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A. Dynamic Equilibrium

Equilibrium, Acids and Bases. A. Dynamic Equilibrium. equilibrium theories and principles apply to a variety of phenomena in our world. eg) blood gases in scuba diving CO 2 in carbonated beverages buffers in our blood.

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A. Dynamic Equilibrium

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  1. Equilibrium, Acids and Bases A. Dynamic Equilibrium • equilibrium theories and principles apply to a variety of phenomena in our world eg) blood gases in scuba diving CO2 in carbonated beverages buffers in our blood • reactions are often which means that not only are the reversible… products formed but the reactants can be reformed • we use the double arrow to show this relationship eg) A + B ⇌C + D

  2. the forward and reverse reaction will proceed at different rates…it depends on the concentration of the reactants and products • if we start with only the reactants A and B, the forward reaction will initially be the fastest as it is the only reaction possible forward • as the products C and D are formed, thereaction willand thereaction will slow down reverse speed up • at some point, the rates of forward and reverse reactions become equal

  3. Dynamic Equilibrium forward reaction equilibrium Rate reverse reaction 0 Time

  4. a system is said to be ina state of when: dynamic equilibrium 1. the of the forward and reverse reactions are rates equal observable(macroscopic)properties 2. the of the system, such as temperature, pressure, concentration, pH are constant 3. the system is a system at closed constant temperature

  5. B. Classes of Reaction Equilibria • there are three classes of chemical equilibria: 1. favoured (percent rxn ) reactants <50% <50% A + B ⇌ C + D 2. favoured (percent rxn ) products >50% >50% A + B ⇌ C + D

  6. 3. to the right (percent rxn ) quantitative >99% >99% A + B ⇌ C + D or A + B  C + D

  7. C. The Equilibrium Constant • experiments have shown that under a given set of conditions (P and T) a specific quantitative relationship exists between the equilibrium concentrations of the reactants and products • one reaction that has been studied intensively is that between H2(g) and I2(g) (simple molecules and takes place in gas phase no solvent necessary!) H2(g) + I2(g) ⇌ 2 HI(g)

  8. when different combinations of H2(g), I2(g), and HI(g) were mixed and the concentrations measured, it was discovered that equilibrium was reached in all cases different • even though the equilibrium [ ] are , the end quotient was the same each time (within experimental error)

  9. this led to the empirical generalization known as the which says that there is a between the concentrations of the products and the concentrations of the reactants at equilibrium Law of Equilibrium constant ratio

  10. this law can be expressed mathematically: For the reaction aA + bB ⇌ cC + dD The law is: Kc = [C]c[D]d [A]a [B]b equilibrium constant where: Kc = A, B = C, D = a, b, c, d = coefficients reactants products balancing

  11. is constant for a reaction at a given …if you change the temperature, Kc also changes Kc temperature • it is common to ignore the units for Kc and list it only as a numerical value (since depends on the powers of the various [ ] terms) • when determining Kc use only the species that are in or gas aqueous ***unless all states are the same, then use them all

  12. the the value of Kc, the greater the tendency for the reaction to favor the higher forward direction (the products) • if Kc is then the reaction is favoured greater than 1, products • if Kc is then the reaction is favoured less than 1, reactants • Kc indicates the and not the percent reaction rate of the reaction • catalysts will not affect the [ ] at equilibrium… they only increase the rate of the rxn

  13. Example 1 Write the equilibrium law for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas. 2 NO(g) + O2(g) ⇌2 NO2(g) Kc = [NO2(g)]2 [NO(g)]2[O2(g)]

  14. Example 2 Write the equilibrium law for the following reaction: CaCO3(s) ⇌ CaO(s) + CO2(g) Kc = [CO2(g)] *** do not include solids in Kc

  15. Example 3 Write the equilibrium law for the following reaction: 2 H2O(l) ⇌ 2 H2(g) + O2(g) Kc = [H2(g)]2[O2(g)] *** do not include liquids in Kc

  16. Example 4 Phosphorus pentachloride gas can be decomposed into phosphorus trichloride gas and chlorine gas. a) Write the equilibrium law for this reaction. PCl5(g) ⇌ PCl3(g) + Cl2(g) Kc = [PCl3(g) ][Cl2(g)] [PCl5(g)]

  17. b) If the [PCl5(g)]eq = 4.3 x 10-4 mol/L, the [PCl3(g) ]eq = 0.014 mol/L and the [Cl2(g)]eq = 0.014 mol/L then calculate Kc. Kc = [PCl3(g) ][Cl2(g)] [PCl5(g)] = (0.014)(0.014) (4.3 x 10-4) = 0.46

  18. Example 5 Find the [SO3(g)] for the following reaction if Kc = 85.0 at 25.0C. 2 SO2(g) + O2(g) ⇌2 SO3(g) 0.500 mol/L 0.500 mol/L ??? Kc = [SO3(g) ]2____ [SO2(g)]2[O2(g)] 85.0 = [SO3(g) ]2 (0.500)2(0.500) [SO3(g) ]2= 10.625 [SO3(g) ] = 3.26 mol/L

  19. Practice Question

  20. Practice Question

  21. Practice questions • Workbook time!!! - page 1

  22. D. Graphical Analysis • a graph of vs. can be used to see when equilibrium has been reached…as soon as the concentrations , you can read this time off the graph concentration time don’t change any more

  23. SO3(g) 75 SO2(g) Concentration (mol/L) 50 O2(g) 25 10 20 30 0 Time (s) Example 1 Consider this rxn: 2 SO2(g) + O2(g)⇌ 2 SO3(g) At what time does equilibrium get reached and what is the value for Kc?

  24. Kc = [SO3]2 [SO2]2[O2] = (75)2 (50)2(25) = 0.090 Equilibrium is reached at approximately 20 seconds.

  25. Practice • Workbook – p.4

  26. E. Le Châtelier’sPrinciple – Video Ted-Ed • states that when a chemical system at is disturbed by a the system adjusts in a way that Le Châtelier’s principle equilibrium change in property of the system, opposes the change • this takes place in a three-stage process 1. initial equilibrium state 2. shifting non-equilibrium state 3. new equilibrium state

  27. Particle Model of Matter • Review: add heat = movement increases in particles = more collisions = more chemical reactions • Use this to explain the following instead of memorizing it!!!

  28. a system can be affected by a change in concentration, temperature and or volume (pressure) 1. Concentration Changes increase • an in the [ ] of the products or reactants favours the other side of the equation • a in the [ ] of the products or reactants favours decrease the same side of the equation

  29. eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g) ***Haber-Bosch process ↑ [N2(g)] will shift the equilibrium ↑ [NH3(g)] will shift the equilibrium  [NH3(g)] will shift the equilibrium to the products to the reactants to the products • changes in concentration have on the value of no effect Kc

  30. 2. Temperature Changes • energy is treated like a or reactant product eg) reactants + energy ⇌ products reactants ⇌ products + energy • if cooled, the equilibrium shifts so more heat is produced (same side) • if heated, the equilibrium shifts away from the heat so it cools down (opposite side)

  31. a change in temperature is the only stress that the value of Kc!!!!!!! will change • if the shift is towards the side, Kc will product increase • if the shift is towards the side, Kc will reactant decrease

  32. Practice Question

  33. 3. Volume and Pressure Changes • with gases, volume and pressure are related (volume , pressure )  ↑ • the concentration of a gas is related to volume (pressure)…volume , concentration ↓ ↑ http://michele.usc.edu/java/gas/gassim.html • an caused by a in volume causes a shift towards the side of the equation with increase in [ ] drop fewer moles of gas eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g) 4 moles 2 moles will shift to NH3(g)

  34. if the number of moles are the on both sides of the reaction, a change in volume (pressure) has same no effect • changes in volume and pressure have on the value of no effect Kc

  35. Practice Question

  36. 4. Colour Changes • in many equilibrium systems, the reactants will have a different colour than the products • predictions can be made about the equilibrium shift and the resulting change in colour

  37. Example Use the following reaction to predict the equilibrium shift and resulting colour change when the stresses are applied 2 CrO42(aq) + 2 H3O+(aq) ⇌ 2 Cr2O72(aq) + 3 H2O(l) yellow orange products, orange a) a crystal of Na2CrO4(s) is added b) a crystal of K2Cr2O7(s) is added reactants, yellow c) a few drops of concentrated acid is added products, orange d) water is removed products, orange e) a few crystals of NaOH(s) are added reactants, yellow

  38. Practice • Practice – p.2

  39. all of the changes that can happen to systems in equilibrium can be shown graphically: Example State what change to the equilibrium takes place at each of the labelled parts of the graph:

  40. NH3(g) N2(g) Concentration (mol/L) H2(g) A B C D Time (min) Manipulations of An Equilibrium System N2(g) + 3 H2(g)⇌ 2 NH3(g) + energy Stress Equilibrium Time A addition of H2(g) B addition of inert gas, addition of catalyst C decrease in volume D increase in energy

  41. Practice Question

  42. F. ICE Tables • we can use a table set-up to calculate the equilibrium concentrations and/or Kc for any system • you must be able to calculate all before you can use the equilibrium law equilibrium [ ]

  43. Example 1 Hydrogen iodide gas decomposes into hydrogen gas and iodine gas. If 2.00 mol of HI(g) is place in a 1.00 L container and allowed to come to equilibrium at 35C, the final concentration of H2(g) is 0.214 mol/L. Find the value for Kc. 2 HI(g) ⇌ H2(g) + I2(g) 0 2.00 mol/L 0 Initial –0.214 mol/L x 2/1 +0.214 mol/L +0.214 mol/L x 1/1 Change = –0.428 0.214 mol/L 1.572 mol/L +0.214 mol/L Equil.

  44. Kc = [H2(g)][I2(g)] [HI(g)]2 = (0.214)(0.214) (1.572)2 = 0.0185

  45. Example 2 In a 500 mL stainless steel reaction vessel at 900C, carbon monoxide and water vapour react to produce carbon dioxide and hydrogen. Evidence indicates that this reaction establishes an equilibrium with only partial conversion of reactants to products. Initially, 2.00 mol of each reactant is placed in the vessel. Kc for this reaction is 4.20 at 900C. Calculate the concentration of each substance at equilibrium. CO(g) + H2O(g) ⇌ CO2(g) + H2(g) 2.00/0.5L = 4.00 mol/L 0 2.00/0.5L = 4.00 mol/L 0 I –xmol/L +xmol/L x 1/1 –xmol/L +xmol/L x 1/1 C E 4.00 x mol/L xmol/L 4.00 x mol/L xmol/L

  46. Kc = [CO2(g)][H2(g)] [CO(g)][H2O(g)] 4.20 = (x)(x) (4.00 x)( 4.00 x) 4.20 = x2 (4.00 x)2 ***Note, this is a perfect square so to solve for x simply square root both sides of the equation, then solve 2.05 = x (4.00 x) 2.05(4.00 x) = x 8.05  2.05x = x 8.05 = 3.05x x = 2.69 mol/L

  47. 4.00 – 2.69 = 1.31 mol/L [CO(g)] = 4.00 – x = [H2O (g)] = 4.00 – x = [CO2(g)] = x = [H2(g)] = x = 4.00 – 2.69 = 1.31 mol/L 2.69 mol/L 2.69 mol/L

  48. Example 3 Gaseous phosphorus pentachloride decomposes into gaseous phosphorus trichloride and chlorine gas at a temperature where Keq = 1.00  103. Suppose 2.00 mol of PCl5(g) in a 2.00 L vessel is allowed to come to equilibrium. Calculate the equilibrium [ ] of each species. PCl5(g)⇌ PCl3(g) + Cl2(g) 2.00mol/2.00L = 1.00 mol/L 0 mol/L I 0 mol/L C –xmol/L x 1/1 +x mol/L +x mol/Lx 1/1 xmol/L E 1.00 – xmol/L xmol/L

  49. Kc = [PCl3(g)][Cl2(g)] [PCl5(g)] 1.00  10-3 = (x)(x) (1.00 - x) ***at this point, you would have to use the quadratic formula to solve for x • when the concentrations are greater than the equilibrium constant, we can make an that greatly simplifies our calculations 1000 X approximation • if Kc is very small, the equilibrium doesn’t lie very far to the right and  x is a very small number

  50. ***in this example 1.00 – x can be assumed to be 1.00 since x is really small, so… 1.00  10-3 = (x)(x) (1.00 ) x2 = 1.00  10-3 x 1.00 x = 0.0316 ***now you can calculate the [ ]eq for each species …substitute x into the equilibrium values in the ICE table [PCl5(g)]eq = 1.00 mol/L – 0.0316 mol/L = 0.967 mol/L [PCl3(g)]eq = 0 + 0.0316 mol/L = 0.0316 mol/L [Cl2(g)]eq = 0 + 0.0316 mol/L = 0.0316 mol/L

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