Chapter 18 Lecture - Chemical Thermodynamics PDF

Chapter 18 Chemical Thermodynamics CHY 102 Chapter Outline Section 18.1 Entropy and Spontaneity Section 18.2 Entropy Changes Section 18.3 Entropy and Temperature Section 18.4 Gibbs Free Energy Section 18.5 Free-Energy Changes and Temperature Section 18.6 Gibbs Free Energy and Equilibrium ...

Chapter 18 Chemical Thermodynamics CHY 102 Chapter Outline Section 18.1 Entropy and Spontaneity Section 18.2 Entropy Changes Section 18.3 Entropy and Temperature Section 18.4 Gibbs Free Energy Section 18.5 Free-Energy Changes and Temperature Section 18.6 Gibbs Free Energy and Equilibrium  18.1: Entropy and Spontaneity Entropy (S) is a measure of the degree of disorder or randomness in a system Entropy of a system increases whenever there is: A phase change from a more condensed to less condensed phase (solid à liquid, liquid à gas) Ø Solids are highly organized, liquids less so, gas particles move randomly An increase in temperature within a given phase Ø More kinetic energy increases the motion of the particles An increase in the number of gas particles or dissolution of a solid in an aqueous solution Ø Increase in number of particles provides more ways for the system’s energy to be distributed  18.1: Entropy and Spontaneity Example - Determine the sign of ΔS for the following processes: a. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔS = + The number of particles increases, and a gas is formed b. CO2(g) → CO2(s) ΔS = − A gas becomes a solid c. ΔS = + A solid becomes a gas d. condensation appearing on bathroom mirror after a shower ΔS = − A gas becomes a liquid  18.1: Entropy and Spontaneity Entropy increases when there are more energetically equivalent ways to arrange the components of that system. Entropy was defined mathematically by Boltzman in the 1870s k = Boltzman constant = 1.38 x 1023 J/K S = k ln W W = the number of energetically equivalent arrangements (microstates) possible for the system What are microstates? Consider a box with a divider down the middle containing two particles on the left side A on top B on top A and B both A and B both B on bottom A on bottom on top on bottom Since there are four possible arrangements for this system, W=4  18.1: Entropy and Spontaneity If, in the last example, the divider were to be removed: 16 microstates 1 macrostate There are now 16 possible arrangements, therefore W=16. For any system, we can calculate W as: X = # of locations n W=X n = # of particles In the above example, there are two particles and four possible locations in which the particles can be, so W = 42 = 16 Equivalent microstates form a macrostate, and the most probable macrostate is the one containing the highest number of microstates. Each circle represents a microstate. Which macrostate is most probable? B A A. A and B together A B A B B. A and B in separate locations A B B A A C. A in top left, B in bottom area B D. All of the above are equally probable. A B B B A A Each circle represents a microstate. Which macrostate is most probable? B A A. A and B together 3 A B A B B. A and B in separate 6 locations A B B A A C. A in top left, B in bottom 1 area B D. All of the above are equally probable. A B B B A A  18.1: Entropy and Spontaneity In general, the tendency towards disorder is a statistical probability; a system will spontaneously move to a more spread-out arrangement Ø A more disordered system is more probable than an ordered one The criteria for spontaneity are not solely based on the entropy change for the system. The entropy change of the surroundings must also be considered, and how the system and surroundings together affect the total entropy of the universe Sfewer moles < Smore moles Ø Processes that decrease the entropy of a system can still be spontaneous The Second Law of Thermodynamics states that 2NH3 à N2 + H2 spontaneous processes always result in an overall increase in entropy, S, of the universe  18.2 Entropy Changes The third law of thermodynamics states that the entropy of a pure, perfectly ordered, crystalline substance at absolute zero is zero: Scrystal (0 K) = 0 At T = 0K, there is no thermal energy and the particles in a perfectly ordered crystal would exist in a single microstate: Scrystal = k ln W = k ln 1 = 0 Standard molar entropies, S°, are calculated based on the third law of thermodynamics, and these can be used to calculate standard changes in entropy, ΔS°: DS rxn  = å m[S  (products)] - å n[S (reactants)] Where m and n are the coefficients of the products and reactants from the balanced chemical equation, and standard thermodynamic conditions are 298 K, 1 atm gas pressure, and 1 M solute concentration. Standard Entropies at 298 K of Selected Substances Compound J Compound J S S (state) mol × K (state) mol × K CO2(g) 213.8 HCl(g) 186.9 CaO(s) 38.1 Hg(l) 75.9 CaCO3(s) 91.7 Hg(g) 175.0 H2(g) 130.7 I2(s) 116.1 H2O(l) 70.0 I2(g) 260.7 H2O(g) 188.8 O2(g) 205.2  18.2 Entropy Changes Example - Predict the sign of ΔS and calculate the standard entropy change associated with the reaction of gaseous hydrogen and gaseous oxygen to form water vapor. Step 1: Write out the balanced chemical equation 2 H2(g) + O2(g) → 2 H2O(g) Prediction: The number of moles of gases decreases in the reaction, so ΔS will be negative (−). Step 2: Calculate the standard change in entropy using standard entropy data from the table DS rxn  = å m[S  (products)] - å n[S (reactants)] DS rxn  = 2S  ( H 2 O(g)) - [2S  ( H2(g)) + S  (O2(g))] æ J ö é æ J ö æ J öù DS  = 2 mol ç 188.8 ÷ - ê2 mol ç 130.7 mol × K ÷ + 1 mol ç 205.2 mol × K ÷ ú rxn è mol × K ø ë è ø è øû J DS rxn  = -89.0 K  18.3: Entropy and Temperature An overall change in entropy is equal to the change in entropy of the system plus the change in entropy of the surroundings: ΔS univ = ΔS sys + ΔS surr For a process to be spontaneous, ∆Suniv must be positive The entropy change for the system may be negative, but the process still be spontaneous  18.3: Entropy and Temperature For an isothermal process (i.e., no heat transfer between system and surroundings), and at constant pressure the entropy change of the surroundings is related to the thermal energy change of the system by: -DH sys ΔS surr = T Example – Calculate ΔSsurr and ΔSuniv for the reaction of gaseous hydrogen and gaseous oxygen to form 2 mol of water vapor at 298 K Step 1: Write the balanced chemical equation 2 H2(g) + O2(g) → 2 H2O(g) Step 2: Calculate DH rxn  Ø Look up DH f values in Appendix A.1 of textbook: H2O(g) = -241.8 kJ/mol, H2 and O2 are in their elemental state = 0 kJ/mol DH rxn  = å m[ DH f (products)] - å n[ DH f (reactants)] DH rxn  = 2 ( DH f (H2O(g)) - [2 ( DH f (H2(g)) + DH f(O2 )(g))] æ kJ ö é æ kJ ö æ kJ ö ù DH  rxn = 2 mol ç -241.8 ÷ - ê2 mol ç 0 ÷ + 1 mol ç 0 mol ÷ ú è mol ø ë è mol ø è øû DH rxn  = -483.6 kJ Step 3: calculate ΔSsurr -DH sys ΔS surr = T -( -483.6 kJ) 1000 J DS surr  = ´ 298 K 1 kJ J DS surr  = 1.62 ´ 103 K Step 4: Calculate ΔSsys and ΔSuniv Ø We calculated ΔSsys for this reaction in the previous example; ΔSsys = -89.0 J/K ΔS univ = ΔS sys + ΔS surr J J ΔS univ = -89.0 + 1620 K K J ΔS univ = 1530 K  18.4 Gibbs Free Energy Gibbs free energy, G, is a state function that an be used for predicting if a process is spontaneous at a given temperature: G = H - TS For a chemical reaction occurring under standard conditions but nonstandard temperature, the change in Gibbs free-energy is: DG  = DH  -T DS  ∆G° = - à Reaction is spontaneous in the forward direction ∆G° = + à Reaction is spontaneous in the reverse direction ∆G° = 0 à Reaction is at equilibrium Example - Determine ΔG° for the reaction of hydrogen and oxygen to form water given ΔS° = -89J/K and ΔH° = -483.6 kJ. 2 H2(g) + O2(g) → 2 H2O(g) æ Jö æ 1 kJ ö DG  = DH  -T DS  DG  = -483.6 kJ - 298 K ç -89.0 ÷ ç 1000 J ÷ DG  = -457.1 kJ è Kø è ø  18.4 Gibbs Free Energy The standard free energy of formation, DG f , is the free-energy change associated with the o formation of a substance from its component elements, in their standard states. DG rxn  = å m[ DG f (products)] - å n[ DG f (reactants)] Example - Calculate DG rxn  for the combustion of propane. Compound ∆Gf° (kJ/mol) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) C3H8(g) -23.4 CO2(g) -394.4 Solution: Use standard free energy of formation values to calculate ∆Grxn° H2O(l) -237.1 DG rxn  =[3DG f (CO2(g)) + 4DG f(H2O(l))] - [ DG f(C3H8(g)) + 5DG f(O2(g))] é æ kJ ö æ kJ ö ù é æ kJ ö æ kJ ö ù DG  rxn = ê3 mol ç -394.4 ÷ + 4 mol ç -237.1 ÷ - ú ê1 mol ç -23.4 ÷ + 5 mol ç 0 mol ÷ ú ë è mol ø è mol ø û ë è mol ø è øû DG rxn  =( -2131.6 kJ) - ( -23.4 kJ)= -2108.2 kJ   Determine DG rxn for the following reaction: 2 A(g) + B2(g) → 2 AB(g) A. −400 kJ B. −250 kJ C. −200 kJ D. −100 kJ E. −50 kJ Compound(state) 𝚫𝑮∘𝐟 (kJ/mol) A(g) 0 B2(g) −100 AB(g) −150   Determine DG rxn for the following reaction: 2 A(g) + B2(g) → 2 AB(g) DG rxn  = å m[ DG f (products)] - å n[ DG f (reactants)] A. −400 kJ B. −250 kJ 𝑘𝐽 𝑘𝐽 𝑘𝐽 Δ𝐺°!"# = 2 𝑚𝑜𝑙 −150 − 2 𝑚𝑜𝑙 0 + 1 𝑚𝑜𝑙 −100 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 C. −200 kJ D. −100 kJ 𝚫𝑮°𝒓𝒙𝒏 = −𝟐𝟎𝟎 𝒌𝑱 E. −50 kJ Compound(state) 𝚫𝑮∘𝐟 (kJ/mol) A(g) 0 B2(g) −100 AB(g) −150  18.5 Free-Energy Changes and Temperature Standard values are measured at 298K. ∆H and ∆S values only vary slightly with temperature, so standard values can be used for them at temperatures other than 298K One way ∆G can be calculated at temperatures that are not 298K is by calculating ∆H° and ∆S°, then using ΔG = ΔH° – TΔS° We can also determine the temperature at which a reaction becomes spontaneous, i.e., when ∆G=0: ΔG = ΔH – TΔS = 0 DH T= DS T will provide the minimum temperature at which the forward reaction is spontaneous.  18.5 Free-Energy Changes and Temperature Example - Determine the minimum temperature for the spontaneous vaporization of liquid mercury using data from the table below Hg(l) → Hg(g) Compound ∆S° (J/mol・K) Solution Hg(g) 175.0 J J J Hg(l) 75.9 DS = 175.0  - 75.9 DS  = 99.1 mol × K mol × K mol × K Compound ∆H° (kJ/mol) Hg(g) 61.4 𝑘𝐽 𝑘𝐽 𝑘𝐽 Δ𝐻°!"# = 1 𝑚𝑜𝑙 61.4 − 1 𝑚𝑜𝑙 0 = 61.4 Hg(l) 0 (elemental state) 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 kJ 61.4 DH mol T= = = 620 K DS J æ 1 kJ ö 99.1 ç K è 1000 J ÷ø ∴ The vaporization of mercury becomes spontaneous at temperatures above 620 K.  18.6 Gibbs Free Energy and Equilibrium For other nonstandard conditions, the change in free energy for a reaction can be calculated as: ΔG = ΔG° + RT ln Q Reaction Quotient Gas constant Temperature (QC for aqueous 8.314 J mol-1 K-1 (K) QP for gas) Recall that Q occurs under nonequilibrium conditions. The result of the above equation tells us the direction in which the reaction will proceed: Ø ΔG = negative à reaction proceeds spontaneously Ø ΔG = positive à reaction does not proceed spontaneously Ø ΔG = 0 à reaction is at equilibrium At equilibrium, Q = K and ΔG = 0, so the above equation becomes: ΔG° = -RT ln K K < 1 à The equilibrium position lies closer to the lower- energy reactants When Q = 1, ΔG > 0 à The reverse process is spontaneous K = Q & ΔG = 0 à No net change in forward or reverse direction K > 1 à The equilibrium position lies closer to the lower- energy products When Q = 1, ΔG < 0 à The forward process is spontaneous  18.6 Gibbs Free Energy and Equilibrium Example – a) Calculate ΔG for the Haber process at 365 K for a mixture of 1.5 atm N2, 4.5 atm H2, and 0.75 atm NH3, b) In which direction is the reaction currently proceeding? N2(g) + 3 H2(g) ⇌ 2 NH3(g) a) Step 1: Start by calculating ΔG° from DG fvalues found in Appendix A.1. æ kJ ö DG  = 2 mol ç -16.4 ÷ - [0 + 0]= -32.8 kJ è mol ø Step 2: Calculate Q (PNH )2 Qp = 3 (PN )(PH )3 2 2 (0.75)2 -3 Qp = = 4.1 ´ 10 (1.5)(4.5)3 Step 3: Plug your calculated values for ΔG° and 𝑄 into the ∆G equation ΔG = ΔG° + RT ln Q kJ æ J 1 kJ ö DG = -32.8 + ç 8.3145 ´ ÷ (365 K) ln(4.1 ´ 10-3 ) mol è mol × K 1000 J ø DG = -49.5 kJ/mol b) Since ∆G is negative, the system is not at equilibrium and the reaction is occurring spontaneously in the forward direction Given the ∆G° value calculated for the Haber process in the last example, what is the equilibrium constant for the Haber process at 298 K. A. 5.6 x 105 B. 1.01 C. 4.74 x 108 D. 1.02 x 103 Given the ∆G° value calculated for the Haber process in the last example, what is the equilibrium constant for the Haber process at 298 K. ΔG  - ΔG° = -RT ln K K=e RT A. 5.6 x 105 B. 1.01 DG  -( -32.8 kJ) 1000 J Calculate exponent: - = ´ = 13.24 RT æ J ö 1 kJ C. 4.74 x 108 ç 8.3145 mol × K ÷ (298 K) è ø D. 1.02 x 103 Now solve for 𝐾: K = e 13.24 K = 5.6 × 105