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# Physical Chemistry II CH 433

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This 290 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 433 at North Carolina State University taught by Stefan Franzen in Fall. Since its upload, it has received 31 views. For similar materials see /class/223997/ch-433-north-carolina-state-university in Chemistry at North Carolina State University.

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Date Created: 10/15/15

Chemistry 433 Lecture 8 Statistical Thermodynamics Thermochemistry Considerations Exact Differentials NC State University The rotational partition function The rotational partition function can be derived from rotational energy levels The sum over levels has the form q 0 2J 1exp JJ1 If the levels are sufficiently closely spaced relative to thermal energy then the sum can be written as an integral gm 0 2J 1eXp 3 JJ1d1 The rotational partition function Then we make the substitution let u JJ1 J2 J Then du 2J1dJ so that I w 2 21 T qro 1 exp is the rotational partition function for a diatomic molecule in the high temperature limit For most gas phase molecules at room temperature the high temperature limit is valid The vibrational partition function The vibrational partition function is a exp 20 exp mm exp exp exp 2 h00 1 X kT tia 3i l Review of the translation partition function We have seen that the translation partition function can also be derived using a Gaussian integral It is 32 gm 2 V We can also write it as qtrans 13 A Where L is called the thermal wavelength 1 mmJ h2 The molecular partition function The molecular partition function is the product of the Individual partition functions rotation vibration and translation q V 7 2 e mggans 8gib 8got 4831 abcd Z e38taransz e38gib2 e138 ot2 eB8Cejlec a b C d qtrans q vib q rot q elec K 2 qror Trot Krot Average energy and heat capacity We have seen that the internal energy can be written as Brno 68 Many books write the energy as E physics and statistical mechanics however in thermodynamics we often write it as U They are the same thing 3an we we have seen that the heat capacity is the derivative Ul the energy with respect to temperature 6U CV 39 a Tiv Heat capacity for a diatomic or polyatomic molecule The derivative can be rewritten as 1 2L L 21 as W or war was Therefore the heat capacity can 70 be Written as V Recalling that Q qNN we can also write the internal energy and heat capacity in terms of the molecular partition function M Zazlnq U 68 C NM 882 The contribution of rotation The contributions of the translational rotational and vibrational partition functions to the internal energy can be made separately The fact that they can be written as lo39arithms helps wreatl The rotational contribution to the energy and heat capacity for a diatomic is nQ t nKmt InB UIB NNkT UOtRT ifNNA 2 CV Nk52aln B Nk CWR ifN NA Consequences of including rotation We have seen that the internal energy and heat capacity of a monatomic ideal gas are Um gRT cvm R These were derived from the translational partition function see lecture 4 In this lecture we have formally incmuw The contribution from rotations for a diatomic molecule The internal energy and heat capacity of a diatomic molecule are Um 2RT cm 2R Rotation of polyatomic molecules Diatomic molecules have two rotational degrees of freedom and polyatomic molecules have three rotational degrees of freedom The partition function for a polyatomic can be written as qmt 87c2lAkT T8n2lBkT128n2leT12 h2 h2 h2 KAKBKC 12 83 It is immediately obvious that the contribution from rotations is Urot 32RT and CVrot 32 R The total values are Um 3RT cvm 3R The contribution of vibrations The vibrational partition function has a different form than either the rotational Ol39 tranSI39dLIUll39dl une IIIIpOIL dIIL unfel ellue is that the energy of vibrational modes is not zero at absolute zero The zero point energy is 12hv To properly account for this zero point energy we will need to introduce a reference internal energy U0 which is the zero point energy The vibrational partition function is mal b exp 7 q 1 exp Shoo lnqV b In 1 exp ma The contribution of vibrations The first term is the zero point energy n 3903 6B 2 The second term is the population of higher vibrational levels a In 1 exp Bhoo exp Bhoohoo Vib lt95 1 exp Shoo To see how big this term is let s consider the function in terms kT ie 18 We will make a substitution x hvkT If vibrational energy levels are larger than thermal energy hv gtgt kT then x gtgt 1 The contribution of vibrations The second term is the population of higher vibrational levels b exp xx kT V 1 exp x The function is plotted to the 08 rivht Note that the axis units of RT are appropriate for a molar energy kJmol and kT in the above expression is appropriate 02 for a single molecule J D The pressure Based on the definition of pressure volume work dU PdV so that ev lZ ll The indexj comes from the fact that we are going to average over an ensemble collection of s stems We are assuming that the number of molecules and volume of each system is the same Using our previous definition for the calculation of a molecular property the average pressure is ltPgt 2 pjN V B1 1N V Little pj is the probability The pressure Substituting in we find 8U e BEjNV GUJ P pl Wil a v 39 mt a v Since the derivative is with res ect to volume the ressure depends only the translational partition function The derivative of the partition function with respect to volume is ac 5U j crimp Vile The average pressure can then be written as P cxlmlS 3l The statistical formula for pressure The pressure can be expressed sol trms f h partition function ltPgtkita le With this definition in hand we can also write the thermodynamic function of enthalpy in terms of the partition function aan vaan H UPV 68 6V Differential relationships for enthalpy have defined a relationsp alpy and Internal energy H U PV The infinitesimal cnange In the State function n reSUItS In HdHUdUPdPVdV Therefore dH dU PdV VdP Now we substitute dU Sq 8w into this expression dH 5q 8w PdVVdP Since SW PdV dH 8q VdP At constant pressure dP 0 and we have dH qu A note on using current flow in calorimetry measurements Energy is measured in Joules Electrical energy is used to delivery heat in calorimetry applications In electrostatics the units of energy are Joules Coulombs Volts J CV Coulomb is a unit of charge and volt is a unit of potential A charge moving through a potential is a little IlKe a waterran The problem here is that we need a dynamic description since moving charge is not static C A note on using current flow in calorimetry measurements As charge moves through the potential energy heat is Released Power Amperes Volts W IV Power has units of energy per unit time So as the current flows through a wire at a certain rate heat energy is added to the system at that rate C A note on using current flow in calorimetry measurements As charge moves through the potential energy heat is Released Power Amperes Volts W IV Power has units of energy per unit time So as the current flows through a wire at a certain rate heat energy is added to the system at that rate C O A note on using current flow in calorimetry measurements As charge moves through the potential energy heat is Released Power Amperes Volts W IV Power has units of energy per unit time So as the current flows through a wire at a certain rate heat energy is added to the system at that rate C o A note on using current flow in calorimetry measurements As charge moves through the potential energy heat is Released Power Amperes Volts W IV Power has units of energy per unit time So as the current flows through a wire at a certain rate heat energy is added to the system at that rate A note on using current flow in calorimetry measurements As charge moves through the potential energy heat is Released Power Amperes Volts W IV Power has units of energy per unit time So as the current flows through a wire at a certain rate heat energy is added to the system at that rate A note on using current flow in calorimetry measurements Over a period of me c M energy adwd w c arm is Energy Amperes Volts time I M This equation is used in the text to describe a number of calorimetry applications The standard state The standard state of a substance is its pure form a 1 bar of pressure The standard state can have any temperature but the temperature should be specified For example the standard enthalpy of vaporization of water Is the enthalpy of vaporization at 373 K the boiling point and 1 bar of pressure Biological implications Obviously the food we eat release heat in our bodies This heat is used both to maintam body temperature and no processes that build up our bodies anabolic processes We often talk about calories in the food we eat The calorie on a cereal box is equal to 1000 calories in the chemical nomenclature 1 calorie 4184 Joules Therefore the Intake required for an average man of 12 MJday is about 3000 calories per day in the sense of diet Thermodynamic cycles The energies and enthalpies of ionic solids are dominated b Coulombic interactions The lattice enthal can be calculated from Coulombic interactions Using the calculated lattice enthalpy and experimental data one can obtain the enthalpy of formation of an ionic cullu by means of the BornHaber cycle This is illustrated on the next slide for a generic salt of a monovalent ion MX39 The reactions needed to obtain MX39 in the solid phase to M X39 in the va or hase are Viven below The overall process of separating the ions in an ionic solid into the constituent ions is written as VX39 S 9 V g X39 g Lattice enthalpy The BornHaber cycle The thermodynamic cycle is illustrated The scheme shows that if one knows all of the energies except one diffiCUlL LU measure quantity one can calculate it using Hess law Lattice Enthalpy MX39 gt M X39 Ionization Enthalpy Potential Election of Formation Afflnlty Enthalpy of Vaporization 12 X2 g M s gt M 3 Bond Dissociation gt Xg The BornHaber cycle This reaction is composed of the following steps MX39 s 9 M s 12 X2 g Dissociation enthalpy M S 9 M g IntnaIpy or vaporization 12 X2 g 9 X g Bond dissociation M 9 M e39 Ionization potential X e39 9 X39 Electron affinity Thus we can express the above in terms of a thermodynamic cycw hat allows us to dc e dissociation energy using known experimental quantities This is necessary for ionic solids because it is essentially impossible to directly measure the dissociation enthalpy The dissociation enthalpy is equal to the enthalpy of formation but opposite in sign The above cycle and the reasoning applied will allow us to determine with some the forces between ions due to Coulombic interactions Dependence of internal energy on cnanges In v0Iume and temperature We have seen from the first law that the internal energy change in the system l5 equal LU LIIU vvum uune dnu the heat transferred Work depends on volume changes and heat transfer leads to changes in temperature Thus it is logical that the initial description of internal energy depends on volume and temperature Thus for example the depend ence of U on volume at constant temperature is l 8U U U aVTclv And similarly the dependence of U on temperature at constant volume is r E U U8TVdT Dependence of internal energy on cnanges In v0Iume and temperature Putting these together we have 39 w w u UlavldvlarldT which can be written as the total derivative w w dUlavl wlarlvdt This called the total derivative The idea is that it includes the variables in a space Here the space is VT space We shall see that these are not the only possible variables that can describe the dependence 0 u Dependence of internal energy on cnanges In v0Iume and temperature The I artial derivatives are slo es in the s ace w w MT V arlv CV We have already seen CV and we know it as the heat capacity At constant volume However nv is new nv is the change in the internal energy when the volume of a substance is changed at constant temperature lfthe intermolecular interactions are zero ideal gas then nv will be zero However real interactions between molecules can give rise to nonzero xv Joules set out to measure nv in an expansion experiment but was not successful Expansion coefficient and isothermal compressibility In order to I roceed with the next level of analsis we intro duce the expansion coefficient 1 0 vaTp The expansion coefficient is a measure of the change in molar volume ie also inverse density of a material with temperature Many substances expand as the temperature is increased hence the name expansion coefficient The isothermal com ressibilit39 is 1 6V K vaPT a measure of the degree to which a substance wi news a higher density smaller molar volume at high pressure Heat ca aoit relationshi s The heat ca acit at constant I ressure and constant volume are related by GP CV The above expression applies only to an ideal gas We have seen the utility of that expression in that it applies to monatomic diatomic and polyatomic gasses For liquids and solids we have the formula 2 CID CV The JoulesThompson coefficient The differential of the enthalpy can be expressed as dH HcPdP CPdT Where p is the JoulesThomson coefficient The enthalpy change be written dH 6Ppo 6T dT 0T 6PT 6PH0Tp CP The derivation of the above expression relies on the permutation relation S zllS le flsi Therefore Modern measurement of the JOUIesThompson coerrlcient It is relatively difficult to perform measurements under constant H isenthalpic or adiabauc COIIuiLiUIIS ereoe the modern measurements are at constant temperature In other words 6PT T is measured Based on the expression on the previous slide aT w T 6PH0Tp CF Chemistry 433 Lecture 24 The Chemistry of Climate Change CO2 Equilbria NC State University The short term carbon cycle Photosynthesis removes carbon atoms from the atmosphere and turns them into food for green plants This process requires energy derived from the sun s radiation At night when plants turn off their photosynthesis and undergo respiration carbon is replacw wok nto he air lw en the plant dies and begins to decay much of the trapped carbon within the plant is released back into the atmosphere There is a small percentage that remains as fixed carbon Even on a time scale of years this cycle is mostly reversible meaning that at steady state there is little carbon fixed by these processes They are important however and if the biomass on earth decreases eg defo te no capture of 002 will decrease Carbon Cycle Calcium carbonate in sedimentary rock Roughly 75 of the carbon injected into the atmosphere by nonorgan means usually volcanoes was no way quotno deposits of calcium carbonate limestone These deposits make up the largest reservoir in the carbon cycle Limestone is formed from bicarbonate HCO ions weathered and dissolved in the ocean The ions along with the skeletal remains of marine life accumulate on the ocean floor Limestone formation involves a series of chemical reactions all of which have a net effect of removing carbon dioxide from the atmosphere Weathering of limestone deposits by rain tends to return carbon atoms to shortterm reservoirs thereby replenishing the concentration of atmospheric carbon dIOXIde 602 release Weathering Midocean ridge Carbonate metamorphism Copyright 2004 Pearson Prentice Hall Inc The CarbonateSilicate Geochemical Cycle CaCO3 SiO2 CaSiO3 CO2 General Considerations Fixation of CO2 in organic form requires an energy source In biology that energy source is the sun s radiation The formation of CaCO3 is thermodynamically favorable However the process is slow The calcium content of seawater is about 380 mgL The solubility product of calcium ca e KSID Ca2CO321 5 x 10399 at 298 K The concentration of CO3239 depends on pH because of the two acid equilibria co2 H20 9 HCO339 H 9 co3z 2H The Balance between the Atmosphere and the Ocean The most common method of CO sequestration CO2 Atmosphere Concentration 1945 270ppm 2005 350ppm 2050 550ppm How much of the carbon ends up on the ocean floor a Chemical Equilibria and the Solubility of Carbon Dioxide A Chemist s view the ideal ocean We can define an ideal or inorganic ocean as follows a body of water in contact with gas eous carbon dioxide and containing dissolved strong electrolytes dissolved carbon dioxide bicarbonate ions and carbonate ions all at equilibrium Note that this ocean does not include any biological activity so it can serv only to provide the inorganic background for the realocean Procedure for treating coupled equilibria First step List the species We consider the process of carbon dioxide in the atmosphere in equilibrium with aqueous carbon dioxide which can form carbonic acid Carbonic acid has two dissociations First we create a list of chemical species present in the system at equilibrium We will use 2 11 JUS 112U1 mg Hcogltaqgt Step 2 create a SpeciesbyElement Matrix ureate a matrix Wlth specnes along the top and elements along the side H co2 H20 C02aq C02g Hco H 1 0 2 0 0 1 C 0 1 0 1 1 1 O 0 3 1 2 2 3 Each entry in the matrix corresponds to the elemental composition of each species The original organization is arbitrary but in the end you will want to have a diagonal unit matrix row reduction Row Reduction Procedure The row reduction process can be described as follows 1 Any row can be multiplied by any number positive negative integral a fraction Any row can be added to any other before or after multiplication so if you multiply by 1 this means you can add or subtract any row from any other You can repeat these operations as many times as you wish You can interchange any two rows the species heading the row moves with the row This is the final form that we want to obtain H CO2 H20 C02aq C02g HCO H 1 0 0 2 2 1 CO2 0 1 0 1 1 1 H20 0 0 1 1 1 0 This matrix is in echelon form Note that the species have changed in the left hand column We will go through the steps used in this case Row reduction procedure Multiply the carbon row by 3 H C02 H20 C02aq C02g HCO H 1 0 2 0 0 1 C O 3 O 3 3 3 O 0 3 1 2 2 3 Row reduction procedure Add the C row to the 0 row H C02 H20 C02aq C02g HCO H 1 0 2 0 0 1 30320 1 0 1 1 1 0 U U 1 1 1 U Row reduction procedure Multiply the 0 row by 2 H C02 H20 C02aq C02g HCO H 1 0 2 0 0 1 30320 1 0 1 1 1 0 U U 394 A A u Row reduction procedure Add the 0 row to the H row H C02 H20 C02aq C02g HCO H 1 o o 2 2 0 30320 1 0 1 1 1 U U 1 1 1 U Step 3 Row Reduce the Matrix to Echelon Form H co2 H20 H 1 0 0 co2 0 1 0 1go 0 0 1 C02 aQ 2 1 1 C02g 2 1 1 Hco 1 1 0 The species forming the unit matrix red square are the independent species We can form independent net reactions from the columns blue rectangles The Chemical Significance of the Row Reduction In the original speciesbyelement matrix each column is a vector in a space in which the elements are basis vectors Row reduction changes the basis vectors to species in the s stem at e uilibrium ie each column ex resses the species labeling the column as a combination of s ecies in the s stem ie each nontrivial column is a net reaction The process of row reduction assures that the column vectors are independent Each column expresses each species as a combination of other species C02aq 2H aq CO 2aq H200 C02 g 2HaQ CO aq H200 HCOE aq ng aq Haq These chemical reactions are a complete set of independent net reactions As such they provide a basis for a complete consideration of the equilibrium state They may be added and subtracted with the cancellation of species to obtain alternative sets of independent net reactions but each set will contain ewe unee et reactions and wi pwvme a basis for the same consideration of the equilibrium state An alternative set of independent net reactions The following are linear combinations from the set found by rowreduction C02aq C02g CO2 aq H200 Haq HCO aq HCOE aq ng aq Haq For each independent net reaction there is an equilibrium constant In the Henry s Law constant the pressure is in bar activities we M Ma w molalities K2 m rnCOZ 45gtlt1O7 In Hcog H 11 K3 mCngmH mHC03 56gtlt10 Add any constraints to the system In the study of the ocean we can take the concentration of hydrogen ions and the partial pressure of carbon dioxide as known for a given State or tne atmospnere and tne ocean based on the known pH and partial pressure of 002 H63x10 9 PCO238gtlt104bar we can calculate the expected value of other species from the equilibria We do not need a constraint for H20 since it does not appear in the equilibrium constants Solve for concentrations We therefore have three equations snd three unknowns and we can solve for the molalitiw v t cabon WWW bearing species 002 380 x 104 30 127 x105 HCO339 45 x 107 002 H 907 x 104 003239 56 x 103911 HCO339 Ht 806 x 106 We also know that Ht OH39 1014 Charge balance condition In order to consider other states of the system it is necessary to consider charge conservation The hydrogen ion concentration is far below the concentrations of the negative ions There is no charge balance without additional cations We therefore determine the molality of cations required to balance the charges due to 003239 aq HCO339 aq and H aq cb OH39 HCO339 2CO32 H 92x104 The majority of these compensating ions are calcium These ions enter the ocean through riverine fluxes ie weathering Determining total amounts of 002 The total mass of the earth s oceans is mocean quotV quot quot3921 kg Natmeq represents the number of moles of 002 from the atmosphere that have dissolved in the ocean Natmeq 127 x 105 moa13 x 1021 kg 165 x 1016 moles The number of moles of 002 in the atmosphere is Force AreaPressure 515 x 1014 m2105 Pa natmCO2 XCO2FOFC6 N g ms2 x MW kgmo MW is the mol weight of the atmosphere 0029 kgmol nath02 38 x103945 x 10191032998 67 x 1016 kg Or natmcozl Nair aq 025 The ocean is saturated in CaCO3 Now to return to the question of the ocean as a sivnificant sink for anthro ovenic carbon dioxide When the species were listed CaCO3s was not included In order for it to form the system must be saturated with respect to its formation It is reported that the surface of the ocean is 232 x 103 molal in Ca2 The solubility product for calcium carbonate 492 X 09 0821003 1 Ksp in the ocean the ion product is 232 x103 x 80 x106 185 X 108 which is greater than the solubility product ie for the simple system considered ere the solution is supersaturated Saturation and States of Disequilibrium The concentration of the CaCO3 is not necessarily in e uiibrium In other words there can be an excess or deficit of Ca2 and 003239 in the solution above the CaCO3 solid lfthere is an excess of ions in solution the solution is supersaturated were s a dew we te solution is subsaturated KSlo Ca2CO3239 lfthe system is not in equilbrium then there will be a driving force AG to attain equilibrium AG AGO RTan The three possible states are AG gt 0 Q Ca2CO3239 gt K Supersaturated AV 7 c u 7 Loa2CO3239 K Saturated AG lt 0 Q Ca2CO3239 lt K Subsaturated The ocean as a 002 sink So could CaCOgs be a significant sink for anthro ovenic carbon dioxide It has been observed that there are no large deposits of CaCOgs on the bottom of the deep ocean so the answer is probably no There are several possible explanations for this and all possibly contribute to failure of the precipitate to form or its rediss deep ocean 1 002 redissolves in the deep ocean 2 Ionic activities are reduced high salt concentration 3 Surface free energy presents a barrier to precipitation Pressuredependent equilibrium There are large pressures up to nearly 1 kbar in the deep ocean One scenario is that calcium carbonate forms at the surface and sinks to depths where it redissolves because of effect of pressure on the dissolution reaction CaCO3 gt Ca2 C032 AV 62 cm3mol molar volume of reaction The molar volume water of solvation is less around the ion that around the solid or the bulk For exam le at 8000 m P 800 atm lnKPK AV PR lnKPK 0062 L800 atm008206298 K 20 K800 atm e UK e U49z x 109 56 x 108 Millero in Geochimica et Cosmochimica Acta 59 661 1995 Sedimentation and the Snow Line The formation of calcium carbonate as a sediment in the oceans has bee occurring for billio v w h process leads to the formation of limestone sedimenary rock One can imagine CaCO3 forming white particles and settling to the bottom This is like snow in the ocean since the particles build up a layer on the ocean floor However in the deep ocean the pressure shifts the equilibrium so that this snow melts before it reaches the bottom Below about 5000 m there is no limestone on the ocean floor This part of the deep ocean is known as the abyssal plan Ocean acidification Furthermore if calcium carbonate were precipitating in an inorganic ocean the carbonate forming reactions below would be drawn to the right constantly forming hydrogen ions and decag e HCO 2 co2 H C02aq H20 2 co2 2H unless there were some mechanism consuming hydrogen ions at the surface Thus the pH of the oceans is falling in the short term The oceans are becoming more acidic Carbon recycling in the ocean We have seen that the equilib o xpi the large quantity ca 40 times the amount in air of carbon dioxide in the sea However living organisms in the sea die and fall toward t dvpths As y as so they are oxidized and produce carbon dioxide and their skeletal remains often contain calcium carbonate At great depths the disso c dioxide has over time become significantly super saturated relative to the atmosphere because the upwelling required for equilibration c wow cgug a buildup of carbon dioxide at depth before a steady state is reached Ocean Carbon Recycling At depth JaL U3 s C02aq H200 gt 2HCO aq the surface C02aq gt C02g HCOE aq gt C02 2 Haq C02 aq 2H aq gt C02 g H200 Where does humanmade 002 go The ocean takes u onl 25 of the carbon dioxide output by humans As a result the pH of the ocean is decreasing The pH of the ocean has decreased from 82 to 81 UI tIIU iIIUUOLrIClI IUVUIULIUII On a longer time scale 002 is fixed in the ocean as CaCO3 However CaCO3 precipitation is slow and the ocean supersaturated In caICIum carponate Chemistty 433 Lecture 15 Applications of Free Energy NC State University Thermodynamics of glycolysis PeaotTon kJmoi DrgTucose ATP eDegTuooseesephosphate ADP AG 7T6 7 DrgTucoser rphosphate eDeTmotoseesephosphate on T 7 Defmotosee seoTphosphate ATP gt DeTmotoseeT Brdiphosphate ADP a 7T4 2 DefructoserT Brdiphosphate egiyoeraioehyoeesephosphate hate AG giyoeraioehyoeesephosphate phosphate NAD gt phosphogTycerate NADT T My 5 3 ADP gt sephosphogiyoerate ATP AG phosphogTycerate AG pymyate NADT T h gt Taotate NAD u lt E 939 a m o g a 1 a I lt 1 a co2 acetaTdehyde NADT T h gt ethanoT NAD Phosphorylation of glucose DrgTucose ATP eDegTuooseesephosphate ADP AG 7T6 7 The reaction can be decomposed into two reactions DrgTucose phosphate sDegTucoseesephosphate H20 Asa MA 3 ATTgt H20 gtADT3gt phosphate AG 73T 0 The sum ofthe two reactions resuTts Tn an oyeraii negatTye free energy change uhderstahdard conditions in this manner the stroh T spontaneous hydroTysis ofATP Ts coupTed to the otheTWTse unspohtaheous gTucose phosphoryTatioh ThTs reaction Ts typTcaT ofthe roTe pTayed oyATP Ththe ceTT Note that the vaTues TOTAG assume a concentration oTT M CTearTy the concentrations Th the ceTT are oiteh quTte different from the standard state and thTs WTTT haye profound consequences forthe dTT eCtTOT T of spontaneous hge The role of enzymes ATT of the reactions Th the gTycoTytic pathway are cataTyzed by enzymes Porexarnpie the reaction considered on the previous sTTde Ts oataiyzeo o kinetics ofthe reaction but not the thermodynamics eWTTT consider he T39OTe ofcataTysts in the second hanofthe course Notice that A6 for certain steps Ts posTtTye Porexarnpie DrgTucoser rphosphate eDeTmotoseesephosphate AG M 7 Ts oataiyzeo by phosphogiucose Tsomerase The equiTibhum constant forthis process Ts Tlt expteraPTT expteT7008 3T3TOT 0 5 The concentration ofDrfruc oser rphosphate at equiTibrTum WTTT be Tess than that of DrgTucoser rphosphate Question oTyen that K 0 5 forthe reaction DrgTucoser rphosphate eDeTmotoseesephosphate n a t uhdei stai idai d conditions T e forTi iTtTaT concentrations ofT M A T M B 0 5 M C 0 667 M DOT5M Question oTyen that K 0 5 forthe reaction DrgTucoser rphosphate eDeTmotoseesphosphate n a 0 under standard conditions T e forTT iTtTaT concentrations ofT M A TM B 05M K 5 C0667M X 1K 1 0333 D gluc G phos1 x133 D 0T5M D fruc G phos1x 67 Intracellular conditions are not equilibrium conditions e sulesedueht steo lh a selles etleaetlehs ls hlghly soehtaheeus ltth thlswllltend tel deplete the oleduet telthe oleyleus lesetleh Thus mule etthe oleduelWllltehd tel be telmed by L Chatellers ollhelole We can observe thls duahtltatlyely by eehsldellhe the yalue ell o the leselleh duetleht SinEE A G AG R T o the yalue utAG ma hell he zele lh uthekuvds the eeuoled selles h eh but ale oelsed set atthe eyelall etetteet eh a selles etleaetlehs s tel oleduee a het soehtsheeus change Sample Problem in Metabolism The ehzyme aldulase estalyzesthe eehyelsleh ettluetese lErdlphusphate FDP tel dlhydruxyactune ohesohate DHAF and glyceraldehyder rphusphuate GAP UndevphysluluglcalBundltluns the eeheehtlatlehs etihese oeeles lh led bland eells elythleeyles ale FDP35 W DHAP13D W and GAP15 uh Wlllthe eehyelsleh eeeul soehtsheeusly uhdelthese eehdltlehs7 Sulutlun The standard tlee Enevgy change telthe lesetleh ls FDPgt DHAP GAP AGD 233 id and DHAPGAPFDP13EI y laws y l jj y ltd 57 y m6 c W lh SEED Jmul a al Jmellqolu lows 57 y W 434 Jmul EIle 43 ldmul The lesetlethll eeeul soehtsheeusly uhdelthe eehdltlehs etthe eell Protein folding example Two state model r 39 F b ku Unfolded Folded K FlU K Fraction folded ff Fraction unfolded 1ff Thermodynamic model for twostate equilibria K ff1ff ff KK1 K erAGDRT ff 1 eAGDRT 1 eAHDRTeVASDR Theternperature at yyhleh the protelrl ls 5m rolded or DNA l5 SEWn hyhrldlzed Carl be deflrled asTm the melt temperature Tm AG 0 or T AH As Equl ibrium melt curves Protellls or DNA H S 39 Dkllmul ZDDJanl x 39 Mostly folded lylostly rlyorlolzeo Fraction Folded ZED 250 Jim 20 Jon Tehlpelzlllle In this case Tm 300 K AHDIAS Van t Hoff plots Is Slope AH R 7 Fli Data 2 u 4 as m lxlo K 1 The standard method for obtaining the reaction enthalpy is a plot of In Kvs 1T Thermodynamics of DNA hybridization A combination ofspectroscopy and calorimetry was used to determine the free energies ofmelting ofshort oligo nucleotides Based on these measurements the he energy ofa helix can be determined based on10 sets of nearestneighbor pairs shown on the next slide In addition to these values we need to know the he energy ofthe initiation ie the rst base pair The overall free energy is then calculated from AG AGEl initiation ZAG nearest neighbors AG AHEl AS 5 7AA 5 7AA 7 3 en 3 T gt e79 7381 einia T T 57A 5eAr 37T aaiTA 753 eaan 7997 A A 5 7 577A 3 A aaiAT 738 7251 e7i5 T c 5 A 5eAc 3 T aaiTG e54 7272 7732 o A 57c 5ecA 376 aaiGT e79 7243 e55n T Breslauer et al PNAS 93 3746 1986 6 AG AHEl AS 5 7A 5 er LT gt 37m 757 7325 7359 A 576 576A 37c eLCT 757 7234 53a T o 57c 57cc 376 gt 76c ei5i 7498 eiim c c 576 5eoc 37c e346 eiau 7454 eiizi c 57c 5ecA 376 aaiGT eiau emu eiim Breslauer et al PNAS 93 3746 1986 Sample problem Determine the melt temperature for the oligonucleotide 5 ATAGCA3 gt 5 ATAGCA 3 39 3 TATCGT 5 3 TATCGT 5 Solution AG AG initiation EAG neare5t neighbors a oAT uTA 9A6 use 00A AGGCIniIAGTAAGATAGTCAGCGAGGT 209 763 738 767 7130 769 Note that We use GC initiation iftnere is a Single GC base pair Only uSeAT initiation oftne Strands are all Aand T Sample problem Determine the melt temperature for the oligonucleotide 5 ATAGCA 3 5 ATAGCA 3 3 TATCGT 5 3 TATCGT 5 Solution conto AG 46 8 w Notice that tne tne free energy orinitiation i5 positive initiation i5 tne cnaln together To calculate the melt temperature We need the enthalpy ofreactiona Well AT TA AG sc CA AH AH TA AH AT AH TC AH CG AH G 7360 7251 7326 746 Z 4644KJ Sample problem Given that AG 168 kJmol and AHEl 1644 kJmol forthe hexamer determine the melt temperature C B 48 C C 52 C D 58 C Sample problem Given that AG 168 kJmol and AH 1644 kJmol for the hexamer determine the melt temperature 42 39C B 48 ElC C 52 DC AS AHD AGE 164416Bl298 x1000 39 u 4953 JmuIK D 58 C The melt temperature occurs when AG 0 T AHE39IASEl 1644004953 331 K 58 C Noncovalent forces in proteins What holds them together Hydrogen bonds mm Saltbridges quot In quot Dipoledipole interactions Mfk Hydrophobic effect M 391 Van der Waals forces quotM L What pulls them apart Conformational Entropy DipoleDipole Interactions Dipoles often line up in this manner Example oc helix Electrostatic Interactions Coulomb s Law V q1q2sr Example of a hydrogen bond NH39OC Example of a Salt Bridge Main Chain Main Chain Glutamate Hydrogen bonding in water Hydrophobic interactions Contributions to AG o Sample problem The thermal unfolding ofa serine protease SP from a thermophilic bacterium was studied and gave the following Thermodynamic data AG 154 kJmol and AHEl 750 kJmol forthe process SP random coil gtSP native Determine the melt temperature A 82 ElC B 92 ElC C 102 C D 112 Sample problem The thermal unfolding ofa serine protease SPfrom thermophilic bacterium was studied and gave the following Thermodynamic data AG 154 kJmol and AHEl 750 kJmol forthe process P random coil 439SP native Determine the melt temperature A 82 ElC AS Al l r Aeaw 775 0 l5 41298 x1000 B 92 C 7200 JrnoHK C 102 C The melttemperature occurs when AG 0 D 112 ElC AS 7750007200 375 K 102 Cl Kinetics Catalysis I Enzymes Catalysis involves lowering of the energy barrier Reactants Products Energy AHjF Nuclear Coordinate A Catalyst pIUVIucS an CIILCI I Iatlvc rcauuun pat Ivvay with a lower activation energy or activation enthalpy Types of catalysis Homogeneous catalysis the catalyst s u the same phase as the reactants Exam le acid or base catalsis Heterogeneous catalysis the catalyst is in a different phase from the reactants Example metal complexes surfaces zeolites Enzymatic catalysis the catalyst is a protein that has a substrate binding site and controlled reaciv path Zeolites an Important Class or catalysts Database of zeolite structures httpwwwizastructureorgdatabases Example search for ZSMS Unit cell parameters a 20090A b 197383 c 13142 1 alpha 900 beta 900 gamma 900 volume 521128 A3 Basis for heterogeneous catalysis in zeolites Zeolites are crystalline solids made up of SiO4 building blocks These tetrahedral units join together to form several different ring and cage structures The characteristic that separates zeolites from allsilica minerals is the substitution of aluminum into the crystalline framework The substitution of aluminum generates a charge imbalance which is compensated by a proton The acid site formed behaves as a classic Bronsted acid or protw Watg uciu sue The highly acuc sues combined with the high selectivity arising from shape selectivity and large internal surface area makes the zeolite an ideal industrial catalyst H Zeolites shape selective catalysis The alkylation of benzene with propylene is an important petrochemical process because the product cumene is a chemical intermediate used to synthesize phenol and acetone Classical industr p cwsses are based on quotolid phosphoric acidquot catalysts with problems of handling safety corrosion and waste disposal These can be avoided by using zeolite catalysts Zeolites shape selective catalysis The medium pore size zeolite ERBl has greater reactivity fv cumene formation than larger pore size catalysts Zeolites shape selective catalysis Calculated energy sur39race ror cumene in BEA Diffusion of cumene in zeolite BEA ZieglerNatta catalyst for pOIymerization of ethylene ZieglerNatta catalysts are an important class of mixtures of chemical compounds remarkable for their ability to effect the polymerization of olefins hydrocarbons containing a double ca oncarbon bond to polymers of high molecular weights and highly ordered stereoregular structures These catalysts were originated in the 19505 by the German chemist kari Ziegler tor the pOIymerIzatIon of ethylene at atmospheric pressure Ziegler employed a catalyst consisting of a mixture of titanium tetrachloride and an alkyl derivative of aluminum Giulio Natta an Italian chemist extended the method to other olefins and developed w m m of the Ziegler catalyst based on his findings on the mechanism of the polymerization reaction ZieglerNatta catalyst for pOIymerization of ethylene TiCl3 can arrange itself into a number of crystal structures The one that we39re interested in is called ocTiCI3 It looks something like this Cl I Cl Cl I C ll 1L I C I C I l I v Cllllh 39 1L J CI I l I CI CI CldCl I Illn T CI IIIum T CI Clln CI I I I ClCll ClCl IllI I 9 T CI I C ITi u ITi CI CI cClCl Cl Cl Cl As we can see each titaniumatom is coordinated to six chlorine atoms with octahedral geometry ZieglerNatta catalyst for pOIymerization of ethylene At the surface of the crystal a titanium atom is surrounded on one side by five chlorine atoms but on the other side by empty space This leaves titanium a chlorine short Titanium as one of the transition metals has six em t orbitals resultin from one 45 and five 3dorbitals C39 13 in the outermost electron shells All The surface Ti atom has an CI 3 empty orbital shown as an GIL I NC em 5 uare in the icture quotTiquot DtY q I3 Cll Cl Cl ZieglerNatta catalyst for pOIymerization of ethylene Titanium wants to fill its orbitals But first AC2H52C enters the picture 1t UUI lclLCb Ul Ie Ul iLb eLI IyI gluupS LU the impoverished titanium but kicks out one of the chlorines in the process We still have an empty orbital In El Cl H3CHZCLClCH3 D l ID ClTIICI HscHZC CHchs Cl Cl ZieglerNatta catalyst for pOIymerization of ethylene The the aluminum is coordinated though not covalently bonded to the CH2 carbon atom of the ethyl group it just donated to the tchll quotum and to one of the chlorine atoms adjacent to the titanium Cl There is still a vacant site H3CH2C39quot CH3 where Cl CH2 D polymerization x can occur IllIll Ti39 Cl Cl CI ZieglerNatta catalyst for pOIymerization of ethylene This process forms the active polymerization catalyst which happens to be insoluble unlike the 2 components that make up the complex so we have what is commonly termed a heterogeneous catalyst also known as a solid solution Cl H CH I CH 3 H3CH2C 39NN 3 Cl CH Illh x u T Cl lclquot H CI 6 l C ZieglerNatta catalyst for pOIymerization of ethylene Upon binding ethylene forms bonds with the Ti atom and the carbon of the ethylene ligand CI I H CH3 CH c H3CH2C lC CI H C ZieglerNatta catalyst for pOIymerization of ethylene The growing polymer chain is initiated CH3 H2C C39 H CH I C 3 H3CH2C l 39339 c llITi ZieglerNatta catalyst for pOIymerization of ethylene The vacant site is CH3 available for the H20 nexr ethylene molecule to H CH3 bind Enzymatic Catalysis Michaelis Menton Kinetics Alcohol Deh droenase Serine Proteases MichaelisMenton kinetics The rate of an enzyme catalyzed reacuon in which substrate S is converted into products P depends on the concentration of the enzyme E ev though the enzyme does not undergo any net change ka kb ES lt gtES gtPE k a MichaelisMenton rate equa ons ka kb E S lt gt ES gt P E k kaE5 k55 ka55 ka39ES kit 5 Steps in the MichaelisMenton mechanism Step 1 Bimolecular formation of the enzyme E and and substrate S E S ES rate of formation of ES kaES Step 2 Unimolecular decomposition of the complex ES E S ucc of decomposition of ES ka ES Step 3 Formation of products and release from the enzyme ES P E rate of formation of P kbES The rate law of interest is the formation of the product in terms of E and S The enzyme substrate complex can be eliminated The enzyme substrate complex is formed transiently and can be approximated using the steady state approximation kaE5 ka3955 kitEs z o The result of this approximation is kaE5 ES 39 k kb Pseudofirst order MichaelisMenton kinetics In an experiment we know the total enzyme concentration EO and not the unbound enz me E The total concentration of enzyme EO E ES k 50 E5 5 ES a k39 k leallallgca to a b kaEo5 ES ka39 kb ka MichaelisMenton parameters The rate of formation of product can be written dP kb5 7 EO Where k m where KM is the Michaelis constant and kb is the maximum turnover number The Michaelis constant is k3 kb K 77 ka Limiting conditions of enzyme reactivity Maximal rate If there is excess substrate present the rate is limited by the rate at which the ES complex raIIs apart Ine rate or rormatlon or prooucts IS a maximum and Vmax kbE0 is called the maximum velocity 0 Second order regime If S ltlt KM then the rate of formation of products is dPdt kbKM EOS The rate depends on S as well as Eo A plot of 1k yields kb and KM but not the rate constants ka and Ka39 Ine latter rate consrants can pe Obtained from stoppedflow experiments Catalytic rate constant Kcat Catalytic rate The rate constant kb is also called the catalytic rate constant or kcat Since this is an intrinsic rate constant it is not an observable However it can be calculated from observables kcat VmaXE0 Turnover number When the enzyme is saturated ie when S gtgt KM then kcat is also called the turnover number Integrated form of the MichaelisMenten equation 3 KMIn Vmaxt OI 310 t Kmln 310 S Vmax KMIn 2 23912 V maX The halftime 112 is the time at which half of the original substrate is consumed General expression for reaction velocity Based on the previous analysis the velocity at an arbitrary substrate concentration is 1 039 39 39 LineweaverBurke Plots The MichaelisMenton expression is nonlinear The LineweaverBurke plot is linearized plot of data V Vmax Vmax Vmax o This expression has the norm of an equation no a line y intercept slope x 0 Such plots are not necessary today with common nonlinear fitting programs Inverse Reaction Rate vmaXv LineweaverBurke Plots 400 JO 0 O 200 100 Transition State Stabilization The original idea of the enzyme having maximum complementarity to the TS was put forward by Linus Pauling in 1946 It wasn39t until the early 7039s that the idea was put on a more solid grounding As put forward by Lienhard and Wolfenden the idea is as s KniF kn ES ESF gt EP I Ks I Kt C k C Ki ES 4 ES gt EP Transition State Stabilization Defining the egulb u w sw w w wwwtiw w swnts W SHS Kt ESES from TS theory AG RT ln Kf and kobs kBTheAGRT Thus kn kBThKnat and KC KBThKC4E where c means catalyzed and n means uncatalyzed From the scheme you can see that K5 K3 Knit Kt hence KtKs KeirKniE however kCkn KeirKniE Therefore the observed mc c u cce t kCkn KtKs gtgt 1 Therefore the transition state geometry Si must bind more tightly than the substrate S in its equilibrium geometry Transition State Analogs The transition state stabilization hypothesis was tested by designing socalled transition state analogs molecules which mimick the real TS as closely as possible One of the first enzymes examined quotas p oli c accase inc 0 O H H C00 30039 l l l H H H The compound on the right is a planar TS state analog This molecule quotas found to be a good inhibitor with Ki some two orders of magnitude smaller than Km The Role of Entropy In a seminal paper Page and Jencks showed that the loss in entropy in going from a bimolecular to a unimolecular reaction ie E S ltgt ES could account for as much as 108 of the observed rate enhancement In other words this m ch free energ w l cme fr m the intrinsic binding energy The entropy loss arises from the loss of translational and rotational degrees of freedom when the SUbStl aLc lb UUUI Iu IIIC configurational entropy IS S kB an where W is the number of degrees of freedom available to a molecule Inhibition An inhibitor is any compound that causes a decrease in the catalytic rate We will consider noncovalent ligands that can bind to the enzyme The general scheme is shown below 5 kc I inhiitr E 4 ES gt E P Inhibition occurs if I I kiEIS lt kCES l s 1 k EIHEIS 39gt EI P competitive inhibition Competitive inhibition results from the direct competition between the I and S for the substrate binding site There is an additional equilibrium constant Elli EIE I K 1 El The VCIULILy under these conditions turns out to be lmax V ocM 5 0 1 K Noncompetitive Inhibition Noncompetitive inhibition arises when I can bind at site that is not the same as the substrate binding site There is an additional equilibrium constant I E E f 1 El Here the complex IE indicates that t c WW woo ot bind in the same site as the substrate The velocity under these conditions is V 5 Vmax 1 ocM5 17 Distinguishing competitive and non competitive inhibition The tradition method is to use LineweaverBurke analysis On a LB plot competitive inhibition will give rise to lines of varying slope but with the same intercept On the other hand noncompetitive inhibition will give rise to lines with ifferent intercer ts Noncompetitive 1 OKMI OLKM 06 V Vmax Vmax Vmax ocKM 5 ocM 1 Vmax Vmax Vmax Competitive l I Reaction Rate vmax Plots of competitive and non competitive inhibition 5 Noncompetitive Competitive Enzymatic catalysis Alcohol dehydrogenase The enzyme alcohol dehydrogenase EL 1111 lb also known as aldehyde reductase This enzyme belongs to the oxidoreductase class of enzymes Alcohol dehydrogenase catalyzes the oxidation of ethanol to acetaldehyde with the reduction of NAD to NADH The alcohol can be in the of a primary secondary cyclic secondary or a hemiacetal to produce these products aldehyde ketone and NADH This reaction occurs during the glycolysis pathway in the mitochondria of animal cells Cofactors include zinc or iron that act on primary and secondary alcohols or hemiacetals Enzymatic catalysis Alcohol dehydrogenase Zinc functions as a Lewis acid it deprotonates the alcohol substrate in order to facilitate hydride transfer The hydride is transferred through space from the OH group of the substrate to the C4 position of the nicotinamide ring Scheme 1 Zn 343 Zn 548 Zn s43 Bow Hwo 439 5gb mH o 7 o quotH70 H c H NADG H CHNADEG H NADH R R R R 037 T 073 p10 Substrate binding site The hydrophobic pocket Val294 Pro295 lie318 Zinc oxygen involved in the dehydrogenation reaction shown in white Application of MichaelisMenten Kinetics In the presence of excess NAD the rate for the reaction CH3CHZOH NAD gt CH3CHO NADH H Can be determined as a function of ethanol CH3CH20H Rate molL M391S391 3025 0007 006 39 39 0015 011 0031 016 0068 021 01 023 02 028 a 04 028 mm Bendinskas et al J Chem Ed 82 1068 2005 Application of MichaelisMenten Kinetics In the presence of excess NAD the rate for the reaction CHscHzOH NAD gt CH3CHO NADH H Can be determined as a function of ethanol CH3CH20H molL 0007 0015 0031 0068 01 02 UJ 04 Rate Mis1 006 011 016 021 023 028 ULU 028 Rate of Ethanol Consumption M434 0 m m l 0 N o I 0 A 01 0 A o I Vmax 03 M391 s1 KM 005 M l I l l 01 02 03 04 Ethanol A Lineweaver Burke plot is a double reciprocal I5L 1CH3CH20H 1Rate mol L M391S391 16 1428 1557 557 V 5 323 525 147 477 100 455 8 50 357 e 33 345 4 39 2 5 3 2390 4390 6390 8390 1CO 1amp0 143910 1Ethanol A Lineweaver Burke plot is a double reciprocal I0L 1CH3CH20H 1Rate M39IISquot 16 1428 1667 Wmax 33 Ms 66 v w 12 KMVmax 009 M 323 625 147 477 100 450 3 50 357 e Vmax 03 M391 3391 33 345 4 KM 003 M 2 5 3 5 7 2390 4390 6390 8390 100 1amp0 110 1Ethanol Serine Proteases Trypsin is one of the three principal digestive proteinases the other two being pepsin and chymotrypsin Trypsin and chymotrypsin are both serine proteases that are quite similar They have a catalytic triad of AspHisSer Trypsin continues the process of digestion begun in the stomach in the small intestine where a slightly alkaline environment about pH 8 promotes its maximal enzymatic activity Trypsin hydrolyzes peptides containing arginine and lysine Ch motr sin h drol zes e tides containin39 t rosine phenylalanine tryptophan methionine and leucine Trypsin is the most discriminating of all the proteolytic enzymes in terms of the restricted number of chemical bonds that it will attack Chemists use trypsin widely as a reagent for the orderly and unambiguous cleavage of such molecules Structural classification serine proteases Class All beta proteins Fold Trypsinlike serine proteases barrel closed n6 S8 greekkey duplication conSIsts or two domains or tne same TOICl Superfamily Trypsinlike serine proteases Families Prokaryotic proteases 9 Eukaryotic proteases 41 Viral proteases 4 beta sheet in the first domain is openeo rather than forms a barrel Viral cysteine protease of trypsin fold 3 Zymogens protease precursors Most proteases are synthesized in an inactive form This form is known as the zymogen A protein cleaveage step is required to active the protease This type of control is important for the transport of enzymes capable of protein degradation quot if r 47 1 Chymotrypsin Chymotrypsinogen with inhibitor Prokaryotic structural examples Trypsin from Streptomyces griseus Mechanistic overview Substrate binding General base catalysis by imidazole to activate the Ser OH Nucleophilic catalysis by Ser OH to form a tetrahedral adduct Stabilization of the tetrahedral transition state by hydrogen bonding to the quotoxyanion holequot and by the electrostatic environment provided in part by Asp102 General acid catalysis of the departure of the leaving group to form the acylenzyme covalent intermediate and departure of the P1 leaving group amine or alcohol Reverse of the above to hydrolyze the acylenzyme Beginning with the imidazole activating wter by general base catalysis facilitate its nucleophilic attack on the carbonyl Se aeiepiseieaeememenism Asp Asp His sz 2 j7 1 SQ CHZ Catalytic triad 39 quot Q N x m 0 quotW ii ill Ml39l l39 llilili k I R Substrate A l polypepiide l 1 3 I H v Telrnhedrul intermediate OACIdbase catalySIS cum le 0Covalent catalySIs quot oProximityorientation 5 effects M 23th 0Also not depicted c113 here electrostatic 3 Eggmmm Nil cl Q 2 am A catalysis transition i H U i H M AcyI enzymeinlermediile J Hlx His g 57 A5 Hr II2 l 5 IL N 095 I T I 39 CH2 ii 0 lx I New Cvlermmus l i olclaaved polypeptide r I i I chain I Telmhedral intermediate Adiquot enzyme warghl 1999 xmni lnny and Snnr inc All right inruwnri Substrate binding to Chymoltpymotrypsin free enzyme 0 A 102 c SP His57 o HN N HO S er195 Hydrophobic pocket o n H H Active snte N N Oxyanion hole GIy 93 Ser39gs Substrate a polypeptide AAn C CH NH C H NH AA R1 u The oxyanion hole The OWaSiei i l emfGEWeases u b I Oxyanion hole Aim Robenus JDquot Kmul J Alden RA and Birkm LL may I i 4502 1572 Copyright 1999 John Wiley and Sons Inc All rights vaunted Role of oxyanion hole in serine protease mechanism OEIectrostatic catalysis OPreferential binding of transition state The catalytic triad The key experiment that elucidates the role of as artate in the As HisSer catalytic triad is the mutation of aspartate 102 to asparagine Since the aspartate residue is essential there has been a great deal of interest in understanding the charge relay hypothesis h39m iquot Q Acetic ACld Imidazole Methanol Asp His Ser Using Density Functional Theory to model the catalytic triad The role of the aspartate can be modeled by determining the charge on oxygen and the potential energy for removal of hydrogen in from the serine oxygen by calculation PES Potential Energy Surface g gCharge on O Systematically change this group to H20 39OH etc Calculated potential energy surfaces for deprotonation of the serine hydroxyl l l 80 CH3OH 2 CHSOHImHZO g 60 CH30H ImCHSCONHCH3 8 CH3OHImCH300239 5 g CH30HImH20 g 40 CHSOHlm lJJ a 2 E 20 D I o l 10 12 14 16 18 O H Bond Distance A Modified MichaelisMenten scheme for serine proteases The appropriate reacti0n scheme Ior a serine proteaSe involves the release of two intermediates ie the N and C terminus of the cleaved peptide k1 k2 k3 ESltgtES gtEA gtEP k1 l P1 no distinguish UCLWEEI IdLCb k2 and k3 one uses esters that form a stable 4coordinate intermediate For these k3 lt k2 See Ferscht Enzyme Kineticsquot Chapter 5 Key points 1 The Arrhenius expression for the rate constant K AeEaRT 2 The assumption of transition state theory TST activated com lex is in e uiibrium with reactants 3 Relationship of TST rate constant and Arrhenius rate const 4 A catalyst lowers the barrier for a reaction It provides an alternative reacwn pathway but does not alter the pwducts 5 Homogeneous vs Heterogeneous catalysis 6 Examples a zeolites shape selective catalysns b ZieglerNatta polymerization catalyst c Alcohol deh dro enase cl Serine protease Chemistry 433 Lecture 11 Second Law Applications NC State University Summary of entropy calculations In the last lecture we derived formula for the calculation of the entropy change as a function of temperature and volume changes These are summarmed m we LaUIc ucluvv Constant temperature AS annV2V1 Constant volume As novlnT2T1 Constant pressure AS nCplnT2T1 Sometimes we only have pressure information and the entropy change can be rewritten as follows P V P1V1 P2V2 gt F 7 AS 7 ln 7 ln Isothermal compression Calculate the entropy for an irreversible compression of oxygen gas The initial pressure of the gas is 1 bar in a volume of 100 L The final pressue u the u the gas us 10 bar and the temperature is 400 K Solution Note that you need to obtain the number of moles The problem does not ask you for a molar entropy Write down the ex ression for the entro AS 17 lnj 7 ln We need either the ratio of volumes or the ratio of pressures We are given the pressures so we can use those P2P1 10 bar1 bar 10 Isothermal compression We obtain the number of moles using the ideal gas law RV1 1O5Pa01 m3 RT 831 JmoK400 K 1 bar100 L 00831 L barmol K400 K 300 moles I Now we can substitute into the entropy expression AS 17 Mg 30 mo831 Jmol Kn 10 249 JK Entropy of mixing When two substances can mix there is a spontaneous tendency for this occur We quantify this using the entropy state function If we consider two containers sep w b a stopcock t N2 gas is in one and Br2 gas is in the other we know from experience that the gases wi mix once the stopcock is opened Entropy of mixing For each gas we can describe the mixing as a volume change The N2 gas is originally on the right side contained in volume V at pressure P After opening the stopcock the available volume is 2V and the partial pressure of P1 x1P The same is true for Br2 lts initial pressure is P2 and final pressure is P2 x2P We treat the entropy as the sum of two expansions ie an expansion for each gas AmixS AexpS l AexpSZ AWXS n1R ln n2R ln P P ArmXS nRx1lnx1 x2lnx2 Note that the mole fraction applies to the final composition Entropy of phase transition The entropy of phase transition can be calculated using the enthalpy and temperature of the transition qPfus AfusH A188 Tm Tm Example using the data in the Table calculate the entropy of val orization forthe following com ounds Compound AvapH Tb Solution Use the following relation kJmol K C6H6 308 3532 A S M C2H6 147 1846 Vap TVap CCI4 300 3500 CH4 818 1117 Br2 297 3324 H28 187 2128 Entropy of phase transition The entropy of phase transition can be calculated using the enthalpy and temperature of the transition qPfus AfusH Ams Tm Tm Example using the data in the Table calculate the entropy of val orization forthe following com ounds Compound AvapH Tb AvapS kJmol K JmoIK Note the similarity in C6H6 308 3532 872 the values for the CzHe 147 1846 796 entropy of vaporization 834 This is known as Br24 295 3324 885 Trouton S rule39 H28 187 2128 877 Conformational entropy The entropy of a polymer or a protein depends on the number of possible conformations This concept was realized first more than 100 years ago by Boltzmann The entropy is proportional to the natural logarithm of the number of possible conformations W S R In W For a polymer W MN where M is the number of possible conformations per monomer and N is the number of monomers For a typical polypeptide chain in the unfolded state M could be a number like 6 where the conformations include different luv angles and side chain angleb un tlle OLhUI lldnu VVIIUII Lhc protein is folded the conformational entropy is reduced to W 1 in the theoretical limit of a uniquely folded structure Thus we can use statistical considerations to estimate the entropy barrier to protein folding Multinomial distribution The total number of ways that a group of N objects can be arranged in M different categories is M The flUJn tLDle dl UUUIIIUlellt dppllcs tU a 5tU dLlUll VVhUlU were are more than two groups In general if there are M different groups that we can place N objects in we have as the total number of ways W MN 111 nm where H is the product operator indicating that there are M terms mUItipllcu together in the denomllldLUl THU ValucS of the nm are the numbers of objects in each of the groups Clearly the nm must sum up to the total number of objects M 2 mm N m 1 Permutation of Letters We have already seen that the permutations of the indices ijk etc gives rise to N different combinations The permutation of indices in the sum over Boltzmann factors lead to the factor of N in the partition function Q qNN for indistinguishable particles In this application we are assuming that all of the indices are uque am therefore the number of ways of arranging them is given by W N 1111 ie there is only one index in each group Suppose we asked how many ways there are of arranging the letters in the word MINIMUM There are three Ms and two ls with a total of seven letters In this case then there are four groups the group of M N l and U The tota numberofways is 7 4 56 7 W 3211 2 420 Application to a game of chance This is discussion is based on a game called YAHTZEE On each turn you role five dice You obtain points based on the configuration of the dice The following point scheme is made in the instructions 50 all five dice have the same number YAHTZEE 30 full house two of and kind with three of kind 40 long straight five sequential numbers 12 three of a kind with any two ot paired We define a configuration as 123456 where 1 is number of ones etc rolled on a particular cast of the dice Consider how many ways there are to get a YAHTZEE of all ones In this case each of the five die must have the same number The configuration is 500000 and the number of ways is given by 55OOOO 1 Application to a game of chance It is obvious that there is only one way to get all ones But what about a full house three of one kind and two of another We must be specific since there are a number of full houses possible how many Let us take two twos and three sixes The configuration is 020003 and the number of ways this can be aChievcd Io uzluzL0003 IO There are two possible long straights The configuration is 111110 and the number of ways this can be achieved is 511110 1205 24 The overall probability of obtaining any configuration is given by W for that configuration compared to the total number of configurations M For YAHTZEE is 65 7776 Therefore the Probability of getting a long straight is 2247776 0006 Problem solving We can identify the following main types of problems that involve entropy change Isothermal expansioncompression reversibleirreversible Temperature change Equilibration Mixing Phase Transition Statistical or Conformational Entropy Adiabatic trick question if q 0 then AS 0 Whenever you are solving an entropy problem remember to consider both system and surroundings The system is always calculated along a reversible path Equilibration We have seen a simple example where there are two metal blocks both made of the same material However this need not be the case For any arbitrary materials 1 and 2 in contact we need to know the initial temperatures and heat capacities to calculate the final temperature 71 Q2 Cp1T1 Teq Cp2T2 Teq Now we solve for the equilibrium temperature Teq Cp1T1 Cp1Teq Cp2T2 Cp2Teq Cp1T1 Cp2T2 Cw Cp2Teq 0pm OWE T 6 quot Cpl1 Cpl2 Equilibration Calculating the entropy Once you have obtained the equilibrium temperature the entropy is easily calculated from Te Te AS1 nCp1 ln A82 nCp2 ln In such problems you are assuming that the two objects in thermal contact are a closed system The overall entropy for heat flow should be positive since heat flow from a hotter to a colder body is a spontaneous process Example The coffee cup problem A hiker uses a amminum WNW wp If the mass of the cup is 4 grams and the ambient temperature ie the cups temperature is 10 C and the hiker pours 50 ml of coffee with a temperature of 90 oC into the cup A Calculate the equilibrium temperature B Calculate the entropy change Gas exchange in a green leaf The cells of the spongy layer are irre ular in shag e and loosely packed Their main function seems to be the temporary storage of sugars and amino acids synthesized in the palisade layer They also aid in the exchange of gases between the leaf and the environment During the day these cells give off oxygen and water vapor to the air spaces that surround them They also pick up carbon dioxide from the air spaces The air spaces are interconnected and open to the outside through pores called stomata singular stoma Entropy 02 mixing in a the stomata of a leaf During photosynthesis O2 is produced in the thylakoid membrane of green leaves The gas meets the atmosphere in the stomata of the leaf Assuming that the following mole fractions exist calculate the molar entropy of mixing Insude x02 in uo and xN2 in uo Outside x52 out 02 and x112 out 08 Assume that the stomata closes with an equal volume of outside air and inside gas present in an enclosed space Solution Assume that the initial state is premixed gas both inside and outside Cal xygen wpound and nvge compound 2 The final composition is xf1 xi1 x012 05 022 035 xf2 rxi2 x022 05 082 065 Entropy 02 mIXIng In a the stomata of a leaf Write down the total entropy change as products final mixed composition minus reactants premixed w om inside wol outside There is an entropy of mixing of each of the two reactants that must be subtracted from the entropy of the final mix Note that there is half as much of each of the reactant gases as the product AStotal ASfinamix ASmixin ASmixout Xofln X01 X02In X02 094 Xi1 Xi2n X12 Since the problem aSKs for me molar entropy we really need AS nal mixn so we can write the formula as Entropy 02 mixing in a the stomata of a leaf Write down the total entropy change as products final mixed composition minus reactants premixed w om inside wol outside There is an entropy of mixing of each of the two reactants that must be subtracted from the entropy of the final mix Note that there is half as much of each of the reactant gases as the product X01 2 X02 2 2 831 JmOl K005 0421 Jmol K AStorem R Xf1n Xf1 szln Xf2 2 ln x01 In X02 Xi1 XI39Z This is a small entropy Part of the reason is that the gases were already mixed If they had been completely unmixed the entropy would have been 54 JmolK Phase Transition Compare the entropy of sublimation of water to the entropy of vaporization at 0 00 Which is larger Why Solution The entropy of a phase transition is given by Athase T phase ASphase From the information in Atkins page 61 we find that AvapH 4507 kJmol and AsubH 5108 kJmol Using these values and the temperature 73 45070 Jmo Asvap 273 K 1651 Jmo K 51080 Jmo Assub 273 K 1871 Jmol K Conformational entropy of a protein Estimate the conformational entropy of myoglobin Myoglobin has 150 residues with 6 possible conformations per residue Assume there is a unique conformation for the folded structure Solution The definition 8 R an is known as the statistical entropy R is the gas constant and W is the number of possible conformations for a strucwre and W M For the unfolded protein W 6150 and S 150R ln6 For the folded protein W 1 and S 0 There conformational entropy is AfoldS AunfoldS and is therefore 2233 Jmol or 22 kJmol The Levinthal Paradox The Levinthal paradox assumes that all of the possible conformations will be sampled with equal probability until the proper one N native is found Thus the funnel surface looks like a hole in a wolf course The I aradox states that if a protein samples all 6quotquot conformations it will take a time longer than the age of the universe to find the native fold N for a polypeptide where M 100 and if it takes 103911 seconds to sample each possible conformation M Conformational ntropy The Pathway Model Imagine that the a unique pathway winds through 1 the surface to the hole gt The path starts atA and 2 the folding goes through 3 a unique and welldefined set of conformational changes Here the entropy must decrease rapidly since M the number of degrees of Conformat39onal gt freedom in the folding pathway is quite Entropy small compared to 6quot On this diagram the configurational entropy is given by the width of the funnel and the rela by the height relative to the bottom folded state The vertical axis is energy NOT free energy Evidence for folding pathways One piece Of evidence IOl folding pathays comes 0le trapping disulfide intermediates This method was pioneered by Creighton using BPTI and has only been used on other proteins as Reduced SD51544 EU51533 SD511433 f t a p 55514 33 Native Creighton et a1 Prog Biophys Mol Biol 33 231 1978 Beyond 2 Kim et al showed that some ofthe previous data and interpretations were wrong The major 2DS species contains the two native DS39s 3051 and 555 The third disulfide is formed quite slowly because it is quite buried It is possible to isolate a stable species with only the first two disulfides formed and the third remaining in the reduced form Studies with a mutant in which the third DS was replaced by 2 Ala and which folded at a similar rate to the wild type support the idea that the trapped disulfide species have partial nativelike structure These observations and others like it can be used to the idea of a pathway into question The folding funnel shown here represents the change gt in energy for a large number of folding paths 3 that lead to the native LIJ 9 configuration There are no energy barriers This implies that all paths have an equal probability leading Conformatlonal gt to the folded state The funnel Entropy shown here has no energy barriers and all paths lead directly to th n tiv tate Th thi funnel i n i t nt with tw t t folding behavior Dill and Chan Nature Struct Biol 1997 41019 Barriers and misfolding The energy surface does not have to be a smooth 1 trajectory Ihere can be gt barriers that will trap 9 intermediate states These 0 states may then be observed UJ to determine aspects of the folding trajectory Whenever an intermediate is observed quot there will be a question as to Conforn lational Whether thlS IS part OT a pathway or whether a funnel description is more applicable Note that the funnel provides the possibility for misfolding This will typically result in multiple minima in the energy landscape Dill and Chan Nature Struct Biol 1997 410 19 Chemistry 433 Lecture 29 Application to kinetics in myoglobin NC State University Photoyss in heme protehs Carb noxide bound to heme can be photolyzed U E Deoxy Heme Excitation by light is a method for initiating ligand dynamics or enzyme intermediates for kinetic studies Teng Srajer Moffat Nature Struct Biol 199417o1 MbCO The peptide backbone is shown as a ribbon that follows the cxhelical structure of myoglobin The structure shown is at equilibrium Conformational substates are called A states MbCO The photoproduct Iron moves out or me neme plane when CO is photolyzed CO moves to a docking site and is parallel to the heme plane Conformational substates are called B states Deoxy Mb The deoxy structure has no CO ligand The protein backbone has shifted to permit a water to enter the distal pocket This IUIIII is often referred to an 5 state Porphne orbitals Nodes in Porphne orbitals Absorption spectra for MbCO and deoxyMb 150 B MbCO g 100 Deoxy Mb 50 Q 0 l l xl 400 450 500 550 600 Wavelength nm Ligand recombination is a sum of single exponentia processes at room temperature SC CIDgee kgemkm CDME kbi HisFePCO k escape HisFeP CO kgeminate 11v kbimolecular Difference spectrum from nanosecond transient absorption spectroscopy HiSFePCO The gemnate and bimoecuar yea St qngee ltkgemkescgtr cpbie kbiz k gem kesc k gem k 980 k gem pgem 980 phi k deoxy Mb The expermenta kmetics are bEXponenta however the y may appear smge exponenta f geeikgemkesct bieikbit His FePCO HisFeP CO kgeminate hv kbimolecular HisFePCO 20 40 60 80 Time x10393 seconds 100 Parameters Qge 004 Dbi 096 kobs Sl kbi Sl The expermenta kmetics are bEXponenta however the y may appear smge exponenta 39 101 His FePCO HisFeP CO kgeminate hv kbimolecular HisFePCO f geeikgemkesct bieikbit 0 20 40 60 80 Time x10393 seconds 100 Parameters cDge 004 Dbi 096 kobs Sl kbi Sl St The experimental khetCs are bexponenta and 775 IS the rapc7 gem are phase 02 00 St geeltkgemkescgtt CIDbie kbi O 5 10 15 Time X106 seconds Parameters qbge 004 Dbi 096 kobs Sl kbi Sl The experimental khetCs are bexponenta as seen on the 0gt39me plot 10 I r Geminate 03 Phase Blmolecular hase 06 04 02 Z geeiacgeerkesc bieikbit 00 4 l I I 1 10398 10 7 10396 10395 10 4 10 3 10 2 1039 Time seconds Parameters qbge 004 Dbi 096 kobs 5X105 51 kbi Sl The bmoeCuar rate constant is a pseudo rst order rate constant 10 I r 03 Para meters kbi kbiTCO 0396 Bimolecular kbi 106 M391 5391 Phase 04 ICO 01 mM 1 mM 02 004 l l I 1039 10 7 10396 10 5 10 4 10 3 10 2 10391 Time seconds AXa ligand replacement in the H936 ca vty mutant of Mb Heme H39 Gly 93 quot IS 93 n Proximal cavity myoglobin mutant of Mb C 0 recombination kinetics depend an axial ligand but escape kinetics do not 250 K 00 Kinetics 08 06 AA 04 02 00 IIIIIIIII IIIIIllIl II mml II ml in7 10396 10395 10394 10393 1039 Timeis CO rebinding rate de ends on axial livand Livands are imidazole 4methyl imidazole or 1methyl imidazole Franzen and Boxer JBC 1997 272 9655 Quest0n H93G ca vZy mutant Kinetic data on the H93G cavity mutant Ligand c1322em kobs X106 s39l kbi X106 s39l Wild type 0039 1040 1000 4Methyl Imidazole 0074 1080 1000 Imidazole 013 1150 1000 lMethyl Imidazole 085 6660 1000 Determine the bimolecular yield of the H93G with 4methyl imidazole A 0074 B 0470 C 0826 D 0926 Quest0n H93G ca vZy mutant Kinetic data on the H93G cavity mutant Ligand c1322em kobs X106 s39l kbi X106 s39l Wild type 0039 1040 1000 4Methyl Imidazole 0074 1080 1000 Imidazole 013 1150 1000 lMethyl Imidazole 085 6660 1000 Determine the bimolecular yield of the H93G with 4methyl imidazole A 0074 B 0470 C 0826 phi pgem 1 D 0926 Dbl 1 CDgem Quest0n H936 ca vZy mutant Kinetic data on the H93G cavity mutant Ligand c1322em kobs X106 s39l kbi X106 s39l Wild type 0039 1040 1000 4Methyl Imidazole 0074 1080 1000 Imidazole 013 1150 1000 lMethyl Imidazole 085 6660 1000 What is the geminate rate constant for the H93G4Me Im mutant A 697 x 103 5391 B 859 x 104 5391 C 472 x 105 5391 D 100 x 106 5391 Quest0n H936 ca vZy mutant Kinetic data on the H93G cavity mutant Ligand c1322em kobs X106 s39l kbi X106 s39l Wild type 0039 1040 1000 4Methyl Imidazole 0074 1080 1000 Imidazole 013 1150 1000 lMethyl Imidazole 085 6660 1000 What is the geminate rate constant for the H93G4Me Im mutant A 697 x 103 5391 B 859 x 104 5391 C 472 x 105 5391 D 100 x 106 5391 q k gem k gem gem k gem k bi k obs 1 k gem CIgemk obs Analyzing cavity mutant data Kinetic data on the H93G cavity mutant Ligand Ingem kobs X106 s39l kbi X106 s39l Wild type 0039 1040 1000 4Methyl Imidazole 0074 1080 1000 Imidazole 013 1150 1000 lMethyl Imidazole 085 6660 1000 d kgem k gem k D k Dbi 9617 1 i b gem k gem k bi k obs gem gem o S q 1 q39396quot Ligand bi 1ng x105 s39l kescapeX106 s39l Wild type 0961 0416 0998 4Methyl Imidazole 0926 0859 0999 Imidazole 0870 1495 1000 lMethyl Imidazole 0150 560 1000 Lecture 27 Kinetic The Arrhenius rate constant Transition state rate constant The Arrhenius equation The empirical observation is that Ea In k In A R7 for many reactions This means that a plot of In k vs 1T ives a straiht line A is the preexponential or frequency factor Ea is the activation energy Also k Ae EaRT Experimental determination of Arrhenius parameters We can plot nk vs 1T to determine the activation energy dink ECquot Rd17 A plot of In k vs 1T yields a slope of EaR and an intercept of In A Energy Activated complex theory Intermediate Products Nuclear Coordinate Reactants The diagram depicts a reaction coordinate The intermediate is the activated complex The transition state The activated complex is a distorted structure that is interuim bn tor the reactants and that of the products At the peak of the potential energy surface between the reactants and products lies the transition state The fundamental assumption of activated complex theory is that the transition state Is in equilibium with the reactants and products A B H Ci The assumption of equilibrium between the reactants and the transition state Since the formation of the activated complex Ci occur in equilibrium with the reactans we can express the equilibrium constant as Kl Ci AB and the rate constant is given by the product of a fre uenc factor kBTh for the formation of the complex times the equilibrium constant AH AH The relationship of kinetics and thermodynamics The I rinci e of microsco ic reversibilit re uires that kf k r Using definitions from thermodynamics and from transition state theory AGVRT AG RT e f e AGRT Note that at constant temperature the prefactor kBTh is the same for each of the rate constants e The math behind the comparison Using the definition of free energy AG AH TAS V eac ee e my we AsRe AHRT ASOIR AH RT e e eAsrlRe AHrlRT e which can be separated into AsR AHRT eASOIR 9 f and e AHOIRT e f eASR e AHRT The connection of entropy and enthalpy These equations imply simply that ASquot A8 As AHO AH AH Note that the relationship between the enthalpies can be seen graphically in the energy diagram that we started with Note t and have the same meaning Relationship to the Arrhenius parameters The transition state rate constant k kBThe39AGRT is k kBTheAsR eAHtRT and the Arrhenius rate constant is k AeEaRT which leads to the identification A kBTheASR The frequency factor depends on the exponential of the activation entropy Ea AH where Ea is the activation enthalpy Debye Huckel Theory Inicatmlhr Solutions of electrolytes are nonideal at relatively low concentrations The activities of ions in solution is relatively large compared to neutral compounds ons interact through a Coulombic potential that varies as 1r r is the distance petween IonS Neutral solutes interact through London dispersion forces that vary as 1r6 The greater the charge on the ions the larger the deviations from ideality For example per mole of CaCl2 dissolved the deviation from ideal behavior is larger than for NaCl due to the 2 charge of calcium ion These considerations lead to the concept of an ionic atmosphere Ionic solutions We consider a general salt CVAV which dissociates into n cations and n anions per formula unit CVAV s gt VCZ aq VAZ39 aq Where vz vz O by electroneutrality We write the chemical potential of the salt in terms of the chemical potentials of its constituent ions according to H2 VH V H where the subscript 2 refers to the ionic solute Ionic solutions As for neutral solutes we have 2 u2 RT In 82 but for ionic solutes we have that II II RT In 8 and u u RT In a v In a v In a In a2 which implies that a2 an aquot39 We can use this development of introduce the mean ionic activity a i an a 39where v v v Ionic solutions We cannot define the activity coefficients of individual ions but we can determine the mean activity coefficients by the same means used to determine the activity coefficients of other substances The mean activity coefficients are defined based on single ion activity coeffiCIents a myand a my where m and m are the molalities of the individual ions given by mvm and mvm Ionic solutions In analogy with the definition of the mean ionic activity ai we define a mean ionic molality mi by rni n mn rnn and a mean ionic activity coefficient gi by Y n Yn y m Given these definitions we can write ain minyin Activity coeffi i n At low concentrations of ionic solute the mean activity coefficient goes as In vi 1173zz1c0 2 Here CO is 1 molL and where 2 and z are the charge ofthe positive and negative ions involved in a chemical reaction This is the concentration in the standard state This standard concentration cancels the units of ionic strength Inistrnuh The ionic strength is calculated using 2 2 I C ZC ZC where 2 and z are the charge of the positive and negative ions respectively and c and c are their concentrations It is important to note that 2 and 2 ions that make up the solution are not necessarily the same charge as those involved in the chemical reaction In other words excess ions in solution contribute to the ionic atmosphere Origin of the DebyeHuckel Theory Where does the factor of 1173 come from In 1925 Debye and Huckel derived the following form for the activity coefficient in ionic solutions 2 K61 ln Vi SneoerkT This expression is based on Coulomb s law and on the concept of a Debye length lK The concept of Debye length This expression is based on Coulomb s law and on the concept of a Debye length 1K The Debye length can be thought ofthe approximate radius of the ionic atmosphere of an ion For a 1 1 electrolyte l 304 pm K Vc moiL When the concentration ofthe electrolyte is 001 M the Debye length is approximately 3000 pm or 3 nm Limitations of the DebyeHuckel approximation The general formula for K is 262NA1000Lm 3 K gogrkT IcmolL Debye Hiickel theory is valid only in the limit of low concentrations The theory breaks down when the concentration ofthe electrolyte is greater than about 100 mM Chemistry 433 Lecture 11 Second Law Applications NC State University Summary of entropy calculations In the last lecture we derived formula forthe calculation of the entropy change as a function of temperature and volume changes These are summarized in the table below Constant volume Constant pressure Sometimes we only have pressure information and the entropy change can be rewritten as follows P V Pi Vi Psi2 2 3 7 i AS nRInM annP2 Isothermal compression Calculate the entropy for an irreversible compression of oxygen gas The initial pressure ofthe gas is 1 ari volume of 100 L The nal pressure ofthe ofthe gas is 10 bar and the temperature is 400 K Solution Note that you need to obtain the number of moles The problem does not ask yuu for a molar entropy Write down the expression forthe entropy i AS nRInM annPz We need either the ratio of volumes or the ratio of pressures We are given the pressures so we can use those P2IPi 10 bar1 bar 10 Isothermal compression We obtain the number of moles using the ideal gas law 105 Pa01m3 RT 831 JImoI K400 K 1 bar1oo L 00831 L barmoi K4oo K 300 moles n Now we can substitute into the entropy expression AS nR In 30 mola31 JmoI KIn 10 249 JK Entropy of mixing When two substances can mix there is a spontaneous tendency for this occur We quantify this using the entropy state function Ifwe considertwo containers separated by a stopcock It N2 gas is in one and Br2 gas is in the other we knowfrom experience that the gases will mix once the stopcock is opened Entropy of mixing For each gas we can describe the mixing as a volume change The N2 gas is originally on the right side contained in volume V at pressure P After opening the stopcock the available volume is 2V and the partial pressure of P1 xyP The same is true for Brz Its initial pressure is P2 and nal pressure is P2 xZP We treat the entropy as the sum of two expansions ie an expansion for each gas Amixs Aexpsl Aexpsz P P AWS n1R In3 an I432 AWS n1Rnx1 an Inx2 AWS nRxylnx1 lenxz Note that the mole fraction applies to the fl al cornposition Entropy of phase transition The entropy of phase transition can be calculated using the enthalpy and temperature ofthe transition AMH T ms qF Jus Afuss TM Example using the data in the Table calculate the entropy ofvaporization for the following compounds u 3532 A sALpH 1m van Tap Entropy of phase transition The entropy ofphase transition can be calculated using the enthalpy and temperature ofthe transition A2155 qF39Jus AmsH Tm Tm Example using the data in the Table calculate the entropy ofvaporization forthe following compoun s Note the 52m2lar2ty In thevaluesforthe entropy ofvaponzauon Th25 25 known as Trouton s rule Conformational entropy The entropy ofa polymer or a protein depends on the number ofpossible conformations T 39s concept was realized rst more than 100 years ago by Boltzmann The entropy is proportional to the natural logarithm ofthe number of possible conformations W S R In W For a polymer W MN where M is the number of possible conformations per monomer and N is the number of monomers For a typical polypeptide chain in the unfolded state M could be a number like 6 where the conformations include different any angles and side chain angles On the other hand when the protein is folded the conformational entropy is reduced to W 1 in the theoretical limit ofa uniquely folded structure Thus we can use statistical considerations to estimate the entropy barrier to protein folding Multinomial distribution The total number ofways that a group of N objects can be arranged in M different categories is MN The multinpmial coef cient applies to a situation where there are rh39ore than two groups In general ifthere are different groups that we can place N objects in we have as the total number ofways W N2 11 nml where 1391 is the product operator indicating that there are M terms multiplied together in the denominator The values ofthe nm are the numbers of objects in each ofthe groups Clearly the nm must sum up to the total number of objects M Z nquot N m 1 Permutation of Letters We have already seen that the permutations ofthe indices ijk etc gives rise to N different combinations The permutation ofindices in the sum over Boltzmann factors lead to the factor of N in the partition function Q qNN for indistinguishable particles In this application we are assuming that all ofthe indices are unique and therefore the number ofwa s of arranin them is iven b l W 12121212 ie there is only one index in each grou Suppose we asked how many ways there are of arranging the letters in the word MINIMUM There are three Ms and two ls wit a to al of seven letters In this case then there are four groups the group ofM N I and U The total numberofwaysis W 7 4567420 3222121 2 Application to a game of chance This is discussion is based on a game called YAHTZEE On each turn you role ve dice You obtain points based on the con guration of the dice The following point scheme is made in the instructions 50 all ve dice have the same number YAHTZEE 30 full house two of and kind with three of kind 40 long straight ve sequential numbers 12 three ofa kind with any two not paired We de ne a con guration as 123456 where 1 is number of ones e c ro e on a pa icu ar ca ofthe dice Consider how many ways there are to get a YAHTZEE of all ones In this case each ofthe ve die must have the same number The con guration is 500000 and the number of ways is given by 550000 1 Application to a game of chance ittsooytousthattheretsoniyonewa togetaii ones But what apout a full house three uf urie kind anu we ufanuther We rnusthe specific SlflCE there are a number offull houses posstoie how man 7 Let ustake two twos ano three sixes The conrtguratton ts tutztututua ano the numper orways thts e achieved is SlDlzl l l lsl WEI2x6 tEI There are two posstpie iong stratghts The conrtguratton ts n M an the numper or ways thts can pe 4 achieved is 5it ltlt lt lEIl tZDS 2 The oyeraii propaptitty or optatntng any conrtguratton ts gtyen pyvvrort at conrtguratton compareo tot e totai numper o conrtgurattons MN ForYAHTZEE ts 55 7775 The erore the Propaptitty or getttng a iong stratght ts 2247775 n EIEIE Problem solving We can toenttiy the roiiowng rnatn types or propiems that tnyoiye entropy change isothermai expansioncompression reyerstpieitrreyerstpie Temperature change Eoutitpratton Mixing Phase Transttton Stattsttcai or Conformational Entropy Aotapattc trick ouestton tro Eltheri AS El Wheneyer you are soiytng an entropy propiem rememper to constoer poth system ano surrounotngs The system ts aiways caicuiateo aiong a reyerstpie path Equilibration We haye seen a strnpie example where there are two rnetai blocks poth maoe or the same rnatertai materialst ano 2t we neeo to nowthe tntttai temperatures ano heat capactttes to caicuiate the rtnai temperature at T 02 OttTt 7 Tu a Co27T Tu Now we soiye for the eoutitprtum temperature Tm CrtTtaCttTu 2sz Cute CttTt Ctzn0ttctz1t T CrtTnCnTz a Cut Co Equilibration Calculating the entropy Once you haye ootatneo the eoutitprtum temperature the entropy ts eastiy caicuiateo from Te Te AS r70 lri AS2 770 72 lri in such roblerns ou are assumtng thatthe two optects tn therrnai contactare a cioseo system the y i o entropy rorheat riow shouio pe post ye stnce heat riow rrom a hotter to a coioerpooy ts a spontaneous process it Example The cuffee cup prublem Ahikeruses an aiurntnurn cuffee cup lfthe mass ufthe cu tsA grams and the ambienttemperature cup stemperature isrlU ac anuihe hiker peutssu rni ufcuffee h aiempetaiute etau ac tnict the cup A caicuiaieiheegutitpttumiempetaiute a caicuiate iheenitctpy change Gas exchange in a green leaf The ceiis orthe spongy iayer are trreguiar tn shape ano iooseiy packed Thetr rnatn runctton seems he temporary storage of sugars ano amtno actos synthestzeo tnthe paitsaoe iayer They aiso ato tn the exchange orgases to the outstoe through pores caiieo stomata singular stoma Entropy O2 mixmg in a the stomata of a leaf Durtng photosynthests o2 ts proouceo tn the thylakoid memorane orgreen ieayes The gas meets the atmosphere tnthe stomata of the leaf Assumtng that the roiiowtng moie rracttons exist caicuiate the moiar entropy of mixing instoe XDZ m u 5 ano M m u 5 Outstoe x32 M u 2 ano N2 M a a Assume that the stomata closeswith an equal volume of outstoe atr ano tnstoe gas present tn an encioseo space Soiutton Assume that the tntttai state ts premixed gas poth tnstoe ano outstoe Call o gen compouno t ano nttrogen compouno 2 The final compostttort ts E5Ll22E35 EISrLlE2EE Entropy O2 mixing in a the stomata of a leaf Write down the total entropy change as products nal mixed nal mix Note that there is half as much of each ofthe reactant gases as the product Astute As name nR x ln x xuln x nR 2 Asmxim Asmxioul nR Xmln xm xozln x02 Tx ln xM x ln xq Since the problem asks for the molar entropy we really need ASmLWn so we can write the formula as ASWW R x ln xM len er Entropy O2 mixing in a the stomata of a leaf Write down the total entropy change as products nal mixed nal mix Note that there is half as much of each ofthe reactant gases as the product X01 2 in x Xi an x92 7m xm m x 831 J mol K005 0421 JmoI K This is a small entropy Part ofthe reason is that the gases were already mixed lfthey had been completely unmixed the entropy would have been 54 JmolK Phase Transition Compare the entropy of sublimation of water to the entropy of vaporization at 0 ElC Which is larger Why Solution The entropy ofa phase transition is given by AH Asphm T phase Phase From the information in Atkins page 61 we nd that AWH 4507 kJmol and AWH 5108 kJmol Using these values and the temperature of 273 K we nd 45070 Jmol Asvap W 1651 JmoI K Assub 510237 33ka 1871 JmoI K Conformational entropy ofa protein Estimate the conformational entropy of myoglobin Myoglobin has 150 residues with 6 possible conformations per residue Assume there is a unique conformation for the folded structure Solution The de nition S R InW is known as the statistical entropy R is the gas constant and Wis the number of possible conformations for a structure and W MN For the unfolded protein W 615 and S 150R ln6 For the folded protein W 1 and S 0 There conformational entropy is AmmS AWEMS and is therefore 2233 Jmol or 22 kJmol The Levinthal Paradox A N Conformatlonal Entro in a golf course The paradox states that ifa protein samples all 6M conformations it will take a time longer than the age of the universe to nd the native fold N for a polypeptide where M 100 and if it takes 1039M seconds to sample each possible conformation The Pathway Model lmtahgine thatdtheha uni ue 1 pa way wrn s roug 6 esu ace 0 e oe gt thhe th startshat 2 and changes Here the entropy must decrease rapidly since N the number of degrees of Conformatlonal gt 39eedom in the folding pathway is quite Entropy small compared to 6 On this diagram the con gurational entropy is given by the width ofthe funnel and the relative by the height relative to the bottom folded state The vertical axis is energy NOT free energy Evidence for folding pathways One piece of evidence for folding pathways comes from trapping disulfide intermediates This method was pioneered by Creighton using BPTI and h t 39oteins as only been used on 0 her p w 3051514 305L538 mama33 I M E 3 easmug Reduced S l Native Creighton etal Prog Biophys Mol Biol 33 231 1978 B eyo nd 2 Kim et al showed that some ofthe previous data and interpretations were wrong The major 2DS species contains the two native DS39s 3051 and 555 The third disul de is formed quite slowly because it is quite buried It is possible to isolate a stable species with only the rst two disul des formed and the third remaining in the reduced form Studies with a mutant in which the third DS was replaced by 2 Ala and which folded at a similar rate to the wild type support the idea that the trapped disul de species have partial nativelike structure These observations and others like it can be used to the idea of a pathway into The Folding Funnel The folding funnel shown here represents the change gt in energy for a large number of folding paths that lead to the native an equal probability leading hconforr atlonal gt to the folded state The funnel Entropy shown here has no energy barriers and all paths lead directly to the native state Thus this funnel is consistent with two state folding behaVIor D111 and chanNalure Strucf B101 1997410719 Barriers and misfolding The energy surface does not have to be a smooth T trajectory There can be barriers that will trap intermediate states These in ermediate is observed there will be a question as to Conformational 4 gt whether this is part of a pathway Entropy or whether a funnel description is more applicable Note that the funnel provides the possibility for misfolding This will typically result in multiple minima 39n the energy landscape D111 and Chan Nature sum Biol 1990410719 folding trajectory Whenever an 39 t Microscopic Treatment of the Equilibrium Constant Lecture The chemical potential The chemical potential can be expressed in terms of the partition function oanj 1quotRT aNj To see this we first expand anj starting with the fact that Qj qjNJle In Q len qj In N len qj len N N and then take the derivative with respect to N aan q 6N nqj an 1 1 n j J j Consideration in a theoretical chemical reaction This result ini s h a RTIn For the reaction I VAA VBB lt gt VYY39I39 VZZ we have amp i i or VylnNyVzlnNZ VAnNA VBnNB 0 V V V V NnyZZ quqZZ V V V V NAANBB qAA BB Relation to the equilibrium constant We can express the concentration dependence in the equilibrium constant as where pj is the number density of species j pj NjV PM PEApr which shows that the equilibrium constant can be expressed in terms of molecular partition func20nTsj NYV vYNZV V2 qYV vyqZV v2 6 NAVVANBV VB qAVVAqBVVB KCT In considering the molecular partition function we must consider the kinematic contributions and electronic contributions In other words if we write the molecular partition function as q q transq rotq vibq 6166 the molecular motions translation rotation and vibration are kinematic contributions to the available energy space The electronic partition function is somewhat different since it represents the binding energy of a molecule with respect to constituent atoms Until now we have simply stated that qelec gelec the electronic degeneracy This is equivalent to ignoring the energetic contributions to chemical bonds and treating molecules as translating rotating and vibrating collections of nuclei that are held together by bonds However when we deal with chemical reactions there are by definition changes in bonding We can accommodate this by writing D RT qelec gelece 0 where DO represents the binding energy The binding energy D0 is equal to the equilibrium energy De shown in the figure minus the zero point energy D0 De hyZ The zero point energy is shown as the red stripe at the bottom 20 For CO this energy is 0 calculated to be 1067 cm391 and De is 96545 cm391 Thus the zero point energy is typically a small correction CO Potential Energy 20 40 60 Energy cm 391 x103 80 391007 I I I I I 100 105 110 115 120 125 Internuclear Distance A By making this subtraction we also quotremovequot the zero point energy from the vibrational partition function The vibrational partition function that we have considered up to now is e hmZkBT qvib 1 e hmkBT The term in the numerator represents the zero point energy When this term is incorporated into qelec the vibration partition function becomes 1 qw39b 1 e f zcokBT In this form we can state that for a singly degenerate vibration qvib 1 at T O K and the magnitude of the vibrational partition function increases with temperature Using this separation the significance of the kinematic partition functions is clear These represent the temperature dependence of occupation of various levels translation rotation and vibration respectively The electronic partition function on the other hand represents the energy of stabilization of the molecule with respect to its constituent atoms In raCI qelec approaches Innnlty as T 9 O K What sense does this make Well if we consider qelec as a contribution to an equilibrium constant we can think of temperature as a parameter that determines the relative stability of the molecule The bound state of the molecule will be most favored at T O K As temperature increases there is some thermal tendency for the molecule to dissociate even though this is small at most temperatures of interest to chemists In this sense qelec is quite different from the kinematic partition functions Vibrational temperature To simplify the writing of the partition function we can define a vibrational temperature Vib We define em hmkB 1 QVib 1 e wbT Thus the vibrational partitionc function has a simple form Note that the vibrational temperature represents the at significant population of higher vibrational levels occur Rotational partition function and rotational temperature In the high temperature limit the rotational partition TunCtIon is q N 872kT rot N h2 B where B is called the rotational constant The rotational spectrum line spacing is 23 Thus we can define a rotational temperature mt such that 872k rot h2 N L Qroz N amt Symmetry number For molecules with an axis of symmetry there are fewer unique rotational states accessible The partition function is therefore reduced by the symmetry number 6 which corresponds to the multiplicity of the symmetry axis For example for a diatomic molecule the symmetry number is 2 For the rotation about the axis of symmetry in ammonia is 3 Qroz z 69 rot The translational partition function As discussed previously the translational partition function can be written as q v trans 3 A where A is the thermal wavelength 2 A h 27tka Notice that the V in the translational partition function cancels the V in the number density Thus the contribution of the translational partition function to equilibrium is the thermal wavelength This can still be significant if moles of gas are created or destroyed in the chemical reaction 10272008 Chemistry 433 Lecture 21 Two component systems NC State University Total derivative for two components We considerthe thermodynamics of twocomponent systems The ideas discussed here are easily generalized to multicomponent systems For a solution consisting of n1 moles of component 1 and n2 moles of component 2 the Gibbs energy is a function T and P and the two mole numbers n1 and n2 The dependence on these variables is indicated by writing G GTPn1n2 The total derivative of G is given by an Elwyn lman l39mdnz as 6P m 012 as Gibbs energy at fixed composition If the composition of the solution is fixed then we have dn1 dn2 0 and the last two terms are zero In this case the functional form of the Gibb39s energy is exactly the same as we have seen previously dGg 73 ad dF where 6G 6G er S and lap V The chemical potential in a mixture The chemical potential is defined as lJl 1T1P1 2 Gm for component 1 and an analogous eduation holds for component 2 In generalthere may a greater numberof components and each will have an associated chemical potential that is the derivative of the Gibbs energy with respect to the mole numberofthat component It is also evident that the chemical potential is a molar Gibbs energy for one component and for more than one component it is a partial molar Gibbs energy This is an intensive property and is just the Gibbs energy per mole m General Gibbs energy for a mixture Fora binaw solution the Gibbs energy is G SdT VdP ttldn14ttzdn2 At constant T and P we have dG ujdn1 itan A general expression forthe Gibbs energy is G ll39ln1 P2quot Fora one component system G un consistent with the statements made previously that p is a molarGibbs energy Otherthermodynamic quantities have associated partial molarvalues The easiest to see physically isthe partial molarvolume Vmy 6 VBnl Partial molar volume For a two component mixture the volume is v vmn1 vzymn2 For example when lepropanol and water are mixed the final volume V ofthe solution is not equaltothe volumes of pure le ro anol and water The mixture oftwo components that can interact in a noneidealfashion leads to a solution volume that is greater or lessthan that ofthe pure components The partial molar volumes allow this to be quantified Otherthermodvnamic quantities can also be expressed as partial molar derivatives In general forthe jth component we have din 75 WP 10272008 GibbsDuhem Equation Starting with G pqnj uznzwe can differentiate to obtain dG du1n1 du2n2 ujdn1 i12an Comparison with the above equation dG ujdn1 uzdn2 leads to du1n1 duznz 0 lfwe divide both sides by n1 nzwe have duiXi dues 0 where x and xzare moie fractions These last two equations are two forms ofthe Gibberuhem equation The Gibberuhem equation is important because it tells that ifwe knowthe chemical potential ofone component as a function of composition we can determine the other GibbsDuhem Equation For example the chemical potential of substance 1 in a two component mixture is 391 IIIl RTIn Xi Where 0 g x1 g 1 The superscript is the IUPAC notation for a property of a pure substance We can differentiate with respect to x1 and substitute into the GibbsDuhem equation to obtain X X db etc 301th 7 Rd y 1n x x dx dx whims Ila and since dx1 dx2 we have dx Rial uz u RTlnxZ Thus we have shown that one can derive the chemical potential of substance 2 from substance 1 The expression u ujquot RT In x implies that we can determine the chemical potential of any substance based on the knowledge of the chemical potential of the pure substance and the mole fraction x One of our goals in the study of nonideal solutions will be to prove this Recall that iftwo phases are in equilibrium their chemical potentials are equal We can use this fact to our advantage At any given tempemture a liquid has a vapor pressure This meansthatthe chemical potential ofthe vaporabove the liquid must equal the chemical potential ofthe liquid itself This is just anotherway of sayingthatthe liquid and its vaporare in chemical equilibrium Whamquot Ifthe pressure ofthe vapor phase is low we can consider it to be ideal Thus we have fin Wail p1 in RT In P where we have simply stated the chemical potential ofthejth component ofthe liquid relative tothat of its standard state of 1 bar of pressure pf in Chemical potential of a solution For pure componentj the equation becomes uj U u vap ujquot T RT In Pf Thus uf uj RT In PJPJ This is a central result for the study of liquid This result uses information from the vapor phase chemical potential above the liquid to give us information on the chemical potential in the liquid Ideal Solutions Ideal solutions obey Raoult s law Pi lPl We choose as ourfirst example a binary solution imagine thatthetwo types of molecules in a binary solution are randomly distributed throughout the solution Raoult39s law states thatthe partial pressure of liquid 1 above the liquid is equaltothe mole fraction ofthe liquid in a solution time the partial pressure ofthe pure liquid This holds for ideal solutions An ideal solution isapproached by binary solutions of moleculesthat have simiiar properties leg benzene and toluenel 10272008 Geometric considerations We can understand Raoult39s law by means of a physical picture of the surface of a liquid The two types of molecules can only enter the vapor phase if they reach the surface The red and blue spheres represent molecules ortwo species in a binary solution Since the surface area is approximately proportional to the mole fraction assumingthat the molecules have the same size then the ability ortne molecule tojump into the vapor phase is proportional to its concentration mole fraction at the surface We can considerthe red and blue spneres to be two molecules in an ideal binary solution We represent the vapor pressure or each ortne pure liquids by Pm39 or Paw Vapor pressure in a twoscomponent mixture Interpreting Raoult s law We can rewrite Raoult39s law as X mlm This leads to an expression for the chemical potential 1 p RT1nX Calculated ideal vapor pressure plot Themtal vapor pressure over an ideal salullan ls my a p r p1 xlgt39 r x1P139 lrxllP39x1Pl39 P r x1e F We can make a plotofvapor pressure vs campaslllan for benzene and tolueneas follows We plotxMWPMW39rrom e Um 1 where new Eutorrand llkewlse we plot XWM from 1 to u notrngthatxwme 171 Note that e lEUmrr Both of the vapor pressures ouoteo here are t he benzene but we could have chosen toluene and the dlagram would have the same information Calculated ideal vapor pressure plot toluene Vapor Pressure Torr 02 04 08 08 10 xbenzene 10272008 Calculated ideal vapor pressure plot Vaporpressurermmpasltlan olagram afan loeal salutlan or benzene ano toluene The eooauons oseo to generate the llnes 1 XIP1 vvnere 1ls benzene Q ano z lsmluene 2 Vannr Pressure Tm Thinking in terms of the vapor pressure In a blnary salutlan the campasltlan onne vapor ls noune same vaparwlll be largerthan the malefractlan afbenzene ln salutlan Composition of the vapor To calculate the malefractlan mth vapor we use Daltan s lavv Daltan s lavv states thanhe parllal pressore nere reao vapor pressore as well ora gas ls eooal m the malefractlan vor campanentl n the gas phasetlmes themtal pressore P vamsl Natethatwe nave a waym calculate bath p ano Pwmframthe equatlans apove p xlgt ano PM p x1Pl39r m so v IvPW Considering the composition of the va por We can expano on the plat apoverorpenzene andmluene pv ploulnga vaparcurvethat represents thetatal vapor pressure as a functlan onne mole fractlan n the vaporpnase Nate thatthe vlaletmtal llne lnlheflgure apove representsthe vapor pressure nctlan onne malefractlan n the lloolo pnase all vi P ano Pl39 Vanar Pressule rrorr Derivation of the vapor curve Pm X2P27Y2 and Xflpmal39 P1 lP2quotP1 Plural Peml 39 Fi Pf Psz Plural 39 Plural P2 P2quot Ps wo VP PINEquot Pv39ivz P MP W Pl Pz llP2quotP1 l2 1 Pz llP2quot Psz quot Ps39m k Pm Ps Pz llPZ llsyzl Ps yzl Pm Pl Pz llP2 lP1quotP2 l2 Vapor Pressure Torr Plot of the vapor curve llqulcl llnE 100 vapor curve 50 o 02 04 06 08 10 xbenzene Ynemens 10272008 Explanation of the liquid and vapor curves The plot shown is a pressurercomposition oiagrarn This can be shown by iabeiingthe regions orthe e upper region at sor oientiv high a pressure is the liquid region above the liquid line The E r with vapor That is to savthat two phases coexist in this region Vapor Pressure Torr R o r The twophase region N o a m 0 liquid liquid vapo apor an o 02 04 06 0 8 10 Xbenzene Ybentene Explanation of the two phase region ihthe hooro region the oornposrtroh is MW anothe rrnpheo xww xgmi or more generally x1 aho x 17x1 in the vaporpressurermmpasilian oragrarn bv examiningpmnls x aho V Vapor Pressure Torr Thetie line An Nmo o o 0 on o 02 O 4 O 6 08 1 D Xhenzene Yhenterie Calculating composition in each phase We can oehhethe mole fractions orirooro aho vapor respectively as Hr or n n x 7i 4 and 7 if him n y nanfquot n Where N am hm arethetotai hornber orrnoies rhthe homo aho vapor phases respectively The overall rnoie rraotroh is given by Usinglhe fact that zn t he a xm r we We can calculate the reiatrve arnooht orhooro aho vapor apuir u menu 7 Illustration of the lever rule This eooatroh hasagraphroai rhterpretatroh We akalan expahsroh or the region arooho P on the behzehetoioehe vapor pressurermmpasilian piot mom coo ybenz vapor 035 040 045 050 055 060 Zhenzene 10272008 Example calculation using the lever rule E laqu Equallun or nl qnV Pls called the lever rule ll 11 7 05 l andthereldre 21 05 andthe yapdr pressure is pm mu run we can calculate bath yum and xmm thatwill giye the limits ulthe llE llne Xi leir Pmi39llvcm39r Pwi39l lion 7 GDlHBD r 6m D 120 333 vim apneaPW llalllsnlmu 06 In this case giyen that 2M 05 as shdwn in the ligure we have 6 r I 5 l I 5 7U 333 I 167 Graphical interpretation of the lever rule thicethat there is an inyerse relatidnship between the distance aldngthetielinetda phase and theamdunt dtthatphase Pulnl with liquid liethere is a negligible amduntdvyapdr cdmpared td the liquid whenthe cdmpdsitidn isvl Temperatureccomponent phase diagrams We can display the cdmpdsitidn dtthe sdlutidn and yapdr phases tdtal pressure rdr example mpientpressure may be 760 td r lllhls r sureiscdnsidered Pmmlurasul i n x17 l wul39PllP139PllDrxll7snlurr39PllP139Pl r gnituded 39 a P determined rdmthe Clauslusr clapeyrdn eduatidn 1dr the two species rl The pressure P andtemperature r are reverence temperature intd the eduatidn apdyetd yield xlwhlch determines the liquid cdmpdsitidn curye nample hemene and methannl H Mdlecule Methandl 3372 353 men rdrrlexplamlu Dunnm 1dr penzene 1 men TurrlexleSlD DDZerlTl 1dr methandl 1 llrexpBJDlDDDZB1TllllExp4 ZBlD DZerlan ExpBJDlDDDZBBJTm x Calculating the vapor curve Aswas descriped abovelurlhe yapdrpressurecdmpdsitidn diagram we can relate yllu x1 andtherepy dptain a yapdr curye P ml The regidn between the liquid cdmpdsitidn curye and the yapdr cdmpdsitidn curye is a two phase regidn and the leyer rule appliesthere aswas seen abovelurlhe lane Vzpm so yapdr prESSUFEsEDmpDSllan phase diagram E 5 na o39z a4 as n Malellacllun 310 Fractional distillation rhetemperature cdmpdsitidn diagram shdws huwlracllunal distillatidn wdrks nt einitial cdmpdsitidn dttheliduidisA nd cdnd pdintc Ev pdiling the sdlutidn with the cdmpdsitidn atpdintc the yapdr has cdmpdsitidn n We can cdntinuethis prdcess ntil we haye essentially pure Lprupanul lthe lettmdst pdint dnthephase diagram Saluiipn remperaluie l c i no 02 04 us us 10 Mai lra llorl Practical note In practice the process ofdistillation can be done in a single apparatus using a column of beads such thatthe temperature is lowered gradually ascendinglhe column The composition changes gradually as the Vapor moves uptlie column The beads create a nucleus forcondensation ofthe vaporso that an equilibrium between liquid and Vapor can be maintained 10272008 Stirling s Approximation In confronting statistical problems we often encounter factorials of very large numbers The factorial N is a product NNlN22l Therefore In N is a sum N In N 221 In m where we have used the property of logarithms that logabc loga logb logc The sum is shown in figure below The sum of the area under the blue rectangles shown below up to N is In N As you can see the rectangles begin to closely approximate the red curve as m gets larger The area under the curve is given the integral of In X N lnNi11nm sf lnxdx v du To solve the inte al use integration by parts u dv uv 7 fr Here we letulnXand dVdX ThenvXand dudXX N N N dx Ilnxdxxlnx 7 x7 0 0 0 Notice that XX l in the last integral and X In X is 0 when evaluated at zero so we have N N flnxdxNlnN f dx 0 0 Which gives us Stirling s approximation In N N In N N As is clear from the figure above Stirling s approximation gets better as the number N gets larger Let s try a few numbers N N In N NlnN N Error 10 363 x106 151 1302 138 50 304 x1064 1484 1456 188 100 933 x10157 3637 3605 088 150 571 x10262 6050 6016 056 My calculator overheats at 200 That is all right since we have shown that the result is converging In thermodynamics we are often dealing very large N ie of the order of Avagadro s number Clearly for these values Stirling s approximation is excellent Chemistry 433 Lecture 6 Heat Capacity State and Path Functions NC State University Question Which expression correctly gives the work of expansion v2 gt v1 V2 A nRTIn V2 B nRTIn V2 C 17 In V2 D nR n Question Which expression correctly gives the work of expansion v2 gt v1 V2 A nRTIn V2 B nRTIn V2 C 17 In V2 D nR n The measurement of heat We must carefully distinguish between heat and temperature When we add heat to the system its temperature increases We can use measurement of the temperature to determine how much heat has been added However we need to know the heat capacity of the system in order to do this Heat supplied HeatcapaCIty Temperature rise The heat capacity is called C If we perform a heat exchange at constant volume then we designate LhU neat Gapdb39lty as UV If the process occurs at constant pressure we call the heat capacity CP 7 Calorimetry The science of heat measurement is called calorimetry A calorimeter consists of a container in a heat bath A physical or chemical process occurs in the container and heat is added or removed from the heat bath The temperature increases or decreases as resu u own3 the heat capacity of the bath we can measure the amount of heat that has been added or removed from the system Energy in the form of heat ows into the bath Calorimetry In the studies of biological systems there are two important types of calorimetry 1 Differential scanning calorimetry DSC 2 Isothermal titration calorimetry ITC In DSC the temperature is increased at a conSIant neatlng rate and the heat capacity is measured DSC is used for determining the parameters associated with phase transitions eg protein unfolding denaturation DNA hybridization etc In ITC the temperature is held constant while one compon is added to another The heat of interaction eg binding is measured using this method ITC is widely used to determine the enthalpy of binding eg for proteinprotein and proteindrug interactions among other types of biological appHca ons Molar and Specific Heat Capacities we use molar heat capacities for pure substances As the name implies the units are JmolK for the molar heat capacity We write the molar heat capacity at constant volume as Cvm For mixtures we cannot use a molar heat capac n we use the specific heat capacity which is the heat capacity per gram of material with units of JgK Calculating the internal energy change We have seen that the internal energy depends only on temperature For example for a change of pressure and temperature at constant volume we saw that AU qV since the work is zero for a constant volume plUbeSS Thus at constant volume AU qV CVAT But in fact when we consider the origin of the internal energy in the kinetic theor of gases we realize that AU gnRAT CVAT and therefore CV gnR and CVm 2 R Nloo The heat capacity at constant pressure For a constant ressure ste we saw in the last lecture that 7P 7V PAV By analogy wit the constant volume process qp CPAT and therefore CPAT CVAT PAV CPAT CVAT nRAT 0 CV 17 so that 0 nR and CRm 2 R Nlcn Definition of the enthalpy Based on these considerations we can see that there is a new state function the energy at constant pressuw This state function is known as the enthalpy H The enthalpy change Is AH qp CPAT and we can rewrite the relationship from the previous slide as AH AU PAV We have also defined the relationship between the internal energy and the enthalpy Te PAV term represents the work of expansion or compression done against the atmosphere during a chemical reaction We use enthalpy instead of internal energy under normal conditions oecause It InCIuoes this work automatically Heat Capacity for a Diatomic Molecule For a diatomic molecule there is contribution from rotations as well as translations This means that as heat is added to the sstem the rotational levels can be I o ulated in addition to an increase in molecular speed The kinetic theory of gases considers only the speed An approximate rule is that we obtain a COHtl lbuuun tu Lllc llCdt GapaClLy UV Of 12nR fOl39 each degree of freedom We saw that for a monatomic gas the heat capacity was CV 32nR A diatomic gas has two rotational degrees of freedom and so the heat capacity is approximately CV 52nR What does this say about CP Well the relationshi between CP and CV holds for all gases so CP 72nR for a diatomic ideal gas Question What is the internal energy of a monatomic gas AU 2R BU 2RT CU 2nR DUnRT Question What is the internal energy of a monatomic gas AU 2R BU 2RT CU 2nR DUnRT Since the molar internal energy was not specified and simply the energy D is the best answer Question If the heat capacity of a diatomic gas is 52nR what is the internal energy of a diatomic ideal gas AU R BU CU DU 2 Q 2nRT 2nR 2nRT Question If the heat capacity of a diatomic gas is 52nR what is the internal energy of a diatomic ideal gas AU R BU CU DU 2 Q 2nRT 2nR 2nRT Adiabatic Processes If a process occurs in an isolated system then no heat can be transferred between the system and surroundings In this case the heat transferred q is zero ie q 0 Therefore AUW We call such processes adiabatic Actually this special case is of great importance For example when a column of air rises in the atmosphere it expands anu cools adiabatically Expressed in differential format dU8W CVdT PdV Hr w hv h finiin fh inrnl nr in VI terms of the heat capacity and the work in pressurevolume terms Adiabatic Processes Using the form on the previous page we can derive the relationship WWW m Imus wcl Vptre dU 8W CVdT PdV CVdT fdv dT CV T 17 V 72 V2 dT dV R T V1 v T2 i CVInf nRn V1 Adiabatic Processes Using the form on the previous page we can derive the relationship between the volume change and temperature T2 39 H v2 T1 CV T2 V1 nRCV T39V2 23 T2 for an ideal monatomic gas 2 25 T2 for an ideal diatomic gas 2 This expression is great practical value since you can r i h m r r fir i ri Thi hnmenn leads to rain over mountains and cooling that affects ecosystems at high elevation Question Which statement is true for an adiabatic compression A AU q B AU q w C AU w D AU q w Question Which statement is true for an adiabatic compression A AU q B AU q w C AU w D AU q w Question Which statement is true for an adiabatic expansion A The temperature increases as gas expands B The temperature decreases as gas expands C The temperature remains constant D The work done is equal to the hea transferred Question Which statement is true for an adiabatic expansion A The temperature increases as gas expands B The temperature decreases as gas expands C The temperature remains constant D The work done is equal to the hea transferred Path Functions We have seen that work and heat are path functions The magnitude of the work and heat depends notjust on the final values of the T and P but also on the path taken We can summarize the paths and their implications in the table below Path Condition Result Isothermal AT 0 w q Constant V AV 0 w 0 AU CVAT Constant P AP 0 w PAV qp CpAT Adiabatic q 0 AU w State Functions At present we have introduced two state functions Internal Energy AU Intnalpy AH State functions do not depend on the path only on the value of the variables N can make the analogy with elevation e pmential energy at an elevation h which we call Vh does not depend on how we got to that elevation If we compare Vh1 In Kalelgn to Vnz on Mt Mitchell the difference Vh2 Vh1 is the same regardless of whether we drive to Mt Mitchell through Statesville or Asheville The work we do to get ie how much gas we use in a car is a path function Question Enthalpy is a state function therefore it depends only on A the temperature and pressure B the elevation C the work of expansion in a reaction D the heat transferred in a reaction Question Enthalpy is a state function therefore it depends only on A the temperature and pressure B the elevation C the work of expansion in a reaction D the heat transferred in a reaction Lecture 25 Kinetics Experimentaltechniques Definition of Rate First Order Processes Exponential Decay Experimental Techniqr s 0 Monitor reaction progress using pressure conductivity spectrophotometry etc in real time o Quench reaction after a given time by rapid cooling or solvent trapping o Initiate process by flow stoppedflow rapid mixing or flash photolysis 77me scales for expenmenta measurement of reaction kni39 s o Spectroscopy is a key tool for monitoring reaction progress on rapid time scales 0 Flash I hotosis is a convenient method for initiating a timedependent process with real time acquisition as rapid as ten femtoseconds 0 Rapid migtlting owed reactions can be monitored at xed distances with 100 us time resolution 0 Stoppedflow has a 1 ms mixing time o Quenching method suitable for slow reactions De nition of r For any given reaction e g A B Rate of consumption of A is vA dAdt Rate of formation of C is vc dCdt Sign convention reactants are consumed and therefore dReactantdt lt 0 o In general for any species vj1njdJdt where nj is the stoichiometric coefficient Rates ofappearan e For a chemical reaction dJdt nJVJWh e J is the stoichiometric coef cient 0 Example A ZB 3C gt D 2E o If dAdt v then the rate ofdisappearance is dBdt 2v dCdt 3v and the rate of appearance is dDdt v and dEdt 2v Rates ofappearan o For a chemical reaction dJdt HJVJWHG J is the stoichiometric coef cient 0 Example A ZB 3C gt D 2E o va dAdt 2v dBdt 3v dCdt then v dDdt and 2v dEdt using the correct sign convention 11132008 The rate is proportional to concentratioquot s Example The rate law is v kAB where each reactant is museu to the first power The coefficient k is called the rate constant The rate law can be determined by the isolation method The reaction is run in an excess of all but one reactant to determine the dependence on concentration Units of the rate const 2 The units of rate constant are always su that they convert into a rate expressed as concentration divided by time a e A 3 mol L391 s391 L mol391 s391mol L391mol L391 Rate k A2 B mol L391 s391 L2 mol392 s391mol2 L392mol L391 Quest0n What are the units of k in vA kA A mol L391 s391 B L mol391 s391 C mol L391 D s391 Quest0n What are the units of k in vA kA A mol L391 s391 B L mol391 s391 C mol L391 D s391 rstorder KneLf39 s o A first order rate process dAdt kA 014 7 Ir dl A Am 2 dA Ir dl 1 A i M At ln AJD kl rstorder Kinet39 s o A first order rate process dAdt kA Al kt quot A Al We 11132008 Firstorder lane 39 s o The rst order dAdt kA o The solution is AADekt 0 Known as single exponential kinetics Halflife o krl2 lnA02A O ln12 is the de nition ofa halflife 112 ln2k rate constant is give Halflife and rate const t The halflife for 23EU is 45 x 109 years What is the rate constant for radioactive decay or 238U A 15 x 103910 var1 B 75 x 10399 year391 C 225 x 109 year1 D 15 x 10399 var1 Halflife and rate const t The halflife for 238U is 45 x 109 years What is the rate constant for radioactive decay or 238U A 15 x 103910 year1 B 75 x 10399 year391 C 225 x 109 year1 D 15 x 10399 var1 Exponen tia kineti 3 3 7 epoZ E 05 expo 5 we expmt e a 027 An egtponenuai process rs typical of rst order kinetics Decav ot radioacuve nuclei rs me texlbook example Exponen tia kineti 9 1o E W i epoZ 6 9 7 expo E we epoQ E 027 8 1 1 1 1 U 2 4 a 1 Time on lhe vans we can plot populauon or concentrauon These are two wavs of saving lhe sarne mg Exponential kineti C m e E 399 i epoZ k 39 6 5 7 expo E oar exprzt r E 027 e 8 1 1 1 1 U z 4 a 1 Time An egtponenua1 process nas a 1e lJrne tnat corresponds to c 1kwherellt15 one rate constant 11132008 Exponen tia kineti m as os ova D2 E 5 2 c 8 c 8 4 a Eemnds We focus on the exponenha otocess wtth a tate constant ofk 15quot Exponen tia kineti Concenn39attm At 4 5 Tune The tntttat rate ts the stooe at Dme zeto Thts ts gtven ov the bme ttne tn the gure It ts obtatned ttom the dertvauve Exponen tia kinetf r 1 alUx C 10 as E 06 HERE Slope 04 rate 1m E 02 8 3 I t t 2 4 6 True The tnttat rate ts the stooe atttme zeto Thts ts gwe r1 bv the btoe ttne tn the gure It ts obtatned totn the derwahve Exponen tia kineti g Slapeequot rkequot e g 3 1eez7182 0 0367malLS g uzr Z 6 4 Ttme At t 1secono the stooe ts agatn gtven bv the derwahve at 1 Second NOW therate15 SmaHer Concertratton At At t 1secono the stooe ts agatn gtven bv the derwauve t at 1 Second Now the rate 15 smaHe k 1oet secono mzno Concertratton At At t 1secono the stooe ts agatn gwen bv the derwahve at 1 Second NOW therate15 SmaHer 11132008 Exponen tia kineti 0 711111 4111 m MP T W T e HERE xtapee1e2ez7 82 1M6 0135malLx Concentration At 4 s Time At t 2 seconds the sobe is again given by the derivative at 2 seconds The conuhues to decrease as A is used up Exponen tia kineti 0 d eel 7 Slaper e e ke e 62 ns HERE Sloperleze27182 e rate 0135malL5 n 27 RATE CONSTANT 1 squot e a 1 Concentration At 4 Time At t 2 seconds the siope is again given by the derivauve at 2 seconds The conuhues to decrease as A is used up Question A A first order process occurs with a rate constant s 1 If the initial concentration AEl 1 M what is the concentration after 30 seconds A 0 5 M B 005 M C 0005 M D 00005 M Question A first order process occurs with a rate constant 01 s If the initial concentration AEl 1 M what is the concentration after 30 seconds A 0 5 M B 005 M C 0005 M D 00005 M Approach to equ bf m A Rate equauons are 39 dU Jdt 4891 MB 39 d51dt KM 39 MB Simuitaneoussoiuuons of k1 k1 these equauons ieads to a rate constant Of k K Equiiibrium rsabbroached B as the sum of the forward and reverse rate constans Approach to equiIb m Let X be the deviatron from equiiibrium X M U l 5 then A A 4 X and B 5 e X Rate equauons are 39dWdt MA 39 MES kid113 X 39 MOB T X ah awem kA1rkB1q 0 Therefore rdxdt k kgtlt xx expir m w 11132008 o A a B and B a A o L 1 constant for such a process can be related for the forward and reverse rate constants k1 k 1 0 From before km k1 1671 TL K PthCIpe of microscopic EVE1539 fty Question A protein folds with a rate constant of 100 s39l At Eh melt temperaturequot the free energy of the folding transition is AGO 0 What is the observed rate constant for folding at this temperature B 200 s391 C 50 s391 D 0 s391 Question A protein folds with a rate constant of 100 s39l Att melt temperaturequot the free energy of the folding transition is AGO 0 What is the obsened rate constant for folding at this temperature A 100 s391 B 200 s391 C 50 s391 D 0 s391 Summay of rstorder proq ses 1 Firstorder processes have an exponential timercourse 2 The rate constant k can be related to a 1e time I ora halftime 112 K lr K mu 112 3 The unils of the rate constant are s391 4 The approach in equilibrium is the sum of forward and reverse rate constanis 5 The rate v is the instantaneous change slope 5 The uniis for the rate are moleslitersec Ms39l 11132008 Chemistry 431 Lecture 5 The First Law of Thermodynamics NC State University 912008 Background Thermodynamics is a macroscopic science that could e used with reference to atoms s century thermodynamics was deve oped to un erstand engines eg steam engines and other transformations of energy 0 3 E o m 3 539 m E r 3 m m E However our approach is to make constant reference to the microscopic world that we know exists based on many experiments Thus we will use an approach related to statistical thermodynamics We are concerned with the average properties of a collection ofa large number of atoms or molecules The Conservation of Energy Energy in thermodynamics is the capacity to do work The work occurs against an opposing force and can be de ned as a force operating through a distance When chemical change used to generate work there is otten heat generated as well We will show in this lecture that the sum ofthe heat generated and the work done is a constant amount of energy Thus the rst law of thermodynamics is an expression ofthe conservation of energy Internal Energy Heat Exchanged Work Done PressureVolume Work Students otten wonder why we begin with pressurevolume work After all there are lots of kinds of work starting with the basic de nition that work is a force acting through a distance However the work performed by an engine involves the ex ansion ofa gas to drive a iston and it is this work that is of practical interest in the design of steam and internal combustion engines It is also pressurevolume work that leads most directly to an understanding of the relationship between heat and work In order to study the transfer of energy in the form ofheat and work we de ne the system the engine or mechanical device we are interested in and the surroundings Basic Definitions System and Surroundings An open system is a system that can exchange both energy and matter with its surroundings A closed system is a s stem that can exchange ener but not matter with its surroundings An isolated system is a system that exchange neither matter nor energy with its surroundings If a system is isolated we say that it is adiabatic Greek for not going through Heat does not go through the walls of container of an isolated system The opposite is diathermic Work and Heat We considera chemical reaction that generates both work and heat For example Zns 2 HClaq gt ZnCI2aq H2g Note that one mole ofH2 gas is reduced for each e uivalent 0 Zn consumed lfthe reaction takes place in a container that is tted with a piston the production ofgas will cause the 39ston to rise 399 work acti n 0 heat taking place Work and Heat lfwe constrain the piston so that it cannot move then the pressure in the container will rise This will result in an increase in the temperature PV nRT Thus more heat will be released into the surroundings in this case Although we have not accounted for the difference quantitatively we know that energy must be conserved and so the amount of heat released must equal the work that would have been done M I heat taking place 9 12008 Exothermic and Endothermic Reactions A reaction that releases heat is exothermic A reaction that absorbs heat is endothermic The reaction we considered on the preceeding slide is exothermic Endothermic reactions are ch less 39equent We will see later that entropy must account for the fact that such reactions can be spontaneous An example of an endothermic reaction is the dissolution of sodium nitrate in water This reaction is the basis of instant cold packs that are included in rst aid kits The measurement of work Work is de ned as the result of a force acting through a distance This easily seen for gravitation If an ob39ect of mass m is litted to a higher elevation in the gravitational eld ofthe earth the work done is w mgh Idwf2FdhmgrdhmghZ h quot1 quot1 The units of work are kg m2s2 which are also called Joules Equivalent sets of units are listed below J1Nm1Pam31kgmzs 2 Just as there is a sign associated with work ofli ing or letting something fall in a gravitational eld there is a si associated with work done and heat exchanged by the system The Sign Convention We can summarize the sign convention in the following table Work done w by system Heat flows a Alternatively we can depict the convention as follows wlt0 wgt0 qlt0 qgt0 The Sign Convention We can summarize the sign convention as follows 1 Choose the correct answer A Work done by the system is positive B Work done by the system is negative 2 Choose the correct answer A Heat absorbed by the surroundings is positive B Heat absorbed by the surroundings is negative Internal Energy Now that we have given a symbolic form to both heat q and work w we can state the rst law in terms of symbols AU w q The symbol U represents the internal energy ofthe system This is the energy that is conserved The rst law states that any change in the internal energy AU results either from We can also state the rst law in differential form dU 6w Sq In this case there is an important difference between the differential written with d and that written with 6 rdrtne reyerstpie ase and we call it w p c y w Jaywer PaulrFQV dv Patvzrvt aw m PM rndst pe a srnaiier pressdre expanSion tnan rpr tne reyerstpie Graphical Comparison The pide edrye ts an tsdtnerrnai patn T T from the initial State to the final State reP L andT 3EEllt And P L andT3nntlt t t t t t 100 lnitlal i mnmnw 80 PM i sanstantv Pressure atrn Graphical Comparison The pide curve ts an tsdtnerrnai patn T T from the initial State to the final State Here EIEIa L andT 3EIEIilt And P atrn L an T3ttntlt t t E 100 CenetantT E 80 i ConstantV e 60 Step 7 CunstantP a 40 Final 9 20 PtvtT D o 0 20 2 5 1 u 1 5 Volume L 912008 Graphical Comparison Graphical Comparison The plue curve ls an lsotherrnal path TT from the lhltlal state to the flhal state ere The plue curve rs an lsothermal path I T from the initial State to the final State E 1003 lnlllal LEM E 100 inmal If ALMW E 30 FVV T 7 Constantv 30 Pl V 7 Constantv 9 so P step 7 Constant P 93 60 P Stew 7 Constant P a 40 A Pr Vn Ta a 40 Pr Vn Ta a A g 20 Pl Vth 9 20 Pl Veri D D r r r r r r r r r oo 05 1390 1395 2390 25 on 05 10 1395 2390 25 Volume L Volume L Graphical Comparison Work of Egtltpansion he reo ano Vlolet lrnes represent the lrreVerslhle path the initial State to the final State Note thatt e pressure rall nrstto the nnal external pressure reo llne Then the expansron occurs at constant pressure Vlolet llne e plue curve rs an lsothermal path I T from the initial State to the final State ereP nu amV 4 L arldT 3EEllt a L ano sun k g 2 1004 l innan LN e 1004 l inlet 4 F 30 PquotV T 7 constantv fa 30 7 Constantv 9 60 PStep 7 Constant P g 60 F39Sien 7 ConstantP P V T P V T a 40 r r n a 40 r r Final g 20 Pl VlT 9 20 Pl vl Tl D m r r r r r r r r r r 0 0 0395 10 15 0 2395 0 O 10 15 V 25 Volume L Volume L Work of Egtltpansion Work of Egtltpansion The red and Vlolet llnes represent the lrreVerslhle path The red and Vlolet llnes represent the lrreVerslhle path fro the initial State to the final State Note that the from the initial State to the final State Note that the pressure rall nrstto the nnal external pressure reo llne et l pressure rall nrst to the nnal external pressure reo llne the exuanslon occurs at constant pressure Viol the expansron occurs at constant pressure Vlolet llne lrle r r r r r r r r r r r r 5 10 l gut3T i Comm 10 3 CcnstantT E 807 tr l 7 Cans an g 80 v l 4 7 Constantv 9 e eo PStep 7 CunstantP P Vl T a a 40 39 l Ftnat 9 9 20 PerTt a D o u 2 5 1 u 1 5 20 Volume L 912008 Work of Egtltpanslon The red and Vlolet llhes represent the lrreVerSlhie path from the lhltlai State to the flhai State Note thatthe pressure rall rlrst to the rlhal external pressure reo llhe Thenthe EXDahSlOH occurS at conStant ureSSure Vloiet ilne 39E 100 7 l 539 i rant r 7 5 3 F V T39 7 Constant v 9 so PStep 7 Constant P PM T a 40 39 Fmal 9 20 Pl vlTl a t l t l t t o o 0 5 2o 2395 1390 15 Volume L Work of Egtltpanslon The red and Vlolet llhes represent the lrreVerSlhie path from the lhltlai State to the flhai State Note that the pressure rall rlrstto the rlhal external pressure reo llhe Theht e ekpahsloh occurs at cohstahtpressure lwolet ilhe E 100 39Fm ii39 7 Constant 30 l 7 Constantv 9 60 PStep 7 ConstantP P V T a 40 39 Flnal 9 20 Pl wt D r r r t o o o 5 25 t 1 a 1395 20 Volume L Graphical Comparison graphlcaiiy asthe area uhoerthe curves For the reVerSloie work thlS lSthe area uhoerthe olue curve For the lrreVerSlhie work thlS iS the rectahole under the Vlolet ilhe 2 1004 l innan Low 2 10 l inmal 4 F 30 PquotV T 7 Constantv fa 30 Constantv 9 60 PStep 7 ConstantP g 60 F39Slen 7 ConstantP PVlT PVhT a 40 l l MI a 40 r 1 Final 9 20 PllVth 9 20 17s fillvtlTi D l m t t t t t t t t t t no 0 10 1395 390 2395 00 10 15 V 25 Volume L Volume L Graphical Comparison graphlcaiiy asthe area uhoerthe curves rorthe reVerSloie work thlS lSthe area uhoerthe olue curve For the lrreVerSlhie work thlS lSthe rectahole under the Vlolet ilhe Graphical Comparlson Slhce the work iS the integral lh PVV space ltcah represehteo graphlcaiiy asthe area uhoerthe curves For the reVerSloie WorkthlS lSthe area underthe t1 ue curve For the lrreVerSlhie WorkthlS lSthe rectanuie under the Violet ilne r r r r r E 100 i Emc39T 7Censtantr 80 39 l 7 Constantv 9 so F39Step 7Constantrgt PVT 40 rl rtnal 9 20 let D 00 05 20 2395 Graphical Comparison Clearly the work or ekpahsloh iS greatest ror the reVerSloie ath Thls iS alwa strue There are rh y posslole hone reVerSloie paths and the s sterh perrorrhs less work along anV ofthoSe uathS than lt doeS alone the reVerSlhie oath r r t t r E 100 n 3 Constant T E 80 quot l 7Constantv e 60 F39Step 7 ConstantP PVT a 40 39 Ftnal 9 20 PlVlTl D l on 10 15 2390 25 VolumeL 912008 Work of Compression The two patns snown peiow represent tne reyersrpie pioe ano srngiestep rrreyersrpie reo ano yroiet patnsrora cornpressron Forthe srngiestep patn tne external pressor rnost egoai tne nnai internal pressure ano itis strii a constant r A 100 g 80 Final Fslev L pr V r Pp Vi Te 3 so 7 a 40 g 20 cc r r r r r r O 0 D 5 1 O i 5 2 D 2 5 Volume L Work of Compression Thus the rrreyersrpie patn regorres tne sorroonorngsto oo rnore work on tne systern wgt u tnan tne reyersrpie patn Again we can see tne work grapnrcaiiy astne area onoer the curve 1003 r l r r r Pressure aim l l 00 05 20 25 I 1 1 O i 5 Volume L Work of Compression rnostne rrreyersrpie patn regorres tne sorroonorngs to go rnore work on tne systern wgt tnan tne reyersrpie patn A 100 7 r 3 so 7 Final P Step 39 7 11 pp Vrr Tr Pm Vn Te 2 60 a 40 inihai 3 20 r P Tr CL t r on 05 l r 20 25 r l KO i 5 olume L Isothermal Processes For an roeai gas tne internal energy rstne kinetic energy There is no potential energy roran roeai gas so tnrs rnost be true Therefore 3 A U eTNRT Um 7R7 For an rsotnerrnai processr2 T1 constant so A Ur magan 3an u The internal energy cnange is always zero ror an rsotnerrnai process ir we return to tne nrst iawwe see tnat AU w g u 07w The systern apsorps neat rrorn tne sorroonorngs wrtn an energy equal to tne work oone Isochoric and Isobaric Processes isocnonc rneans constantyoiorne The reo line in tne preceeorng ngores is a constant yoiorne process in socn rocessw srnce ov u Recall tnat 5w rPdV Therefore AU qvfor an rsocnonc process The v sopr scnpt rneans tnat v is neio constant isopanc rneans constant pressure in a constant pressure expansion sucn as we saw in rne preceemng rrarneswe know now to calculate tne work w VP oven we know tnatAU rortne rsocnonc process is rov Srnce tne eno pornt or tne rsocnorano rsopar is tne sarne as tnat ror tne isotherm we know tnat oyeraii AU u Therefore o7 oV PmAv The systern apsorps neat rrorn tne sorroonorngs wrtn an energy equal to tne work oone The relationship of qV and qp We can oecornpose tne internal energy ror tne constanty stant P steps into two parts AU1 and AUZ Since tne oyeraii AU u AU1 AU2 rqv ope PAy we can soiye for new obtain ue um PAV r E 100 n i Constantr E 80 i Constantv a eo step i CunstankP 5 my 40 r a Fl al 9 20 van D l l 2 o a 2390 2 5 1 u 1 5 Volume L Chemistry 433 Lecture 14 Gibbs Free Energy and Chemical Potential NC State University The internal energy expressed in terms of its natural variables We can use the combination of the first and second laws to derive an expression for the internal energy in terms of its natural variables If we consider a reversible process dU 8g 8w 5w PdV definition of work Sq TdS second law rearranged Therefore dU TdS PdV This expression expresses the fact that the internal energy U has a T when the entropy changes and a slope P when the VOIUme Changco vvc use expreSSiUIl LU UCI IVC the P and T dependence of the free energy functions The Gibbs energy expressed in terms of its natural variables IU the natural variables IUI Lhc Uluua unalgy vvc ucglll with the internal energy dU TdS PdV and substitute into dH dU PdV VdP to find dH TdS VdP S and P are natural variables of enthalpy and using dG dH TdS SdT we find dG SdT VdP T and P are natural variables of G Once again we see why G is so useful Its natural variables are ones that we commonly experience T and P The variation of the Gibbs energy with pressure We shown that dG VdP SdT This differential can be used to determine both the pressure and temperature de endence of the free enerw At constant term erature SdT O and dG VdP The integrated form of this equation is P2 AGf WW 1 P For one mole of an ideal gas we have agE 2 AGm RTL P RTlnP1 Note that we have expressed G as a molar quantity Gm Gn The variation of the Gibbs energy with pressure We can use the above expression to indicate the free energy at some pressure P relative to the pressure of the standard state P 1 bar GmT G T RT In T3 G0T is the standard molar Gibbs free energy for a gas As discussed above the standard molar Gibbs free energy is the free energy of one mole of the gas at 1 bar of pressure The Gibbs free energy increases logarithmically with pressure This is entirely an entropic effect Note that the 1 bar can be omitted since we can write RTln 1 garRTln P RTln 1RTln P The pressure dependence of G for liquids and solids If we are dealing with a Ii uid or a solid the molar volume is more or less a constant as a function of pressure Actually it depends on the isothermal compressibility K 1NaV6PT but K is very small u s a number of the order 10394 atm391 for liquids and 10396 atm391 for solids We have discussed the fact that the density of liquids is not strongly affected by pressure The small value of K is another way of saying the same thing For our ur oses we can treat the volume as a constant and we obtain GmT G T VmP 1 Question What is the isothermal compressibility K 1VaVaPT for an idealgas A K 1N B K 1P C K 1T D K nRTP Question What is the isothermal compressibility K 1VaVaPT for an idealgas A K 1N B K 1P C K 1T D K nRTP Question Compare the following two equations for the pressure dependence of the Gibbs free energy Which one applies to the formation of diamond trom graphite A GmT G T RTln 1 gar B GmT Gem leP 1 Question Compare the following two equations for the pressure dependence of the Gibbs free energy Which one applies to the formation of diamond trom graphite A GmT G T RTln 1 gar B GmT Gem leP 1 Systems with more than one component U to this I oint we have derived state functions for ure systems The one exception is the entropy of mixing However in order for a chemical change to occur we must have more than one component present we need generaze the methods to account for the presence of more than one type of molecule In the introduction we stated that we would do this using a quantity called the chemical potential The chemical potential is nothing more than the molar Gibbs free energy of a particular component Formally we write it this way Hi an I moles of i changes with all other Rate of change of G as number of TPj i variables held constant Exam le a as hase reaction Let s consider a gas phase reaction as an example We will use a textbook example N204 9 2 N02 9 We know how to write the equilibrium constant for this reaction P502 PN2O4 At constant T and P we will write the total Gibbs energy as K d6 llVozano2 N204an204 d6 ZHNozdn N204dn We use the reaction stoichiometry to obtain the factor 2 for N02 Definition of the Gibbs free energy change for chemical reaction We now define AFXHG Ame dnTP This is AWG but it is not AmnGO Note that we will use AWG and AG interchangably If we now apply the pressure dependence c one compoc 0 P GmT G T RTln m to a multicomponent system p 0 I W pT RTln 1 bar These two expressions are essentially identical The chemical potential pi is nothing more than a molar free energy Question How should one think of the chemical potential p of component j A It is the potential energy of that component l5 It IS a monar rree energy of tnat component C It is potential entropy of that component D It is a molar entropy of that component Question How should one think of the chemical potential p of component j A It is the potential energy of that component l5 It IS a monar rree energy of tnat component C It is potential entropy of that component D It is a molar entropy of that component Single component GmT G T RTln 1 gar Multiple components each have a p pT pT RTn P 1 bar

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