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General Chemistry II CHEM 152  Unit 2 Week 6
Week 6 Reading Assignment Chapter 14 – Sections 14.6 through 14.9 (calculations - Le Châtelier)
What to Expect ,[object Object],[object Object],[object Object],[object Object]
Q...the Reaction Quotient ,[object Object],[object Object],[object Object],[object Object]
Reaction Quotient Example ,[object Object],[object Object],[object Object]
Reaction Quotient Example ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Q vs. K ,[object Object],[object Object],[object Object]
Problem ,[object Object],[object Object],[object Object],[object Object],[object Object]
Problem ,[object Object],[object Object],[object Object],[object Object]
Problem ,[object Object],[object Object],[object Object],0 ∞ 1 K  0.21 Q  2.5
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.
Butane-Isobutane Equilibrium butane isobutane Let us consider the following equilibrium Shifting Equilibrium
Butane   Isobutane ,[object Object],[object Object],Shifting Equilibrium To answer this question we can calculate the “reaction quotient”  Q  and compare it to K:
Butane  Isobutane ,[object Object],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
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.
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?
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
Changing Volume or Pressure N 2 O 4      2NO 2 ,[object Object],[object Object],[object Object],K does not change
Another Reaction H 2  + I 2      2HI What would happen in this case if similar “stresses” are applied to this equilibrium?
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
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.
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
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
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)
Le Châtelier’s Principle ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Equilibrium Calculations Let ’ s now learn how to calculate equilibrium constants or the concentrations of reactant and products at equilibrium.
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
One [Product] eq  Known ,[object Object],[object Object],I C E +  2A B 4C initial molarity 1.00 1.00 0 change in concentration equilibrium molarity 0.50
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
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
None [Product] eq  Known ,[object Object],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?    
H 2 (g)  +  I 2 (g)    2 HI(g) K = 55.3 ,[object Object],[object Object],[object Object],[object Object],Step 1.  Set up  a  table to define equilibrium    concentrations. [H 2 ] [I 2 ] 2[HI] Initial  Change Equilib Equilibrium Concentrations + 
[object Object],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
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.    
Nitrogen Dioxide Equilibrium N 2 O 4 (g)     2 NO 2 (g) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Equilibrium Concentrations 
[object Object],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
Equilibrium Concentrations ,[object Object],[object Object],[object Object],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
Approximations to Simplify the Math ,[object Object],[object Object],K = ax 2 Conc - bx K = ax 2 Conc
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
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
The approximation is valid!! Equilibrium Concentrations  [I 2 ] 2[I] initial 0.500 0 change - x +2 x equilibrium 0.500-  x 2 x
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
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?
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
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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Lect w6 152_abbrev_ le chatelier and calculations_1_alg

  • 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.
  • 4.
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 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.
  • 14.
  • 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.
  • 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.
  • 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.
  • 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.
  • 32.
  • 33.
  • 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.
  • 36.
  • 37.
  • 38.
  • 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

Editor's Notes

  1. Update for Tro.
  2. Tier 1
  3. Tier 1.5
  4. Tier 2
  5. Tier 1.5
  6. Tier 1.5
  7. Tier 1.5