) [NO][O2]1/2 1 1/2 K"c = ––––––––– = ––– [NO2]2 Kc = 2.14 x 10–14 = 1.46 x 10–7 Modifying the Chemical Equation (cont’d) [NO2]2 Kc = ––––––––– = 4.67 x 1013 (at 298 K) [NO]2 [O2] What will be the equilibrium constant K"cfor the new reaction?

  15. Modifying the Chemical Equation (cont’d) • For the reverse reaction, K is the reciprocal of K for the forward reaction. • When an equation is divided by two, K for the new reaction is the square root of K for the original reaction. • General rule: • When the coefficients of an equation are multiplied by a common factor n to produce a new equation, we raise the original Kc value to the power n to obtain the new equilibrium constant. • It should be clear that we must write a balanced chemical equation when citing a value for Kc.

  16. Quick Check • So… • If we reverse a reaction, to find the new Kc we would… • If we double a reaction, to find the new Kc we would… • If we third a reaction, to find the new Kc we would…

  17. Example 14.2 The equilibrium constant for the reaction: ½ H2(g) + ½ I2(g) HI(g) at 718 K is 7.07. (a) What is the value of Kc at 718 K for the reaction HI(g) ½ H2(g) + ½ I2(g) (b) What is the value of Kc at 718 K for the reaction H2(g) + I2(g) 2 HI(g)

  18. Try It Out N2(g) + 3H2(g) ↔2NH3(g) For the equation above the Kc= 3.6 x 108 What is the value ofKc at the same temp for: NH3↔ ½ N2 + 3/2 H2 5.3 x 10-5

  19. Try It Out! N2(g) + 3H2(g) ↔2NH3(g) For the equation above the Kc= 3.6 x 108 Determine the value for Kc for the reaction: 1/3 N2 + H2↔2/3 NH3 7.1 x 102

  20. (PG)g(PH)h Kp = (PA)a(PB)b Equilibria Involving Gases • In reactions involving gases, it is often convenient to measure partial pressures rather than molarities. • In these cases, a partial pressure equilibrium constant, Kp, is used. Kc and Kp are related by: Kp = Kc (RT)Δn(gas) where Dn(gas) is the change in number of moles of gas as the reaction occurs in the forward direction. Dn(gas) = mol gaseous products – mol gaseous reactants

  21. Remember “R” .0821 atm · L mol · K Units in problem have to be Kelvin and atm!

  22. An Example 2H2S (g)↔ 2H2 (g)+ S2 (g) Kp= 1.2 x 10 -2 Kc = ? (@ 1065 deg C) 1.1 x 10-4

  23. Example 14.3 Consider the equilibrium between dinitrogentetroxide and nitrogen dioxide: N2O4(g) 2 NO2(g) Kp = 0.660 at 319 K (a) What is the value of Kc for this reaction? (b) What is the value of Kp for the reaction 2 NO2(g) N2O4(g)? (c) If the equilibrium partial pressure of NO2(g) is 0.332 atm, what is the equilibrium partial pressure of N2O4(g)?

  24. Suppose we need: N2O(g) + 3/2 O2(g) 2 NO2(g) Kc(1) = ?? N2O(g) + ½ O2(g) 2 NO(g) Kc(2) = 1.7 x 10–13 2 NO(g) + O2(g) 2 NO2(g) Kc(3) = 4.67 x 1013 The Equilibrium Constantfor an Overall Reaction and we’re given: • Adding the given equations gives the desired equation. • Multiplying the given values of K gives the equilibrium constant for the overall reaction. • (To see why this is so, write the equilibrium constant expressions for the two given equations, and multiply them together. Examine the result …)

  25. Let’s Try One… (#25) What is the value of Kp at 298 K for the reaction ½ CH4 (g) + H2O (g) ↔ ½ CO2 (g) + 2H2 (g), given the following data at 298 K? CO2 (g) + H2 (g) ↔ CO (g) + H2O (g) Kp = 9.80 x 10-6 CO (g) + 3 H2 (g) ↔ CH4 (g) + H2O (g) Kp = 8.00 x 1024

  26. Try It Out! (#27) • Determine Kc at 298 K for the reaction 2CH4 (g) ↔ C2H2 (g) + 3 H2 (g), given the following data at 298 K. CH4 (g) + H2O (g) ↔ CO (g) + 3 H2 (g) Kp= 1.2 x 10-25 2C2H2 (g) + 3O2 (g) ↔ 4CO (g) + 2H2O (g) Kp = 1.1 x 102 H2 (g) + ½ O2 (g) ↔ H2O (g) Kp= 1.1 x 1040

  27. Example: CaCO3(s) CaO(s) + CO2(g) Equilibria Involving PureSolids and Liquids • The equilibrium constant expression does not include terms for pure solid and liquid phases because their concentrations do not change in a reaction. • Although the amounts of pure solid and liquid phases change during a reaction, these phases remain pure and their concentrations do not change. [CaO] [CO2] Kc = –––––––––– [CaCO3] Kc = [CO2]

  28. Write the Kc expression for: C(s) + H2O (g) ↔CO(g) + H2(g) MgCO3(s) ↔ MgO(s) + CO2(g) 2H2(g) + O2(g)↔ 2H2O(l) H2S(g) + I2(s)↔ 2HI(g) + S(s)

  29. Example 14.4 The reaction of steam and coke (a form of carbon) produces a mixture of carbon monoxide and hydrogen, called water-gas. This reaction has long been used to make combustible gases from coal: C(s) + H2O(g)CO(g) + H2(g) Write the equilibrium constant expression for Kc for this reaction.

  30. Equilibrium Constants: When Do We Need Them and When Do We Not? • A very large numerical value of Kc or Kp signifies that a reaction goes (essentially) to completion. • A very small numerical value of Kc or Kp signifies that the forward reaction, as written, occurs only to a slight extent. • An equilibrium constant expression applies only to a reversible reaction at equilibrium. • Although a reaction may be thermodynamically favored, it may be kinetically controlled … • Thermodynamics tells us “it’s possible (or not)” • Kinetics tells us “it’s practical (or not)”

  31. 2H2(g) + O2(g) ↔2 H2O(g) Kp = 1.4 x 10 83 @ 298 K A term in the denominator of the equilibrium approaches zero and makes the value of the equilibrium constant very large. Very large Kc or Kpfavors forward reaction

  32. C(s)+ H2O(g) ↔CO(g) + H2(g) Kp = 1.6 x 10-21 @ 298 K To account for a small numerical value of an equilibrium constant, the numerator must be very small. Very small Kc or Kp favors reverse reaction

  33. SO….. A reaction is most likely to reach a state of equilibrium in which significant quantities of both reactants and products are present if the numerical value of Kc and Kpis neither very large nor very small, roughly in the range from 10-10 to 1010

  34. The Reaction Quotient, Q • For nonequilibrium conditions, the expression having the same form as Kc or Kpis called the reaction quotient, Qc or Qp. • The reaction quotient is not constant for a reaction, but is useful for predicting the direction in which a net change must occur to establish equilibrium. • To determine the direction of net change, we compare the magnitude of Qc to that of Kc.

  35. When Q is smaller than K, the denominator of Q is too big; we have “too much reactants.” When Q =K, equilibrium has been reached. When Q is larger than K, the numerator of Q is too big; we have “too much products.” The Reaction Quotient, Q

  36. A better graphic

  37. CO(g) + 2 H2(g)↔ CH3OH (g) If the Kc for this reaction = 14.5, then what direction will the reaction proceed if the initial concentrations are CO = 0.100 M H2 = 0.100 M CH3OH = 0 M Left to right

  38. Try It Out! (#49) • If a gaseous mixture at 588 K contains 20.0 g each of CO and H2O and 25.0 g each of CO2 and H2, in what direction will a net reaction occur to reach equilibrium? Explain • CO (g) + H2O (g) ↔ CO2 (g) + H2 (g) Kc = 31.4 at 588 K

  39. Stop – Homework Time • You should be able to complete problems 20, 22, 24, 26, 28, 30, 32, 50, & 52

  40. Today… • Turn in: • Gizmos - basket • Our Plan: • Review/Quiz • LeChatelier’s POGIL • Notes • Demonstration • Homework (Write in Planner): • Investigation 13 Pre-Lab

  41. Quick Check • For the reaction 2NO(g) ↔ N2 (g) + O2 (g), Kc = 9.92 at 2273 K. • Write the equilibrium expression (Kc) for this reaction. • What is Kc for this reaction: 2N2(g) + 2O2 (g) ↔ 4NO(g) (0.0102) • At equilibrium, the concentration of nitrogen and oxygen are both 1.5 x 10-3 M. What is the concentration of NO? (4.76 x 10-4 M) • What is Kp for the reaction? (9.92)

  42. Stresses on Equilibrium • Chemists make a living preventing equilibrium. They want to “knock it out of whack” to maximize a certain product. • Stresses can be applied to shift equilibrium and maximize the amount of product, as you saw in the POGIL.

  43. Le Châtelier’s Principle • When any change in concentration, temperature, pressure, or volume is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change. • Analogy: Begin with 100 men and 100 women at a dance. • Assume that there are 70 couples dancing, though not always the same couples (dynamic equilibrium). • If 30 more men arrive, what happens? • The equilibrium will shift, and shortly, more couples will be dancing … but probably not 30 more couples. • Real Example – Deer populations

  44. Changing the Amounts ofReacting Species (concentration) • At equilibrium, Q = Kc. • If the concentration of one of the reactants is increased, the denominator of the reaction quotient increases. • Q is now less thanKc. • This condition is only temporary, however, because the concentrations of all species must change in such a way so as to make Q = Kc again. • In order to do this, the concentrations of the productsincrease; the equilibrium is shifted to the right.

  45. Concentration • An increase in concentration will cause the reaction to shift to the opposite side until a new equilibrium is achieved.

  46. Important Note • Remember that this is a change in concentration so changing the amount of a pure solid or liquid which is already in equilibrium should not make the reaction consume the additional solid or liquid.

  47. Example • Example: 2HgO (g) ↔2Hg (l) + O2 (g) ∆ H = +43.4 kcal • Add HgO, what happens to equilibrium? • Add O2, what happens to equilibrium? • Add Hg, what happens to equilibrium?

  48. Example 14.7 Water can be removed from an equilibrium mixture in the reaction of 1-octanol and acetic acid, for example, by using a solid drying agent that is insoluble in the reaction mixture. Describe how the removal of a small quantity of water affects the equilibrium. CH3(CH2)6CH2OH(aq) + CH3COOH(aq) CH3(CH2)6CH2OCOCH3(aq) + H2O(soln) H+ p. 591 – As water is removed, the reverse reaction is slowed and the forward, water-forming reaction is stimulated. Not all of the water that is removed can be replaced, however, because a new equilibrium could not tolerate a constant concentration of H2O at the same time that the other three concentrations change. Thus, in the new equilibrium the amount of water is somewhat less than in the original equilibrium, the amount of octyl acetate is greater, and the amounts of both 1-octanol and acetic acid are less.

  49. Changing External Pressure or Volume in Gaseous Equilibria • When the external pressure is increased (or system volume is reduced), an equilibrium shifts in the direction producing the smaller number of moles of gas. • When the external pressure is decreased (or the system volume is increased), an equilibrium shifts in the direction producing the larger number of moles of gas. • If there is no change in the number of moles of gas in a reaction, changes in external pressure (or system volume) have no effect on an equilibrium. Example: H2(g) + I2(g) 2 HI (g) What would happen if pressure is increased?

  50. Example • Example: 2HgO (g) ↔2Hg (l) + O2 (g) ∆ H = +43.4 kcal • Add pressure, what happens to equilibrium? • Increase volume, what happens to equilibrium?

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