# Suppose human’s memory decays exponentially as a first order - PowerPoint PPT Presentation Download Presentation Suppose human’s memory decays exponentially as a first order

Suppose human’s memory decays exponentially as a first order Download Presentation ## Suppose human’s memory decays exponentially as a first order

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1. Suppose human’s memory decays exponentially as a first order reaction and the half life is one day. Three days have passed since you came to class last Friday. If you have not reviewed what you learned, what is the percentage of chemistry you remember from last Friday? IF you need the equations for first order reaction, here they are r = k [A]

2. Chapter 14 Chemical Equilibrium

3. phase equilibrium Liquid Gas

4. Equilibrium Two opposite processes have reached the same rate. Note: equilibrium is dynamic!

5. Reactants  Products Reactants  Products Reversible Reaction Reactants ⇌ Products

6. kf N2O4(g) ⇌ 2NO2(g) kr Both directions are elementary reactions What are the differential rate laws? rf = kf[N2O4] rr = kr[NO2]2

7. kf N2O4(g) ⇌ 2NO2(g) kr Suppose we start with a flask of pure N2O4 gas rf = kf[N2O4] rr = kr[NO2]2 At equilibrium, the forward and reverse reactions are proceeding at the same rate. rf = rr

8. kf N2O4(g) ⇌ 2NO2(g) rf = kf[N2O4] = rr = kr[NO2]2 kr Once equilibrium is achieved, the concentration of each reactant and product remains constant.

9. kf N2O4(g) ⇌ 2NO2(g) kr rf = kf[N2O4] rr = kr[NO2]2 kf[N2O4] = kr[NO2]2 K =

10. a A + b B ⇌ c C + d D K: equilibrium constant. [ ]: concentration at equilibrium! K: independent of concentration, dependent upon temperature. recall d=m/V K: no unit.

11. Write the equilibrium expression for K for the following reactions: (a) 2O3(g) ⇌ 3O2(g) (b) 2NO(g) + Cl2(g) ⇌ 2NOCl(g) (c) Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) a A + b B ⇌ c C + d D

12. At one temperature: one K, infinite number of equilibrium concentrations. K is independent of concentrations. Recall d = m/V.

13. What Does the Value of K Mean?

14. a A + b B ⇌ c C + d D • If K>>1, the reaction is product-favored; product predominates at equilibrium. • If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.

15. N2(g) + 3H2(g) ⇌ 2NH3(g) Haber process

16. Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g) • The following equilibrium concentrations were observed for the Haber process • at 127 °C: [NH3] = 3.1 x 10−2 mol/L, [N2] = 8.5 x 10−1 mol/L, • [H2] = 3.1 x 10−3 mol/L. • Calculate the value of K at 127 °C for this reaction. • Calculate the value of the equilibrium constant at 127 °C for the reaction • 2NH3(g) ⇌ N2(g) + 3H2(g) c) Calculate the value of the equilibrium constant at 127 °C for the reaction given by the equation ⇌ NH3(g)

17. reverse a A + b B ⇌ c C + d D c C + d D⇌a A + b B x n na A + nb B ⇌ nc C + nd D

18. For practice 14.2, page 622 Consider the following chemical equation and equilibrium constant at 25 °C: 2COF2(g) ⇌ CO2(g) + CF4(g) K = 2.2 x 106 Compute the equilibrium constant for the following reaction At 25 °C: 2CO2(g) + 2CF4(g) ⇌ 4COF2(g) K’ = ?

19. Equilibrium constant of a composite reaction 2 NOBr (g) ⇌ 2 NO (g) + Br2(g) Br2 (g) + Cl2 (g) ⇌ 2 BrCl (g) + 2NOBr (g) + Cl2 (g) ⇌ 2NO (g) + 2BrCl (g) Sum of reactions  Product of equilibrium constants

20. HF (aq) ⇌ H+ (aq) + F− (aq) K1 = 6.8 x 10−4 H2C2O4 (aq) ⇌ 2H+ (aq) + C2O42− (aq) K2 = 3.8 x 10−6 Determine the equilibrium constant for 2HF (aq) + C2O42− (aq) ⇌ 2F− (aq) + H2C2O4 (aq) K3 = 0.12 Method: take linear combination of known reactions to construct target reaction.

21. 2NO(g) + O2(g) ⇌ 2NO2(g) K = 5.0 x 1012 What is the equilibrium constant for NO2(g) ⇌NO(g) + ½ O2(g)

22. N2(g) + 3H2(g) ⇌ 2NH3(g) = Kc a A (g) + b B (g) ⇌ c C (g) + d D (g) ∆n = Sum of the coefficients of gaseous products − Sum of the coefficients of gaseous reactants

23. Example 14.3, page 624 Kp for the following reaction is 2.2 x 1012 at 25 °C, calculate the value of Kc. 2NO(g) + O2(g) ⇌ 2NO2(g) Try For Practice 14.3 on the same page

24. Consider the equilibrium N2(g) + O2(g) + Br2(g) ⇌ 2NOBr(g) Calculate the equilibrium constant Kp for this reaction, given The following information (at 298 K): 2NO(g) + Br2(g) ⇌ 2NOBr(g) Kc, 1 = 2.0 2NO(g) ⇌ N2(g) + O2(g) Kc, 2 = 2.1 x 1030 Answer: Kp = 4.3 x 1028

25. Equilibrium Category based on Phase

26. Homogeneous Equilibrium All species have the same phase N2(g) + 3H2(g) ⇌ 2NH3(g) Heterogeneous Equilibrium Not all species have the same phase Kc, Kp 2CO(g) ⇌ CO2(g) + C(s)

27. The concentration of pure solid or liquid is not included in the equilibrium constant expression for heterogeneous equilibria.

28. 2H2O(l) ⇌ 2H2(g) + O2(g) 2H2O(g) ⇌ 2H2(g) + O2(g)

29. Write the expression for K and Kp for the following processes: • PCl5(s) ⇌ PCl3(l) + Cl2(g) • CuSO4 • 5H2O (s) ⇌ CuSO4(s) + 5H2O(g) blue white

30. Hydrated Copper (II) Sulfate on the Left. Water Applied to Anhydrous Copper (II) Sulfate, on the Right, Forms the Hydrated Compound

31. a A + b B ⇌ c C + d D [ ]: concentration at equilibrium! reaction quotient [ ]: concentration at a particular moment.

32. a A + b B ⇌ c C + d D Q = K, system (mixture) is at equilibrium, rr = rf Q > K → rr > rf → system will shift to left to reach equilibrium Q < K → rr < rf → system will shift to right to reach equilibrium

33. Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g) • At 500 °C, K = 6.01 x 10−2 . Predict the direction in which the • system will shift to reach equilibrium in each of the following • cases: • [NH3]0=1.0 x 10−3 M; [N2]0 = 1.0 x 10−5 M; [H2]0 = 2.0 x10−3 M • [NH3]0=2.00 x 10−4 M; [N2]0 = 1.50 x 10−5 M; [H2]0 =3.54 x10−1 M • [NH3]0=1.0 x 10−4 M; [N2]0 = 5.0 M; [H2]0 = 1.0 x10−2 M b) Q = 6.01 x 10−2 c) Q = 2.0 x 10−3 a) Q = 1.3 x 107

34. value of Kc or Kp equilibrium concentrations or pressures

35. N2O4(g) ⇌ 2NO2(g) N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2 (g). 0.600 atm

36. Try Example 14.8 and For Practice 14.8 on page 632

37. CO(g) + H2O(g) ⇌ CO2(g) + H2(g) At 700 K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000 L flask. x = 0.387 (M)

38. Example 14.11, page 636 Consider the following reaction: I2(g) + Cl2(g) ⇌ 2ICl(g) Kp = 81.9 A reaction mixture at 25 C initially contains PI2 0.100 atm, PC12 0.100 atm, and PICl = 0.100 atm. Find the equilibrium partial pressures of I2, Cl2, and ICl at this temperature.

39. Try Example 14.9 and For Practice 14.9 on page 634.

40. H2(g) + F2(g) ⇌ 2HF(g) 3.000 mol H2 and 6.000 mol F2 are mixed in a 3.000 L flask. K at this temperature is 115. Calculate the equilibrium concentration of each component. x = 2.14 (M) or 0.968 (M)

41. Example 14.10, page 634 N2O4(g) ⇌ 2NO2(g) Kc = 0.36 at 100 °C. A reaction mixture at 100 °C initially contains [NO2] = 0.100 M. Find the equilibrium concentration of NO2 and N2O4 at this temperature.

42. H2(g) + I2(g) ⇌ 2HI(g) Kp = 1.00 x 102. Suppose HI at 5.000 x 10−1 atm, H2 at 1.000 x 10−2 atm, and I2 at 5.000 x 10−3 atm are mixed in a 5.000-L flask. Calculate the equilibrium pressure of each component. x = 3.55 x 10−2 (atm) or −7.19 x 10−2 (atm)

43. Example 14.12, page 638 2H2S(g) ⇌ 2H2(g) + S2(g) Kc = 1.67 x10−7 at 800 °C A 0.500 L reaction vessel initially contains 0.0125 mol of H2S at 800 °C. Find the equilibrium concentration of H2 and S2.

44. a A + b B ⇌ c C + d D • If K>>1, the reaction is product-favored; product predominates at equilibrium. • If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.

45. Try Example 14.13 on page 638.

46. value of Kc or Kp equilibrium concentrations or pressures

47. Example 14.5, page 628 Consider the following reaction: CO(g) + 2H2(g) ⇌ CH3OH(g) A reaction mixture at 780 C initially contains [CO] = 0.500 M and [H2]= 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant?