2EquilibriumDefinitionsEquilibrium is a process in which two opposing processes occur at the same time and at the same rate such that there is no net change.The phenomena of equilibrium occurs in chemical systemsSuch systems are said to be reversible, which means that a process occurs in one direction but the reverse process can also occur at the same time and at the same rate.
3Rate of forward reaction = rate of reverse reactions At Equilibrium:Closed system – no matter/energy/pressure changesNo macroscopic changesReactants and products both present (and usually in different amounts)[reactant] = constant, [product] = constantCan be approached from both sidesRate of forward reaction = rate of reverse reactions
4Dynamic Equilibriumdynamic equilibrium = a balance between forward and reverse processes occurring at the same rate & this is denoted by a double arrow** Dynamic equilibrium will not occur if any of the chemicals, reactants, or products escape or are removed from the container.
5Party Analogy 30 people at a house party 8pm: 10pm: 16 people in the kitchen14 people in the living room10pm:Different people but same number in each room
6Dynamic Equilibrium CO2(g) CO2(aq) Example: Closed bottle of pop CO2 gas leaving dissolved state and entering gas stateCO2 gas ALSO, leaving gas state and entering liquid stateNo visible changeCO2(g) CO2(aq)
7Equilibrium Double Arrow equilibrium is symbolized with an equation containing a forward (→) and a reverse (←) arrow combined into:N2O4 (g)2NO2 (g)
8Equilibrium Double Arrow forward reaction = in an equilibrium equation, the left-to-right reactionreverse reaction = in an equilibrium equation, the right-to-left reactionCO2(g) CO2(aq)ForwardReverse
9Drinking Bird Equilibrium https://www.youtube.com/watch?v=Bzw0kWvfVkAAt 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
103 Types of Equilibrium 1. Solubility Equilibrium (dissolving process) 2. Phase Equilibrium (change of state)3. Chemical Reaction Equilibrium (reactants ⇆ products)
11Types of Equilibrium #1solubility equilibrium = a dynamic equilibrium between a solute and a solvent in a saturated solution in a closed systemI2(s) I2(aq)
12Solubility Equilibrium Saturated solution = a solution containing the maximum quantity of a soluteBeyond the solubility limit, any added solute will remain solid and not dissolve
13Solubility Equilibrium kinetic molecular theory states that particles are always moving and collidingeven if no changes are observedDissolution = the process of dissolving
14(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.
15Digesting a Precipitate Allow precipitates to sit for long periods of time before filteringThe longer you wait the more pure the crystal, also the larger the crystalIf precipitate forms quickly, impurities maybe trapped in the precipitate
16Types of Equilibrium #2phase equilibrium = a dynamic equilibrium between different physical states of a pure substance in a closed systemclosed system = a system that may exchange energy but NOT matter with it’s surroundingsH2O(l) H2O(g)H2O(s) H2O(l)
18Types of Equilibrium #3chemical reaction equilibrium a dynamic equilibrium between reactants and products of a chemical reaction in a closed systemreversible reaction = a reaction that can achieve equilibrium in the forward or reverse direction
19Chemical Equilibrium Equilibrium is reached when the rate of the forward reaction equals the rate of thereverse reaction.H2(g) + I2(g)2HI(g)(H = -13kJ/mol)The H value always refers to the forward reaction. 13 kJ of energy is liberated for every mole of HI formed.For the whole reaction:H2(g) + I2(g)2HI(g)(H = -26kJ)
20Chemical Reaction Equilibrium In a Closed System N2O4(g) 2 NO2(g)
22Reversible ReactionsThe same dynamic equilibrium composition is reached whether we start from pure N2O4(g), pure NO2(g), or a mixture of the two, provided that environment, system and total mass remain the same.
23Calculating the Equilibrium Constant The equilbrium constant, Keq, is the ratio of equilibrium concentrations at a particular tempKc for solution-phase systems or Kp for gas- phase systemsKeq = [C]c[D]d for the eqn[A]a[B]b aA+bB cC+dDNote: The equilibrium constant depends ONLY on the concentration of gases (not liquids/solids)
24Questions: Equilibrium Law Expression 1. Write the equilibrium law expression for the following:a) 2NO2(g) ↔ N2O4(g)b) 2HI(g) ↔ H2(g) + I2(g)2. A reaction vessel contains NH3, N2 and H2 gas at equilibrium at a certain temperature. The equilibrium concentrations are [NH3] = 0.25mol/L, [N2] = 0.11mol/L and [H2] = 1.91 mol/L. Calculate the equilibrium constant for the decomposition of ammonia.K =[N2O4(g)][NO2(g) ]2K =[H2(g)] [I2(g) ][HI(g) ]22NH3(g) ↔ N2(g) + 3H2(g)K =[N2(g)] [H2(g) ]3[NH3(g) ]2K =[0.11] [1.91 ]3[0.25 ]2K = 12.3
25Questions: Equilibrium Law Expression 3. Nitryl chloride gas, NO2Cl, is in equilibrium at a certain temperature in a closed container with NO2 and Cl2 gases. At equilibrium, [NO2Cl] = mol/L and [NO2] = mol/L. If K = 0.558, what is the equilibrium concentration of Cl2? 4. Write a balanced equation for the reaction with the following equilibrium law expression:K =[NO2(g)]2[NO (g) ]2 [O2 (g) ]
26Heterogeneous Equilibria homogeneous equilibria = equilibria in which all entities are in the same phaseReactants and products are all gas or all aqueousheterogeneous equilibria = equilibria in which reactants and products are in more than one phaseReactants and products are in different phases
27Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N2O4 (g) NO2 (g)Kp =NO2P2N2O4PKc =[NO2]2[N2O4]
28Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.CaCO3 (s) CaO (s) + CO2 (g)Kc =[CaO(s) ][CO2(g)][CaCO3(s)][CaCO3(s)] = constant[CaO(s) ] = constantKc[CaO(s)][CaCO3(s)]= [CO2(g)]Constant values are incorporated into the equilibrium constant and are NOT included in the equilibrium expressionKc = [CO2(g)]The concentration of solids and pure liquids are considered to be constant and are not included in the expression for the equilibrium constant.
29PCO2 = Kp PCO does not depend on the amount of CaCO3 or CaO CaCO3 (s) CaO (s) + CO2 (g)PCO2= KpPCO2does not depend on the amount of CaCO3 or CaO
30Equilibrium favors the reactant side N2O4 (g) NO2 (g)equilibriumequilibriumequilibriumStart with NO2Start with N2O4Start with NO2& N2O4Equilibrium favors the reactant side
31CHECKPOINTThe reaction at 200C between ethanol and ethanoic acid produces ___________________ and __________________.Write the equation for this reactionDetermine the equilibrium constant expression for the reaction
32Calculating Equilibrium Concentrations (when given one concentration) Sample Problem:When ammonia is heated it decomposes:2NH3(g)↔ N2(g) + 3H2(g)When 4.0 mol of ammonia is introduced in a 2.0L container and heated. Theequilibrium 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 valuesSTEP 2: Setup an ICE TableSTEP 3: Determine the value of XSTEP 4: Use x value to determine the other quantities
36Reversible ReactionsFor a given overall system composition, the same equilibrium concentrations are reached whether equilibrium is approached in the forward or the reverse directionWhat about Keq will it be the same in fwd/rev?
37Equilibrium Tubes Heat + N2O4 (g) 2NO2 (g) BrownColourlessThe effects of temperature on equilibriumENDOTHERMIC RxnVery ColdColdHot
38NO2 is one of the chemicals in smog! In the summer on hot, windless days an orange haze is seen over the horizon, this is NO2In the winter, the smog doesn't go away, it is just less noticeable. The cooler temperatures lead to more N2O4 and less NO2 which we can't see as well!N2O4 (g) NO2 (g)BrownColourless
39Qualitative 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 dilateIf body T too low shiver, surface blood vessels constrictIf blood CO2 levels ↑ breathe deeper & fasterIf blood sugar levels ↑ insulin released to remove excess glucose
40Le 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.
41Le 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 productshttps://www.youtube.com/watch?v=dIDgPFEucFM
42Le Châtelier’s Principle System starts at equilibrium.A change/stress is then made to system at equilibrium.Change in concentrationChange in temperatureChange in volume/pressureSystem responds by shifting to reactant or product side to restore equilibrium.
43Le Châtelier’s Principle Change in Reactant or Product ConcentrationsAdding 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.
44Le Châtelier’s Principle Change in Reactant or Product ConcentrationsIf H2 is added while the system is at equilibrium, the system must respond to counteract the added H2That is, the system must consume the H2 and produce products until a new equilibrium is established.Equilibrium shifts to the right.Therefore, [H2] and [N2] will decrease and [NH3] increases.N2 (g) H2 (g) ↔ NH3 (g)
45Change in Reactant or Product Concentrations N2 (g) + 3H2 (g) NH3 (g)AddNH3Equilibrium shifts left to offset stress
46Change in Reactant or Product Concentrations aA + bB cC + dDChangeShifts the EquilibriumIncrease concentration of product(s)leftDecrease concentration of product(s)rightIncrease concentration of reactant(s)rightDecrease concentration of reactant(s)left
47Le Châtelier’s Principle Effect of Temperature ChangesThe 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.
49Effect of Temperature Changes 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.
50Gas Law – Boyle’s Law Relationship: Pressure & Volume As pressure on a gas increases, the volume of the gas decreases
51Le Châtelier’s Principle Effects of Volume and PressureAs 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.
53Effects of Volume and Pressure A (g) + B (g) C (g)ChangeShifts the EquilibriumIncrease pressureSide with fewest moles of gasDecrease pressureSide with most moles of gasIncrease volumeSide with most moles of gasDecrease volumeSide with fewest moles of gas
54Le Châtelier’s Principle Adding a Catalystdoes not shift the position of an equilibrium systemsystem will reach equilibrium sooner
55Le Châtelier’s Principle Adding a Catalystlowers 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
57Le Châtelier’s Principle Adding Inert Gasespressure of a gaseous system at equilibrium can be changed by adding a gas while keeping the volume constantIf 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
59No 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.
60No Sweat!The carbon dioxide dissolves in body fluids forming the following equilibrium system:
61No 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.
69The Equilibrium Law Expression & The Equilibrium Constant, K At constant temperature, regardless of initial concentrations the concentrations of reactants and products always give a constant value KaA + bB cC + dDK =[C]c[D]d[A]a[B]bProductsReactants
70Equilibrium 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.
71Recall: Equilibrium achieved from any combination of reactants and products
72N2O4 (g) 2NO2 (g) K = [NO2]2 [N2O4] = 4.6 x 10-3 Regardless of initial concentrations, at a given temperature, the relationship of the equilibrium concentrations of reactants and products always yields a CONSTANT value, KN2O4 (g) NO2 (g)K =[NO2]2[N2O4]= 4.6 x 10-3constant
73Ways Different States of Matter Can Appear in the Equilibrium Constant, K MolarityPartial Pressuregas, (g)YESaqueous, (aq)---liquid, (l)solid, (s)
74Questions: Equilibrium Law Expressions Write the equilibrium law expressions for the following reactions:NH4Cl(s) ↔ NH3(g) + HCl(g)2H2O(l) ↔ 2H2(g) + O2 (g)2NaHCO2(s) ↔ Na2CO3(s) + H2O(g) + CO2(g)K =[NH3 (g) ] [HCl(g)]K =[H2 (g) ]2 [O2(g) ]K =[H2O (g) ] [CO2(g) ]
75Equilibrium 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 reactionK is independent of the initial concentration of the reactants and products, but on the concentrations at the equilibrium
76What 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.ProductsReactants
78Question: Magnitude of K Consider the reaction: H2(g) + I2(g) ↔ 2HI(g) + heatAt 448⁰C, K=50.5. Would you predict the value of K to be higher or lower at 300⁰C?At 448⁰C K >> 1 = PRODUCTS favouredHeat lowered = rxn shifts to PRODUCT sideAt 300 ⁰C K > 50.5
79Equilibrium 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 entitiesGet the equilibrium law expression from the net ionic equation
80Equilibrium 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
82The 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
83The Reaction Quotient, Q reaction quotient, Q = a test calculation using measured concentration values of a system in the equilibrium expressionthink of Q as being similar to KK is calculated using concentrations at equilibriumQ may or may not be at equilibrium
84The Reaction Quotient, Q aA + bB cC + dDQ =[C]c[D]d[A]a[B]b
85The 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.
86ICE Table Initial, Change, Equilibrium I = initial concentration of reactants and products before reactionC = change in the concentrations of reactants and productsthe start and the point at which equilibrium is achievedE = concentrations of reactants and products at equilibrium.
87Solving Equilibrium Problems with ICE Example1: For the above reaction [N2]i = 0.32mol/L and [H2]i = 0.66mol/L. At a certain T and P, [N2]eq = 0.20mol/L. What is the value of K under these conditions?
88Example 2At 150°C, K for the reaction I2(g) + Br2(g) ↔ 2IBr(g) is found to be 1.20x102 . Starting with 4.00mol of each of iodine and bromine in a 2.00L flask, calculate the equilibrium concentrations of all reaction components.
89Example 3Unlike 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, QWhen the reaction 2HI(g) ↔ H2(g) + I2(g) takes place at 445°C, the value of K is If [HI]=0.20mol/L, [H2]=0.15mol/L and [I2]=0.09mol/L, is the system at equilibrium? If not, in which direction will it shift to reach equilibrium?
90Example 4For the reaction H2(g) + F2(g) ↔ 2HF(g) , K is 1.5x102 at SATP. Calculate all equilibrium concentrations if 4.00mol of H2(g) , 4.00mol of F2(g) and 6.00mol of HF(g) are initially placed in a 2.00L reaction vessel.
91Calculations 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.
92Assumption“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 Kinitial conc. divided by KIf # is greater 100 then drop the x in the denominator
93When the 100 Rule assumption fails We must use the quadratic equation
94Example 5If 0.50 mol of N2O4(g) is placed in a 1.0L closed container at 150C, what will be the concentrations of N2O4(g) and NO2(g) at equilibrium? (K = 4.50) N2O4(g) ↔ 2NO2(g)
97Remember Solubility?Solubility = the concentration of a saturated solution of a solute in solvent at a specific temperature and pressureSolubility is a specific maximum concentrationDegree of Solubility:UnsaturatedSaturatedSupersaturated
98SolubilityUnsaturated solution = a solution containing less than maximum quantity of a soluteSaturated solution = a solution containing the maximum quantity of a soluteSupersaturated solution = a solution whose solute concentration exceeds the equilibrium concentration
101The Solubility Product Constant Solubility product constant (Ksp) = the value obtained from the equilibrium law applied to a saturated solutionSimilar to Keqno unitsAt a specific temp.Example: AgCl(s) ↔ Ag+(aq) + Cl-(aq)Ksp =[Ag+(aq) ] [Cl-(aq) ] = 1.8x10-10 at 25⁰C
102Equilibrium exists between a saturated solution and excess solute. DissolvingPrecipitationSaturated SolutionExcess Solute
103Solubility vs. Solubility Product Solubility = the amount of a salt that dissolves in a given amount of solvent to give a saturated solutionmol/L or g/100mLSolubility Product = the product of the molar concentrations of a the ions in the saturated solutionKsp has no units
104Table of Ksp Appendix C8 (page 802) Usually only for low solubility ionic compoundsHigh solubility compounds form solutions that do not tend to be saturated & no equilibrium is established
105Calculating Solubility using the Ksp Value Example 1:Calculate the molar solubility of cobalt (II) hydroxide at 25⁰C if Ksp = 1.1x10-15 at this temperature.
106Calculating Ksp using Solubility values Example 2:Calculate Ksp for silver chromate (Ag2CrO4) if its solubility is 0.29g/L at 25 ⁰ C.
107Predicting 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 formTrial Ion Product = the reaction quotient applied to the ion concentrations of a slightly soluble salt
108Using Q to Predict Solubility Q is greater than Kspsupersaturated solutionPrecipitate will fromQ is equal to KspSaturatedPrecipitate will not formQ is less than Kspunsaturated
110Calculations 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 mol/L AgNO3 solution, will a precipitate form?
111Homework:Page 486 #1,2,4Page 488 #5Worksheet: Extra Solubility ProblemsQuiz on Thursday April 18 ICE problem + solubility
112Common Ion EffectCommon Ion Effect = A reduction in the solubility of a salt caused by the presence of another slat having a common ion
113The Laws of Thermodynamics Energy & Equilibrium:The Laws of Thermodynamics
114Thermodynamics Thermodynamics = the study of energy transformation 3 fundamental laws of thermodynamicsLaws used to understand why certain changes occur but others do not
115First 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.
116First Law of Thermodynamics Remember:Total energy of the universe = system surroundingHess’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
117Enthalpy 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 bondIt 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.
118Bond EnergyBond 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 breakEnergy is released when product bonds form
119Bond EnergyA bond that has a higher bond energy (i.e. Requires more energy to break) is more stable.
120Enthalpy & Entropy Changes Together Determine Spontaneity Endothermic = + ∆HExothermic = - ∆HExothermic reactions tend to proceed spontaneously
121Spontaneous Reactionspontaneous reaction = one that, given the necessary activation energy, proceeds without continuous outside assistanceExample: a sparklerNeeds light from a flame for activationOnce lit, the available fuel combusts quickly and completely, releasing large amounts of energy as heat and light
122Entropyenthalpy is not the only factor that determines whether a chemical or physical change occurs spontaneouslyentropy, S = a measure of the randomness or disorder of a system, or the surroundings
123Entropy Increase entropy = increase randomness = +∆S When entropy increases in a reaction, the entropy of the products, Sproducts, is greater than the entropy of the reactants, Sreactants, yielding an overall positive change in entropy, S.
124Entropy decrease entropy = decrease randomness = -∆S When entropy decreases in a reaction, the entropy of the products, Sproducts, is less than the entropy of the reactants, Sreactants, yielding an overall negative change in entropy, S.
131Enthalpy, Entropy, and Spontaneous Change Changes in the enthalpy, ∆H, and entropy, ∆S, of a system help us to predict whether a change will occur spontaneouslyExothermic reactions (-∆H) involving an increase in entropy (+∆S) occur spontaneously, because both changes are favouredEndothermic reactions (+∆H) involving a decrease in entropy (-∆S) are not spontaneous because neither change is favoured
132Enthalpy, 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
133Free Energy -∆G = spontaneous +∆G = nonspontaneous free energy (or Gibbs free energy), G = energy that is available to do useful workIn 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
134Second 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 universeMathematically, Suniverse > 0
135Second Law of Thermodynamics a system’s entropy, Ssystem, can decrease (the system becomes more ordered), so long as there is a larger increase in the entropy of the surroundings, Ssurroundings, so that the overall entropy change, Suniverse, is positive.
136Problem?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 fluidsbuilding highly ordered structures such as nests, webs, and space huttles.
137Not 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
138Oh 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.
139Predicting 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: