Principles 2 Week 11

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Monday March 25, 2024 Day 48 1^st Law of Thermodynamics: A Review of Calorimetry Reaction Spontaneity: What is it?
Textbook Readings 6.4: The First Law of Thermodynamics 6.5: Constant Volume Calorimetry 6.7: Constant Pressure Calorimetry 18.2: Spontaneous and Nonspontaneous Processes	Course Lectures 6.6 pdf Video Calorimetry 17.1 pdf Video Thermo: First Law
The First Law Of Thermodynamics: Conservation of Energy and Calorimetry	Spontaneous Chemistry
Objectives 1. State the 1st Law of Thermodynamics and give real world examples of the law. 2. Apply the 1st Law of Thermodynamics to chemical situations including constant pressure and bomb calorimetry. Determine the heat of a reaction. 3. Define "spontaneity" and "non-spontaneity" from a chemical point of view. 4. Give examples of how spontaneous processes sometimes require an initial "push." 5. Identify the key changes that are required for a process to be spontaneous.
Homework Problems 41.1 The First Law of Thermodynamics states that energy is conserved; meaning that energy must be completely accounted for. At some point during the day, you will eat something and your body, a biochemical machine, will convert the food (potential energy) into other forms of energy. List as many of these other energy forms as you can. 41.2 The First Law of Thermodynamics (a.k.a. "conservation of energy"), is the underlying principle at work in calorimetry experiments. A reaction takes place and either produces heat (exothermic) or absorbs heat (endothermic). In constant pressure calorimetry, energy is inventoried in two ways: i. The heat absorbed (or released) by the calorimeter: q_calorimeter = C_cal x ΔT_calorimeter (Note that C_calincludes contributions of the cup, stirring rod, thermo probe and other calorimeter factors. C_cal is usually measured uniquely for each calorimeter and doesn't require "mass",) ii. The heat absorbed (or released) by the solution (often approx. water) q_solution = m_solution x c_solution x ΔT_solution Note: c_water = 4.184 J/g^oC Consider the following: 60.0 mL of 0.500 M HCl is mixed with 45.0 mL of 1.75 M NaOH in a calorimeter with known calorimeter constant (125. J/^oC) . The initial temperature of the calorimeter and solutions before mixing is 22.60 ^oC. After mixing, the temperature rises to 28.55 ^oC. a. Calculate the amount of heat energy gained by the calorimeter. b. Calculate the amount of heat energy gained by the solution assuming it to be mostly water c. Calculate the total heat gained in the experiment (1st Law of Thermo) d. Determine the ΔH_rxn using the limiting reactant. 41.3 For many years, a large boulder sits at the top of a tall mountain. One day, a reckless hiker leans up against the boulder which is just enough for it to begin rolling down the mountainside. Is the rolling boulder a "spontaneous" process? Explain. 41.4 For several months, a convenience store keeps their supply of charcoal briquettes in piles outside the store's front entrance. One day a reckless hiker holds a lit match to the pile and it begins to burn taking the store with it. Is the burning pile of briquettes (...and store) a spontaneous process? Explain 41.5 Compare your answers to 41.3 and 41.4. How are they alike? How are they different? 41.6 Spontaneous chemical reactions can occur slowly or quickly. In fact, the speed of the reaction has nothing to do with whether the reaction is spontaneous or not. a. Give an examples of both fast and slow spontaneous chemical reactions. b. Re-write your answers in part "a" as non-spontaneous reactions. 41.7 As discussed in the first video above, spontaneous processes are often exothermic. However, they can also be endothermic. Give an example of an endothermic, spontaneous chemical reaction. 41.8 "Perpetual motion machines" claim to be able to go "forever" and to be a source of free, unlimited energy. They violate the Laws of Thermodynamics and require an outside source of energy to fool the unsuspecting victim. What force makes all perpetual motion machines impossible? What is required for a supposed perpetual motion machine to appear perpetual? (Additional video information available here). Click and drag the region below for correct answers 41.1 In our body, food energy is converted into motion and sound. To my knowledge, we don't produce light energy (no one glows in the dark but if you do it's a superpower). Extra food energy also contributes to the body's fat stores; chemical potential energy to be used at a later time. 41.2 a. 0.743₇₅ kJ b. 2.61₃₉₅₄ kJ c. 3.35₇₇₀₄ kJ d. -111.₉₂₃ kJ/mol (note "-" sign) 41.3 Yes, the boulder rolling down the hill is spontaneous. It does require an initial "kick" or "activation energy", but once the boulder begins rolling it continues. 41.4 Yes. Like the boulder, the pile of charcoal requires activation energy to get it started burning. Once started, the burning takes place with no help from us. 41.5 Both are spontaneous process requiring activation energy. The boulder converts its energy into sound, kinetic energy and ultimately heat. The burning charcoal converts its PE into light, sound, kinetic energy (hot air rises) and of course, a lot of heat. 41.6 Metal rusiting is a very slow but spontaneous process. Charcoal also spontaneously converts into diamond ... very slowly. The combustion of gasoline in a car's engine is spontaneous and occurs very quickly on a millisecond time scale. 41.7 Ice melting at room temperature. Ice requires heat to melt (endothermic) and yet an ice cube melts spontaneously under the correct temperature circumstances. 41.8 Friction is a perpetual motion machine's downfall. There is no way to escape friction that will gradually convert what ever energy is in the machine to waste heat. If someone claims to have a perpetual motion machine, they are trying to fool you. The internet is full of examples (just google the topic) of clever deceptions which must always have a "secret energy source" operating behind the scenes.

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Tuesday March 26, 2024 Day 49 Entropy: Disorder and Energy Clumps
Textbook Readings 18.3: Entropy and the Second Law of Thermodynamics	Course Lectures 17.3 pdf Video Entropy
Introduction to Entropy	Lumpy Energy, Stirling Engines and the Death of the Universe
Objectives 1. Describe entropy in terms of randomness and disorder. 2. Describe entropy in terms of state availability. 3. Describe entropy in terms of lumpy energy. 4. Differentiate and provide examples of high & low entropy situations
Homework Problems 42.1 Entropy is often described as a measure of the randomness or disorder of a system. For example, a clean & tidy apartment is low entropy whilst a messy apartment is thought to have high entropy. As pointed to in the first video above, comparisons such as these aren't entirely correct and instead we should be thinking about the number of "states" available in each situation. If you are a perfectionist, how many ways can your clean & tidy apartment exist? If you are a perfectionist, how many ways can your apartment be messy? Which of the two represents more entropy? 42.2. Which has greater entropy? The Sun or the moon? Why? 42.3 In the second video above, an alternative definition for "entropy" is offered up (5:50). What is it? 42.3 Give examples of low entropy energy sources available to us here on Earth. Which of them have the sun as their ultimate source? Are there any earth-bound energy lumps that don't have the sun as their source? 42.4 What does the phrase "heat death of the universe" mean?_{(2nd video 8:00)} 42.5 What is the "arrow of time?" _{(2nd video 10.15)} Click and drag the region below for correct answers 42.1 There is only ONE way for the apartment to be "clean" and that's with everything in its proper place. There are an infinite number of ways for the apartment to be messy. All it takes is one out of place item, and the perfectionist begins the cleaning process. There are more ways for an apartment to be messy. Thus, this "macrostate" is the one with the greatest entropy. 42.2 The sun has more entropy because it's bigger and the atoms that are the sun are moving around with great speed. Simply, there are more possible "States" for the atoms/ions in the sun than there are for the cold rock that is the moon. 42.3 Lumpy energy. Low entropy is where we have energy concentrated in a single place. As the universe "winds down", the energy that's currently clumped in places like the sun will gradually become more an more evenly distributed throughout the universe leading to something called the universe's "heat death." 42.4 Refer to the video and 42.3 for this answer. 42.5 The arrow of time is the direction time takes as entropy in the universe increases.

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Wednesday March 27, 2024 Day 50 2nd Law of Thermodynamics: Macrostates and Microstates
Textbook Readings 18.3: Entropy and the Second Law of Thermodynamics	Course Lectures 17.4 pdf Video* Macrostates and Microstates Microstates/Macrostates Demonstration Spreadsheet
What is the Second Law of Thermodynamics?	The Second Law of Thermodynamics: Heat Flow, Entropy, and Microstates
Objectives 1. Provide alterate definitions of the 2nd Law of Thermodynamics. 2. Describe how the number of states available to isolated systems affects their entropy. 3. Calculate/predict entropy changes given heat and temperature information. 4. Calculate the absolute entropy "S" of a system using the Boltzman equation for entropy.
Homework Problems 43.1 The 2nd Law of Thermodynamics has different, equivalent forms. Here are several: i. The entropy of the universe is always increasing. ΔS_universe> 0 ii. No cyclic process can convert heat into work at 100% efficiency. iii. Local processes can experience a decrease in entropy but only if entropy increases by an equivalent or larger amount somewhere else. iv. Heat spontaniously flows from "hot" to "cold". Consider each of the following spontaneous process and choose from among the 2nd law statements above best describes what's going on. a. Current auto engines are only on average ~35% efficient. Auto engineers are are working on ways to trap the waste heat and make the engine more efficient. b. Even in my Yeti coffee mug, the temperature of my coffee slowly drops through the morning. c. I decrease entropy in my house by vacuuming the carpet with an electric vacuum cleaner but near the power plant that furnishes the electricity, things are getting dirtier (more entropy). d. The "heat death of the universe" 43.2 A heat engine uses 150. kJ of heat energy and performs 45. kJ of work at 25.0 ^oC. a. How much heat is released into the surroundings? b. What is the percent efficiency of this heat engine. c. The "surroundings" gain heat energy and their ideal/reversible entropy change can be calculated as: Q Δ S_surroundings = ----------- T Where "Q" is the heat gained (+) or lost (-) by the surroundings and T is the temperature in Kelvins. Calculate the entropy change experienced by the surroundings for the heat engine above. Include the correct sign (+/-) and units in your answer. 43.3 An endothermic chemical reaction consumes 30.5 kJ of heat energy that's removed from the surroundings @ 25.0 ^oC. Calculate Δ S_surroundings . 43.4 Consider the following situation. You (Jade) have decided to move into a 2 bedroom appartment with two other friends (Wade & Cade) and the looming question is how to populate the bedrooms. Here are the four distribution possibilities: _{(Macrostate 1)} Bedroom A: 0 People Bedroom B: 3 People _{(Macrostate 2)} Bedroom A: 1 Person Bedroom B: 2 People _{(Macrostate 3)} Bedroom A: 2 People Bedroom B: 1 Person _{(Macrostate 4)} Bedroom A: 3 People Bedroom B: 0 People Each of these general situations is known as a "Macrostate". Continuing, for each of the macrostates there is a possible distribution of people. Let's consider the second macrostate listed above: _{(Macrostate 2)} Bedroom A: 1 Person Bedroom B: 2 People Here's how the individuals can be assigned: _{(Microstate 1)} Bedroom A: Jade Bedroom B: Cade, Wade _{(Microstate 2)} Bedroom A: Cade Bedroom B: Wade, Jade _{(Microstate 3)} Bedroom A: Wade Bedroom B: Jade, Cade These possibilities were most easily determined by recognizing that each of the three people could appear individually only once in room A. Each of the 3 arrangements above is considered a "Microstate" and there were 3 total microstates possible. a. How many total microstates (a.k.a. "W") are possible for Macrostate 1 above b. How many total microstates (a.k.a. "W") are possible for Macrostate 2 above c. How many total microstates (a.k.a. "W") are possible for Macrostate 3 above d. How manytotal microstates (a.k.a. "W") are possible for Macrostate 4 above 43.5 Entropy (S) is determined using the Boltzman Equation: S = k_b ln W where W is the number of possible microstates and k_b = 1.38 x 10^-23 J/K Consider the 2 BR apartment problem above except now for 4 people Zade has joined the group a. What are all of the possible macrostates in this situation? b. For each macrostate, determine the number of microstates possible. c. Use the Boltzman equation above to determine the entropy for each macrostate. d. The macrostate with the highest number of microstates is considered to be the most probable. What macrostate is the most probable in this situation? d. Maximum entropy is associated with macrostates that are most uniformly distributed. Compare your entropy results from part "c" and determine if the largest entropy is associated with the most evenly distributed arrangement of people within the apartment. 43.6 Use the 2-Volume spreadsheet to determine: a. the number of ways 16 particles can be arranged in 2 connected volumes. b. the entropy of the most probable macrostate. c. four ways that 16 particles can be arranged in the most probable macrostate. Click and drag the region below for correct answers 43.1 a. ii b. iv c. iii d. i, iii and iv ....depending on how deeply you think about this. 43.2 a. 105 kJ of heat released into the surroundings. b. 30% c. + 352 J/K (Note that entropy is usually reported in J/K and not kJ/K units) 43.3 -102 J/K 43.4 a. 1 microstate possible i. BR#1: No one BR#2: W, J and C b. 3 microstates possible: i. BR#1: W BR#2: J, C ii. BR#1: J BR#2: W, C iii. BR#1: C BR#2: J,W c. 3 microstates possible: i. BR#1: J, C BR#2: W ii. BR#1: W, C BR#2: J iii. BR#1: J,W BR#2: C d. 1 microstate possible i. BR#1: W, J and C BR#2: No one 43.5 a. Macrostate i: BR#1: No one BR#2: 4 people Macrostate ii: BR#1: 1 person BR#2: 3 people Macrostate iii: BR#1: 2 people BR#2: 2 people Macrostate iv: BR#1: 3 people BR#2: 1 person Macrostate v: BR#1: 4 people BR#2: no one b. Macrostate i: 1 microstate possible BR#1: 0 BR#2: J W C Z Macrostate ii: 4 microstates possible BR#1: J BR#2: W C Z BR#1: W BR#2: J C Z BR#1: C BR#2: J W Z BR#1: Z BR#2: J C W Macrostate iii: 6 microstates possible BR#1: J W BR#2: C Z BR#1: J C BR#2: W Z BR#1: J Z BR#2: W C BR#1: C W BR#2: Z J BR#1: C Z BR#2: J W BR#1: W Z BR#2: J C Macrostate iv: 4 microstates possible BR#1: W C Z BR#2: J BR#1: J C Z BR#2: W BR#1: J W Z BR#2:C BR#1: J C W BR#2: Z Macrostate v: 1 microstate possible BR#1: J W C Z BR#2: 0 c. Macrostate i: S = 0 J/K Macrostate ii: S = 1.913 x 10^-23J/K Macrostate iii: S = 2.473 x 10^-23 J/K Macrostate iv: S = 1.913 x 10^-23 J/K Macrostate v: S = 0 J/K d. Macrostate iii is the most probable (greatest number of possible combinations or microstates) and has the greatest entropy. e. See answer to part "d" 43.6 a. Add them all up and there are 65536 ways to arrange 16 particles in 2 containers. Aren't you glad you weren't asked to write them all down? :) b. The most probable state is the one with 8 particles in one volume and 8 particles in the other volume. There are 12870 ways to do this (microstates) and S = 1.306 x 10^-22 J/K. c. Pick 4 (many are possible) V#1: 1 2 3 4 5 6 7 8 V#2: 9 10 11 12 13 14 15 16 V#1: 9 2 3 4 5 6 7 8 V#2: 1 10 11 12 13 14 15 16 V#1: 1 9 3 4 5 6 7 8 V#2: 2 10 11 12 13 14 15 16 V#1: 1 2 9 4 5 6 7 8 V#2: 3 10 11 12 13 14 15 16

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Thursday March 28, 2024 Day 51 Entropy and Heat Flow
Textbook Readings Entropy and the Surroundings Univ. of Texas	Course Lectures 18.2 pdf Video Spontaneity Delta H and Delta S 18.3 pdf Video Chemical Spontaneity 18.4 pdf Video Threshold Temperatures
Calculating Entropy Changes for Surroundings	Chemical Spontaneity
Objectives 1. Describe the effects of positive and negative ΔH_rxn and ΔS_rxn values on reaction spontaneity. 2. Describe the effect of large/small temperature values on reaction spontaneity. 3. Predict reaction spontaneity given 4. Predict reaction spontaneity given temperature, ΔH_rxn and ΔS_rxn values. 5. Determine threshold temperatures in situations where temperatures affect spontaneity.
Homework Problems 44.1 A powerful statement of the 2^nd law is that for all spontaneous changes, the entropy of the universe must always increase. Mathematically we can write this as: ΔS_universe> ΔS_surroundings + ΔS_system As we learned yesterday, the surrounding's entropy change is determined using the following relationship: Q Δ S_surroundings = ----------- T Combining these relationships gives: Q ΔS_universe> ----------- + ΔS_system T Now, if the "system" is a chemical reaction, then the heat "Q" is simply -ΔH_rxn ...and the relationship becomes ... - ΔH_rxn ΔS_universe> ----------- + ΔS_rxn T Note that the "-" sign in front of ΔH_rxn is required since the heat lost/gained by the by the surroundings is heat that is gained/lost by the system. Reactions are spontaneous ONLY if ΔS_universe> 0 a. What combination of ΔH_rxn (+ or -) and ΔS_rxn (+ or -) guarentees a positive & spontaneous value for ΔS_universe? b. What combination of ΔH_rxn (+ or -) and ΔS_rxn (+ or -) guarentees a negative & non-spontaneous value for ΔS_universe? c. An endothermic reaction has a positive ΔH_rxn . As such it DOESN'T help make ΔS_universepositive because of the negative sign. in the equation above. How can temperature be used to minimize the effect of the positive ΔH_rxn ? d. In some cases, ΔS_rxn < 0 (i.e. negative). This DOESN'T help achieve a positive ΔS_universe. However, if ΔH_rxn is negative (exothermic reaction), the negative signs cancel and the term works in favor of a positive ΔS_universevalue. How can temperature be used to enhance this effect and thus overcome the negative ΔS_rxn term? 44.2 Consider each of the following situations and determine if a spontaneous reaction will result. Cite temperature requirements for temperature as needed. a. ΔH_rxn > 0 (endothermic) ΔS_rxn > 0 b. ΔH_rxn > 0 (endothermic) ΔS_rxn < 0 c. ΔH_rxn < 0 (exothermic) ΔS_rxn > 0 d. ΔH_rxn < 0 (exothermic) ΔS_rxn < 0 44.3 When ammonium nitrate, NH₄NO₃, spontaneously dissolves in water the temperature of the solution formed decreases. What are the signs of ΔH_rxn, ΔS_rxn and ΔS_universe ? 44.4 For the following reaction, determine ΔS_universe at room temperature (25^oC) and state whether the reaction is spontaneous under these conditions. I_2(s) → 2I_(g) ΔH_rxn° = +213.6 kJ/mol and ΔS_rxn° = +245.2 J/molK. IMPORTANT: ΔH_rxn° has units of kJ/mol and ΔS_rxn° has units of J/molK 44.5 As you can see from the numbers below, the following reaction has a negative ΔH value (favoring spontaneity) and a negative ΔS value (doesn't favor spontaneity). What is the reaction's threshold tempearture and under what temperature conditions is it spontaneous? 2 H₂S_(g) + 3 O_2(g) → 2 SO_2(g) + 2 H₂O_(g) ΔH°_rxn = -1036 kJ/mol and ΔS°_rxn = -153.2 J/molK 44.6 Given the information below, determine the temperature conditions that insure the reaction is spontaneous. NH₄Cl_(s) → NH_3(g) + HCl_(g) ΔH°_rxn = 176.79 kJ/mol and ΔS°_rxn = 285.44 J/molK Click and drag the region below for correct answers 44.1 a. Always spontaneous when ΔH_rxn is negative and ΔS_rxn is positive b. Never spontaneous when ΔH_rxn is positive and ΔS_rxn is negative c. To minimize the effect of a positive ΔH_rxn (endothermic) value, increased temperatures are required (makes the fraction smaller) d. To minimize the effect of a negative ΔS_rxn value, low temperatures makes the positive fraction ΔH_rxn /T, more significant 44.2 a. Spontaneous at high temperatures only. b. Never spontaneous c. Always spontaneous d. Spontaneous at low temperatures only. 44.3 ΔH_rxn > 0 (endothermic) ΔS_rxn > 0 ΔS_universe > 0 (because it's a spontaneous process) 44.4 ΔS_universe = - 471 J/K ... reaction is non-spontaneous at room temperature 44.5 This exothermic reaction is spontaneous at temperatures below 6762 K 44.6 Reaction is spontaneous above temperatures of 619 K.

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Friday March 29, 2024 Day 52 Entropy Comparisons and Calculations
Textbook Readings 18.4 Entropy Changes Associated with State Changes Thermodynamic tables	Course Lectures 18.3 pdf Video Chemical Spontaneity 19.2 pdf Video Factors Affecting Entropy
Entropy and Phase Change	How to Calculate Changes in Entropy
Objectives 1. Describe how the entropy of various physical phases compare. 2. Qualitatively determine entropy changes for reaction where gases are products and reactants. 3. Qualitatively compare entropy for large/small atoms or molecules. 4. Use standard tables to determine entropy and enthalpy changes for chemical reactions.
Homework Problems 45.1 Carefully examine each of the following reactions and their gaseous reactants and/or products. Use these observations to determine if ΔS_rxn is positive or negative Lastly, use the stated ΔH_rxn values to determine under what conditions the reaction is spontaneous. a. 2 CO_(g) + O_2(g) → 2CO_2(g) ΔH_rxn = -566 kJ/mol b. 2 NO_2(g) → 2 NO_(g) + O_2(g) ΔH_rxn = +113.1 kJ/mol c. 2 H_2(g) + O_2(g) → 2 H₂O_(l) ΔH_rxn = -483.6 kJ/mol d. CO_2(g) → C_(s) + O_2(g) ΔH_rxn = +393.5 kJ/mol 45.2 Put the following Noble Gases in order of increasing entropy: Ar, He, Ne, Kr 45.3 Put the following molecular alcohols in order of increasing entropy: Butanol: CH₃CH₂CH₂CH₂OH Ethanol: CH₃CH₂OH Propanol: CH₃CH₂CH₂OH Methanol: CH₃OH 45.4 Compare the methane molecule (CH₄) to the methanol molecule (CH₃OH). What types of motion is each molecule capable of? Which molecule has more energy "states" Which molecule has more entropy? 45.5 Protein molecules consist of thousands of atoms covalently bonded together. How would you describe the entropy of a protein molecule? 45.6 Thermodynamic tables are used to determine ΔH_rxn and ΔS_rxn values at 25^oC using the following recipies: ΔS_rxn = Σ nS_products - Σ nS_reactants ΔH_rxn = Σ nΔH^o_products - Σ nΔH^o_reactants For the following reaction, use the thermodynamic tables to calculate ΔS_rxn , ΔH_rxn and ΔS_universeat 25^oC. Is the reaction spontaneous under these conditions? NH₄Cl_(s) → HCl_(g) + NH_3(g) 45.7 For the following reaction, use the thermodynamic tables to calculate ΔS_rxn , ΔH_rxn and ΔS_universeat 25^oC. Is the reaction spontaneous under these conditions? H₂O_(g) → H₂O_(l) 45.8 For the following reaction, use the thermodynamic tables to calculate ΔS_rxn , ΔH_rxn and ΔS_universeat 25^oC. Is the reaction spontaneous under these conditions? 2 H₂S_(g) + 3 O_2(g) → 2H₂O_(l) + 2 SO_2(g) Click and drag the region below for correct answers 45. 1 a. ΔS < 0 (negative) ΔH < 0 (negative, exo) Rxn. is spont. at low temps b. ΔS > 0 (positive) ΔH > 0 (positive, endo) Rxn. is spont. at high temps c. ΔS < 0 (negative) ΔH < 0 (negative, exo) Rxn. is spont. at low temps d. Impossible to determine ΔS from general observations. 1 mol of gas on both sides. 45.2 Low Entropy _{(Small atomic size)} : He … Ne … Ar …. Kr High entropy_{(large atomic size)} 45.3 Low Entropy_{(Low Molecular Weight)}: CH₃OH … CH₃CH₂OH … CH₃CH₂CH₂OH … CH₃CH₂CH₂CH₂OH High Entropy _{(High Molecular weight)} 45.4 CH₄: Molecular rotation (spinning) C - H bond stretching (vibration) H - C - H bending CH₃OH Molecular rotation (spinning) C - H bond stretching (vibration) H - C - H bending O - H bond stretching C - O - H bond bending O - H twirling CH₃OH has more motion modes and has more energy states … and therefore more entropy compared with methane CH₃OH has more entropy (239.9 J/mol-K for CH₃OH vs. 186.3 J/mol-K for CH₄) 45.5 Protein molecules have soooooo many ways to move. They're very floppy and can wiggle and jiggle in so many ways Protein molecules have a lot of entropy because of all the ways they can move. 45.6 ΔS_rxn = 285.1 J/mol-K ΔH_rxn = 176.2 kJ/mol ΔS_universe = -305.88 J/mol-K Reaction is Non-spontaneous at this temperature 45.7 ΔS_rxn = -118.8 J/mol-K ΔH_rxn = -44.0 kJ/mol ΔS_universe = +28.85 J/mol-K Reaction is spontaneous at this temperature 45.8 ΔS_rxn = -390.8 J/mol-K ΔH_rxn = -1124 kJ/mol ΔS_universe = +3.38 kJ/mol-K Reaction is spontaneous at this temperature

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