Principles 2 Week 12

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Monday April 1 Day 53 Introduction to Gibbs Free Energy: What is it?
Textbook Reading 18.6 Gibbs Energy	Course Lectures 17.2 pdf Video* Chemical Free Energy
Objectives: 1. Identify positive and negative Gibbs energy values with spontaneous and non-spontaneous processes correctly. 2. Associate Gibbs energy changes with energy input and output for non-spontaneous and spontaneous processes respectively. 3. Use Gibbs energy and thermodynamic efficiencies to determine maxiumum available work and waste heat values.	Introduction to Gibbs Free Energy
Homework Problems 46.1. Gibbs energy is the maximum energy available to do meaningful work in frictionless and reversible situations. In reality, the amount of work available is a fraction of the Gibbs energy we might calculate for a reaction. Consider the combustion of methane given by the following chemical equation and ΔG value. CH_4(g) + 2 O_2(g) → 2 H₂O_(g) + CO_2(g) ΔG = -900.8 kJ/mol a. Is this reaction spontaneous? How do you know? b. A methane powered delivery truck carries 120. gallons of methane (CH₄ ...instead of gasoline). If 1 gallon of methane weighs 3.54 pounds, determine the total Gibbs energy available for the combustion of all 120. gallons of fuel. (Hint: Use ΔG above and watch your units) c. Your textbook claims that internal combustion engines are only ~30 % efficient at best. Use this figure to determine the maximum amount of useable energy available from the the combustion of 120. gallons of methane fuel. d. How many kJ of heat energy are released as the truck consumes all 120 gallons of fuel? e. If all of the waste heat was used to boil water, how many pounds of water could be vaporized? (Useful Information: For water at 100 ^oC, H_vap = 40.7 kJ/mol) 46.2 Consider the following situation: 3 MnO_2(s) → Mn₃O_4(s) + O_2(g) ΔG = 115 kJ/mol a. Is this reaction spontaneous? How do you know? b. As a non-spontaneous reaction, energy must be delivered to the reaction to make it happen. Ideally, how much energy would be required to transform 6 moles of MnO_2(s)into products? c. If we're able to deliever energy to the reaction with 45% efficiency, how much energy would we actually need to transform 6 moles of MnO_2(s)into products? d. Rewrite the chemical reaction and ΔG above in their spontaneous form. 46.3 Which of the following statements is NOT correct? a. When ΔG for a reaction is negative, the reaction is spontaneous b. When ΔG for a reaction is positive, the reaction is non-spontaneous c. When ΔG for a reaction is zero, the reaction is at equilibrium d. All of the above are correct. Click and drag the region below for correct answers 46.1 a. The reaction is spontaneous because it experiences a drop in Gibbs Energy. That is, ΔG < 0 (negative) b. 1.08₁₉ x 10⁷ kJ is ideally the amount of energy available from the fuel. c. 3.24₅₈ x 10⁶ kJ more likely available given the inefficiencies of the truck's engine. d. The other 70% is released as heat: 7.57₃₆ x 10⁷ kJ e. 7,390 pounds of water could be vaporized! 46.2 a. This reacction is non-spontaneous as written since it represents and increase in Gibbs Energy That is, ΔG > 0 (positive) b. Ideally, 3 moles of MnO_2(s) would require 115kJ of energy. Therefore, 6 moles of MnO_2(s) will require 230. kJ of energy c. 511 kJ would be required as our energy delivery is only 45% efficient. d. Mn₃O_4(s) + O_2(g) → 3 MnO_2(s) ΔG = -115 kJ/mol 46.3 d. All of the above are correct . ΔG is used to identify spontaneous processes (ΔG < 0) from non-spontaneous process (ΔG > 0). This begs the question: What's the significance is ΔG = 0? In this case the reaction neither produces nore requires energy to take place. In fact, the reaction is visually static even though both forward and reverse reactions are taking place. In otherwords, when ΔG = 0 , we have equilibrium! :)

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Tuesday April 2, 2024 Day 54 Determining Gibbs Free Energy Changes
Textbook Readings 18.6 Gibbs Energy	Course Lectures 19.1 pdf Video* Gibbs Free Energy
Gibbs Free Energy Gibbs Free Energy	Using Gibbs Free Energy . Using Gibbs Free Energy
Objectives 1. Know the significance of the phrase "Goldfish Are Horrible Without Tartar Sauce" 2. Use the ΔG = ΔH - TΔS to suggest combinations of entropy and enthalpy that give rise to spontaneous and non-spontaneous reactions. 3. Describe the temperature dependance of ΔG = ΔH - TΔS and how it affects reaction spontaneity. 4. Calculate ΔH & ΔS for chemical reactions from standard thermodynamic data and then determine ΔG for the reaction at specific temperatures. 5. Determine threshold temperatures for endo and exothermic reactions.
Homework Problems. 47.1. Consider the following reaction and its thermodynamic information: NH₄NO_3(s) → NH₄⁺_(aq) + NO₃^-_(aq) ΔH_f^o (kJ/mol) -365.56 -132.51 -205.0 S_f^o (J/mol-K) 151.08 113.4 146.4 a. Calculate ΔH_rxn for this reaction using ΔH_rxn = (Σn_prod* ΔH_f^o _prod ) - (Σn_reac* ΔH_f^o _reac ) b. Calculate ΔS_rxn for this reaction using ΔS_rxn = (Σn_prod* S_f^o _prod ) - (Σn_reac* S_f^o _reac ) c. Use the ΔG = ΔH - TΔS equation and your results from parts "a" and "b" to determine ΔG (kJ/mol) for this reaction at 25^oC. d. Is this reaction spontaneous or non-spontaneous as written. e. If the reaction is spontanteous, is it entropy driven, enthalpy driven or both? 47.2 Determine ΔG for the following reaction given the thermodynamic data provided and say whether the reaction is spontaneous as written or not. 2 N_2(g) + 3 O_2(g) → 2N₂O_3(g) ΔH_f^o (kJ/mol) 0 0 83.72 S_f^o (J/mol-K) 191.5 205.0 312.2 47.3 Exothermic reactions with decreasing entropy (i. e ΔH < 0 & ΔS < 0) will only be spontanteous at low temperatures (cooler environments facitate heat transfer from the reaction into the surroundings). Endothermic reactions with increasing entropy (i. e ΔH > 0 & ΔS > 0) will only be spontanteous at high temperatures (warmer environments act as a heat source for the endothermic heat reactant). In each if these two cases, there is a "threshold temperature" that defines the temperature ranges for spontaneity. This threshold temperature can be determined by setting ΔG = 0 and solving for T: ΔG = 0 = ΔH - TΔS Consider the following reaction and determine the range of temperatures (^oC) required for the reaction to be spontaneous. CH_4(g) + N_2(g) + 163.8 kJ → HCN_(g) + NH_3(g) ΔS_rxn = 16.1 J/mol-K 47.4 Consider the following and determine the range of temperatures (^oC) required for the reaction to be spontaneous SBr_4(g) → S_(g) + 2 Br_2(l) Given: ΔH^o = +115 kJ .....and..... ΔS^o = +125 J/K Click and drag the region below for correct answers 47.1 c. ΔG = -4.360 kJ/mol d. Reaction is spontaneous as written e. Reaction is endothermic :( Entropy increases for this reaction :) and so is "entropy driven" 47.2 ΔG = +278.8 kJ/mol Reaction is non spontaneous and requires work to occur. 47.3 This reaction will be spontaneous at temperatures above 9.9 x 10³ ^oC 47.4 This endothermic reaction is spontaneous at temperatures above 647 K

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Wednesday April 3, 2024 Day 55 Determining Gibbs Free Energy Changes from Standard Gibbs Energies
Textbook Readings 18.6 Gibbs Energy	Course Lectures 19.1 pdf Video* Gibbs Free Energy
Objectives 1. Calculate ΔG^o_rxn using standard thermodynamic ΔG^o_f values. 2. Interpret ΔG^o_rxn values to determine reaction spontaneity.	Gibbs Free Energy
Homework Problems As you already know, ΔG^o_rxn can be calculated via the equation: ΔG^o_rxn = ΔH^o_rxn - T ΔS^o_rxn Goldfish are Horrible without Tartar Sauce However, standard free energies have been tabulated in thermodynamic tables making the determination of ΔG^o_rxn much simpler under standard conditions: ΔG^o_rxn = Σ nΔG^o_products - Σ nΔG^o_reactants Note that and are determined in exactly the same way. In the problems below, use the provided information or thermodynamic tables to determine ΔG^o_rxn. 48.1 Consider the following reaction and Gibbs energy information: 2 C₂H_2(g) + 5 O_2(g) → 4 CO_2(g) + 2 H₂O_(l) ΔG_f^o (kJ/mol) 209.2 0 -394.4 -237.2 a. Determine ΔG^o_rxn for the reaction. b. Is the reaction spontaneous as written? If not, rewrite the reaction in its spontaneous form. 48.2 Consider the following reaction and Gibbs energy information: CaCl_2(s) + CO_2(g) + H₂O_(l) → CaCO_3(s) + 2HCl_(g) ΔG_f^o (kJ/mol) -748.8 -394.4 -237.2 -1129.1 -95.3 a. Determine ΔG^o_rxn for the reaction. b. Is the reaction spontaneous as written? If not, rewrite the reaction in its spontaneous form. 48.3 Consider the following reaction: 3N_2(g) + 4 H₂O_(g) → N₂O_4(g) + 2N₂H_4(l) a. Use the Thermodynamic Tables to determine ΔG^o_rxn for the reaction. b. Is the reaction spontaneous as written? If not, rewrite the reaction in its spontaneous form. 48.4 As we know, when ΔG^o_rxn < 0 (negative), the forward reaction is spontaneous as written. Additionally, when ΔG^o_rxn > 0 (positive), the forward reaction is nonspontaneous as written BUT the reverse reaction IS spontaneous. Thus, ΔG^o_rxn = 0 defines the line between reaction spontaneity and non-spontaneity. What chemical situation is defined by ΔG^o_rxn = 0? 48.5 A reaction is known to be spontaneous, but only at low temperatures. What can be said about ΔG^o_rxn , ΔH^o_rxn , and ΔS^o_rxn ? Click and drag the region below for correct answers 48.1 a. ΔG^o_rxn = -2470.4 kJ/mol b. Reaction is spontaneous as written. 48.2 a. ΔG^o_rxn = +60.7 kJ/mol b. Reaction is NOT spontaneous as written Spontaneous: CaCO_3(s) + 2HCl_(g) → CaCl_2(s) + CO_2(g)+ H₂O_(l) ΔG^o_rxn = -60.7 kJ/mol 48.3 ΔG^o_rxn = +1312.8 kJ/mol b. Reaction is NOT spontaneous as written Spontaneous: N₂O_4(g) + 2N₂H_4(l) → 3N_2(g) + 4 H₂O_(g)ΔG^o_rxn = -1312.8 kJ/mol 48.4 Equilibrium. When ΔG^o_rxn = 0 for a chemical reaction it is in equilibrium with both forward and reverse reactions taking place simultaneously and with the same rate. 48.5 At low temperatures, ΔG^o_rxn < 0 (negative) but at high temperatures ΔG^o_rxn > 0 (positive) Regardles, this must be always be true for this reaction: Δ H^o_rxn < 0 (negative) ΔS^o_rxn < 0 (negative)

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Thursday April 4, 2024 Day 56 Non-Standard Gibbs Energies
Textbook Readings 18.9: Gibbs Energy Change for Non-Standard States	Course Lectures 19.3 pdf Video Non Standard Gibbs Energy
Free Energy Changes, Nonstandard Conditions Free Energy Changes, Nonstandard Conditions	Calculating Free Energy: Non-Standard Pressures Calculating Free Energy: Non-Standard Pressures
Objectives 1. Calculate ΔG⁰_rxn using tabulated thermodynamic information. 2. Formulate the reaction quotient (Law of Mass Action) for chemical reactions 3. Differentiate between standard conditions (1 M, 1 atm, and 25^oC) from non-standard conditions. 4. Calculate ΔG_rxn under non-standard conditions using ΔG⁰_rxn, T and Q.	Practice Problem: Free Energy and Nonstandard Conditions
ΔG⁰_rxntells us a lot. Determined either via the "goldfish" equation or by tabulated standard free energies, ΔG⁰_rxn tells us whether the reaction is spontaneous or not. Also, the ΔG⁰_rxn value gives us the maximum amount of work we can expect from the reaction under perfect conditions ...or...rhe minimum amount of energy required to make a non-spontaneous process happen. Our thermodynamic tables are formulated only for standard conditions: Concentrations: 1.00 M Pressures: 1.00 atm Temperature: 298.15 Kelvin However, most chemical reactions occur under non-standard conditions. In these cases we can calculate ΔG_rxn using the following formulation: ΔG_rxn = ΔG⁰_rxn + RT lnQ where... ΔG⁰_rxn is determined from standard thermodynamic tables R is the the gas law constant 8.314 J/mol-K T is the temperature (Kelvins) Q is the reaction quotient obtained via the Law of Mass Action (Note that the ^o symbol is used to identify values calculated under standard conditions) Homework Problems 49.1 Consider the following reaction: 4 NH_3(g) + 5 O_2(g) → 6 H₂O_(g) + 4 NO_(g) a. Use the thermodynamic data tables to calculate ΔG⁰_rxn at 298.15 K. b. Calculate ΔG_rxn at 298 K for the following mixture of reactants and products: 2.0 atm NH_3(g) 1.0 atm O_2(g) 1.5 atm H₂O_(g) 1.2 atm NO_(g) c. Standard conditions would be when all pressures equal to 1 atm. Compare your answers to "a" and "b". Which of the two situations is "more" spontaneous? 49.2 Consider the reaction of nitrogen monoxide and chlorine to form nitrosyl chloride: 2 NO_(g) + Cl_2(g) → 2 NOCl_(g) a. Use the thermodynamic data tables to calculateΔG⁰_rxn at 298.15 K. ΔG⁰_f = 66.2 for NOCl (not shown in table) b. Calculate ΔG_rxn at 298 K for for the following mixture of reactants and products: 0.40 atm NO 0.50 atm Cl₂ 0.10 atm NOCl c. Standard conditions would be when all pressures equal to 1 atm. Compare your answers to "a" and "b". Which of the two situations is "more" spontaneous? Click and drag the region below for correct answers 49.1 a. ΔG⁰_rxn = -955.6 kJ/mol b. ΔG_rxn= -954.6 kJ/mol c. Reactions are spontaneous when ΔG_rxn< 0 (negative) Reactions are more spontaneous in situations where ΔG_rxn is even MORE negative. Being more negative, "a" is the more spontaneous possibility. 49.2 a. ΔG⁰_rxn = -42.8 kJ/mol b. ΔG_rxn= -48.0 kJ/mol c. Reactions are spontaneous when ΔG_rxn< 0 (negative) Reactions are more spontaneous in situations where ΔG_rxn is even MORE negative. Being more negative, "b" is the more spontaneous possibility.

Friday April 5, 2024 Day 57 Gibbs Energy and Equilibrium
Textbook Readings 18.9: Gibbs Energy Change for Non-Standard States 18.10: Gibbs Energy and Equilibrium	Course Lectures 19.4 pdf Video Gibbs Energy and Equilibrium
Gibbs Free Energy and Equilibrium	Gibbs Free Energy and the Equilibrium Constant
Objectives 1. Use standard Gibbs energies to determine equilibrium constants (and vice versa) 2. Describe how large/small equilibrium constants relate to reaction spontaneity and Gibbs energy changes.
The connection between equilibrium and thermodynamics: We know that ΔG_rxn = ΔG⁰_rxn + RT lnQ and that at equlibrium... ΔG_rxn = 0 and Q = K_eq Thus we have 0 = ΔG⁰_rxn + RT lnK_eq and upon rearranging: ***** ΔG⁰_rxn = - RT lnK_eq ***** WHOO HOO! Thermodynamic Gibbs energies can be used to determine equilibrium constants. Homework Problems 50.1 Nitrogen dioxide, NO₂, dimerizes to form nitrogen tetroxide according to the reaction: 2 NO_{2 (g)} → N₂O_{4 (g)} a. Using the thermodynamic tables to calculate the standard molar Gibbs energy for the reaction, ΔG^o_rxn, at 298.15 K. b. Use your ΔG^o_rxn to determine a value for the equilibrium constant, K_eq , at 298.15 K. 50.2 For the following reaction ∆H_f^o = -197.8 kJ/mol and ∆S_f^o = -187.9 J/mol K. Assume that ∆H_f^o and ∆S_f^o are independent of temperature and calculate the equilibrium constant for this reaction at 262 K. 2SO_2(g) + O_2(g) → 2SO_3(g) 50.3 The solubility product constant (K_sp) of lead(II) iodide is 1.4 × 10^-8 at 25°C. Calculate ΔG^o_rxn for the dissociation of lead(II) iodide in water. What does this value of ΔG^o_rxn suggest about the spontaneity of this reaction? PbI_2(s) → Pb²⁺_(aq) + 2I⁻_(aq) 50.4 The solubility product constant (K_sp) of lead(II) chloride is 1.17 × 10^-5 at 25°C. Calculate ΔG^o_rxn for the dissociation of lead(II) chloride in water. What does this value of ΔG^o_rxn suggest about the spontaneity of this reaction? 50.5 Identify which of the two lead salts (problems 50.3 and 50.4) are most soluble based on their K_sp values. Do the ΔG^o_rxn values support your solubility conclusion? Click and drag the region below for correct answers 50.1 a. ΔG^o_rxn = -2.80 kJ/mol b. K_eq = 3.09 50.2 K_eq = 4.2 x 10²⁹ 50.3 ΔG^o_rxn = +45 kJ/mol Quite non-spontaneous and favors the solid which isn't very soluble. 50.4 ΔG^o_rxn = +28 kJ/mol Again, non-spontaneous favoring the reactants (solid) which isn't very soluble. 50.5 Both salts are 1:2 salts and a direct comparison of K_sp values let's us determine that PbCl₂ is a more soluble salt than PbI₂. The positive ΔG^o_rxn values indicate both reactions are non-spontaneous. Hence, both salts are considered sparingly soluble. However, the ΔG^o_rxn value for PbCl₂ is closer to zero and is therefore more more spontaneous than PbI₂ whose ΔG^o_rxn value is further from zero. Thus, PbCl₂ is more soluble based on ΔG^o_rxn considerations.

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