FA24 CH301 Unit 4 Review Notes-9 PDF

Fall 2024, CH 301, Unit 4: Thermodynamics – List of Question Types (QTs) 1. Thermodynamics Terms and Characteristics 2. Thermodynamic Signs 3. First Law: Internal Energy Theory 4. Calculating Internal Energy (ΔU = q + w) 5. Heat Capacity and Specific Heat Capacity 6. Ca...

Fall 2024, CH 301, Unit 4: Thermodynamics – List of Question Types (QTs) 1. Thermodynamics Terms and Characteristics 2. Thermodynamic Signs 3. First Law: Internal Energy Theory 4. Calculating Internal Energy (ΔU = q + w) 5. Heat Capacity and Specific Heat Capacity 6. Calculating work (w) for Gases (w = –PΔV) 7. First Law: Enthalpy Theory 8. Heating Curve – Description 9. Heating Curve – Calculation 10. Energy Diagrams of Reactions 11. Bomb Calorimetry – Theory and Experiment 12. Bomb Calorimetry – Calculation 13. Enthalpy and Bond Energies – Calculation 14. Terms and Concepts: Formation Reactions 15. Hess’ Law: Enthalpy of Formation – Calculation 16. Hess’ Law: Reaction Enthalpy – Calculation 17. Second Law: Entropy and Microstates 18. Absolute Molar Entropy – Ranking 19. Predicting Changes in Entropy 20. Second Law: ΔS of Surroundings 21. Gibbs Free Energy from Table Values – Calculation 22. Gibbs Free Energy: Signs for Chemical Reactions 23. Gibbs Free Energy: Signs for Phase Changes 24. Gibbs Free Energy: Transition Temperature – Calculation 25. Gibbs Free Energy: Compound Stability 4.1 Thermodynamics Terms and Characteristics These are vocabulary questions. Know the definitions to get them right. Open, Closed, and Isolated Systems Extensive vs Intensive Properties o Open systems – like an open water bottle – allow transfer of matter Extensive Properties: an additive property of a system. and energy. o Closed systems – like a closed water bottle – do not allow transfer e.g., volume, mass, of matter, but do allow transfer of energy. moles, energy o Isolated systems do not allow transfer of matter nor energy. Bomb Example: If you take two buckets with 1 L water each and combine calorimeters are isolated systems due to being insulated. them, now you have 2 L water. Very simple idea that combining two things adds them together. State Functions vs Path Functions Intensive Properties: a property that defines a system. This is not additive. If you ratio two extensive properties, you get out an State Functions: are variables that define how a system starts from how intensive property. it ends. We calculate final-initial with state functions. State functions ask only final- e.g., temperature, initial, like ΔH, ΔG, ΔS, ΔU density, pressure Example: If you take two buckets of water, each at 25°C, and Example: If you take a cup of water from your room at about 25°C, fly combine them, the temperature is still 25°C. It does not add to to the sun reaching about 5600°C, fly to the North Pole reaching -40°C, make it 50°C. Pressure and density are intensive because they are then return to your room, there was no net change in the temperature. a ratio of two extensive properties (P = force/area, ρ = mass/vol). Path Functions: follow the mechanism, or the path, for a given process. Practice Exam Problem: Path functions are q and w Which of the following statements is correct? (heat and work). a) Volume is an extensive property and temperature is an Example: If you push a box around the room first to the door, then to intensive property. correct the window, then back to the starting point, the amount of work (w) b) Mass is an extensive property and moles is an intensive you put in depends on how far (and how hard) you pushed it each time. property. It depends on the path you took. We use PV work (w = -PΔV) rather than c) Both energy and temperature are extensive. force-distance work (w = Fd), but the idea is the same. d) Both pressure and density are extensive. 4.2 Thermodynamic Signs Signs are the most important fundamental concept in thermo!! Many questions are missed because of getting signs wrong. What you need to know: BE THE SYSTEM. Generally: If something LEAVES the system, the sign is (-) Pay attention to wording!! Examples: If something ENTERS the system, the sign is (+) “Done on surroundings” is the same as “leaving the system” Specifically: “Done on the system” is the same q and w are negative when energy leaves the system as “entering the system” q and w are positive when energy enters the system ∆G is negative = spontaneous; ∆G is positive = non-spontaneous ∆S is positive = increasing disorder; ∆S is negative = increasing order ∆H is negative = exothermic; ∆H is positive = endothermic Heat vs Work: Practice Exam Problem Heat is chaotic energy, whereas work is a Which one of the following statements is true? concerted flow of energy that can be easily a) Work is the concerted flow of energy and is positive controlled/guided when it is done on the system. correct Both are PATH functions, meaning the b) Heat is the concerted flow of energy and is positive methods you take to heat something, change when it is done on the system. its volume, etc affect the outcome c) Work is the concerted flow of energy and is positive when it is done on the surroundings. d) Heat is the concerted flow of energy and is positive when it is done on the surroundings. 4.3 and 4.7 First Law: Internal Energy and Enthalpy Theory These are theory questions with lots of words in T/F statements. Know the first law, that ΔUuniv = 0 means that energy is conserved Know that ΔU = q + w Know that U is all the sources of potential and kinetic energy in a system Know that calorimetry allows ΔUsys = – ΔUsurrondings to measure unknown ΔU reactions Know that ΔH = ΔU + PΔV In a closed system, V is constant so ΔU = qv because no PΔV work In an open system, P is constant so ΔH = qp Practice Exam Problem Which of the following statements is/are true concerning the first law of thermodynamics? I) The internal energy (U) of the universe is conserved II) The internal energy of a system plus that of its surroundings is conserved III) The change in internal energy (ΔU) of a system and its surrounding can have the same sign a) I, III d) III only b) I only e) I, II correct c) I, II, III f) II only 4.4 Calculating Internal Energy (ΔU = q + w) This is a simple calculation: ΔU = q + w For which you are given q and w. The only difficulty is SIGNS. entering leaving Practice Exam Problem w (+) w (-) A CD player and its battery together do 500 kJ of work, and the SYSTEM battery also releases 250 kJ of energy as heat and the CD player q (+) q (-) releases 50 kJ as heat due to friction from spinning. What is the change in internal energy of the system, with the system regarded as the battery and CD player together? a) – 750 kJ Explanation: Example: You charge your phone battery to increase charge by b) + 200 kJ “do work” means w = – 500 kJ 5 kJ, but the phone gives off 1 kJ of heat in the process. What is c) – 800 kJ correct “releases energy” means the change in internal energy? d) – 200 kJ q = – 250 kJ – 50 kJ = – 300 kJ ΔU = q + w = –500 – 300 = – 800 kJ Solution: Solve ΔU = q + w by first assigning signs to q and w. The SYSTEM is always the chemical reaction – here, the chemical reaction in the battery is the system. You INCREASE charge of the battery, meaning w = + 5 kJ “Gives off heat” means q = – 1 kJ ΔU = q + w = 5 – 1 = 4 kJ 4.5 Heat Capacity and Specific Heat Capacity What you need to know: 𝑞 Definition of heat capacity from q = mCΔT → C = 𝑇 and know that it is the amount of heat (J) to raise the temperature by one degree. Know that C increases with size or amount. It also increases with IMF strength. 𝐽 𝐽 Famous examples: CH2O = 4 ; metals are 0.3 to 0.4. 𝑔∙°𝐶 𝑔∙°𝐶 Be able to do simple math in comparing heat content. If heat added to two compounds is the same (q1 = q2), then: 𝑚1 𝐶1 𝛥𝑇1 = 𝑚2 𝐶2 𝛥𝑇2 Example: For the same mass of H2O (C = 4 J/g∙°C) and Cu (C = 0.4 Practice Exam Problem J/g∙°C, what is the ΔT for water if the same amount of heat added Consider the following specific heats: to copper increased its temperature by 10°C? Copper 0.384 J/g∙°C Lead 0.159 J/g∙°C Solution: Since qH2O = qCu, we can say that Water 4.18 J/g∙°C 𝑚𝐻2𝑂 𝐶𝐻2𝑂 𝛥𝑇𝐻2𝑂 = 𝑚𝐶𝑢 𝐶𝐶𝑢 𝛥𝑇𝐶𝑢 Glass 0.501 J/g∙°C If the same amount of heat is added to identical masses And, since the masses are the same: of each of these substances, which substance attains the 𝐶𝐻2𝑂 𝛥𝑇𝐻2𝑂 = 𝐶𝐶𝑢 𝛥𝑇𝐶𝑢 highest temperature change? Now, solve the relative calculation. a) Water c) Glass (4)𝛥𝑇𝐻2𝑂 = (0.4)(10) b) Copper d) Lead correct 0.4 ∙ 10 𝛥𝑇𝐻2𝑂 = = 1°𝐶 Explanation: 4 Given the equation, q = mCΔT, if mass and q are constant Or, since the only changing values are C and ΔT for the two (the same for each species), then as C↑, ΔT↓. Therefore, species, simply realize that as C↑, ΔT↓, and water’s C is ten times the species with the lowest specific heat capacity will have the value of copper’s, therefore the ΔT for water is ten times less. the greatest change in temperature. 4.6 Calculating Work (w) for Gases (w = -PΔV) This is a simple calculation: W = -PΔV = -ΔngRT Example: What is work for the chemical reaction below at 300K, and is it done on the system or the surroundings? For a chemical reaction in which there is a change in # of 2 H2 (g) + O2 (g) ⇌ 2 H2O (l) gas molecules, it is easy to determine PΔV work. Solution: Solve using w = -ΔngRT, and remember these: ONLY include At T = 300 K, Example: What is the work done if a piston decreases and gases! RT ≈ 2.5 kJ from 4 L to 2 L with 2 atm of pressure, and is it done on the system or the surroundings? Convert into Joules using Δng = nproducts − nreactants = 0 − 3 = −3 1 L∙atm = 100 J. w = −Δng RT = −(−3)(8.314)(300) = 3 ∙ 2.5 kJ = +7.5kJ Solution: Since this is not a chemical reaction, solve Positive work is entering the system, so 7.5 kJ of work is directly using w = -PΔV. done on the system. 𝑤 = −𝑃𝛥𝑉 = −2 𝑎𝑡𝑚 (𝑉𝑓𝑖𝑛𝑎𝑙 − 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 ) = −2 𝑎𝑡𝑚 (2𝐿 − 4𝐿) = 4 𝐿 ∙ 𝑎𝑡𝑚 = 400 𝐽 Practice Exam Problem A syringe increases in volume by 3 liters at a constant pressure Therefore 400 J is done on the system because ΔV is (-). of 10 atm. Calculate the work in Joules (1 L∙atm = 100 J) a) 3000 J of work on the system. b) 3000 J of work on the surroundings. correct c) 300 J of work on the system. d) 300 J of work on the surroundings. Explanation: w = -PΔV = -10 atm * 3 L = -30 L∙atm = -3000 J. Negative sign means work leaves the system = done on surroundings. 4.8 Heating Curve – Description Know the equations and constants to use in each region of the curve: Heating Curve of Water To solve, simply be able to tell which regions T (°C) (constants) are part of the calculation. V Example: Water is heated from 75°C to steam at 110°C. Which constants are used? IV Start at 75°C, in Region III → Cwater 100 Phase change at 100°C in Region IV → ΔHvap III Heat to 110°C in Region V → Csteam Boiling II 0 Point I Freezing point q Practice Exam Problem A mixture of ice and water is heated to 50°C. Equations / Constants in each region above: What constants are needed to calculate the amount of heat introduced into the system? I. Solid warming: q = mCsolidΔT a) Cice, ΔHfus, Cwater II. Solid melting: q = mΔHfusion b) Cwater Don’t make this problem c) Cice, ΔHfus, Cwater, ΔHvap, Cgas III. Liquid warming: q = mCliquidΔT d) ΔHfus, Cwater correct hard!! You need to do ZERO calculations! Explanation: IV. Liquid vaporizing: q = mΔHvap Starting with ice + water means it’s at 0°C – the only temp where ice and water can exist V. Gas warming: q = mCgasΔT at the same time at equilibrium – first we phase change ice to water (fully) using ΔHfus, then we warm the water to 50°C using Cwater. 4.9 Heating Curve – Calculation You will calculate the amount of heat it takes to warm H2O from one temperature to another. To do this, know the five regions of the heating curve of water and their constants: Example 1: How much energy is required to melt 10 g ice? Heating Curve of Water Answer: This involves only Region II, so: T (°C) q = mΔHfus = (10 g)(340 J/g) = 3400 J Example 2: How much energy is required to heat 10 grams V water from liquid at 0°C to liquid at 100°C? IV Answer: Since there is no phase transition, this only 100 involves Region III, so: III q = mCwaterΔT = (10 g)(4 J/g°C)(100°C) = 4000 J Boiling II Example 3: How much energy is required to heat 10 grams 0 Point ice from -20°C to water at 0°C? I Answer: First we warm ice (Region I), then transition Freezing from solid → liquid (Region II), so: point qI = mCiceΔT = (10 g)(2 J/g°C)(20°C) = 400 J q qII = mΔHfus = (10 g)(340 J/g) = 3400 J qtot = 400 J + 3400 J = 3800 J Equations / Constants per region: You will be given the values: I. Solid warming: q = mCiceΔT Cice = 2 J/g∙°C II. Solid melting: q = mΔHfusion ΔHfus = 340 J/g Practice Exam Problem How much heat is required to heat 0.1 g of ice at -30°C to steam at 100°C? III. Liquid warming: q = mCwaterΔT Cwater = 4 J/g∙°C Use the constants provided on the left. IV. Liquid vaporizing: q = mΔHvap ΔHvap = 2260 J/g a) 80 J c) 306 J correct V. Gas warming: q = mCsteamΔT Csteam = 2 J/g∙°C b) 193 J d) 74 J Explanation: First we warm ice to 0°C (I), then we convert to water (II), then heat water to 100°C (III) where it then boils to steam (IV). qI = (0.1)(2)(30) = 6 J qII = (0.1)(340) = 34 J qIII = (0.1)(4)(100) = 40 J qIV = (0.1)(2260) = 226 J qtot = 6 + 34 + 40 + 226 = 306 J 4.10 Energy Diagrams of Reactions This is the easiest question on the test. Be able to interpret the graph in order to determine if energy is entering (+) or leaving (-) the system and how much it changed by. products reactants reactants products Energy enters so ΔU Energy leaves so ΔU is positive. is negative. Practice Exam Problem: Explanation: What is the change in enthalpy (ΔH) for the ΔH = Hproducts - Hreactants reaction from A → B given the reaction Products (red) = 300 kJ coordinate diagram below? Reactants (blue) = 250 kJ ΔH = 300 – 250 = 50 kJ Increase in energy = endothermic a) -50 kJ; endothermic b) -150 kJ; exothermic c) 50 kJ; endothermic correct d) 300 kJ; exothermic 4.11 Bomb Calorimetry – Theory and Experiment Understand the basics of calorimetry and be able to answer T/F questions. Basic bomb calorimeter set up: Practice Exam Problem Which of the following statements about calorimetry Insulated O2 supply Ignition box experiments is TRUE? calorimeter and wires container a) Water is often used to measure the temperature change because it has a small heat capacity. b) Bomb calorimeters provide a measure of enthalpy Thermometer by measuring heat at a constant pressure. Stirrer c) The water in a bomb calorimeter must be constantly stirred. Correct d) Calorimeters are sufficiently insulated that the heat Sample change calculated from ΔT is completely accurate. (System) Explanation: Water a) FALSE – water is used because it has a high heat capacity (and is readily available…) b) FALSE – Bomb calorimeters measure internal energy by measuring heat at a constant volume. Igniting the bomb calorimeter starts the c) TRUE – If the water is not stirred, the ΔT will be reaction of the sample inaccurate. d) FALSE – Calorimeters are not isolated systems. The The water serves as a heat sink in order to heat lost to the calorimeter should be accounted for measure the temperature change to provide a more accurate calculation. The setup requires constant stirring to ensure accurate ΔT Bomb calorimeters measure heat change at constant volume (ΔU = qV) 4.12 Bomb Calorimetry – Calculations This is a classic plug and chug. You will be asked the energy content of something put in a calorimeter. q = mCΔT + CΔT and qsys = – qsurr → Don’t worry about signs. The question will only ask for energy content of a sample. Practice Exam Problem Dr. Laude explodes 10 gummy bears in a bomb calorimeter. The temperature rise for 2,500 grams of water is 20°C. The bomb If given multiples (moles or # calorimeter housing has a heat capacity of 2500 J/°C. How much items), then convert to PER energy is in one gummy bear? (The heat capacity of water is 4 J/g°C) mole or PER item!! a) 250 kJ c) 20 kJ b) 25 kJ correct d) 200 kJ Explanation: Pay attention to the parts of the equation: Given: Solution: 10 gummy bears 𝑞 = 𝑚𝑤 𝐶𝑤 𝛥𝑇 + 𝐶𝑐𝑎𝑙 𝛥𝑇 𝐽 𝐽 q = mCΔT + CΔT mwater = 2500 g = (2500 𝑔) (4 𝑔°𝐶 ) (20°𝐶 ) + (2500 ) (20°𝐶 ) °𝐶 ΔT = 20°C Ccal = 2500 J/°C = 200,000 𝐽 + 50,000 𝐽 water calorimeter Cwater = 4 J/g°C = 250,000 𝐽 = 250 𝑘𝐽 But that’s for 10 gummy bears, so for 1 gummy bear is 250 kJ / 10 = 25 kJ 4.13 Enthalpy and Bond Energies – Calculation Classic plug and chug calculation: Practice Exam Problem Using bond enthalpy data, calculate the change in Given a balanced chemical reaction and table of bond energies, which lists all the energies as positive values, plug BE x # bonds enthalpy for the following reaction. into the equation to solve for change in enthalpy of the reaction: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) Various bond energies: ∆H°rxn = ∑nBEreactants – ∑nBEproducts C–H 413 kJ∙mol-1 H–H 436 kJ∙mol-1 C–C 346 kJ∙mol-1 O=O 498 kJ∙mol-1 C=O 799 kJ∙mol-1 O–H 463 kJ∙mol-1 C≡C 835 kJ∙mol-1 C≡O 1072 kJ∙mol-1 Bond breaking is Bond forming is endothermic (+) exothermic (–) a) -802 kJ∙mol-1 correct c) -1420 kJ∙mol-1 Things to remember: b) 1420 kJ∙mol-1 d) 203 kJ∙mol-1 This is the only reactant – product calculation. The minus sign in the equation accounts for the fact that bond Explanation: forming is exothermic, so the products’ BE values are 1. Draw the Lewis structures for each molecule in the rxn. negative. H | O=O H—O—H This calculation does not account for any intermolecular H—C—H + O=C=O + H—O—H forces, so it is only accurate for gas phase reactions. | O=O It is best to draw the 2D Lewis structure so you can count H the number of bonds. An example of this is provided on 2. Using the bond energies provided and the formula, the right. calculate ∆H°rxn (rounding for calculator-free math): ∆H°rxn = ∑nBEreactants - ∑nBEproducts = [4(C—H) + 2(O=O)] – [2(C=O) + 4(O—H)] IMPORTANT: = 4(413) + 2(498) – 2(799) – 4(463) Pay attention to the SIGNS when you are doing the math! ≈ 4(400) + 2(500) – 2(800) – 4(450) For calculator-free math, you should be = 1600 + 1000 – 1600 – 1800 careful how much you round. You may = –800 kJ∙mol-1 need to round less if your answer is too close to multiple answer choices. 4.14 Terms and Concepts: Formation Reactions Hess’ Law says enthalpy is a state function, which means that it Here’s what you need to know about formation doesn’t matter how you get there; how the system change ends reactions, with examples: minus how it begins will give you the same answer, regardless of A heat of formation reaction is a reaction that: the path. Produces one mole of a substance You will be introduced to four ways to calculate ∆Hrxn: from reactants are elements in their standard Experimentally with a calorimeter (4.11 and 4.12) state (e.g. H2 as a gas, S as a solid, C as graphite) Bond energies (4.13) under standard conditions: room temperature Heats of formation (4.15) (298 K) and 1 atm, as designated by the ° sign. Mixed reaction enthalpies (4.16) By far, the most common approach is to solve using heats of C (graphite) + O2(g) → CO2 (g) formation provided in thermodynamic data tables (covered in H2 (g) + ½ O2 (g) → H2O (g) question type 4.15). H2 (g) + ½ O2 (g) → H2O (l) Practice Exam Problem Calculations with heats and free energies of formation Which of the following is a heat of formation reaction? (∆Hf° and ∆Gf°) are superior to bond energy (BE) a) 2 H2 (g) + O2 (g) → 2 H2O (g) calculations because they account for all phases, not b) H2O (g) → H2 (g) + ½ O2 (g) just gas phase like BE. They are also specific energies to c) H+ + OH– → H2O (l) a substance rather than average energies as with BE. d) H2 (g) + ½ O2 (g) → H2O (l) correct Explanation: a) Not a formation reaction; it forms 2 moles of products. b) Not a formation reaction; does not meet either criteria. c) Not a formation reaction; reactants are not in their standard states. d) Is a formation reaction; reactants are in their standard states and forms only one mole of product. 4.15 Hess’ Law: Enthalpy of Formation – Calculation Classic plug and chug calculation: Practice Exam Problem Calculate the standard reaction enthalpy for the reaction Given ∆H°f values for all compounds and a balanced chemical of calcite with hydrochloric acid. reaction, plug in ∆H°f x n, where n is the number of moles, into CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g) the equation: The standard enthalpies of formation are: CaCl2 (aq) ∆Hf° = -877.1 kJ∙mol-1 ∆H°rxn = ∑n∆H°f, products – ∑n∆H°f, reactants H2O (l) ∆Hf° = -285.83 kJ∙mol-1 CO2 (g) ∆Hf° = -393.51 kJ∙mol-1 Things to note: CaCO3 (s) ∆Hf° = -1206.9 kJ∙mol-1 The values of ∆H°f to plug in are found in HCl (aq) ∆Hf° = -167.16 kJ∙mol-1 thermodynamic data tables a) -98.8 kJ∙mol-1 Remember that ∆H°f for elements in their standard b) -165 kJ∙mol-1 states is 0 c) -38.2 kJ∙mol-1 d) -15.2 kJ∙mol-1 correct IMPORTANT: Explanation: Plug in the ∆Hf° values to the equation and multiply by their Pay attention to the SIGNS when you coefficient in the reaction (rounding for calculator-free math): are doing the math For calculator-free math, you should be ∆H°rxn = ∑n∆H°f, products – ∑n∆H°f, reactants careful how much you round. You may = [∆H°f,CaCl2 + ∆H°f,H2O + ∆H°f,CO2] – [∆H°f,CaCO3 + 2(∆H°f,HCl)] need to round less if your answer is too = (-877.1) + (-285.83) + (-393.51) – (-1206.9) – 2(-167.16) close to multiple answer choices. = -880 – 285 – 400 + 1200 + 340 = -1565 + 1540 = -15 kJ∙mol-1 4.16 Hess’ Law: Reaction Enthalpy – Calculation This QT is quite different in structure from what you’ve seen before. Practice Exam Problem For this question type, you’ll be given: Calculate the standard reaction enthalpy for the reaction: A few ∆H°rxn values with chemical reactions N2H4 (l) + H2 (g) → 2 NH3 (g) The overall reaction for which you want to calculate ∆H°rxn Given: This is based on Hess’ Law which states that, because enthalpy is a state N2H4 (l) + O2 (g) → N2 (g) + 2 H2O (g) ∆H° = -543 kJ∙mol-1 function, if two or more chemical reactions can be added together to give 2 H2 (g) + O2 (g) → 2 H2O (g) ∆H° = -484 kJ∙mol-1 a specific reaction, their ∆H°rxn can be added to give the ∆H°rxn for that specific reaction. N2 (g) + 3 H2 (g) → 2 NH3 (g) ∆H° = -92.2 kJ∙mol-1 The idea is to rearrange and combine given reactions and ∆H°rxn to yield a) -1119 kJ∙mol-1 the desired reaction and its unknown ∆H°rxn. Typically this is done by some b) -151 kJ∙mol-1 correct combination of c) -243 kJ∙mol-1 Flipping reactions, which “flips” the sign of the enthalpy (x it by -1) d) -945 kJ∙mol-1 Multiplying reactions by factors (e.g. x½, x2, x3, etc), which multiplies the enthalpy by the same factor Explanation: First, identify where molecules in the overall reaction appear in the to perfectly cancel out to get the desired reaction. sub-reactions. Make sure to pay attention to states and coefficients. Then, decide how to change the reactions to cancel what you don’t need. Reactions 1 and 3 have the correct amount of N2H4 (l) and NH3 (g) on the correct sides of the reaction. Leave those Don’t forget to correct the reactions alone for now. corresponding ∆H° when Flip Reaction 2 so that the “2 H2” flip to the right side to you FLIP or MULTIPLY cancel 2 of Reaction 3’s “3 H2” on the left. reactions by some factor! N2H4 (l) + O2 (g) → N2 (g) + 2 H2O (g) ∆H° = -543 kJ∙mol-1 2 H2O (g) → 2 H2 (g) + O2 (g) ∆H° = + 484 kJ∙mol-1 N2 (g) + 3 H2 (g) → 2 NH3 (g) ∆H° = -92.2 kJ∙mol-1 Make sure everything cancels perfectly to give the desired reaction. It does, so no more changes need to be made. Finally, add the ∆H° values to find ∆H°rxn = -151 kJ∙mol-1. 4.17 Second Law: Entropy and Microstates Two parts: ΔSuniv theory and the entropy microstate argument ΔSuniv Theory Microstate Argument Given ΔSuniv = ΔSsys + ΔSsurr for a phase transition at given Entropy (S) is a measure of disorder or the number of microstates (W). temperature, be able to assign signs for ΔSuniv, ΔSsys, ΔSsurr. Microstates are all the ways an atom or molecule partitions its thermal How to do it: motion. Think of this as number of ways to distribute or arrange energy. S = kln(W) so as # microstates ↑, disorder↑ = S↑ ΔSuniv + if process happens – if process doesn’t happen W is a vastly large number, so we only calculate for simple cases As volume increases, # microstates increases (V↑ = W↑) ΔSsys + if s → l → g As complexity of molecule (size) increase, # microstates in trans., vib., – if g → l → s and rot. modes of motion increases 𝛥𝐻𝑠𝑦𝑠 ΔSsurr 𝛥𝑆𝑠𝑢𝑟𝑟 = − so… Practice Exam Problem 𝑇 Which of the following statements about the second law is CORRECT? + if exothermic – if endothermic a) The change in entropy of the universe is positive for spontaneous processes. correct Example: b) Exothermic reactions decrease the entropy of the surroundings. What is the sign of ΔSuniv for water boiling at 25°C? c) At absolute zero, perfect crystals have an absolute entropy that is → Water boils above 100°C, so this doesn’t happen. negative. ΔSuniv = (–) d) As the number of microstates increase, the entropy of a system decreases. What is the sign of ΔSuniv for water freezing at -10°C? Explanation: a, b, and d are found directly on this page. C is a concept from the → Water freezes below 0°C, so this happens. positional entropy question type: perfect crystals have 1 orientation so ΔSuniv = (+) S = kln(W) = Nkln(1) = 0 4.18 Absolute Molar Entropy – Ranking A ranking problem in which you determine relative amount of disorder (relative absolute entropy) for different compounds. Important factors to use in ranking: Practice Exam Problem Most important: phase!! Which of the following compounds has the greatest absolute molar entropy? o Gas > Liquid > Solid a) A mole of hexane (C6H14) in the gas phase correct Size of molecule (MW) within the same phase b) A mole of hexane (C6H14) in the liquid phase o ↑MW = ↑S, generally c) A mole of cyclohexane (C6H12) in the gas phase d) A mole of cyclohexane (C6H12) in the liquid phase Mixtures are more disordered (more ways to arrange molecules) Explanation: Overall, this is comparing hexane to cyclohexane, and o Mixtures > Pure substances liquid to gas. Hexane > Cyclohexane due to ↑floppiness (know Floppy structures > Rigid structures that “cyclo-” means “cyclic” = ring structure). Based o Diamond (sp3 carbon) is more rigid than graphite (sp2 off the molecular formula, hexane also has ↑MW. carbon) Gas > Liquid due to more freedom of movement. o “floppy” structures are those that have greater freedom of movement Pay attention to what order the question is asking!! Least vs Greatest 4.19 Predicting Changes in Entropy This kind of problem presents different physical and chemical processes and asks you to determine sign of ΔSsys. Important factors to consider in ranking: Most important: phase!! Change in moles/# molecules if same phase ΔH = (+) Gas Use coefficients in the reaction given: ΔS = (+) Liquid 2 H2 (g) + O2 (g) → 2 H2O (g) ΔS is (–) because 3 → 2 ΔH = (—) C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) ΔS is (+) because 6 → 7 Solid ΔS = (—) Practice Exam Problem Consider the processes I) NaCl (s) → Na+ (aq) + Cl– (aq) II) CaCO3 (s) → CaO (s) + CO2 (g) III) H2O (l) → H2O (s) Which entropy will increase as the process proceeds from left to right? a) I only d) I, II b) I, II, III e) II only c) III only f) I, II correct Explanation: I) Changes phase from s → aq (aqueous means “in water”, so consider this like a liquid) = positive ΔS. (Also increases in moles.) II) Changes from s → s + g; s → g phase change = positive ΔS. (Also increases in moles.) III) Changes from l → s = negative ΔS. 4.20 Second Law: ΔS of Surroundings From the Second Law of Thermodynamics, we know ∆Suniverse = ∆Ssystem + ∆Ssurroundings Practice Exam Problem ∆Suniverse > 0 for spontaneous processes A hydrogen balloon explodes at room temperature (25°C), releasing 600 kJ of energy into the In other words, for a process to “happen”, the entropy of the surroundings. What is the change in entropy of the universe must increase. surroundings? This simple equation describes how the amount of heat energy a) -2000 J/K determines the change in entropy of a system: b) 2000 J/K correct c) 2 J/K q d) -2 J/K ∆S = T Explanation: Or more specifically defining change in entropy of the We are provided that ∆Hsys = -600 kJ because the surroundings: reaction “releases” 600 kJ of energy, and T = 298K. −∆Hsys ∆Ssurr = Recognize that since this is an exothermic reaction, T the entropy of the surroundings will increase, so ∆Ssurr Note that for an exothermic reaction, ∆Hsys is (–) so ∆Ssurr is (+). = (+), thus a) and d) cannot be the correct answer. In other words, the heat energy from the exothermic process entered the surroundings and produced the increase in entropy. Converting kJ to J and plugging in to calculate ∆Ssurr: −∆Hsys −(−600 kJ) ∆Ssurr = = T 298 K Pay attention to the UNITS! 600,000 J Enthalpy (∆H) is typically in kJ ≈ = 2000 J/K 300 K Entropy (∆S) is typically in J/K 4.21 Gibbs Free Energy from Table Values – Calculation This is a classic plug and chug calculation. Practice Exam Problem Calculate ∆G°rxn for the following reaction at room temperature Given ∆H°f and ∆S°m values from thermodynamic data tables, for using the data below: a chemical reaction: CaC2 (s) + 2 HCl (aq) → CaCl2 (aq) + C2H2 (g) First, find ∆H°rxn = ∑n∆H°f,prod - ∑n∆H°f,reac Then, find ∆S°rxn = ∑n∆S°m,prod - ∑n∆S°m,reac Compound ∆H°f (kJ∙mol-1) ∆S°m (J∙mol-1∙K-1) CaC2 (s) -59.8 69.96 Then, find temperature in Kelvin HCl (aq) -167.16 56.5 Finally, plug in to ∆G°rxn = ∆H°rxn – T∆S°rxn to find ∆G°rxn CaCl2 (aq) -877.1 59.8 C2H2 (g) 226.73 200.94 Important things to remember: a) -279.4 kJ∙mol-1 correct c) -761.9 kJ∙mol-1 T∆S will be in J/K and ∆H will be in kJ. Convert J to kJ or kJ b) -239.4 kJ∙mol-1 d) -463.8 kJ∙mol-1 to J before doing the calculation. Your final answer for ∆G°rxn should be converted to kJ. Explanation: Rounding for calculator-free math: Don’t forget the minus (–) sign in front of T∆S ∆H°rxn = [230 + (-880)] – [2(-170) + (-60)] in kJ/mol Often, ∆G ~ ∆H because T∆S is much smaller. So, see if = 230 – 880 + 340 + 60 you can find a good choice for ∆G without calculating T∆S. = -250 kJ/mol Based on the ∆H°rxn value, the answer could be either a) or b. Finish calculating to determine which one is correct: ∆S°m = [200 + 60] – [2(60) + 70] in J/mol∙K = 260 – 190 = 70 J/mol∙K At room temperature, T = 298 K ≈ 300 K, so: ∆G°rxn = ∆H°rxn – T∆S°rxn = -250,000 J/mol – (300 K)(70 J/mol∙K) = -250,000 J/mol – 21,000 J/mol = -271,000 J/mol = -271 kJ/mol So a) is the closest answer. 4.22 and 4.23 Gibbs Free Energy: Signs for Chemical Reactions and Phase Changes Here’s what you need to know about determining what the sign of ΔH, ΔS, and ΔG are given a chemical system: Chemical Reactions: ΔH and ΔS Phase Changes: ΔH and ΔS Enthalpy, ΔH: ΔH for phase changes has to do with making (-) or breaking (+) Besides context clues (the beaker became cold, the reaction heats etc), you IMFs, just like making/breaking bonds. should know that combustion reactions are exothermic, so ΔH is (-). ΔS for phase changes has to do with freedom of movement. Know how to identify a combustion reaction: a hydrocarbon plus oxygen yields To determine the sign, simply draw out the diagram below. Moving up carbon dioxide and water. is increasing, moving down is decreasing. CxHy + O2 → CO2 + H2O ΔH = (+) Gas Additionally, be aware that the reverse of a reaction has the reverse sign of ΔH. ΔS = (+) Liquid Entropy, ΔS: ΔH = (—) ΔSsys is (+) when system disorder (# microstates) increases. The number of Solid ΔS = (—) microstates is the number of ways you can arrange the molecules – if you have more molecules, or if they can more around more freely, then entropy is greater. ΔSsys is (+) when… Temperature Dependence of ΔG Phase changes in order from solid → liq → gas from left → right. Once you assign signs to ΔH and ΔS for a chemical system, you can # Moles increases from left → right. determine when ΔG will be (-), i.e. when the reaction will be Mixing or dissolving substances. spontaneous, given the equation is: ΔG = ΔH — TΔS Molecule size or complexity increases (generally). ΔGsys ΔHsys ΔSsys System is at higher temperatures. — at low T — — — at all T, always happens — + Practice Exam Problem + at all T, never happens + — What are the signs for change in enthalpy and change in entropy when dry ice sublimes? — at high T + + a) +, + correct c) -, - b) +, - d) -, + Explanation: Submlimation is solid → gas, which “moves up” if you use the Assigning signs properly is diagram on the top right. Changing from s → g absorbs heat (ΔH +) to break one of the most important IMFs, & The molecules can move around more freely (ΔS +). things you need to be able to do in the thermo unit!! 4.24 Gibbs Free Energy: Transition Temperature – Calculation Given a chemical reaction or phase change, be able to calculate the transition temperature to find the temperature at which it becomes spontaneous. There is an equilibrium point between two phases Practice Exam Problem: where ∆Gsys = 0. This suggests that 0 = ∆H – T∆S, so Using the thermodynamic data provided below, at what temperature does carbon dioxide sublimate? ∆𝐻 𝑇= CO2 (s) → CO2 (g) ∆𝑆 For example, we can calculate water’s transition ∆Hrxn = 33.89 kJ/mol temperature for the phase transition from liquid to gas ∆Srxn = 162.67 J/mol∙K given the ∆H and ∆S values: a) -64°C correct c) 200 °C H2O (l) → H2O (g) b) -252 °C d) -25 °C ∆Hsys = 44.01 kJ/mol ∆Ssys = 118.92 J/mol∙K Explanation: Rounding (to solve without a calculator) and plugging in: Remember to convert either J to kJ before solving. 44000 𝐽/𝑚𝑜𝑙 𝑇= ≅ 367 𝐾 = 94°𝐶 ∆𝐻 34000 𝐽/𝑚𝑜𝑙 120 𝐽/𝑚𝑜𝑙∙𝐾 𝑇= = = 212.5 𝐾 ∆𝑆 160 𝐽/𝑚𝑜𝑙∙𝐾 (This value is pretty close to the known boiling point of 212.5 𝐾 – 273 = 60°𝐶 water, 100°C) Since ∆H and ∆S are both positive, this suggests below Which is pretty close to only a) -64°C. 94°C, the boiling of water is nonspontaneous and above 94°C, the boiling of water is spontaneous. 4.25 Gibbs Free Energy: Compound Stability Formation data in the thermodynamic tables allows you to predict whether a substance is stable under standard conditions. Why? If the elements in their standard state react spontaneously, then ∆G°f is negative, and they fall down the free energy hill to make a stable substance. Example 1: C (graphite) + 2 H2(g) → CH4 (g) ∆G°f = -50 kJ/mol Practice Exam Problem: So, methane is more stable than the elements that Based on the following thermodynamic data, which made it. gaseous organic compound is the most stable at room temperature? Ethyne, C2H2 (g) ∆G°f = +209.20 kJ∙mol-1 Example 2: 2 C (graphite) + 2 H2 (g) → C2H4 (G) ∆G°f = +52 kJ/mol Propane, C3H8 (g) ∆G°f = -23.49 kJ∙mol-1 So, the reverse reaction to form the elements is Cyclopropane, C3H6 (g) ∆G°f = +104.45 kJ∙mol-1 favored and ethene is unstable. Methane, CH4 (g) ∆G°f = -50.72 kJ∙mol-1 a) Methane correct c) Cyclopropane Example 3: Does a diamond last forever, given that b) Propane d) Ethyne C (graphite) → C (diamond) has ∆G°f = +3 kJ/mol? Explanation: No. Because ∆G°f > 0, the reverse reaction is favored, A more negative Gibbs free energy of formation, ∆G°f, suggesting graphite is more stable than diamond. value suggests it is easier to form that compound under standard conditions—the formation reaction is This is where kinetics comes in. The rate at which more spontaneous the more negative ∆G°f is. Thus, it diamond decays is so slow that for practical is easiest to form methane, CH4, gas under standard purposes, you can give your diamond jewelry to your conditions, so it is the most stable. great great grandchild without worrying that they can use it to take notes in class.