Ch12 Thermodynamics Powerpoint PDF

Chapter 12. Thermodynamics 2 Ch. 12 Outline 12.1: From Spontaneity to Entropy 12.2: Entropy 12.3: From the 2nd & 3rd Laws of Thermodynamics to Free Energy 12.4: Free Energy of Reactions 12.1: Spontaneous Processes 3 Will a reaction occur “naturally” at a given temperature and pressure, without the e...

Chapter 12. Thermodynamics 2 Ch. 12 Outline 12.1: From Spontaneity to Entropy 12.2: Entropy 12.3: From the 2nd & 3rd Laws of Thermodynamics to Free Energy 12.4: Free Energy of Reactions 12.1: Spontaneous Processes 3 Will a reaction occur “naturally” at a given temperature and pressure, without the exertion of any outside force? In other words, is a reaction spontaneous? A process that DOES occur naturally under a specific set of conditions is called a spontaneous process. A process that DOES NOT occur naturally under a specific set of conditions is called nonspontaneous. Spontaneous Nonspontaneous 3 12.1: Spontaneous Processes A spontaneous process will “just happen” Examples: A mixture of hydrogen and oxygen will react in the presence of a spark. 2H2(g) + O2(g) + spark → 2H2O(g) 4 5 Spontaneity If a reaction is spontaneous in one direction, it will be non-spontaneous in the reverse direction under the same conditions. 2H2(g) + O2(g) → 2H2O(l) Spontaneous 2H2O(l) → 2H2(g) + O2(g) Non-Spontaneous A non-spontaneous reaction is still possible with the continual input of energy. Spontaneity It is important to not confuse spontaneous with fast. The rate of a reaction (Ch. 17) and spontaneity are not necessarily connected. Spontaneous but way too slow 6 Energy and Spontaneity Many spontaneous processes proceed with a decrease in energy. Spontaneous reactions are often exothermic, but NOT ALWAYS. Spontaneous exothermic reaction: Spontaneous endothermic reaction: 2H2(g) + O2(g) + spark → 2H2O(g) HC2H3O2(aq) + NaHCO3(aq) → NaC2H3O2(aq) + CO2(g) + H2O (l) 7 8 Another factor that matters to spontaneity: Dispersal of Matter and Energy When the value opens, the gas spontaneously expands to fill both containers. With an ideal gas, this process results in no change in energy. Heat spontaneously flows from the hotter object to the colder object. But overall there is no change in energy. Entropy Greater, more uniform dispersal of matter and energy is a driving force of a spontaneous process. Entropy = dispersion of matter and energy 10 Ch. 12 Outline 12.1: From Spontaneity to Entropy 12.2: Entropy 12.3: From the 2nd & 3rd Laws of Thermodynamics to Free Energy 12.4: Free Energy of Reactions Boltzmann formula of Entropy S = k lnW k is the Boltzmann constant (1.38 × 10-23 J/K) W is the number of microstates. Microstate (W) – A specific configuration of the locations and energies of the particles in a system. 11 Microstate (W) Microstate (W) – A specific configuration of the locations and energies of the particles in a system. The number of microstates possible is given by: W = nN n is the number of boxes N is the number of particles Consider two particles distributed amongst two boxes. 12 Microstate (W) Now consider four particles distributed among two boxes. There are five possible distributions for this system. Which distribution is most probable? 13 More Microstate (W) Higher Entropy (S) The most probable distribution has the largest number of microstates, therefore the one of greatest entropy and favored. S = k lnW This principle applies to systems with larger numbers of particles. The most probable state will be the one in which the particles are divided evenly throughout the container (greatest entropy): This same principle can also be applied to spontaneous heat flow. 14 15 Entropy and Hess’s Law Entropy is another state function and follows Hess’s Law (chapter 9.3), which states: If a process can be written as the sum of several stepwise processes, the change of the total process equals the sum of the changes of all the steps. ∆S1 ∆S Sinitial ∆S2 The change in entropy for a process is the difference in entropy between the final state and the initial state. ΔSsys = Sfinal – Sinitial ∆S3 Sfinal ΔS = ΔS1 + ΔS2 + ΔS3 Alternatively, the change in entropy for a process is the sum of entropy changes of each stepwise processes. ΔSsys = ΔS1 + ΔS2 + ΔS3 +…. 16 Entropy Changes Practice x + 115.0 +117.0 +2x=24.8 2x = 24.8 -115.0 -117.0 = -207.2 X = -103.6 Step1 Step2 St ep 3 Historical Study of Entropy In 1824, at the age of 28, Carnot published his findings about steam-powered machinery. Carnot died young, at the age of 36. In 1865, Rudolf Clausius’s study of Carnot’s findings led to groundbreaking discoveries about spontaneous heat flow.. Entropy change is expressed as the ratio of the reversible heat (qrev) and the kelvin temperature (T). 𝑞𝑟𝑒𝑣 ∆𝑆 = 𝑇 18 Factors that Influence Entropy-Phase 1) The phase of the substance. Increasing entropy Sgas> Sliquid > Ssolid Solid Liquid Gas Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each. H2O(l), H2O(g), H2O(s) Answer: H2O(s) 0 for a spontaneous process ΔSuniverse < 0 for a nonspontaneous process ΔSuniverse = 0 for a process at equilibrium Third Law of Thermodynamics: The entropy of a pure perfect crystalline substance at zero kelvin is zero. 24 Standard Entropies Ø Third Law of Thermodynamics defines a reference point: at 0 k, entropy of a perfect crystal is zero. 𝑞𝑟𝑒𝑣 Ø Carnot and Clausius defined the entropy change: ∆𝑆 = 𝑇 Ø Scientist did experiments to measure the values of ∆𝑆 from 0 K or almost 0K and integrated ∆𝑆 to yield solute values of S at 25OC. Standard Entropies, S°298 These values are for 1 mole of a substance at 298 K and standard conditions (1 atm, and 1 M for solutions) Entropies are always positive Units of entropy is J/ mol.K ΔS° for Reactions 26 Entropy change of a reaction ∆Srxno can be obtained from S° ΔSrxno = ΣnΔSo (products) – ΣnΔSo (reactants) n is the coefficients of the balanced equation. When calculating ΔS° remember to multiply the standard enthalpies of formation by the coefficients of the balanced equation. ° Calculate DS 298 for the following change. H2(g) O2 (g) H2O (l) *A. -326.6 J/K B. 326.6 J/K C. 520 J/K S° (J mol−1 K−1) 130.7 205.2 70.0 ΔSrxno = (2x70.0) –(2x130.7 + 1 x 205.2) =-326.6 J/K 27 Factors Effecting ΔSsurr Case study: Condensing of water vapor into liquid droplets is spontaneous at low temperature. Condensing of water is exothermic. Heat released from the system will be absorbed by the surrounding and increases its entrope, ΔSsurr is positive. So ΔSsurr is directly proportional to the - ΔHsystem (enthalpy changeof the system) 28 Factors Effecting ΔSsurr Case study: Condensing of water vapor into liquid droplets is spontaneous at low temperature, however, evaporation of liquid water into water vapor is spontaneous at elevated temperature. Why? ∆Suniv = ∆Ssys + ∆Ssurr (for water condensing) When T is low, water condenses spontaneously When T is high, water doesn’t condense spontaneously ∆Ssurr ∆Suniv >0 ∆Ssurr ∆Ssys ∆Suniv 0, the reaction is non-spontaneous in the forward direction If ΔG = 0, the system is at equilibrium Relationship Among ΔG, ΔH, and ΔS ΔG = ΔH – TΔS Spontaneous reactions, those with negative DG, generally have ΔH < 0 Exothermic reaction. A negative ΔH will contribute to a negative ΔG. ΔS > 0 A positive ΔS will contribute to a negative ΔG. Note that a reaction can still be spontaneous when ΔH is positive or ΔS is negative, but not both. 33 ΔG = ΔH – TΔS ∆H ∆S ∆G Spontaneity + - + - + Spontaneous - at low T + at high T Spontaneous at low T Not at high T + + + at low T + at high T Not at low T Spontaneous at high T Nonspontaneous If ΔS is positive, -TΔS will be negative. Then, the higher the T, the more negative the -TΔS , and the more negative the ΔG ΔG Example ΔG = 0 = ΔH – TΔS Þ ΔH = TΔS T = ΔH / ΔS This is the temperature at which ΔG = 0, or by definition, the system is at equilibrium. 35 Direction of Spontaneity Change A reactions has ∆Ho298 = -50 kJ/mol and ∆So298 = -250 J/mol.K. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous? If spontaneous, under what temperature conditions will it become nonspontaneous? A. The reaction is not spontaneous at room temperature. It will become spontaneous when the temperature increased to 400 K or beyond. *B. The reaction is nonspontaneous at room temperature. It will become spontaneous when the temperature decreased to 200 K or lower. C. The reaction is spontaneous at room temperature. It will become nonspontaneous when the temperature increased to 200 K or beyond. ΔG° = ΔH° – TΔS° ΔG° = -50 – 298 (-0.250) = 24.5 kJ/mol ΔG° >0, Nonspontaneous T = ΔH / ΔS = -50/(-0.25) = 200K 36 37 Ch. 12 Outline 12.1: From Spontaneity to Entropy 12.2: Entropy 12.3: From the 2nd & 3rd Laws of Thermodynamics to Free Energy 12.4: Free Energy of Reactions The Standard Free Energy Change, ΔG° Although the Change in Gibbs Free Energy equation is valid under all conditions we will most often apply it at standard conditions. Under standard conditions, ΔG° = ΔH° – TΔS° Pay attention to J vs. kJ in calculations! What is the ∆S° if the ∆H° is +88.00 kJ and ∆G° is +11.21 kJ at 298K? ΔG° = ΔH° – TΔS° A. -258 J B. *+258 J Þ 11.21 kJ = 88.00 kJ -298 ( ΔS°) C. +0.258 J Þ ΔS° = (88.00 – 11.21)kJ /298 = 0.258 kJ 38 Standard Free Energy of Formation, ∆Gf° The Standard Free Energy of Formation (∆Gf°) for a compound is defined as the free energy change for the formation of one mole of a substance from free elements in their standard state at 1 atm and 25 °C. Example: H2 (g) + ½O2 (g) à H2O (l) ∆Gf° = -237.2 kJ/mol Example: write an equation representing ∆Gf° of CaCO3 Ca (s) + C (graphite, s) + 3 O2 (g) à CaCO3 (s) 2 39 40 ∆Gf° can be used to calculate ΔG° For a chemical reaction: mA + nB → xC + yD ΔG°= ΣnΔG°f (products) – ΣnΔG°f (reactants) This equation only works at 298 K, the temperature for which the values of ∆Gf° are tabulated. ΔG°f for any free element in its most stable form at standard conditions is defined as zero. 41 Use the standard free energy data to determine the free energy change for the following reactions, which is run under standard state conditions and 25 °C. Identify it as either spontaneous or nonspontaneous at these conditions. Fe 2O3 ( s) + 3CO( g ) ¾¾ ® 2Fe( s) + 3CO 2 ( g ) D𝐺!" (CO2, g) = -394.36 kJ/mol D𝐺!" (CO, g) = -137.15 kJ/mol D𝐺!" (Fe2O3, s) = -742.2 kJ/mol ΔG° = ΣnΔG°f (products) – ΣnΔG°f (reactants) = [2 (0) + 3 (-394.36) ]–[(-742.2) + 3(-137.15) ] = - 29.43 kJ/mol Spontaneous 42 Additivity of ΔG; Coupled Reactions As with enthalpy and entropy, free energy changes for reactions are additive (that means, follows Hess Law). If Reaction 3 = Reaction 1 + Reaction 2 Then, ΔG3 = ΔG1 + ΔG2 If a reaction is reversed, the sign on DG is also reversed. If a reaction is multiplied by a factor of “n” then DG is also multiplied by a factor of “n”. 43 Additivity of ΔG Practice 1 2 Flip equation 2 , the other way to say, times it by (-1), we get equation 3 : (-1) 2 Þ ∆Gf° = 30 kJ 3 ADP ® ATP Add equation 1 and 3 : Glu + ATP ® G6P + ADP + ADP ® ATP Glu ® G6P ∆Gf° = -17 kJ ∆Gf° = 30 kJ ∆Gf° = 13 kJ Additivity of ΔG Practice Given: 1 P4 ( s) + 5O 2 ( g ) ¾¾ ® P4O10 ( s) ° DG298 = - 2697.0 kJ/mol 2 2H 2 ( g ) + O2 ( g ) ¾¾ ® 2H 2O( g ) ° DG298 = - 457.18 kJ/mol 3 6H 2 O( g ) + P4O10 ( s) ¾¾ ® 4H 3PO 4 (l ) ° DG298 = - 428.66 kJ/mol Determine the standard free energy of formation, DGf° , for phosphoric acid. Here is the equation that represents ∆Gf° of phosphoric acid. # $ # % $ & ¼ x 3 Þ H2O + P4O10 ® H3PO4 % ' % ¼ x 1 Þ & P4 + & O2 ® & P4O10 # + & # # $ & x 2 Þ H2 + ! H " 2 # $ # O2 ® H2O $ + P4 + 2 O2 ® H3PO4 % H2 + P4 + 2O2 ® H3PO4 & ∆Gf° = ¼ (-428.66kJ/mol) = -107.165 kJ/mol ∆Gf° = ¼ (-2697.0kJ/mol) = -674.25 kJ/mol # ∆Gf° = (-457.18kJ/mol) = -342.885 kJ/mol & ∆Gf° = −107.165 + = -1124.3 kJ/mol −674.25 + (-342.885) 44 Additional Study Resources Thermodynamic entropy definition clarification https://www.khanacademy.org/science/physics/thermodyna mics/laws-of-thermodynamics/v/thermodynamic-entropydefinition-clarification Internal Energy https://www.khanacademy.org/science/chemistry/thermody namics-chemistry/internal-energy-sal/v/more-on-internalenergy?modal=1