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Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
Lect w6 152_abbrev_ le chatelier and calculations_1_alg
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Lect w6 152_abbrev_ le chatelier and calculations_1_alg

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    • 1. General Chemistry II CHEM 152 Unit 2 Week 6
    • 2. Week 6 Reading Assignment Chapter 14 – Sections 14.6 through 14.9 (calculations - Le Châtelier)
    • 3. What to Expect
      • Introduce Q
      • Solve Equilibrium Problems
        • ICE Tables
      • Discuss Le Châtelier’s Principle
    • 4. Q...the Reaction Quotient
      • Quotient...what?
      • quotient ( n. ) number resulting from the division of one number by another
      • At EQUILIBRIUM, Q = K
      • For A + B  2 C
    • 5. Reaction Quotient Example
      • Write the reaction quotients for the following reactions:
      • 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g)
      • C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g)
    • 6. Reaction Quotient Example
      • Write the reaction quotient for the following reactions:
      • 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g)
      • Q = [NO 2 ] 4 [O 2 ]
      • [N 2 O 5 ] 2
      • C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g)
      • Q = [CO 2 ] 3 [H 2 O] 4
      • [C 3 H 8 ][O 2 ] 5
    • 7. Q vs. K
      • At equilibrium , Q = K
        • So what happens when Q  K?
        • Q < K or Q > K
    • 8. Problem
      • For the reaction N 2 O 4 (g)  2 NO 2 (g) , the equilibrium constant is 0.21 at 100  C.
      • At the point in the reaction where
      • [N 2 O 4 ] = 0.12 M and [NO 2 ] = 0.55, is the reaction at equilibrium?
      • Q = [NO 2 ] 2 = [0.55] 2 = 2.5
      • [N 2 O 4 ] [0.12]
    • 9. Problem
      • For the reaction N 2 O 4 (g)  2 NO 2 (g) , the equilibrium constant is 0.21 at 100  C.
      • At the point in the reaction where [N 2 O 4 ] = 0.12 M and [NO 2 ] = 0.55, is the reaction at equilibrium?
      • Q = 2.5  0.21
      • In which direction will it proceed?
    • 10. Problem
      • For the reaction N 2 O 4 (g)  2 NO 2 (g) , the equilibrium constant is 0.21 at 100  C.
      • At the point in the reaction where [N 2 O 4 ] = 0.12 M and [NO 2 ] = 0.55, is the reaction at equilibrium?
      • In which direction will it proceed?
      0 ∞ 1 K 0.21 Q 2.5
    • 11. Shifting Equilibrium The equilibrium state for a chemical process can be affected by changes in the concentration of reactants or products, or by varying the temperature and pressure of the system. The direction in which the equilibrium shifts (increasing the concentration of reactants or products) can be predicted.
    • 12. Butane-Isobutane Equilibrium butane isobutane Let us consider the following equilibrium Shifting Equilibrium
    • 13. Butane Isobutane
      • Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M.
      • Now add 1.50 M butane. What are the concentrations of [iso] and [butane] when the equilibrium is re-established? K = 2.5
      Shifting Equilibrium To answer this question we can calculate the “reaction quotient” Q and compare it to K:
    • 14. Butane Isobutane
      • Given that Q= 0.63 and K =2.5 . How can the equilibrium be re-established?
      The system must shift to increase Q until Q = K This is done by increasing the concentration of ISOBUTANE and DECREASING that of BUTANE. Adding reactants to an equilibrium shifts the equilibrium toward products Shifting Equilibrium
    • 15. Le Châtelier’s Principle The outcome is governed by the: Le CHÂTELIER’S PRINCIPLE “ ...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.” Changes in concentration , pressure , and temperature affect the position of chemical equilibrium.
    • 16. Let’s Analyze It N 2 O 4  2NO 2 What would you expect to happen if we double the concentration of NO 2 ? How does K=[NO 2 ] 2 eq / [N 2 O 4 ] eq change?
    • 17. Consider the reaction CaCO 3 (s) + CO 2 (aq) + H 2 O(l)  Ca 2+ (aq) + 2 HCO 3 - (aq) Predict the effect on the equilibrium of: -Removing CO 2 -Adding more CaCO 3 to the system -Adding CaCl 2 to the solution -Adding more water -HCO 3 - is added
    • 18. Changing Volume or Pressure N 2 O 4  2NO 2
      • Predict the effect of:
      • Reducing the volume of the container to one half its initial value;
      • Increasing the volume of the container to two times its initial value.
      K does not change
    • 19. Another Reaction H 2 + I 2  2HI What would happen in this case if similar “stresses” are applied to this equilibrium?
    • 20. V  (P  ) – Reaction shifts to the side with fewer moles of gas V  (P  ) – Reaction shifts to the side with greater moles of gas The value of K DOES NOT CHANGE. Summary
    • 21. Temperature Effects 2NO 2  N 2 O 4 Let’s now analyze the effect of changing temperature. Evaluate the value of K before and after the change.
    • 22. Based on the shifts in the equilibrium with changing temperature, decide whether this is an endothermic or an exothermic reaction. PCl 5 (g)  PCl 3 (g) + Cl 2 (g) Temperature Effects
    • 23. Increase T  More reactants (K decreases)  H = - (exothermic) Temperature Effects Decrease T  More products (K increases) 2NO 2  N 2 O 4 + Heat Treat the Heat as if it were a product
    • 24. Increase T  More products (K increases)  H = + (endothermic) Temperature Effects Decrease T  More reactants (K decreases) Treat the Heat as if it were a reactant Heat + PCl 5 (g)  PCl 3 (g) + Cl 2 (g)
    • 25. Le Châtelier’s Principle
      • Add or take away reactant or product; change volume or pressure:
        • K does not change
        • Reaction adjusts to new equilibrium “position”
      • Change T:
        • Change in K (effect depends on the sign of  H 0 rxn )
        • Therefore change in concentrations at equilibrium
      • Use a catalyst:
      • - Reaction comes more quickly to equilibrium. K not changed; final concentrations not changed.
    • 26. Equilibrium Calculations Let ’ s now learn how to calculate equilibrium constants or the concentrations of reactant and products at equilibrium.
    • 27. All [Products] eq known: H 2 ( g ) + I 2 ( g )  2 HI ( g ) @ 445°C Initial Equilibrium Equilibrium Constant [H 2 ] [I 2 ] [HI] [H 2 ] [I 2 ] [HI] 0.50 0.50 0.0 0.11 0.11 0.78 0.0 0.0 0.50 0.055 0.055 0.39 0.50 0.50 0.50 0.165 0.165 1.17 1.0 0.5 0.0 0.53 0.033 0.934
    • 28. One [Product] eq Known
      • You have the reaction 2 A ( aq ) + B ( aq )  4 C ( aq )
      • with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] eq =0.50 M. What are the equilibrium concentrations? K?
      I C E +  2A B 4C initial molarity 1.00 1.00 0 change in concentration equilibrium molarity 0.50
    • 29. Find the value of K for the reaction 2 CH 4 ( g )  C 2 H 2 ( g ) + 3 H 2 ( g ) at 1700°C if the initial [CH 4 ] = 0.115 M and the equilibrium [C 2 H 2 ] eq = 0.035 M Your Turn +  Construct an ICE table for the reaction For the substance whose equilibrium concentration is known, calculate the change in concentration 2CH 4 C 2 H 2 3H 2 initial 0.115 0.000 0.000 change equilibrium 0.035
    • 30. Find the value of K for the reaction 2 CH 4 ( g )  C 2 H 2 ( g ) + 3 H 2 ( g ) at 1700°C if the initial [CH 4 ] = 0.115 M and the equilibrium [C 2 H 2 ] eq = 0.035 M Answer +3x +x -2x 0.105 0.045 K = (0.035*0.105 3 )/(0.045 2 ) = 0.020 K = [C 2 H 2 ][H 2 ] 3 [CH 4 ] 2 +  2CH 4 C 2 H 2 3H 2 initial 0.115 0.000 0.000 change equilibrium 0.035
    • 31. None [Product] eq Known
      • Imagine that we place 1.00 mol each of H 2 and I 2 in a 1.00 L flask, and they react according with the equation:
      H 2 (g) + I 2 (g)  2 HI(g) If: What are the values for the equilibrium concentrations of all the species in the system?  
    • 32. H 2 (g) + I 2 (g)  2 HI(g) K = 55.3
      • 1.00 1.00 0
      • -x -x +2x
      • 1.00-x 1.00-x 2x
      • where x is defined as the amount of H 2 and I 2 consumed on approaching equilibrium.
      Step 1. Set up a table to define equilibrium concentrations. [H 2 ] [I 2 ] 2[HI] Initial Change Equilib Equilibrium Concentrations + 
    • 33.
      • Step 2. Put equilibrium concentrations into K c expression ( K =[HI] 2 /([H 2 ][I 2 ] ).
      H 2 (g) + I 2 (g)  2 HI(g) K = 55.3 Equilibrium Concentrations [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M Step 3. Solve K expression – In this case, we can take square root of both sides. x = 0.788
    • 34. Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g) Your Turn Find the equilibrium concentrations for NO 2 and N 2 O 4 at 298 K if [N 2 O 4 ] o =0.50 M.  
    • 35. Nitrogen Dioxide Equilibrium N 2 O 4 (g)  2 NO 2 (g)
      • If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations?
      • Step 1. Set up a concentration table
      • [N 2 O 4 ] 2[NO 2 ]
      • Initial 0.50 0
      • Change -x +2x
      • Equilib 0.50 - x 2x
      Equilibrium Concentrations 
    • 36.
      • Step 2. Substitute into K c expression and solve.
      Rearrange: 0.0059 (0.50 - x) = 4x 2 0.0029 - 0.0059x = 4x 2 4x 2 + 0.0059x -0.0029= 0 This is a QUADRATIC EQUATION ax 2 + bx + c = 0 a = 4 b = 0.0059 c = -0.0029 Equilibrium Concentrations
    • 37. Equilibrium Concentrations
      • Solve the quadratic equation for x.
      • ax 2 + bx + c = 0
      • a = 4 b = 0.0059 c = -0.0029
      x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion: x = 0.026 M [N 2 O 4 ] = 0.50 - x = 0.47 M [NO 2 ] = 2x = 0.052 M
    • 38. Approximations to Simplify the Math
        • If there is a large enough difference between the concentration and the K value – SIMPLIFY!
        • Use approximation ONLY when the concentration and the K value are more than a factor of 100 or more apart
      K = ax 2 Conc - bx K = ax 2 Conc
    • 39. For the reaction I 2 (g)  2 I (g) the value of K = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I] ? Equilibrium Concentrations
    • 40. For the reaction I 2 (g)  2 I(g) the value of K = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? Equilibrium Concentrations  [I 2 ] 2[I] initial 0.500 0 change - x +2 x equilibrium 0.500- x 2 x
    • 41. The approximation is valid!! Equilibrium Concentrations  [I 2 ] 2[I] initial 0.500 0 change - x +2 x equilibrium 0.500- x 2 x
    • 42. For the reaction I 2 (g)  2 I (g) the value of K = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? x = 0.00217 0.500  0.00217 = 0.498 [I 2 ] = 0.498 M 2(0.00217) = 0.00434 [I] = 0.00434 M  [I 2 ] 2[I] initial 0.500 0 change - x +2 x equilibrium 0.500- x 2 x  . Approximation is valid
    • 43. Summary Activity 1. Consider the reduction of carbon dioxide by hydrogen to give water vapor and carbon monoxide: H 2 (g) + CO 2 (g)  H 2 O(g) + CO(g) K c =0.10 (at 420 o C) Suppose the initial concentrations of CO 2 and H 2 are the same: 0.050 M. What are the equilibrium concentrations of all the species at 420 o C?
    • 44. Summary Activity 2. If the same reaction as we looked at in Summary Activity 1 was endothermic, what would be the effect on the equilibrium by each of the following: Decrease the Volume Decrease the Temperature
    • 45. Summary Activity 3. Suppose that a mixture of 1.00 mol of HI(g) and 1.00 mol of H 2 (g) is sealed into a 10.0 L flask at 745 K. What will be the concentration of all species at equilibrium? 2HI(g)  H 2 (g) + I 2 (g) K c =0.0200

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