1. Chemical Equilibrium I. A State of Dynamic Balance 1N 2 (g) 3H 2(g) +2NH 3(g) ΔG 0 = -33.1 kJ The reaction is spontaneous under standard conditions The concentrations of the reactants decrease at first… …while the concentration of the product increases But, then, before the reactants are used up, all concentrations become constant

2. Chemical Equilibrium I. A State of Dynamic Balance-when a ________ results in the almost ________ conversion of ________ to ________, the ________ is said to go to __________, but _____ _________ ___ ____ go to __________, most _________ are __________ reaction complete reactantsproducts reaction completionmostreactions donotcompletion reactionsreversible 1N 2 (g) 3H 2(g) +2NH 3(g) 1N 2 (g) 3H 2(g) +2NH 3(g) 1N 2 (g) 3H 2(g) +2NH 3(g) At first, only the reactants are present, so only the forward reaction can occur 1N 2 (g) 3H 2(g) +2NH 3(g)

3. Chemical Equilibrium I. A State of Dynamic Balance-as soon as the ________ ________ begins, the ____________ of the _________ go _____, and the _________ _____ goes _____ as the number of __________ per unit ____ goes _____ forwardreaction concentrations reactantsdown reactionratedown collisions timedown As soon as the products begin forming, the forward reaction rate slows and the reverse reaction begins 1N 2 (g) 3H 2(g) +2NH 3(g)

4. Chemical Equilibrium I. A State of Dynamic Balance-as the _________ proceeds, the ____ of the ________ _________ continues to ________ and the ____ of the ________ ________ continues to ________ until the two _____ are _____, and the system has reached a state of ________ __________ reaction rateforwardreaction decreaserate reversereaction increaserates equal chemicalequilbrium 1N 2 (g) 3H 2(g) +2NH 3(g)

5. Chemical Equilibrium I. A State of Dynamic Balance-at ___________, the ____________ of the ________ and ________ are not _____, but _______, because the ____ of _________ of the ________ is _____ to the ____ of _________ of the ________ equilibriumconcentrations reactantsproducts equalconstant rateformationproducts equalrateformation reactants The Golden Gate Bridge connects San Francisco to Sausalito. If all other roads leading in and out of the two cities were closed… …and the number of vehicles crossing the bridge per hour in one direction equaled the number of vehicles crossing the bridge in the opposite direction… What is true of the number of vehicles in each city throughout the day? Are there the same number of vehicles in each city?

6. Chemical Equilibrium II. Equilibrium Expressions and Constants Cato Maximilian Guldberg 1836-1902 Peter Waage 1833-1900 -while _____ chemical systems have little tendency to _____, and _____ chemical systems _____ readily and ___ to __________, _____ chemical systems reach a _____ of __________, leaving varying amounts of ________ ____________ some reactsome react gocompletionmost stateequilibrium reactant unconsumed -in 1864, Norwegian chemists ______ and _________ proposed the _______ ___________________, which states, at a given ___________, a chemical system may reach a _____ in which a particular _____ of _______ and _______ ____________ has a _______ value Waage GuldbergLaw of Chemical Equilibrium temperature state ratioreactant productconcentrationsconstant

7. Chemical Equilibrium II. Equilibrium Expressions and Constants -the _______ ________ for a _______ at __________ can be written ______________________________, where __ and __ are ________, __ and __ are ________, __, __, __, and __ are the ___________ in the ________ ________, and the __________ _______ __________ is generalequationreaction equilibrium aAaA+bBbBcCcC+dDdD ABreactantsC Dproductsabcd coefficientsbalanced equationequilibrium constantexpression K eq = [C] c [D] d [A] a [B] b -___________ ________ with ___ values __ __ contain more ________ than ________ at ___________, while __________ ________ with ___ values __ __ contain more ________ than ________ at __________ equilibriummixturesK eq >1products reactantsequilibrium mixturesK eq < 1 reactants productsequilibrium

8. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the homogeneous equilibrium for the synthesis of ammonia from nitrogen and hydrogen. 1N 2 (g) 3H 2(g) +2NH 3(g) K eq = [C] c [D] d [A] a [B] b K eq = [C] c [A] a [B] b = [NH 3 ] c [N 2 ] a [H 2 ] b = [NH 3 ] 2 [N 2 ] 1 [H 2 ] 3 The equilibrium is homogeneous because all the reactants and products are in the same physical state (gas)

9. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the equilibrium for the synthesis of Hydrogen iodide from iodine and hydrogen. 1H 2 (g) 1I 2(g) +2HI (g) K eq = [C] c [D] d [A] a [B] b K eq = [C] c [A] a [B] b = [HI] c [H 2 ] a [I 2 ] b = [HI] 2 [H 2 ] 1 [I 2 ] 1 The equilibrium is homogeneous because all the reactants and products are in the same physical state (gas)

10. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the homogeneous equilibrium for the decomposition of Dinitrogen tetroxide into Nitrogen dioxide. 1N 2 O 4 (g) 2NO 2(g) K eq = [C] c [D] d [A] a [B] b K eq = [C] c [A] a = [N 2 O 4 ] a [NO 2 ] c = [N 2 O 4 ] 1 [NO 2 ] 2 The equilibrium is homogeneous because all the reactants and products are in the same physical state (gas)

11. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the homogeneous equilibrium for the reaction of Carbon monoxide and Hydrogen which produces methane (Tetrahydrogen monocarbide) and water. 3H 2 (g) 1CH 4(g) K eq = [C] c [D] d [A] a [B] b K eq = [C] c [A] a = [CH 4 ] c [CO] a = 1CO (g) +1H 2 O (g) + [D] d [B] b [H 2 O] d [H 2 ] b [CH 4 ] 1 [CO] 1 [H 2 O] 1 [H 2 ] 3

12. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the homogeneous equilibrium for the decomposition of Dihydrogen monosulfide into diatomic hydrogen and diatomic sulfur. 2H 2 S (g) 2H 2 (g) K eq = [C] c [D] d [A] a [B] b K eq = [C] c [A] a = [H 2 ] c [H 2 S] a = 1S 2 (g) + [D] d [S 2 ] d [H 2 ] 2 [S 2 ] 1 [H 2 S] 2

13. Chemical Equilibrium II. Equilibrium Expressions and Constants -_________ in which all ________ and ________ are in the same ________ _____ are ____________, but ________ with _________ and ________ in _____ than ___ ________ _____ result in _____________ _________ equilibriareactants productsphysical statehomogeneous reactionsreactants productsmoreonephysical stateheterogeneous equilibria gaseous ethanol liquid ethanol 1C 2 H 5 OH (l) 1C 2 H 5 OH (g) K eq = [C] c [D] d [A] a [B] b K eq = [C 2 H 5 OH (g) ] 1 [C 2 H 5 OH (l) ] 1 K eq =[C 2 H 5 OH (g) ] 1

14. Chemical Equilibrium II. Equilibrium Expressions and Constants -since ______ and _____ ________ and ________ don’t change ___________, (which is really their ______), if the ___________ remains ________, then in the ___________ _______ __________ for a ____________ ___________, the ___________ ________ only depends on the ______________ of the ________ and ________ in the _______ state of matter liquidsolidreactants productsconcentration density temperatureconstant equilibriumconstant expressionheterogeneous equilibrium constant concentrationsreactants productsgaseous

15. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the heterogeneous equilibrium for the decomposition of Sodium Hydrogen carbonate into Sodium carbonate, Carbon dioxide, and water. 2NaHCO 3 (s) 1Na 2 CO 3 (s) K eq = [C] c [D] d [A] a [E] e K eq =[D] d =[CO 2 ] d [H 2 O] e =[CO 2 ] 1 [H 2 O] 1 The equilibrium is heterogeneous because the reactants and products are in different physical states (gas and solid) +1CO 2 (g) 1H 2 O (g) + [E] e

16. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the heterogeneous equilibrium for the decomposition of Calcium carbonate into Calcium oxide and Carbon dioxide. 1CaCO 3 (s) 1CaO (s) K eq = [C] c [D] d [A] a K eq =[D] d =[CO 2 ] d =[CO 2 ] 1 The equilibrium is heterogeneous because the reactants and products are in different physical states (gas and solid) +1CO 2 (g)

17. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the complete, balanced thermochemical equation and equilibrium constant expression for the homogeneous equilibrium for the reaction of hydrazine (Tetrahydrogen dinitride) and Nitrogen dioxide, which produces nitrogen and water. 2NO 2 (g) 3N 2 (g) K eq = 2N 2 H 4 (g) +4H 2 O (g) + [N 2 ] 3 [N 2 H 4 ] 2 [H 2 O] 4 [NO 2 ] 2

18. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the complete, balanced thermochemical equation and equilibrium constant expression for the homogeneous equilibrium for the reaction of Sulfur trioxide and Carbon dioxide, which produces Carbon disulfide and oxygen. 1CO 2 (g) 1CS 2 (g) K eq = 2SO 3 (g) +4O 2 (g) + [CS 2 ] 1 [SO 3 ] 2 [O 2 ] 4 [CO 2 ] 1

19. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the complete, balanced thermochemical equation and equilibrium constant expression for the heterogeneous equilibrium for the reaction of monatomic Sulfur and fluorine gas, which produces Sulfur tetrafluoride gas and Sulfur hexafluoride gas. 5F 2 (g) 1SF 4 (g) K eq = 2S (s) +1SF 6 (g) + [SF 4 ] 1 [SF 6 ] 1 [F 2 ] 5

20. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the complete, balanced thermochemical equation and equilibrium constant expression for the heterogeneous equilibrium for the reaction of magnatite (Fe 3 O 4 ) and hydrogen gas, which produces iron and water vapor. 4H 2 (g) 3Fe (s) K eq = 1Fe 3 O 4 (s) +4H 2 O (g) + [H 2 O] 4 [H 2 ] 4

21. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the homogeneous equilibrium for the synthesis of ammonia and calculate the value of K eq when [NH 3 ] = 0.933 M, [N 2 ] = 0.533 M, and [H 2 ] = 1.600 M. 1N 2 (g) 3H 2(g) +2NH 3(g) K eq = [NH 3 ] 2 [N 2 ] 1 [H 2 ] 3 K eq = [0.933 mol/L] 2 [0.533 mol/L] 1 [1.600 mol/L] 3 K eq =0.399 L 2 /mol 2

22. Chemical Equilibrium II. Equilibrium Expressions and Constants Write the equilibrium constant expression for the homogeneous equilibrium for the decomposition of Sulfur trioxide into Sulfur dioxide and oxygen gas, and calculate the value of K eq when [SO 3 ] = 0.0160 M, [SO 2 ] = 0.00560 M, and [O 2 ] = 0.00210 M. 2SO 3 (g) 1O 2 (g) +2SO 2 (g) K eq = [SO 2 ] 2 [O 2 ] 1 [SO 3 ] 2 K eq = [0.00560 mol/L] 2 [0.00210 mol/L] 1 [0.0160 mol/L] 2 K eq =2.57 x 10 -4 mol/L

23. Chemical Equilibrium III. Le Châtelier’s Principle A. Safety: 1. Hypothesis:What is the effect of temperature on equilibrium? 2. Prediction: 3. Gather Data: The surfaces of the hot plates will be hot enough to cause burns. Use caution. Cobalt(II) chloride hexahydrate is toxic, with an LD 50 = 80mg/kg Avoid ingestion (don’t eat or drink it). Wash hands thoroughly with soap and water before leaving lab. Ethanol is extremely flammable. No open flame. B. Procedure: 1. Pick up a sheet of white construction paper and an artist’s paintbrush.

24. 3. Gather Data: B. Procedure: 4. Use the solution to paint a winter scene on your white construction paper, including a pink-colored field of snow. 3. Add 10 mL of ethanol to the test tube, cap, and shake vigorously until CoCl 2 ·6H 2 O dissolves. If the solution is not light pink, add water dropwise until it turns light pink. Chemical Equilibrium III. Le Châtelier’s Principle 2. With a partner, using a top-loading electronic balance, mass 0.3 grams of CoCl 2 ·6H 2 O, crush it into a fine powder using a mortar and pestle, and place it in a test tube. 5. To simulate the coming of spring, warm your painting over the hotplate in the fume hood. Record your observations.

25. 4. Analyze Data: Chemical Equilibrium III. Le Châtelier’s Principle 4Cl - (aq) 1CoCl 4 2- (aq) +6H 2 O (l) 1Co(H 2 O) 6 2+ (aq) + Hexahydrate Co 2+ ion (pink)chloride ionTetrachlorocobaltate ion (blue) + heat 4Cl - (aq) 1CoCl 4 2- (aq) +6H 2 O (l) 1Co(H 2 O) 6 2+ (aq) + Hexahydrate Co 2+ ion (pink)chloride ionTetrachlorocobaltate ion (blue) 5. Draw Conclusions:

26. Chemical Equilibrium III. Le Châtelier’s Principle Henry-Louis Le Châtelier 1850-1936 -in 1888, ________________________ discovered that there are ways to _______ _________ in order to make _________ more __________ Henry-Louis Le Châtelier controlequilibria reactionsproductive -____________________ states that if a ______ (like a ______ in __________) is applied to a system at __________, the system _____ in the ________ that _______ the _____ Le Châtelier’s Principle stresschange temperature equilibriumshifts directionrelieves stress -________ that reach __________ instead of going to __________ do not ________ as much reactionsequilibrium completion produce

27. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium A. Changes in Concentration Write the equilibrium constant expression for the equilibrium for the reaction of Carbon monoxide and Hydrogen to produce methane and water. Then, calculate the K eq value when [CO] = 0.30000 M, [H 2 ] = 0.10000 M, and [CH 4 ] = 0.05900 M, and [H 2 O] = 0.02000 M. 3H 2 (g) 1CH 4(g) K eq = 1CO (g) +1H 2 O (g) + [CH 4 ] 1 [CO] 1 [H 2 O] 1 [H 2 ] 3 K eq = [0.05900 mol/L] 1 [0.30000 mol/L] 1 [0.02000 mol/L] 1 [0.10000 mol/L] 3 = 3.933 L 2 /mol 2 = first equilibrium position K eq

28. Chemical Equilibrium III. Le Châtelier’s Principle -_________ the ____________ of ___ _________ the _______ of _________ between ___ and ___, _________ the _____ of the _______ _______ A. Changes in Concentration 3H 2 (g) 1CH 4(g) 1CO (g) +1H 2 O (g) + increasingconcentration COincreasesnumber collisionsCO H2H2 increasingrate forwardreaction -the system responds to the ______ of the addition of _______ by forming more _______ to bring the system back into equilbrium stress reactant product

29. Chemical Equilibrium III. Le Châtelier’s Principle A. Changes in Concentration K eq = [CH 4 ] 1 [CO] 1 [H 2 O] 1 [H 2 ] 3 K eq = [0.06648 mol/L] 1 [0.99254 mol/L] 1 [0.02746 mol/L] 1 [0.07762 mol/L] 3 =3.933 L 2 /mol 2 3H 2 (g) 1CH 4(g) 1CO (g) +1H 2 O (g) + 0.99254 M0.07762 M0.06648 M0.02746 M =second equilibrium position K eq

30. Chemical Equilibrium III. Le Châtelier’s Principle A. Changes in Concentration -_________ the ____________ of a ________ causes __________ to _____ to the ____ to _______ the ____ of formation of ______ increasingconcentration reactantequilbrium shiftrightincrease rateproduct 3H 2 (g) 1CH 4(g) 1CO (g) +1H 2 O (g) + 3H 2 (g) 1CH 4(g) 1CO (g) +1H 2 O (g) + -_________ the ____________ of a ________ causes __________ to _____ to the ____ to _______ the ____ of formation of ______ decreasingconcentration productequilbrium shiftrightincrease rateproduct

31. Chemical Equilibrium III. Le Châtelier’s Principle A. Changes in Concentration 4Cl - (aq) 1CoCl 4 2- (aq) +6H 2 O (l) 1Co(H 2 O) 6 2+ (aq) + Hexahydrate Co 2+ ion (pink)chloride ionTetrachlorocobaltate ion (blue) Predict what should happen to the following equilibrium if hydrogen bonding due to the addition of acetone binds water and effectively removes it from the products.

32. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium A. Changes in Volume-_________ the ______ of the _______ container, according to ______, ________ the ________, which in turn ________ the _____ of _________ between the ________ of the ________, _________ the _____ of the ________ _______ 3H 2 (g) 1CH 4(g) 1CO (g) + 1H 2 O (g) + decreasingvolumereaction Boyle increasespressure increasesratecollision particles reactantsincreasingrate forwardreaction -the _____ in the _________ causes the _____ on the system to be _______ as for every __ _____ of _______ _______ _________, only __ _____ of _______ _______ are _________, which, according to ________, occupies __ the ______, which _________ the ________ shiftequilibrium stressrelieved 4molesgaseousreactant consumed2molesgaseous productproduced Avogadro½ volumedecreasespressure

33. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium Use Le Châtelier’s Principle to predict how each of these changes would affect the ammonia equilibrium system. 1N 2 (g) 3H 2(g) +2NH 3(g) a. removing hydrogen from the system __________________________ b. adding ammonia to the system _______________________________ 1N 2 (g) 3H 2(g) +2NH 3(g) equilibrium shifts to the left 1N 2 (g) 3H 2(g) +2NH 3(g) equilibrium shifts to the left

34. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium Use Le Châtelier’s Principle to predict how each of these changes would affect the ammonia equilibrium system. 1N 2 (g) 3H 2(g) +2NH 3(g) c. adding hydrogen to the system _______________________________ 1N 2 (g) 3H 2(g) +2NH 3(g) equilibrium shifts to the right

35. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium How would decreasing the volume of the reaction container affect each of these equilibria? equilibrium shifts to the right2SO 2 (g) 1O 2(g) +2SO 3(g) a. _________________________ stress has no effect on equilibrium1H 2 (g) 1Cl 2(g) +2HCl (g) b. _____________________________ equilibrium shifts to the left2NOBr (g) 1Br 2(g) +2NO (g) c. _________________________

36. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium A. Changes in Temperature-while _______ in _____________ and ________ in _______ cause ______ in _________, they ___ ___ _______ the __________ _______, but a ______ in ___________ causes ______ in both the __________ ________ and the __________ _______ changesconcentration changesvolume shiftsequilibriado notchangeequilibrium constantchange temperaturechange equilibriumposition equilibriumconstant

37. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium A. Changes in Temperature ΔH 0 = -206.5 kJ 3H 2 (g) 1CH 4(g) 1CO (g) + 1H 2 O (g) + -since the _______ for making _______ has a _______ ____, the ________ _______ is _________, and the _______ _______ is __________, so ____ can be thought of as a _______ in the ________ _______ and a _______ in the _______ _______ reaction methanenegativeΔH0ΔH0 forwardreactionexothermic reversereaction endothermicheat reactantforwardreaction product reversereaction +heat

38. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium A. Changes in Temperature-_________ the __________ is like _______ more _______ to the _______ in which _____ acts as a _______ and is _____ ___, in this case, the __________ _______ _______ increasingtemperature addingreactant reactionheat usedup endothermicreverse reaction 3H 2 (g) 1CH 4(g) 1CO (g) +1H 2 O (g) ++heat reactant -__________ shifts to the _____, _________ the ___________ of _______ because _______ is a _______ in the _______ _______ equilibriumleft decreasingconcentration methane reactantreversereaction

39. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium A. Changes in Temperature-_________ the __________ is like ________ _______ from the _______ in which _____ acts as a _______, in this case, the __________ _______ _______ decreasingtemperature removingreactant reactionheat endothermicreversereaction 3H 2 (g) 1CH 4(g) 1CO (g) +1H 2 O (g) ++heat reactant -__________ shifts to the _____, _________ the ___________ of _______ because _______ is a _______ in the _______ _______ equilibriumright increasingconcentration methane productforwardreaction

40. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium In the following equilibrium, would you raise or lower the temperature to get the following results? 1C 2 H 2 (g) 1H 2 O (g) +1CH 3 CHO (g) a. increase the amount of CH 3 CHO______________________________lower the temperature ΔH 0 = -151 kJ 1C 2 H 2 (g) 1H 2 O (g) +1CH 3 CHO (g) +heat b. decrease the amount of C 2 H 2 ________________________________lower the temperature 1C 2 H 2 (g) 1H 2 O (g) +1CH 3 CHO (g) +heat Ethyne (acetylene)Ethanal (ethanaldehyde)

41. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium In the following equilibrium, would you raise or lower the temperature to get the following results? 1C 2 H 2 (g) 1H 2 O (g) +1CH 3 CHO (g) c. increase the amount of H 2 O _________________________________raise the temperature ΔH 0 = -151 kJ 1C 2 H 2 (g) 1H 2 O (g) +1CH 3 CHO (g) +heat

42. Chemical Equilibrium III. Le Châtelier’s Principle-stressors that cause a shift in equilibrium In the following equilibrium, what effect does decreasing the volume of the reaction vessel have? 1CO (g) 1Fe 3 O 4(s) +1CO 2 (g) __________________________________________________________ __________________________________________________________ __________________________________________________________ None. The solids do not change their concentrations, and the number of moles of gaseous reactant equals the number of moles of gaseous product 3FeO (s) + 1CO (g) 1Cl 2(g) +1COCl 2(g) ΔH 0 = -151 kJ In the following equilibrium, what effect does simultaneously increasing the temperature and the pressure have? __________________________________________________________ __________________________________________________________ __________________________________________________________ Cannot predict. An increase in temperature causes a shift in the equilibrium to the left, while an increase in pressure causes a shift in equilibrium to the right. +heat

43. Chemical Equilibrium III. Le Châtelier’s Principle A. Safety: 1. Hypothesis:What is the effect of a change in concentration of reactants and a change in temperature on equilibrium? 2. Prediction: 3. Gather Data: The surfaces of the hot plate will be hot enough to cause burns. Use caution. Cobalt(II) chloride hexahydrate is toxic, with an LD 50 = 80mg/kg Avoid ingestion (don’t eat or drink it). Wash hands thoroughly with soap and water before leaving lab. Concentrated Hydrochloric acid is extremely corrosive. Avoid contact with eyes, skin, and clothing. Goggles, aprons, and gloves mandatory.

44. 3. Gather Data: B. Procedure: 3. Add water dropwise to the test tube until a color change occurs. Record color. ______________ 2. Add 3 mL (60 drops) of concentrated HCl to the test tube. Record color. _____________ Chemical Equilibrium III. Le Châtelier’s Principle 1. With a partner, measure out 2 mL of 0.1 M CoCl 2 solution into a test tube. Record initial color. __________ 4. Add 2 mL of 0.1 M CoCl 2 solution to another test tube. Add concentrated HCl dropwise until the solution turns purple. If the solution turns blue, add water until it turns purple.

45. 3. Gather Data: B. Procedure: 6. Place the test tube in a hot water bath. Record color. ___________ Chemical Equilibrium III. Le Châtelier’s Principle 5. Place the test tube in an ice bath. Record color. ________ 4. Analyze Data: 4Cl - (aq) 1CoCl 4 2- (aq) +6H 2 O (l) 1Co(H 2 O) 6 2+ (aq) + Hexahydrate Co 2+ ion (pink)chloride ionTetrachlorocobaltate ion (blue) A. The equation for the reversible reaction in this experiment is: + heat

46. Chemical Equilibrium III. Le Châtelier’s Principle 4. Analyze Data: A. Use the equation to explain the colors of the solution in steps 1, 2, and 3 In Step 1, the solution is initially a pink color, because the reaction arrives at an equilibrium in which the concentration of the pink-colored 1Co(H 2 O) 6 2+ (aq) is at a higher concentration than the blue-colored 1CoCl 4 2- (aq). In Step 2, the addition of HCl increases the concentration of Cl -, shifting the equilibrium to the right to favor the formation of the blue 1CoCl 4 2- (aq), so the solution turns blue. In Step 3, the increase in concentration of water shifts the equilibrium left, re-establishing a new equilbrium where the concentration of 1CoCl 4 2- (aq) is higher than it was orginally, so the purple color shows more of a balance of pink and blue.

47. Chemical Equilibrium III. Le Châtelier’s Principle 4. Analyze Data: B. Explain how the equilibrium shifts when heat energy is added or removed. In Step 5, since heat acts like a product in the exothermic reverse reaction, removing heat by lowering the temperature causes the equilibrium to shift to the left, increasing the rate of the reverse reaction and causing the solution to turn pink. In Step 6, since heat acts like a reactant in the endothermic forward reaction, adding heat by increasing the temperature causes the equilibrium to shift to the right, increasing the rate of the forward reaction and causing the solution to turn blue. 5. Draw Conclusions:

48. Chemical Equilibrium IV. Using Equilibrium Constants-when a ________ has a _____ ___, the __________ _______ contains _____ ________ than ________ at __________ reactionlarge K eq equilibriummixture moreproducts reactantsequilibrium -when a ________ has a _____ ___, the __________ _______ contains _____ ________ than ________ at __________ reactionsmall K eq equilibriummixture morereactants productsequilibrium A. Calculating Equilibrium Concentrations-__________ ________ can also be used to ________ the __________ ____________ of any ________ in the _______ equilibrium constants calculate equilibrium concentration substance reaction

49. K eq = [CH 4 ] 1 [CO] 1 [H 2 O] 1 [H 2 ] 3 3.933 L 2 /mole 2 = [CH 4 ] 1 [0.850 mole/L] 1 [0.286 mole/L] 1 [1.333 mole/L] 3 =27.7 M 3H 2 (g) 1CH 4(g) 1CO (g) +1H 2 O (g) + 0.850 M 1.333 M ? M0.286 M Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations At 1200 K, the K eq for the following reaction equals 3.933 L 2 /mole 2. What is the concentration of the methane produced, if [CO] = 0.850 M, [H 2 ] = 1.333 M, and [H 2 O] = 0.286 M? [CH 4 ]

50. K eq = [H 2 ] 2 [S 2 ] 1 [H 2 S] 2 2.27 x 10 -3 mole/L= [H 2 ] 2 [0.0540 mole/L] 1 [0.184 mole/L] 2 =0.0377 M 2H 2 S (g) 2H 2(g) 1S 2 (g) + 0.184 M ? M0.0540 M Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations At 1405 K, the K eq for the following reaction equals 2.27 x 10 -3 mole/L. What is the concentration of the Hydrogen gas produced, if [S 2 ] = 0.0540 M, and [H 2 S] = 0.184 M? [H 2 ]

51. K eq = [CO] 1 [CH 3 OH] 1 [H 2 ] 2 10.5 L 2 /mole 2 = [CO] 1 [1.32 mole/L] 1 [0.933 mole/L] 2 =0.144 M 2H 2 (g) 1CO (g) +1CH 3 OH (g) ? M 0.933 M 1.32 M Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations If K eq for the following reaction equals 10.5 L 2 /mole 2, what is the equilibrium concentration of Carbon monoxide, if [H 2 ] = 0.933 M, and [CH 3 OH] = 1.32 M? [CO]

52. Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations from Initial Concentrations Using ICE (Initial, Change, Equilibrium) If the K eq for the following reaction equals 64.0, what are the equilibrium concentrations of I 2, H 2, and HI, if [I 2 ] 0 = 0.200 M, [H 2 ] 0 = 0.200 M and [HI] = 0.000 M? 1H 2 (g) 2HI (g) 1I 2(g) + ? M [H 2 ][I 2 ][HI] Initial Change Equilibrium 0.200 0.000 -1x +2x 0.200 - 1x 2x

53. Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations from Initial Concentrations Using ICE (Initial, Change, Equilibrium) 1H 2 (g) 2HI (g) 1I 2(g) + [H 2 ][I 2 ][HI] Initial Change Equilibrium 0.200 0.000 -1x +2x 0.200 - 1x 2x K eq = [I 2 ] 1 [HI] 2 [H 2 ] 1 64.0= [0.200 – 1x] 1 [2x] 2 [0.200 – 1x] 1 8.00= [0.200 – 1x] 1 [2x] x=0.160 [H 2 ]=0.040 M [I 2 ]=0.040 M [HI]=0.320 M

54. Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations from Initial Concentrations Using ICE (Initial, Change, Equilibrium) If the K eq for the following reaction equals 16.0, what are the equilibrium concentrations of PCl 3, Cl 2, and PCl 5, if [PCl 5 ] 0 = 1.00 M? 1Cl 2 (g) 1PCl 5(g) 1PCl 3(g) + ? M [PCl 3 ][Cl 2 ][PCl 5 ] Initial Change Equilibrium 0.00 1.00 +1x -1x 0.00 + 1x 1.00 – 1x

55. Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations from Initial Concentrations Using ICE (Initial, Change, Equilibrium) [PCl 3 ][Cl 2 ][PCl 5 ] Initial Change Equilibrium 0.00 1.00 +1x -1x 0.00 + 1x 1.00 – 1x K eq = [PCl 3 ] 1 [PCl 5 ] 1 [Cl 2 ] 1 16.0= [x] 1 [1.00 - x] 1 [x] 1 x2x2 =16.0 – 16.0x x2x2 +16.0x-16.0=0 ax2ax2 +bxbx+c=0 x=-b-b±√b 2 – 4ac 2a2a 1Cl 2 (g) 1PCl 5(g) 1PCl 3(g) +

56. Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations from Initial Concentrations Using ICE (Initial, Change, Equilibrium) x=-b-b±√b 2 – 4ac 2a2a x=-16.0±√(16.0) 2 – 4(1.00)(-16.0) 2(1.00) x=0.94(but not -16.9) [PCl 3 ]=0.94 M [Cl 2 ]=0.94 M [PCl 5 ]=0.06 M x2x2 +16.0x-16.0=0 ax2ax2 +bxbx+c=0 x2x2 =16.0 – 16.0x x2x2 =+ 16.0x16.0 x2x2 =+ 16.0x + 64.080.0 (x=+ 8.00) 2 80.0 x=+ 8.00√80.0 x=+ 8.00 8.94 x=0.94 Not a perfect square! 16.0 is an experimentally-determined constant rounded to the nearest tenth!

57. Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations from Initial Concentrations Using ICE (Initial, Change, Equilibrium) If the K eq for the following reaction equals 0.680, what are the equilibrium concentrations of COCl 2, CO, and Cl 2, if [CO] 0 = 0.500 M and [Cl 2 ] 0 = 1.00 M? 1Cl 2 (g) 1CO (g) 1COCl 2 (g) + ? M [COCl 2 ][CO][Cl 2 ] Initial Change Equilibrium 0.000.5001.00 +1x-1x 0.00 + 1x0.500 - 1x1.00 – 1x

58. Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations from Initial Concentrations Using ICE (Initial, Change, Equilibrium) [COCl 2 ][CO][Cl 2 ] Initial Change Equilibrium 0.000.5001.00 +1x-1x 0.00 + 1x0.500 - 1x1.00 – 1x K eq = [Cl 2 ] 1 [CO] 1 [COCl 2 ] 1 0.680= [1.00 - x] 1 [0.500 - x] 1 [x] 1 x=0.340 - 0.340x – 0.680x + 0.680x 2 0.680x 2 -2.02x+0.340=0 ax2ax2 +bxbx+c=0 1Cl 2 (g) 1CO (g) 1COCl 2 (g) +

59. Chemical Equilibrium IV. Using Equilibrium Constants A. Calculating Equilibrium Concentrations from Initial Concentrations Using ICE (Initial, Change, Equilibrium) x=-b-b±√b 2 – 4ac 2a2a x=2.02±√(-2.02) 2 – 4(0.680)(0.340) 2(0.680) x=2.79 or 0.18 [COCl 2 ]=0.18 M [CO]=0.32 M [Cl 2 ]=0.82 M x=0.18 0.680x 2 -2.02x+0.340=0 ax2ax2 +bxbx+c=0