1. Chemical Systems & Equilibrium Unit 4

2. Dynamic Equilibrium  dynamic equilibrium = a balance between forward and reverse processes occurring at the same rate

3. Party Analogy  30 people at a house party  8pm:  16 people in the kitchen  14 people in the living room  10pm:  16 people in the kitchen  14 people in the living room  Different people but same number in each room

4. At Equilibrium:  Closed system – no matter/energy/pressure changes  No macroscopic changes  Reactants and products both present (and usually in different amounts)  [reactant] = constant, [product] = constant  Can be approached from both sides Rate of forward reaction = rate of reverse reactions

5. Dynamic Equilibrium  Example: Closed bottle of pop  CO 2 gas leaving dissolved state and entering gas state  CO 2 gas ALSO, leaving gas state and entering liquid state  No visible change  CO 2(g) CO 2(aq)

6. Equilibrium Double Arrow  equilibrium is symbolized with an equation containing a forward ( → ) and a reverse ( ← ) arrow combined into: N2O4 (g)N2O4 (g) 2NO 2 (g)

7. Equilibrium Double Arrow  forward reaction = in an equilibrium equation, the left-to-right reaction  reverse reaction = in an equilibrium equation, the right-to-left reaction CO 2(g) CO 2(aq) Forward Reverse

8. Drinking Bird Equilibrium  https://www.youtube.com/watch?v=Bzw0kWvfVkA https://www.youtube.com/watch?v=Bzw0kWvfVkA  At rest the vapor and the liquid inside the tube are in an equilibrium  Wet head of bird with water – as the water evaporates from around the head, it takes energy with it, head cools down = vapor inside the head cools and contracts = vacuum = pulls liquid up

9. 3 Types of Equilibrium 1. Solubility Equilibrium (dissolving process) 2. Phase Equilibrium (change of state) 3. Chemical Reaction Equilibrium (reactants ⇆ products)

10. Types of Equilibrium #1  solubility equilibrium = a dynamic equilibrium between a solute and a solvent in a saturated solution in a closed system I 2(s) I 2(aq)

11. Solubility Equilibrium  Saturated solution = a solution containing the maximum quantity of a solute  Beyond the solubility limit, any added solute will remain solid and not dissolve

12. Solubility Equilibrium  kinetic molecular theory states that particles are always moving and colliding  even if no changes are observed  Dissolution = the process of dissolving

13. (a) When the solute is first added, many more ions dissociate from the crystal than crystallize onto it. (b) As more ions come into solution, more ions also crystallize. (c) At solubility equilibrium, solute ions dissolve and crystallize at the same rate.

14. Digesting a Precipitate  Allow precipitates to sit for long periods of time before filtering  The longer you wait the more pure the crystal, also the larger the crystal  If precipitate forms quickly, impurities maybe trapped in the precipitate

15. Types of Equilibrium #2  phase equilibrium = a dynamic equilibrium between different physical states of a pure substance in a closed system  closed system = a system that may exchange energy but NOT matter with it’s surroundings H 2 O (l) H 2 O (g) H 2 O (s) H 2 O (l)

16. Phase Equilibrium

17. Types of Equilibrium #3  chemical reaction equilibrium a dynamic equilibrium between reactants and products of a chemical reaction in a closed system  reversible reaction = a reaction that can achieve equilibrium in the forward or reverse direction

18. Chemical Reaction Equilibrium In a Closed System  N 2 O 4(g) 2 NO 2(g)

19. Reversible Reactions  The same dynamic equilibrium composition is reached whether we start from pure N 2 O 4(g), pure NO 2(g), or a mixture of the two, provided that environment, system and total mass remain the same.

20. Calculating the Equilibrium Constant  The equilbrium constant, Keq, is the ratio of equilibrium concentrations at a particular temp  Kc for solution-phase systems or Kp for gas- phase systems  K eq = [C] c [D] d for the eqn [A] a [B] b aA+bB cC+dD Note: The equilibrium constant depends ONLY on the concentration of gases (not liquids/solids)

21. Questions: Equilibrium Law Expression 1. Write the equilibrium law expression for the following: a) 2NO 2(g) ↔ N 2 O 4(g) b) 2HI (g) ↔ H 2(g) + I 2(g) 2. A reaction vessel contains NH 3, N 2 and H 2 gas at equilibrium at a certain temperature. The equilibrium concentrations are [NH 3 ] = 0.25mol/L, [N 2 ] = 0.11mol/L and [H 2 ] = 1.91 mol/L. Calculate the equilibrium constant for the decomposition of ammonia. K = [N 2 O 4(g) ] [NO 2(g) ] 2 K = [H 2(g) ] [I 2(g) ] [HI (g) ] 2 K = [N 2(g) ] [H 2(g) ] 3 [NH 3(g) ] 2 K = [0.11] [1.91 ] 3 [0.25 ] 2 K = 12.3 2NH 3(g) ↔ N 2(g) + 3H 2(g)

22. Questions: Equilibrium Law Expression 3. Nitryl chloride gas, NO 2 Cl, is in equilibrium at a certain temperature in a closed container with NO 2 and Cl 2 gases. At equilibrium, [NO 2 Cl] = 0.00106mol/L and [NO 2 ] = 0.0108mol/L. If K = 0.558, what is the equilibrium concentration of Cl 2 ? 4. Write a balanced equation for the reaction with the following equilibrium law expression: K = [NO 2(g) ] 2 [NO (g) ] 2 [O 2 (g) ]

23. Heterogeneous Equilibria  homogeneous equilibria = equilibria in which all entities are in the same phase  Reactants and products are all gas or all aqueous  heterogeneous equilibria = equilibria in which reactants and products are in more than one phase  Reactants and products are in different phases

24. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P

25. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = [CaO (s) ][CO 2(g) ] [CaCO 3(s) ] [CaCO 3(s) ] = constant [CaO (s) ] = constant K c = [CO 2(g) ] The concentration of solids and pure liquids are considered to be constant and are not included in the expression for the equilibrium constant. KcKc [CaO (s) ] [CaCO 3(s) ] = [CO 2(g) ]

26. P CO2 = K p CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not depend on the amount of CaCO 3 or CaO

27. N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium Equilibrium favors the reactant side

28. CHECKPOINT The reaction at 200C between ethanol and ethanoic acid produces ___________________ and __________________. 1.Write the equation for this reaction 2.Determine the equilibrium constant expression for the reaction

29. Sample Problem: When ammonia is heated it decomposes: 2NH 3(g) ↔ N 2(g) + 3H 2(g) When 4.0 mol of ammonia is introduced in a 2.0L container and heated. The equilibrium amount of ammonia is 2 0 mol. Determine the equilibrium concentrations of the other two entities. STEP 1: Determine the concentration (initial and equilibrium) for known values STEP 2: Setup an ICE Table STEP 3: Determine the value of X STEP 4: Use x value to determine the other quantities Calculating Equilibrium Concentrations (when given one concentration)

30. Setup ICE Table Determine the concentrations [NH3] initial = 4.0mol/2.0L = 2.0mol/L [NH3] equilibrium = 2.0mol/2.0L = 1.0mol/L

31. Determine the value of X [NH 3 ] (g)equil = 2 0mol / L - 2x [NH 3 ] (g)equil = 1.0mol/L (from calculations in Step 1) 2.0mol/L – 2x = 1.0mol/L -2x = - 1.0mol/L x = 0.5mol/L Use X to determine other quantities

32. constant

33. Reversible Reactions  For a given overall system composition, the same equilibrium concentrations are reached whether equilibrium is approached in the forward or the reverse direction  What about K eq will it be the same in fwd/rev?

34. Equilibrium Tubes The effects of temperature on equilibrium Heat + N 2 O 4 (g) 2NO 2 (g) Colourless Brown Very ColdCold Hot ENDOTHERMIC Rxn

35. NO 2 is one of the chemicals in smog !  In the summer on hot, windless days an orange haze is seen over the horizon, this is NO 2  In the winter, the smog doesn't go away, it is just less noticeable. The cooler temperatures lead to more N 2 O 4 and less NO­ 2 which we can't see as well! N 2 O 4 (g) 2NO 2 (g) Colourless Brown

36. Qualitative Changes in Equilibrium Systems You should be familiar with your own body’s attempt at maintaining equilibrium or “homeostasis”:  If body T too high  sweat, surface blood vessels dilate  If body T too low  shiver, surface blood vessels constrict  If blood CO 2 levels ↑  breathe deeper & faster  If blood sugar levels ↑  insulin released to remove excess glucose

37. Le Châtelier’s Principle  When a chemical system at equilibrium is disturbed by a change in a property, the system adjusts in a way that opposes the change.  In other words: If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

38. Le Châtelier’s Principle Le Chatelier’s Principle: if you disturb an equilibrium, it will shift to undo the disturbance. equilibrium shift = movement of a system at equilibrium, resulting in a change in the concentrations of reactants and products  https://www.youtube.com/watch?v=dIDgPFEucFM https://www.youtube.com/watch?v=dIDgPFEucFM

39. 1. System starts at equilibrium. 2. A change/stress is then made to system at equilibrium.  Change in concentration  Change in temperature  Change in volume/pressure 3. System responds by shifting to reactant or product side to restore equilibrium. Le Châtelier’s Principle

40. Change in Reactant or Product Concentrations Adding a reactant or product shifts the equilibrium away from the increase. Removing a reactant or product shifts the equilibrium towards the decrease. To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product.

41. Le Châtelier’s Principle Change in Reactant or Product Concentrations If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 That is, the system must consume the H 2 and produce products until a new equilibrium is established. Equilibrium shifts to the right. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases. N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)

42. Change in Reactant or Product Concentrations N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress

43. Change in Reactant or Product Concentrations ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left aA + bB cC + dD

44. Le Châtelier’s Principle Effect of Temperature Changes The equilibrium constant is temperature dependent. For an endothermic reaction,  H > 0 and heat can be considered as a reactant. For an exothermic reaction,  H < 0 and heat can be considered as a product.

45. Effect of Temperature Changes

46. Adding heat (i.e. heating the vessel) favors away from the increase: – if  H = + (Endothermic), adding heat favors the forward reaction, – if  H = - (Exothermic), adding heat favors the reverse reaction. Removing heat (i.e. cooling the vessel), favors towards the decrease: – if  H = + (Endothermic), cooling favors the reverse reaction, – if  H = -, (Exoothermic), cooling favors the forward reaction.

47. Gas Law – Boyle’s Law Relationship: Pressure & Volume  As pressure on a gas increases, the volume of the gas decreases

48. Le Châtelier’s Principle Effects of Volume and Pressure As volume is decreased pressure increases. The system shifts to decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. Decreasing the number of molecules in a container reduces the pressure. In a reaction with the same number of product and reactant moles of gas, pressure has no effect.  Only a factor with gases.

50. Effects of Volume and Pressure ChangeShifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas A (g) + B (g) C (g)

51. Le Châtelier’s Principle Adding a Catalyst does not shift the position of an equilibrium system system will reach equilibrium sooner

52. Le Châtelier’s Principle Adding a Catalyst  lowers the activation energy for both forward and reverse reactions by an equal amount, so the equilibrium establishes much more rapidly but at the same position as it would without the catalyst

53. Adding a Catalyst

54. Le Châtelier’s Principle Adding Inert Gases  pressure of a gaseous system at equilibrium can be changed by adding a gas while keeping the volume constant  If the gas is inert in the system, for example, if it is a noble gas or if it cannot react with the entities in the system, the equilibrium position of the system will not change

55. Le Châtelier’s Principle Summary

56. No Sweat!  Chickens cannot perspire.  When a chicken gets hot, it pants like a dog.  Farmers have known for a long time that chickens lay eggs with thin shells in hot weather.  These fragile eggs are easily damaged.  Eggshell is primarily composed of calcium carbonate, CaCO3(s).  The source of the carbonate portion of this chalky material is carbon dioxide, CO2, produced as a waste product of cellular respiration.

57. No Sweat!  The carbon dioxide dissolves in body fluids forming the following equilibrium system:

58. No Sweat!  When chickens pant, blood carbon dioxide concentrations are reduced, causing a shift through all four equilibria to the left and a reduction in the amount of calcium carbonate available for making eggshells.  Solution: Give the chickens carbonated water to drink in the summer.  This shifts the equilibria to the right, compensating for the leftward shift caused by panting.

62. Le Chatelier’s Principle: Warm-up  Page 459 # 4 & 6 AgNO 3 – Hint: check solubility table

64. Graph for Question #4

65. Le Chatelier’s Principle: Warm-up  Page 459 A = ↑ volume B = ↑ temperature C = ↑ [C 2 H 6 ] D = catalyst/inert gas E = ↓ [C 2 H 4 ]

66.  At constant temperature, regardless of initial concentrations the concentrations of reactants and products always give a constant value K aA + bB cC + dD K = [C] c [D] d [A] a [B] b Products Reactants The Equilibrium Law Expression & The Equilibrium Constant, K

67. Equilibrium Law Expression  The molar concentrations of the products are always multiplied by one another and written in the numerator, and the molar concentrations of the reactants are always multiplied by one another and written in the denominator.  The coefficients in the balanced chemical equation are equal to the exponents of the equilibrium law expression.  The concentrations in the equilibrium law expression are the molar concentrations of the entities at equilibrium.

68. Recall: Equilibrium achieved from any combination of reactants and products

69. constant N 2 O 4 (g) 2NO 2 (g) = 4.6 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] Regardless of initial concentrations, at a given temperature, the relationship of the equilibrium concentrations of reactants and products always yields a CONSTANT value, K

70. Ways Different States of Matter Can Appear in the Equilibrium Constant, K MolarityPartial Pressure gas, (g) YES aqueous, (aq)YES--- liquid, (l)--- solid, (s)---

71. Questions: Equilibrium Law Expressions Write the equilibrium law expressions for the following reactions: 1. NH 4 Cl (s) ↔ NH 3(g) + HCl (g) 2. 2H 2 O (l) ↔ 2H 2(g) + O 2 (g) 3. 2NaHCO 2(s) ↔ Na 2 CO 3(s) + H 2 O (g) + CO 2(g) K = [NH 3 (g) ] [HCl (g) ] K = [H 2 (g) ] 2 [O 2(g) ] K = [H 2 O (g) ] [CO 2(g) ]

72. Equilibrium Law Expression  Note: equilibrium constants give no information about the rate of a reaction; they provide only a measure of the equilibrium position of the reaction  K is independent of the initial concentration of the reactants and products, but on the concentrations at the equilibrium

73. What Does the Value of K Mean?  If K >> 1, the reaction is product-favoured; product predominates at equilibrium.  If K << 1, the reaction is reactants-favoured; reactants predominates at equilibrium. Products Reactants

75. Question: Magnitude of K Consider the reaction: H 2(g) + I 2(g) ↔ 2HI (g) + heat At 448⁰C, K=50.5. Would you predict the value of K to be higher or lower at 300⁰C?  At 448⁰C K >> 1 = PRODUCTS favoured  Heat lowered = rxn shifts to PRODUCT side  At 300 ⁰C K > 50.5

76. Equilibrium Reactions in Solution  In addition to gas-phase and heterogeneous reactions, equilibrium reacts can also take place in solution.  It is important to write the reaction components as they ACTUALLY EXIST IN SOLUTION -- represent ions in solution as individual entities  Get the equilibrium law expression from the net ionic equation

77. Equilibrium Reactions in Solution Example: Write the equilibrium reaction and equilibrium law expression for the reaction between zinc metal and copper (II) chloride solution. Cancel out the spectator ions

79. The Reaction Quotient, Q  If a chemical system begins with reactants only, it is obvious that the reaction will initially proceed to the right, toward products.  If, however, reactants and products are both present, the direction in which the reaction proceeds is usually less obvious.  In such a case, we can substitute the concentrations into the equilibrium law expression to produce a trial value that is called a reaction quotient, Q

80. The Reaction Quotient, Q  reaction quotient, Q = a test calculation using measured concentration values of a system in the equilibrium expression  think of Q as being similar to K  K is calculated using concentrations at equilibrium  Q may or may not be at equilibrium

81. The Reaction Quotient, Q aA + bB cC + dD Q = [C] c [D] d [A] a [B] b

82. The Reaction Quotient, Q  Q is equal to K, and the system is at equilibrium.  Q is greater than K, and the system must shift left (toward reactants) to reach equilibrium, because the product-to-reactant ratio is too high.  Q is less than K, and the system must shift right (toward products) to reach equilibrium, because the product-to-reactant ratio is too low.

83. ICE Table  Initial, Change, Equilibrium  I = initial concentration of reactants and products before reaction  C = change in the concentrations of reactants and productsthe start and the point at which equilibrium is achieved  E = concentrations of reactants and products at equilibrium.

84. Solving Equilibrium Problems with ICE Example1: For the above reaction [N 2 ] i = 0.32mol/L and [H 2 ] i = 0.66mol/L. At a certain T and P, [N 2 ] eq = 0.20mol/L. What is the value of K under these conditions?

85. Example 2 At 150°C, K for the reaction I 2(g) + Br 2(g) ↔ 2IBr (g) is found to be 1.20x10 2. Starting with 4.00mol of each of iodine and bromine in a 2.00L flask, calculate the equilibrium concentrations of all reaction components.

86. Example 3  Unlike the previous two examples, it is not always obvious if a system is already at equilibrium, or which way the reaction will shift to reach equilibrium.  In these situations, it is helpful to determine the Reaction Quotient, Q  When the reaction 2HI (g) ↔ H 2(g) + I 2(g) takes place at 445°C, the value of K is 0.020. If [HI]=0.20mol/L, [H 2 ]=0.15mol/L and [I 2 ]=0.09mol/L, is the system at equilibrium? If not, in which direction will it shift to reach equilibrium?

87. Example 4  For the reaction H 2(g) + F 2(g) ↔ 2HF (g), K is 1.5x10 2 at SATP. Calculate all equilibrium concentrations if 4.00mol of H 2(g), 4.00mol of F 2(g) and 6.00mol of HF (g) are initially placed in a 2.00L reaction vessel.

88. Calculations with Imperfect Squares  Our ability to square both sides of the equilibrium law equation greatly simplified the calculation of equilibrium concentrations.  In the absence of perfect squares, a different simplification technique helps us solve the problem.

89. Assumption “The 100 rule”  if the concentration to which x is added or from which x is subtracted is at least 100 times greater than the value of K  initial conc. divided by K  If # is greater 100 then drop the x in the denominator

90. When the 100 Rule assumption fails  We must use the quadratic equation

91. Example 5 If 0.50 mol of N 2 O 4(g) is placed in a 1.0L closed container at 150C, what will be the concentrations of N 2 O 4(g) and NO 2(g) at equilibrium? (K = 4.50) N 2 O 4(g) ↔ 2NO 2(g)

92. Homework  Read section 7.5  Questions

94. Remember Solubility?  Solubility = the concentration of a saturated solution of a solute in solvent at a specific temperature and pressure  Solubility is a specific maximum concentration  Degree of Solubility:  Unsaturated  Saturated  Supersaturated

95. Solubility  Unsaturated solution = a solution containing less than maximum quantity of a solute  Saturated solution = a solution containing the maximum quantity of a solute  Supersaturated solution = a solution whose solute concentration exceeds the equilibrium concentration

97. Solubility Curve of Solids

98. The Solubility Product Constant  Solubility product constant (Ksp) = the value obtained from the equilibrium law applied to a saturated solution  Similar to Keq  no units  At a specific temp.  Example: AgCl (s) ↔ Ag + (aq) + Cl - (aq) K sp = [Ag + (aq) ] [Cl - (aq) ] = 1.8x10 -10 at 25 ⁰ C

99. Equilibrium exists between a saturated solution and excess solute. DissolvingPrecipitation Saturated Solution Excess Solute

100. Solubility vs. Solubility Product  Solubility = the amount of a salt that dissolves in a given amount of solvent to give a saturated solution  mol/L or g/100mL  Solubility Product = the product of the molar concentrations of a the ions in the saturated solution  Ksp has no units

101. Table of K sp  Appendix C8 (page 802)  Usually only for low solubility ionic compounds  High solubility compounds form solutions that do not tend to be saturated & no equilibrium is established

102. Calculating Solubility using the Ksp Value  Example 1: Calculate the molar solubility of cobalt (II) hydroxide at 25 ⁰ C if K sp = 1.1x10 -15 at this temperature.

103. Calculating Ksp using Solubility values  Example 2: Calculate K sp for silver chromate (Ag 2 CrO 4 ) if its solubility is 0.29g/L at 25 ⁰ C.

104. Predicting Precipitation  Instead of using a solubility table…  using Q to determine whether, after mixing, the ions are present in too high a concentration, in which case a precipitate will form  Trial Ion Product = the reaction quotient applied to the ion concentrations of a slightly soluble salt

105. Using Q to Predict Solubility Q is greater than Ksp supersaturated solution Precipitate will from Q is equal to KspSaturated Precipitate will not form Q is less than Kspunsaturated Precipitate will not form

106. Demo: KI + Pb(NO 3 ) 2

107. Calculations involving the prediction of a precipitate (using Q)  Example 3: If 500mL of a 4.0x10-6 mol/L CaCl2 solution is mixed with 300mL of a 0.0040mol/L AgNO3 solution, will a precipitate form?

108. Homework:  Page 486 #1,2,4  Page 488 #5  Worksheet: Extra Solubility Problems  Quiz on Thursday April 18  ICE problem + solubility

109. Common Ion Effect  Common Ion Effect = A reduction in the solubility of a salt caused by the presence of another slat having a common ion

110. Energy & Equilibrium: The Laws of Thermodynamics

111. Thermodynamics  Thermodynamics = the study of energy transformation  3 fundamental laws of thermodynamics  Laws used to understand why certain changes occur but others do not

112. First Law of Thermodynamics “Conservation of Energy”  The total amount of energy in the universe is constant. Energy can be neither created nor destroyed, but can be transferred from one object or place to another, or transformed from one form to another.

113. First Law of Thermodynamics Remember:  Total energy of the universe = system + surrounding  Hess’s Law = the value of ∆H for any reaction that can be written in steps equals the sum of the ∆H values for each of the individual steps

114. Enthalpy Changes & Spontaneity  bond energy = the minimum energy required to break one mole of bonds between two particular atoms; a measure of the stability of a chemical bond  It is also equal to the amount of energy released when a mole of a particular bond is formed.  It is measured in kilojoules (kJ) and is equal to the minimum energy required to break the intramolecular bonds between one mole of molecules of a pure substance.

115. Bond Energy  Bond energy is measured in kilojoules (kJ) and is equal to the minimum energy required to break the intramolecular bonds between one mole of molecules of a pure substance.  Energy is absorbed when reactant bonds break  Energy is released when product bonds form

116. Bond Energy A bond that has a higher bond energy (i.e. Requires more energy to break) is more stable.

117. Enthalpy & Entropy Changes Together Determine Spontaneity  Endothermic = + ∆H  Exothermic = - ∆H  Exothermic reactions tend to proceed spontaneously

118. Spontaneous Reaction  spontaneous reaction = one that, given the necessary activation energy, proceeds without continuous outside assistance  Example: a sparkler  Needs light from a flame for activation  Once lit, the available fuel combusts quickly and completely, releasing large amounts of energy as heat and light

119. Entropy  enthalpy is not the only factor that determines whether a chemical or physical change occurs spontaneously  entropy, S = a measure of the randomness or disorder of a system, or the surroundings

120. Entropy  Increase entropy = increase randomness = +∆S  When entropy increases in a reaction, the entropy of the products, S products, is greater than the entropy of the reactants, S reactants, yielding an overall positive change in entropy, S.

121. Entropy  decrease entropy = decrease randomness = -∆S  When entropy decreases in a reaction, the entropy of the products, S products, is less than the entropy of the reactants, S reactants, yielding an overall negative change in entropy, S.

122. Increase in Entropy

123. Change in Volume of Gaseous Systems

124. Change in Temperature

125. Change in State

126. In Chemical Reactions…

128. Enthalpy, Entropy, and Spontaneous Change  Changes in the enthalpy, ∆H, and entropy, ∆S, of a system help us to predict whether a change will occur spontaneously  Exothermic reactions (-∆H) involving an increase in entropy (+∆S) occur spontaneously, because both changes are favoured  Endothermic reactions (+∆H) involving a decrease in entropy (-∆S) are not spontaneous because neither change is favoured

129. Enthalpy, Entropy, and Spontaneous Change  But what happens in cases where the energy change is exothermic (favoured) and the entropy decreases (not favoured)?  Or when the energy change is endothermic (not favoured) but entropy increases (favoured)?  In these situations, the temperature at which the change occurs becomes an important consideration as well as free energy

130. Free Energy  free energy (or Gibbs free energy), G = energy that is available to do useful work  In general, a change at constant temperature and pressure will occur spontaneously if it is accompanied by a decrease in Gibbs free energy, G -∆G = spontaneous +∆G = nonspontaneous

131. Second Law of Thermodynamics “Law of Entropy”  all changes that occur in the universe. All changes, whether spontaneous or not, are accompanied by an increase in the entropy (overall disorder) of the universe Mathematically, S universe > 0

132. Second Law of Thermodynamics  a system’s entropy, S system, can decrease (the system becomes more ordered), so long as there is a larger increase in the entropy of the surroundings, S surroundings, so that the overall entropy change, Suniverse, is positive.

133. Problem?  Living organisms seem to violate the second law of thermodynamics.  Build highly ordered molecules such as proteins and DNA from a random assortment of amino acids and nucleotides dissolved in cell fluids  building highly ordered structures such as nests, webs, and space huttles.

134. Not really a problem…  Living organisms obey the second law of thermodynamics because they create order out of chaos in a local area of the universe while creating a greater amount of disorder in the universe as a whole

135. Oh no! Thermal Death!  The second law of thermodynamics predicts that the universe will eventually experience a final “thermal death” in which all particles and energy move randomly about.  Life will come to an end because there won’t be any sources of free energy to exploit; stars will stop shining. Waterfalls will stop falling.  All energy will have become randomized. All of the energy that there ever was will still be there, except that it will be uniformly distributed throughout the universe, unable to apply an effective push or a pull on anything. According to the second law, a state of perfect equilibrium is the ultimate fate of the universe.

136. Predicting Spontaneity  The spontaneity of any reaction carried out at constant temperature and pressure can be predicted by calculating the value of G using the following equation, called the Gibbs-Helmholtz equation:

137. ∆G, Spontaneity & Free Energy ∆G = ∆H - T∆S  ∆G = - = spontaneous  ∆G = + = nonspontaneous  Remember: K = ºC + 273

138. Predicting Spontaneity +∆H-∆H +∆S Spontaneity depends on T Spontaneous -∆S nonspontaneous Spontaneity depends on T ∆G = ∆H - T∆S -∆G = spontaneous +∆G= nonspontaneous

141. Third Law of Thermodynamics “Law of Entropy”  The entropy of a perfectly ordered pure crystalline substance is zero at absolute zero.  Mathematically, S = 0 at T = 0 K

142. Calculating Standard Entropy Change  standard entropy = the entropy of one mole of a substance at STAP; units (J/molK)

143. Remember ∆H°  Standard enthalpy change of reaction

144. ∆H° vs ∆S°