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NAME _________________________________________ UNIT 5 (1): THERMODYNAMICS Big Idea 5: The laws of thermodynamics describe the essential role of energy & explain & predict the direction of changes in matter. Enduring understanding 5.A: Two Essential knowledge 5.A.1: Temperature is a measure of the average kinetic energy of atoms and systems with different temperatures molecules. that are in thermal contact will Essential knowledge 5.A.2: The process of kinetic energy transfer at the particulate scale is referred to in exchange energy. The quantity of this course as heat transfer, and the spontaneous direction of the transfer is always from a hot to a cold body. thermal energy transferred from one system to another is called heat. Enduring understanding 5.B: Energy is neither created nor destroyed, but only transformed from one form to another. Essential knowledge 5.B.1: Energy is transferred between systems either through heat transfer or through one system doing work on the other system. Essential knowledge 5.B.2: When two systems are in contact with each other and are otherwise isolated, the energy that comes out of one system is equal to the energy that goes into the other system. The combined energy of the two systems remains fixed. Energy transfer can occur through either heat exchange or work. Essential knowledge 5.B.3: Chemical systems undergo three main processes that change their energy: heating/ cooling, phase transitions, and chemical reactions. Essential knowledge 5.B.4: Calorimetry is an experimental technique that is used to determine the heat exchanged/transferred in a chemical system. Enduring understanding 5.C: Essential knowledge 5.C.1: Potential energy is associated with a particular geometric arrangement of Breaking bonds requires energy, and atoms or ions and the electrostatic interactions between them. making bonds releases energy. Essential knowledge 5.C.2: The net energy change during a reaction is the sum of the energy required to break the bonds in the reactant molecules and the energy released in forming the bonds of the product molecules. The net change in energy may be positive for endothermic reactions where energy is required, or negative for exothermic reactions where energy is released. Enduring understanding 5.D: Essential knowledge 5.D.1: Potential energy is associated with the interaction of molecules; as molecules Electrostatic forces exist between draw near each other, they experience an attractive force. molecules as well as between atoms Essential knowledge 5.D.2: At the particulate scale, chemical processes can be distinguished from physical or ions, and breaking the resultant processes because chemical bonds can be distinguished from intermolecular interactions. intermolecular interactions requires Essential knowledge 5.D.3: Noncovalent and intermolecular interactions play important roles in many energy. biological and polymer systems. Enduring understanding 5.E: Essential knowledge 5.E.1: Entropy is a measure of the dispersal of matter and energy. Chemical or physical processes are driven by a decrease in enthalpy or Essential knowledge 5.E.2: Some physical or chemical processes involve both a decrease in the internal an increase in entropy, or both. energy of the components (ΔH ° < 0) under consideration and an increase in the entropy of those components (ΔS ° > 0). These processes are necessarily “thermodynamically favored” (ΔG ° < 0). Essential knowledge 5.E.3: If a chemical or physical process is not driven by both entropy and enthalpy changes, then the Gibbs free energy change can be used to determine whether the process is thermodynamically favored. Essential knowledge 5.E.4: External sources of energy can be used to drive change in cases where the Gibbs free energy change is positive. Essential knowledge 5.E.5: A thermodynamically favored process may not occur due to kinetic constraints (kinetic vs. thermodynamic control). 439 Learning objective 5.1 The student is able to create or use graphical representations in order to connect the dependence of potential energy to the distance between atoms and factors, such as bond order (for covalent interactions) and polarity (for intermolecular interactions), which influence the interaction strength. [See SP 1.1, 1.4, 7.2, connects to Big Idea 2; Essential knowledge components of 5.A–5.E] Learning objective 5.2 The student is able to relate temperature to the motions of particles, either via particulate representations, such as drawings of particles with arrows indicating velocities, and/or via representations of average kinetic energy and distribution of kinetic energies of the particles, such as plots of the Maxwell-Boltzmann distribution. [See SP 1.1, 1.4, 7.1; Essential knowledge 5.A.1] Learning objective 5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions. [See SP 7.1; Essential knowledge 5.A.2] Learning objective 5.4 The student is able to use conservation of energy to relate the magnitudes of the energy changes occurring in two or more interacting systems, including identification of the systems, the type (heat versus work), or the direction of energy flow. [See SP 1.4, 2.2, connects to Essential knowledge 5.B.1, 5.B.2] Learning objective 5.5 The student is able to use conservation of energy to relate the magnitudes of the energy changes when two nonreacting substances are mixed or brought into contact with one another. [See SP 2.2, connects to Essential knowledge 5.B.1, 5.B.2] Learning objective 5.6 The student is able to use calculations or estimations to relate energy changes associated with heating/cooling a substance to the heat capacity, relate energy changes associated with a phase transition to the enthalpy of fusion/vaporization, relate energy changes associated with a chemical reaction to the enthalpy of the reaction, and relate energy changes to PΔV work. [See SP 2.2, 2.3; Essential knowledge 5.B.3] Learning objective 5.7 The student is able to design and/or interpret the results of an experiment in which calorimetry is used to determine the change in enthalpy of a chemical process (heating/cooling, phase transition, or chemical reaction) at constant pressure. [See SP 4.2, 5.1, 6.4; Essential knowledge 5.B.4] Learning objective 5.8 The student is able to draw qualitative and quantitative connections between the reaction enthalpy and the energies involved in the breaking and formation of chemical bonds. [See SP 2.3, 7.1, 7.2; Essential knowledge 5.C.2] Learning objective 5.9 The student is able to make claims and/or predictions regarding relative magnitudes of the forces acting within collections of interacting molecules based on the distribution of electrons within the molecules and the types of intermolecular forces through which the molecules interact. [See SP 6.4; Essential knowledge 5.D.1] Learning objective 5.10 The student can support the claim about whether a process is a chemical or physical change (or may be classified as both) based on whether the process involves changes in intramolecular versus intermolecular interactions. [See SP 5.1; Essential kn owledge 5.D.2] Learning objective 5.11 The student is able to identify the noncovalent interactions within and between large molecules, and/or connect the shape and function of the large molecule to the presence and magnitude of these interactions. [See SP 7.2; Essential knowledge 5.D.3] Learning objective 5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes. [See SP 1.4; Essential knowledge 5.E.1] Learning objective 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both ΔH° and ΔS°, and calculation or estimation of ΔG° when needed. [See SP 2.2, 2.3, 6.4; Essential knowledge 5.E.2, connects to 5.E.3] Learning objective 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy. [See SP 2.2; Essential knowledge 5.E.3, connects to 5.E.2] 440 Learning objective 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable. [See SP 6.2; Essential knowledge 5.E.4] Learning objective 5.16 The student can use Le Chatelier’s principle to make qualitative predictions for systems in which cou pled reactions that share a common intermediate drive formation of a product. [See SP 6.4; Essential knowledge 5.E.4, connects to 6.B.1] Learning objective 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction. [See SP 6.4; Essential knowledge 5.E.4, connects to 6.A.2] Learning objective 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amoun ts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction ca n produce large amounts of product for certain sets of initial conditions. [See SP 1.3, 7.2; Essential knowledge 5.E.5, connects to 6.D.1] Internal Energy (sum of P.E. and K.E.) can be divided The kinetic energy of chemicals is expressed in the temperature, and motion of the chemicals into 2 really big categories Potential Energy can be converted to defined as Kinetic Energy defined as energy of position of between species relative to an assumed standard energy of motion (possessed by an object in motion) examples of kinetic energy examples of potential energy Chemical Bond Energy P.E. is associated with the chemical energy of bonds. Electromagnetic Spectrum [Light, Thermal, Microwave]; &then there is/are: Sound energy, motion, Between bonded species, there is a bond length or distance and thus the energy possessed by a chemical is associated closely with potential energy … in this case, it’s called, chemical energy There are 3 important terms over which we must be most concerned, with respect to energy: 1) Entropy 2) Enthalpy 3) Free Energy (or Gibbs Free Energy) Very often, we are most concerned with the *change in entropy (ΔS), change in enthalpy (ΔH) and change in Gibbs Free Energy (ΔG). 441 Working Vocabulary: These will be refined over the course of our work …and are simplified, here. Heat: the transfer of energy Work: force acting over distance …. w = F x d or w = -P∆V where gases are involved. Temperature: proportional to the average kinetic energy of the molecule (KEavg) Heat ‘em up, speed ‘em up. Enthalpy (H): *flow of energy (heat exchange) at constant pressure, when 2 systems are in contact with each other. Enthalpy of reaction: ∆Hrxn amount of heat released (negative value) or absorbed (positive value) by a chemical reaction at constant pressure (kJ/molrxn or kJ …[an older unit]) Enthalpy of combustion: Enthalpy of formation: ∆Hf heat absorbed or released when 1 mol of compound is formed from its elements in their standard states in kJ/molrxn Enthalpy of fusion ∆Hfus heat absorbed to melt (overcome IMFs) 1 mol of solid to liquid at a constant temperature (the melting point) ( kJ/molrxn ) Enthalpy of vaporization: Entropy (S): *the measure of the dispersal of matter and energy … the change may be a +ΔS or a -ΔS System: area of the universe on which we are focusing (e.g. the experiment) Surroundings: everything outside of the system Endothermic: +ΔH in kJ/molrxn Exothermic: -ΔH in kJ/molrxn Gibbs Free Energy: criteria for determining the thermodynamic favorability (ΔG = ΔH -TΔS) State Function: a property of matter, INDEPENDENT of pathway (e.g. mass, volume, entropy, ΔH ….) Standard Conditions: Lab conditions of 1 atm, 25°C and 1 M (when solutions are involved). All of this is known by adding the symbol ° to G, H or S. Thus, given ∆H° you automatically know the pressure and temperature conditions which apply to the measurement. 442 I) Thermodynamics is all about energy… it is technically the study of the transformations of energy from one form to another…...And as Dr. Peter Atkins writes; “Energy is the basis of civilization”. The laws of thermodynamics govern chemistry and life. They explain why reactions take place and let us predict how much heat reactions release and how much work they can do. Thermodynamics plays a role in every part of our lives. For example, the energy released as heat can be used to compare fuels, and the energy resources of food are used to assess its nutritional value. (Chemical Principles Quest for Insight p. 235) A) *Heat and work are equivalent means of transferring energy between a system and its surroundings. The total energy of an isolated system is constant. The enthalpy change (ΔH) for a process is equal to the heat released at constant pressure. B) There are, 4 Laws of Thermodynamics Zeroth law of thermodynamics – When two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. … Its big implication is that: * Energy flows from high to low First law of thermodynamics – Energy can neither be created nor destroyed, but it can change forms. In any process, the total energy of the universe remains the same. For a thermodynamic cycle the net heat supplied to the system equals the net work done by the system. The First law is all about the Law of the Conservation of Energy … and it is essentially our “bookkeeper” (Hey! the work bookkeeper has 3 sets of double letters in a row!) It addresses the familiar pieces of chemistry – as covered in the first year course … How much energy is involved in a change? (±∆H… change in enthalpy) Does energy flow into or out of the system? (endothermic / exothermic) What form does the energy finally assume? (EMS chemical or PE KE) Its big implication is that: *The energy of the universe is constant Second law of thermodynamics – The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium … Its big implication is that: *The entropy of the universe is increasing Third law of thermodynamics – The entropy of a perfect crystal at 0 Kelvin is zero It is insanely rare to find a perfect crystal …so entropy values (even for elements) are rarely 0 Its big implication is that: It tells us that the absolute entropy of a substance can be determined at any temperature, higher than zero K …and this is the defense as to why Gibb’s Free Energy (G) and the Enthalpy (H) of an element = 0…. but entropy (S) does NOT equal zero!!!! 443 Law Zeroth SUMMARY: 4 Laws of Thermodynamics Interpretation Implication(s) Energy flows from zones of high When a hot metal is placed in contact with a cooler temperature to zones of lower metal, energy will flow from the hotter to the cooler, temperature until a thermal equilibrium is established and both metals are at the same temperature. Temperature measures average kinetic energy of atoms and molecules. The process of kinetic energy transfer at the molecular level is referred to as heat transfer. First Energy can neither be created nor destroyed. It can only do 2 things …It can be transferred, or be converted to another form. The energy of the universe is constant. For 2 isolated systems in contact with each other, the energy that comes out of one system is equal to the energy that goes into the other system. The combined energy of the two systems remains fixed. E= energy q + heat transfer w work Energy is transferred between systems whether through heat transfer or through one system doing work on the other system. Chemical systems undergo 3 main processes that change their energy: heating/cooling, phase changes, and chemical reactions. Second The entropy of the universe is increasing The universe is constantly increasing the dispersal of matter and energy …. This is an expanding universe. There is no such thing as 100% conversion of chemical energy into useable work. Third The entropy of a perfect crystal = 0 We can deduce entropy at any temperature above 0 K. Check out: 1) Crash Course #17: heat and work: https://www.youtube.com/watch?v=GqtUWyDR1fg 2) State Functions https://www.youtube.com/watch?v=lL2CXpVti34 3) State Functions (more) https://www.youtube.com/watch?v=SEapOFaYDlA 4) Crash Course #18: enthalpy: https://www.youtube.com/watch?v=SV7U4yAXL5I 5) Heat and work (alt) https://www.youtube.com/watch?v=SfHHZxGpwr8 444 C) Systems, Work and Heat 1) System: …the region in which we are interested: such as a flask of gas, a beaker of acid, a reaction mixture, or a muscle fiber (Atkins. p. 236). Everything else, such as a water bath, in which a reaction is immersed (in another beaker) is referred to as the surroundings. The surroundings are where we make observations of the energy transferred into our out of the system. The system and the surroundings jointly make up the universe … but we understand that only a very localized portion of the universe is affected significantly. a) Open system: allows for the exchange of matter and energy between the chemicals and the surroundings …. for example, automobile engines and our bodies! b) Closed system: … possesses a fixed amount of matter, but it can exchange energy with the surroundings. For example, a closed system could be a chemical cold pack used for treating athletic injuries/inflammation. c) Isolated system: … has no contact with its surroundings, so that there is no exchange of matter or energy with the surroundings …. e.g.) a rigid, thermally insulated vessel holding the system …. such as a hot liquid inside a sealed vacuum flask 2) Work… The process of achieving motion against an opposing force… all forms of work can be thought of as equivalent to the work of raising a weight against the pull of gravity. … e.g pushing an electrical current through a circuit. W = Force opposing x distance or work = -P∆V where gases are involved a) unit = Joule where 1 J = 1 kg∙m2∙s-2 (1 J = 1 N∙m in physics) 3) Energy is the capacity to do work … or to create a change a) Internal Energy (E or U) … the total store of energy in a system … b) The absolute value of internal energy cannot be measured (we would need to know the energies of every atom from their electrons to their nuclei)… hence we tend to measure the change in internal energy … 4) When work is done ON a system, (assuming no other changes), the internal energy is increased 5) A system can do 2 types of work: expansion and non-expansion work a) expansion work involves a change in volume of the system e.g. a gas expanding in a cylinder, moving a piston … hence w= -P∆V b) non-expansion work does not involve a change in volume of the system. e.g. the chemical reaction in a battery. 445 6) Heat: The second law is a bit more involved… As written by Patrick Coffey in Cathedrals of Science: …. The second law implies that “While work can always be converted to heat, only a portion of the heat of a system, even under ideal circumstances, can be converted into work in a cyclic process – like a steam engine. The first law is sometimes stated as “You can’t get something for nothing” and the second law as “You can’t even break even” a) Heat and temperature are very different and to grasp the 2nd Law … we need to understand a bit more about “heat”. i) temperature is a property that reflects the random motions of the species of a substance. ii) HEAT is the energy transfer (frictional heating) between two objects due to a difference in temperature. While we often talk about an object being contained by an object in reality. It is not. We call the transfer q … as in q = mc∆T yeah, “q” = heat iii) Heat is the sibling of work, and heat is often referred to as a form of energy (but “heat” per se, is not on the electrochemical spectrum [infrared energy is … but not, heat] Heat is equivalent to the amount of energy that is transferred between the reacting chemicals and surroundings. We limit the work produced during a chemical reaction, by measuring the progression of the reaction in rigid, non-flexible containers (a calorimeter). This limits the work by limiting the movement of the container atoms, over a distance. So, in our measurements (calorimetry) we tend to refer to the energy exchange as heat, (which represents the change in internal energy) …and this is at the heart of the confusion. Intrinsically we recognize that “heat” is a transfer (a verb) not a specific type of energy. e.g. We need to "heat up" a cup of coffee. It’s really heating up out here”. And, most experts don’t even like talking about it as a verb. See: http://hyperphysics.phy astr.gsu.edu/hbase/thermo/heat.html …. According to the late Dr. Mark Zemansky: Don't refer to the "heat in a body", or say "this object has twice as much heat as that body". The First Law of Thermodynamics identifies both heat and work as methods of energy transfer which can bring about a change in the internal energy of a system. After that, neither the words work or heat have any usefulness in describing the final state of the system - we can speak only of the internal energy of the system. b) Essentially the 2nd Law states that there is no such thing as 100% conversion of chemical energy into useable work (energy) c) Thus, one big implication of the 2nd Law is that …. While the energy of the universe is constant … not all of it remains useable… as the entropy of universe is increasing. Hence: Energy is constant…Entropy is NOT 446 II) Reading: Edited from Peter Atkin’s Chemical Principles: A Quest for Insight (p 287) The first law of thermodynamics tells us that, if a reaction takes place, then the total energy of the universe (the reaction system and its surroundings) remains unchanged. But the first law does not address the questions that lie behind the word, “IF”. Why do some reactions have a tendency to occur, whereas others do not? That is, why are some reactions thermodynamically favorable while others are not? Why does anything happen at all? To answer these deeply important questions about the world around us, we need to take a further step into thermodynamics and learn more about energy, beyond the fact that it is conserved… Hence, the second law of thermodynamics is the key to understanding why one chemical reaction has a natural tendency to occur but another one does not. We apply the second law using the very important concepts of entropy and Gibbs free energy. The third law (which does not play as large a role in AP Chemistry) is the basis of the numerical values of these two quantities. The second and third laws jointly provide a way to predict the effects of changes in temperature and pressure on physical and chemical processes. They also lay the thermodynamic foundations for discussing chemical equilibrium (Big Idea 6). We will from this point on be working to blend …to draw together concepts related to the first law (particularly enthalpy) and work. We need to be aware that molecules can occupy only a series of discrete energy levels and so possess only certain energies. A) Thermodynamically Favored Processes (spontaneous processes) and Entropy Thermodynamically Favored Processes definition those processes which occur without outside intervention may be slow or fast may be physical changes OR chemical reactions e.g cooling a block of hot metal the expansion of gas into a vacuum all about direction of procession not speed e.g it is thermodynamically favored for a diamond to revert back to graphite …. but it is a very slow process …. Kinetics and Thermodynamics are both required for a full explanation some require to be initiated (activation energy) “favored” is concerned with the tendency to occur. Whether the process is realized depends on its rates (kinetics) e.g. it is thermodynamically favored for syrup to flow out of a bottle, but this can be very slow, at exceedingly low temperatures A process is spontaneous if it has a tendency to occur without being driven by an external influence (excluding initiation); spontaneous changes need not be fast. B) A classic list of thermodynamically favored changes might include: a ball rolls down a hill – but not “spontaneously” up the hill (recall… spontaneously is an older term for thermodynamically favored … and indicates the lack of outside intervention) 447 a gas fills a vessel uniformly – it never spontaneously collects at one end of a container an iron nail exposed to moisture and oxygen rusts – but iron(III) oxide does not spontaneously revert back to iron and dioxygen heat transfer occurs from an object of greater temperature (hot) to one of lesser temperature (cooler). It does not flow from cool to hot spontaneously (if it did we would not need to have motors on our refrigerators!) wood burns exothermically and spontaneously to produce carbon dioxide and water …. but carbon dioxide and water, when heated together, do not produce wood. At standard pressure and temperatures below 0°∁ water freezes (spontaneously), in a thermodynamically favored process, and at temperatures superior to 0°∁, ice melts to water. As we focus upon the thermodynamically favored reaction we must develop a more fluid grasp of enthalpy and entropy … we will then tie the two together with Gibbs Free Energy III) Enthalpy at Constant Pressure A) When a chemical reaction occurs open to the atmosphere under conditions of constant pressure (as in an open beaker, or cooking something on the stove top), the energy can evolve as both heat and work. Tro p. 265 edited More often than not though, we are interested only in the heat exchange (e.g. the energy released, that can be used to cook our food) & not in the work (the energy used to expand the surrounding air). 1) Hence: Under conditions of constant pressure we therefore deal with the quantity called *Enthalpy (symbol H) … Enthalpy of a system is the sum of its internal energy and the product of its pressure and volume: H = E + PV a) Since internal energy, pressure, and volume are each state functions, enthalpy is also a state function. The symbol ∆H, represents the change in enthalpy. i) at constant pressure q = ∆H ii) ∆H is different from ∆E (the sum of q and work) … They are similar in that both are state functions - but ∆H is a measure of only the heat exchanged when pressure is constant … whereas ∆E is the measure of all of the energy (heat and work) 2) A +∆H indicates that energy is flowing into a system … this is an *endothermic reaction. A -∆H indicates that energy is flowing out of a system, into the surroundings, this is referred to as an *exothermic reaction. 3) unit = kJ/molrxn 448 TRY THIS! Identify each process as endothermic or exothermic, and indicate the sign of ∆H. Here is a hint, to ensure better success … always think of it, from the point of view of the system (the chemicals) … Identify the chemicals first … and then ask “What must happen the chemicals (thermally-speaking) … must they absorb or release energy…. a) perspiration evaporating from a person’s skin * endothermic +∆H b) water freezing to ice in a freezer * exothermic -∆H c) wood combusting with oxygen in a fireplace * exothermic -∆H d) the breaking of a covalent bond * endothermic +∆H e) the grilling (cooking) of a hamburger * endothermic +∆H f) the formation of raindrops via condensation (H2O(g) → H2O(ℓ) * exothermic -∆H B) Calorimetry: q = mcΔT 1) Measurements for the change in enthalpy tend to occur in something called a bomb calorimeter … which is essentially a means of creating an isolated system. a) The bomb calorimeter uses a constant volume … based upon the relationship that w = -P∆V … when a reaction is carried out at constant volume, then ∆V is 0 so w = 0 At this point q essentially = ∆Erxn Note that the bomb is a sealed container, which keeps the volume constant. It is here that we evolve the concept that the amount of energy (in this case, all as heat) absorbed by the calorimeter = the energy released by the reaction Where: qcal = - qrxn Hence: bomb calorimetry occurs at constant volume and measures ∆E for a reaction. http://catalog.flatworldknowledge.com/bookhub/reader/1790?e=averill_1.0-ch05_s03 4) What about coffee-cup calorimetry? I’m glad you asked… That is known as Constant Pressure Calorimetry… Where, qrxn = -qsolution a) This measures enthalpy changes for chemical reactions which occur in solution. b) Hence: Coffee cup calorimetry occurs at constant pressure and measures ∆H for a reaction. 449 TRY THIS! In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 24.5°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/mL and a specific heat capacity constant of 4.18 J/g°C, calculate the change in enthalpy for the neutralization of HCl by NaOH. *q = mcΔT thus: q = (200.0 g)(4.18)(6.7) Hint 1: q = mcΔT and since this is a reaction, in which the masses are additive and the constants are the same for both the acid and base, you simply need to plug and chug ….but watch out for that mass. Hint 2: Notice the unit is given in kJ/mol ..which is appropriate for a change in enthalpy. ans: 5.6 x 103 kJ/mol IV) Stoichiometry and ∆H A) The change in enthalpy is also called : *the heat of reaction and/or the enthalpy of reaction and it is abbreviated at Hrxn 1) The change in enthalpy is dependent upon * the quantities of reactants which react to produce product. a) The magnitude (size) of ∆Hrxn is for the stoichiometric amounts of reactants and products for the reaction as written …. We should use the stoichiometric ratios and the energy of the reaction when trying to determine the total ∆Hrxn b) for instance: Given : C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ∆Hrxn = -2,044 kJ What is the total ∆Hrxn when 3.000 moles of propane are completely combusted in an excess of oxygen gas, producing only carbon dioxide and water? *6,132 kJ 450 TRY THIS! Now, try something a bit more challenging…. A liquid propane tank in a home grill contains 13.2 kg of propane, C3H8. Calculate the energy exchange, in kJ (the heat) associated with the complete combustion of all of the propane in the tank. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ∆Hrxn = -2,044 kJ Hint 1* Convert the kg of propane to moles of propane … This means you should convert to grams and then using the mole mass for propane, convert to moles … this can be done using your wonderfully superior dimensional analysis skills. Hint 2* Use the stoichiometric relationship between propane and the ∆Hrxn to determine the total heat evolved. ans: ∆Hrxn = -6.12 x 105 kJ * kJ = 13.2 kgpropane| 1,000 grams| 1 molepropane| -2,044 kJ| 1 kg 44.09 g 1 molpropane TRY THIS! Ammonia reacts with oxygen gas according to the reaction equation: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) ∆Hrxn = -906 kJ Calculate the energy exchange (the heat) in kJ associated with the complete reaction of 155 g of NH3 in an excess of oxygen gas. *kJ = 155 g NH3 | 1 mol NH3| -906 kJ | = -2060 kJ or -2.06 x 103 kJ 17.03 g 4 mol ans: ∆Hrxn = -2.06 x 103 kJ 451 IV) Entropy: A state function which is a measure of disorder of the matter and energy… It captures how the energy is dispersed throughout the microstates of the matter. A) The entropy of an isolated system increases in any spontaneous (thermodynamically favored) process. 1) The common denominator to each of the spontaneous scenarios in IIB) turns out to be *entropy … which is the dispersion of energy (and very, very often, linked to the matter) throughout the matter of the system. Many see entropy as measure of the molecular randomness or disorder … but that focuses only upon the matter A slightly more sophisticated view is the disorder of the energy….(not just the matter …) B) Entropy is closely linked to the second law of thermodynamics …it is another way of interpreting the law … While it is a measure of the energy dispersion, it may be seen also in a roundabout way *indicating the energy NOT available for work … 1) symbol for entropy (S) and symbol for the change in entropy (±∆S) 2) unit: J∙K-1 or J/molrxn•K …. NOTE…Enthalpy is often given in kJ … beware!!!! 3) entropy is a thermodynamic function that describes the number of arrangements (positions &/or energy levels) that are available to a system existing in a given state. These probable positions and/or energy levels are called microstates a) a microstate is a single possible arrangement of the positions and kinetic energies of the molecules when the molecules are in a specific thermodynamic state. i) e.g. A mixture of gases will exhibit a greater entropy than a similar volume of a single gas … because the mixture has a greater “randomness” or ability to exist in a greater number of variations of molecular order or energy level 4) According to Zumdahl “The key concept is that the more ways a particular state can be achieved, the greater is the likelihood (probability) of finding that state. In other words, …. Nature spontaneously proceeds toward the states that have the highest probabilities of existing” (p. 790) … Keep thinking then, that the driving force is …. well …. probability! 452 a) Predicting the entropy change of a system is based on physical evidence (NMSI Entropy and Free Energy) The greater the dispersal of matter and/or energy in a system, the larger the entropy The entropy of a substance always increases as it changes from solid to liquid to gas. When a pure solid or liquid dissolves in a solvent, the entropy of the substance increase (there are exceptions …e.g. carbonates …in water they bring more order, and decrease entropy) When a gas molecule escapes from a solvent, the entropy increases Entropy generally increases with increasing molecular complexity …since there are more moving electrons! e.g KCl < CaCl2 Reactions increasing the number of moles often increase entropy … and those in which the number of moles of product is less than the number of reactants, entropy decreases. So remember … The greater the number of arrangements, the greater the entropy of the system. b) Many at this level argue positional probability as a driving force (the reason) but I urge you to consider it from the point of view of the energy. For instance: Compare a solid to its liquid phase and to its gaseous phase …What can we say about the entropy? *solid < liquid < gas As some would argue, there is a greater positional probability in the gas phase, when compared to the solid phase (and they would be correct … but I think that’s limited) Consider this … The solid essentially is a single state … We do not even speak about molecules when considering solid para-dichlorobenzene (for instance), because discrete entities do not really exist … hence the energy of the system is essentially dispersed and shared by the entire state … When converted to a gas, the energy is dispersed to individualized molecules, comprising millions of microstates … Hence the energy has been dispersed across a greater level of disorder, and is at a greater entropy. When it comes to the formation of solutions, positional entropy works … but again, 453 I urge you to consider it really from the point of view of the dispersion of energy. The tendency to mix is due to the increased volume available to the particles of each component of the mixture. Granted: when two liquids are mixed, the molecules of each liquid has more available volume and thus more available positions. Perhaps so, but the energy of the system is allowed to be dispersed across a greater number of microstates as those solutions are mixed. Using the argument of energy is a powerful means of visualizing entropy, since entropy is all about energy levels. Fascinating: A little reading on fusion and microstates from 1885 …The second chapter is entitled “Equilibrium in Dissociation,” and attempts to dispel the older belief, based on the caloric theory, that heat is a repulsive force which acts in opposition to chemical affinity. The reason compounds eventually dissociate upon heating is not because the repulsion of the added heat finally overwhelms the attractions of the internal bonds, nor because, in keeping with the newer mechanical theory of heat, the increasing violence of the intramolecular vibrations finally break the internal bonds. Rather it is because the net increase in the number of independently moving species formed upon dissociation is more able to effectively dissipate the system’s internal kinetic energy. George Downing Liveing Chemical Equilibrium The result of the Dissipation of Energy 1885 Deighton, Bell and Co. London. As cited in Bulletin for the History of Chemistry Volume 38 No. 1 2013 p. 41-2 5) A gas has more energetically equivalent configurations than a solid, because the gas has more ways to distribute its energy than a solid. a) A solid’s energy is essentially found in its vibrational motion between its molecules ….it is essentially a single microstate …few options of re-arranging the energy…. b) A gas may exhibit more “places” to put the energy … there is rotational energy or translational energy (straight-line motion) c) Ludwig Boltzmann came up with its expression as: S = k lnW where: k = Boltzmann constant (The universal gas constant / Avogadro’s number) thus k = R/Na = 1.38 x 10-23 J/K W = the number of possible individual microstates that can result in an overall arrangement. d) A chemical system proceeds in a direction that increases the entropy of the universe – it proceeds in a direction that has the largest number of energetically equivalent ways to arrange its components (Tro 820)… its matter, but most importantly … its energy. http://www.wikipremed.com/image.php?img=010304_68zzzz131150_30902_68.jpg&image_id=131150 454 6) Recall: The average kinetic energy of the molecules of an ideal gas is directly proportional to the absolute (Kelvin) temperature of the gas. a) Consequence: The higher the temperature, the faster the molecules move and the greater the kinetic energy they possess (K.E. = ½ mv2) …with mass in kilograms. Additionally, as we saw in the Maxwell-Boltzmann distribution graph, systems at higher temperatures have a broader distribution of molecular speeds. b) Real gases can exhibit 3 kinds of complex motion … Vibrational Rotational Translational (a molecule moving in one direction) http://catalog.flatworldknowledge.com/bookhub/4309?e=averill_1.0-ch18_s04 i) solids have the least, rotational and translational motion. http://education-portal.com/academy/lesson/properties-of-water.html gases can exhibit the highest form of all three: http://wateronthemove.wikispaces.com/Forms+of+Water ii) Consequence: These more complex expressions of kinetic energy, are the means by which molecules can store more energy, as temperature increases. 455 TRY THIS: Right from your Honors Notes … Identify the change in entropy as + or a – Basic entropy interpretations: kJ + H2O(l) H2O(g) S = + l<g H2O(g) H2O(s) + kJ S = - g>s kJ + CH3OH(l) CH2OH(g) C6H6(l) C6H6(s) + kJ Slightly more: complex 2 Fe2O3(s) + 3 C(s) 4 Fe(s) + 3 CO2(g) *S = + S = - S = + 2 AgI(s) + F2(g) I2(s) + 2 AgF(s) S = - 4 Li(s) + O2(g) S = - 2 H2O2(l) 2 Li2O(s) S = + 2 H2O(l) + O2(g) 2 KClO3(s) 2 KCl(s) + 3 O2(g) S = + 4 Cr(2) + 4 K(s) + 7 O2(g) 2K2Cr2O7(s) S = - Advanced: Darn It 2 N2O3(g) 2 N2(g) + 3 O2(g) There's no phase change N2(g) + 3 H2(g) 2NH3(g) S = + 2 SO2(g) + O2(g) 2SO3(g) S = + 2 PaI5(s) 2 Pa(s) + 5 I2(s) S = + S = - TRY THIS: Right from your Honors Notes …Again! The answers to the multiple choice are after question #7 of this TRY THIS section 1) Given: C6H4Cl2(s) + kJ → C6H4Cl2(g) Which statement describes this change? (1) It is endothermic, and entropy decreases. (2) It is endothermic, and entropy increases. (3) It is exothermic, and entropy decreases. (4) It is exothermic, and entropy increases. 2) Which process is accompanied by a decrease in entropy? (1) Deposition of carbon dioxide (2) Evaporation of water (3) Allowing a gas to expand into a larger volume (4) Heating a balloon filled with a gas 456 3) At a pressure of 101.3 kilopascals and a temperature of 373 K, energy is lost from a sample of water vapor causing the sample to change from the gaseous phase to the liquid phase. This phase change is represented by the equation: H2O(g) → H2O(l) + kJ Explain, in terms of particle arrangement, why entropy decreases during this phase change. Be sure to use entropy’s definition … and what is going on in the phase change (per the number of microstates) in your answer…. *Entropy is associated with the dispersion of matter and the energy of that matter. As water vapor condenses the molecules decrease their relative position (a loss of positional entropy) to each other, and in the process, some energy is lost from the system to the surroundings. In the phase change, the number of possible microstates (or possible ways the matter and energy may exist) is more limited, than in the gas phase, the entropy is decreasing 4) Systems in nature tend to undergo changes (1) lower energy and less disorder (2) lower energy and more disorder (3) higher energy and less disorder (4) higher energy and more disorder 5) Which of these biological reactions represents one that decreases the entropy of a cell? (1) digestion (2) hydrolysis (3) catabolism (4) dehydration reactions 6) Even though the process is endothermic, snow can sublime. Which tendency in nature accounts for this phase change? (1) a tendency toward greater entropy (2) a tendency toward greater energy (3) a tendency toward less entropy (4) a tendency toward less energy water 7) Study the dissolving of potassium nitrate in water: KNO3(s) + 34.89 kJ K1+(aq) + NO3-1 (aq) Given all you know, thus far identify which best describe the process (1) (1) (3) (4) It is endothermic, It is endothermic, It is exothermic, It is exothermic, physical chemical chemical, physical and the and the and the and the entropy increases entropy decreases entropy decreases entropy increases TRY THIS answers 1) 2 3 see online 2) 1 4) 2 5) 4 6) 1 7) 1 457 Before going on … what do you know (about)…..? ___entropy is about the dispersion of energy amidst the matter ___entropy has the symbol S ___the change in entropy at standard conditions has the symbol ΔS° ___the greater the number of molecular arrangements, the greater th entropy ___entropy changes with phase ___ a substance in its solid phase has a lower entropy than in its gas phase ___ the particles of a gas have a greater entropy because of greater vibrational, rotational, and translational options “putting” the energy in more “places” ___ when moles decrease, concerning reactants to products, entropy decreases ___ a crystal of AlCl3 has a greater entropy than a crystal of NaCl, due to the greater number of electrons ___the units of entropy are often different from the units of enthalpy ___how to determine a change in entropy given a balanced equation? ___the first three laws of thermodynamics? C) Calculating Standard Molar Entropy (∆S°) at Standard Conditions 1) When calculating molar entropy change: For AP Chemistry the granddaddy of all entropy equations is: ∆S°rxn = ∑∆S°products - ∑∆S°reactants a) Entropy is a state function (it is not pathway-dependent). We can calculate the entropy change for a given process, by taking the difference between the standard molar entropy values (S°) of products and reactants. It is an extensive property … so it depends upon the number of moles 2) Absolute values for entropy can be assigned …. And hence, into our story, enters the third law of thermodynamics. a) The entropy of a perfect crystal of an element at 0 K (absolute zero) represents the lowest possible entropy … the internal arrangement of a perfect crystal is absolutely regular at absolute zero … for molecular motion virtually ceases. b) Essentially the above is a re-working of the third law of thermodynamics. i) Caveat: “A perfect crystal at 0 K is an unattainable ideal, taken as a standard but never actually observed” (Zumdahl 803) 3) As the temperature of a perfect crystal is increased, the random molecular motion increases increasing the “disorder” … in the distribution of the energy among slightly differing microstates, increases (due to the more complex motions) Thus entropy increases. a) S for a perfect crystal is 0 at 0 K … the entropy value for a substance at a particular temperature can be calculated by knowing the temperature dependence of entropy. 458 b) Study the table from your text and complete the following observations: Standard Molar Entropies at 298 K Substance S° (J/K∙mol) H2(g) 130.6 N2(g) 191.5 O2(g) 205.0 H2O(g) 188.8 NH3(g) 192.5 CH3OH(g) 237.6 C6H6(g) 269.2 69.9 H2O(ℓ) 126.8 CH3OH(ℓ) 172.8 C6H6(ℓ) Li(s) 29.1 Na(s) 51.4 K(s) 64.7 Fe(s) 27.23 FeCl3(s) 142.3 NaCl(s) 72.3 Observations of the Data: 1) The enthalpy of formation = 0, but standard molar entropies of elements at the reference temperature are NOT zero proof / examples _____________________________ 2) The standard molar entropies of gases are greater than or less than those of liquids and solids. proof: compare _____________________________ 3) Standard molar entropies generally increase or decrease with increasing molar mass proof: compare _____________________________ 4) Standard molar entropies generally * increase with increasing number of atoms in the formula of a substance (it * increases with molecular complexity). proof: compare _____________________________ TRY THIS! Given all of the information below, predict a numerical range or value for the S° (standard molar entropy) for butane. Then using one or more of the observations from the table, explain your reasoning. Substance Name Formula methane ethane propane butane CH4(g) C2H6(g) C3H8(g) C4H10(g) Value for S° J/K∙mol 186.3 229.6 270.3 ???? * The standard molar enthalpy for butane should be somewhere very close to 310 J/K ∙mol. Based upon the data there appears to be an increase of 40 J/K for the increase in 1 carbon and 2 hydrogen atoms (which references the general formula for the alkanes of CnH2n+2). This prediction should be borne out by the observations from the table, because observation 3 & 4 reference in increasing value of S° for more massive molecules and those molecules exhibiting greater complexity. Butane’s greater mass is relatively unassailable. I submit that it also has a greater complexity due to the possible rotational possibilities given the availability of C – C free rotation. This rotation increases the possible number of microstates, relative to the other examples and thus should exhibit a greater S° 459 TRY THIS: This uses that Granddaddy of an equation… ∆S°rxn = ∑∆S°products - ∑∆S°reactants Calculate the entropy change at 25°C, in J/molrxn•K) from the following standard values, for the reaction: Entropy (J/mol•K) SO2(g) 248.1 O2(g) 205.3 SO3(g) 256.6 2 SO2(g) + O2(g) → 2 SO3(g) Hint 1: * This sort of problem is a repeating theme … You need to calculate the entropy (using the coefficients of the balanced equation) Hint 2:* Multiply each mole value by the entropy value for each substance. Get the total for the product and sum total for the reactants. ans: -188.3 J/ mol•K TRY THIS: Calculate ∆S° at 298 K for the reaction: N2(g) + 3 H2(g) → 2 NH3(g) (B&L p. 829) Table of Standard Entropy Values Substance NH3(g) N2(g) H2(g) S° (J/K∙mol) 192.5 191.5 130.6 *∆S°rxn = ∑S°products - ∑S°reactants ∆S°rxn = 2(192.5 J/ K∙mol) - [(191.5 J/K∙mol) + 3 (130.6 J/K∙mol)] 385 - (191.5 + 391.8) = -198.3 J/K∙mol ans: -198.3 J/K∙mol *Note that given our qualitative work the negative value of ∆S is right in line with this response … As we can see that there is a decrease in the number of moles of gas … or rather, a decrease in the number of microstates …. All we have done in the above example is put a number to that decrease in entropy. TRY THIS! Calculate the standard entropy change, ∆S°, for the following reaction at 298 K. Al2O3(s) + 3 H2(g) → 2 Al(s) + 3 H2O(g) Table of Standard Entropy Values Substance S° (J/K∙mol) H2O (g) 188.8 Al(s) 28.32 Al2O3(s) 51.00 H2(g) 130.6 * ∆S°rxn = 3(188.8 J/K mol) + 2 (28.32 J/K mol) – (51.00 J/K mol) + 3(130.6 J/K mol) 566.4 + 56.64 623.04 - - 442.8 442.8 = 180.24 J/K mol ans: 180.24 J/K mol 460 D) Entropy changes as they apply to phase changes: ΔSsurroundings = qrev T or rather … ΔSsurr = ΔH T 1) this assumes: a constant temperature (as in a phase change), where q is the energy transferred as heat T is the Kelvin temperature at which the transfer occurs. The “rev” signifies that the energy must be transferred reversibly in the sense that any a reversible process is one that can take place in either direction …like a phase change! The equation tells us that when a lot of energy is transferred as heat (a large qrev), a lot of disorder is stirred up in the system and we expect a correspondingly big increase in entropy. For a given transfer of energy, we expect a greater change in disorder when the temperature is low than when it is high. The arriving energy stirs up the molecules of a cool system, which the molecules are already moving vigorously. (Think of a sneeze attracting attention in a library …but very little on a noisy street) Atkins p. 289 2) For instance: Calculate the ∆Sfusion for melting 1 mol of ice at 273 K. The constant for melting water is ∆Hfusion = 6.01 x 103 J/mol. (It’s positive because melting is an endothermic change … also since entropy is extensive, it depends upon the # of mols) ∆Sfusion = qrev which = ∆Hfusion = (1 mol)( 6.01 x 103 J/mol) = 22.0 J/K T T 273 3) This sort of problem blends enthalpy and entropy … and there is an issue with units!! You want to have common units … it would be advisable to change the enthalpy value from kJ/mol to J/mol a) Also, be aware that ∆Ssurr is * temperature dependent. b) Because of this dependency, when you are calculating the change in entropy of the surroundings … when work is done on the surroundings (as in an exothermic reaction, from the point of view of the chemicals) … the change in entropy (for the surroundings) is positive. When the enthalpy indicates that the reaction is endothermic, that means that the energy transfer was from the surroundings to the chemical system …. causing a greater order in the surroundings or a negative entropy value (for the surroundings) 461 TRY THIS! When a process is exothermic, the entropy of the surroundings … a) always increases b) always decreases c) increases or decreases depending upon the process Defend your answer:* a) ∆Ssurr is temperature dependent, and in an exothermic reaction the system transfers energy to the surroundings, causing and increasing in temperature, and thus an increase in the disorder (the number of microstates) of the matter and energy distribution of the surroundings. Entropy must increase. OR: ∆Ssurr= ∆H/T and in an exothermic process, ∆H must be negative for the system BUT positive for the surroundings, hence hence the value for ∆Ssurr must be positive. TRY THIS: Calculate ΔSsurr for the following reaction at 25°C and 1 atm (NMSI: Entropy and Free Energy) In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores. Sb2S3(s) + 3 Fe(s) → 2 Sb(s) + 3 FeS(s) ΔH = -125 kJ/mol *ΔH in joules = 125,000 J/mol * T in K = 298 K *ΔSsurr = ΔH/T = 125,000/298 Why is the ΔSsurr a positive value? Hint 1: *While this is not a phase change, you are asked for ΔSsurr so use the equation ΔSsurr = ΔH/T Hint 2:* Enthalpy is given in kJ …but entropy is in J … convert that enthalpic value to Joules… and check that temperature … is it in Kelvin? ans: 419 J/mol∙K * The reaction is exothermic (based upon the change in enthalpy of -125 kJ/mol. This means that the chemicals lost energy while the surroundings gained it. Thus work was done on the surrounding, causing an increase in the entropy …hence ΔSsurr is positive. 462 TRY THIS: Calculate ΔSsurr for the following reaction at 25°C and 1 atm Given the same scenario as above in the first “TRY THIS”… the antimony ore can be reduced using carbon … according to: Sb4O6(s) + 6 C(s) → 4 Sb(s) + 6CO(g) ΔH = 778 kJ/mol ans: -2.61 x 103 J/mol∙K Why is the ΔSsurr a negative value? *The reaction is endothermic, according to the positive value for the change in enthalpy. This means that the reaction chemicals absorbed energy from the surroundings. Hence the surroundings experience a decrease in entropy …signified by a negative value. Assignment Give it a whirl … Get to Trivedi 18.9 Summary: Hold On! Take a minute out and look at what this packet has covered: Aside from the summary a few pages back … take a whirl at asking: 1) Do you know your basic definitions for such important terms as: Enthalpy, Entropy, System? 2) Do you know how to work with and when to work with: q = mcΔT 3) Do you know that q = mcΔT is most closely associated with enthalpy? 4) Do you know the condition(s) under which q = ΔH? 5) Do you know how to calculate ΔH, using the stoichiometry of a balanced reaction? 6) Do you know the 3 types of motion: vibrational, rotational, and translational & how they mesh with entropy? 7) Do you know how to calculate standard molar entropy (∆S°) using a balanced equation and data? 8) Do you know how to calculate the change in entropy when give ΔH? 9) Have you taken a good hard look at the learning objectives at the front of this packet? 463