Physical Chemistry II
Physical Chemistry II CH 433
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This 9 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 179 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 12 Free Energy Functions NC State University 9232008 System and surroundings both play in role in the entropy In an isolated system the criterion dS gt 0 indicates that a process is spontaneous In general we must consider dSW5 forthe system and dSSW for surroundings Since we can think of the entire universe as an isolated system dSmta gt 0 The entropy tends to increase for the universe as a whole lfwe decompose dSt a into the entropy change for the system and that forthe surroundings we have 39 of the entropy change in the surroundings functions will a low us to eliminate consideration 0 t e surroundings and to express a criterion for spontaneity solely in terms ofparameters that depend on the system Free Energy at Constant T and V Starting with the First Law At constant temperature and volume we have SW 0 and U Sq Recall that dS 2 6qu so we have dU STdS which leads to dU TdS S 0 Since T and V are constant we can write this as dU TS lt o The quantity in parentheses is a measure ofthe spontaneity ofthe system that depends on knovm state functions De nition of Helmholtz Free Energy We de ne a new state function A U TS such that dAg 0 We call A the Helmholtz free energy At constant T and V the Helmholtz 39ee energy will decrease until all possible spontaneous processes have occurred At that oint the sstem will be in e uilibrium The condition for equilibrium is dA 0 time Question The statement dA g 0 means A The condition for equilibrium is dA 0 B Processes are not spontaneous if dA lt 0 C dA cannot be greater than 0 DAl ofthe above time Question The statement dA g 0 means A The condition for equilibrium is dA 0 B Processes are not spontaneous if dAlt C dA cannot be greater than 0 DAll ofthe above time 9232008 De nition of Helmholtz Free Energy Expressing the change in the Helmholtz 39ee energy we have AU TAS for an isothermal change 39om one state to another The condition for spontaneous change is that AA is less than zero and the condition for equilibrium is that AA 0 We write AA AU TAS g 0 at constant T and V lfAA is greater than zero a process is not spontaneous It can occurifwork is done on the system however The Helmholtz 39ee energy has an important physical interpretation Noting the qEV TAS we have AA AU qrev According to the rst law AU qEV wEV so wEV versible isothermal A represents the maximum amount of reversible work that can be extracted from the system Question AA wEV means that the Helmholtz 39ee energy is equal to the maximum amount of reversible work that can be extracted from the system This follows 39om the fact that A The reversible heat is equal to TAS BAis a state Jnc ion 0 The reversible work is the maximum wo k DAll ofthe above Question AA wEV means that the Helmholtz 39ee energy is equal to the maximum amount of reversible work that can be extracted 39om the system This follows 39om the fact that A The reversible heat is equal to TAS B A is a state Jnction C The reversible work is the maximum wo k DAll ofthe above Definition of Gibbs Free Energy Most reactions occur at constant pressure rather than constantvolume Using the facts that SqEV gTdS and 6wEv PdV we have dU gTdS PdV which can be written dU TdS PdV 0 The sign applies to an equilibrium condition and the lt sign means that the process is spontaneous Therefore dU TS PV g 0 at constantT and P We de ne a state function G U PV TS H TS Thus G g 0 at constant T and P The quantity G is called the Gibb39s free energy In a system at constant T and P the Gibb39s energy will decrease as the result of spontaneous processes until the system reaches equilibrium where dG 0 Comparing Gibbs and Helmholtz The quantity G is called the Gibb39s free energy In a system at constant T and P the Gibb39s energy will decrease as the result ofspontaneous processes until the system reaches equilibrium where 0 Comparing the Helmholtz and Gibb39s free energies we see that AVT and GPT are completely analogous except thatA is valid at constant V and G is valid at constant P We can see that PV which is exactly analogous to H U PV the relationship between enthalpy and internal energy For chemical processes we seet at G AH TAS g 0 at constant Tand P AA AU TAS g 0 at constant T and V Conditions for Spontaneity We will not use the Helmholtz free energy to describe chemical processes It is an important concept in the derivation of the Gibbs energy However from this point we will consider the implications of the Gibbs energy for physical and chemical processes There are four possible combinations of the sign of AH and AS in the Gibbs free energy change Description of process ontaneous for T gt AHAS Exothermic sontaneous for T lt AHAS Exothermic sontaneous for all T gt0 Question Fora given reaction we have AH gt 0 and AS lt 0 When will the reaction will be spontaneous A never B when T gt AHAS 0 always LIVVII I I AHAS 9232008 Question Fora given reaction we have AH gt 0 and AS lt 0 When will the reaction will be spontaneous A never B when T gt AHAS 0 always u ImIeII I AHAS m Description of process gt h m39c sontaneous for T gt AHAS ontaneous for T lt AHAS Exothermic sontaneous for all T Gibbs energy for a phase change For a phase transition the two phases are in equilibrium Therefore AG 0 for a phase transition For example for water liquid and vapor are in equilibrium at 37315 K at 1 atm of pressure We can write Ame GHzoggt GHzoo where we have expressed G as a molar 39ee energy From the de nition offree energy we have Avame Ava Hm TAvap n The magnitude ofthe molar enthalpy ofvaporization is 407 kJmol and that ofthe entropy is 1089 JmolK SI Avian 65 Id molquotr37315 Kl8 9 J Kquot malquotu Question Which statement is true for a phase transition AAG0andAS0 DAS0andAH 0 Question Which statement is true for a phase transition AAG0andAS0 DAS0andAH 0 Gibbs energy for a phase change However ifwe were to calculate the free energy of vaporization at 36315 K we would nd that it is 11 kJmol so vaporization is not spontaneous at that temperature lfwe consider the tree energy ofvaporization at 38315 K it is 108 kJmol and so the process is spontaneous AG lt 0 State Function Summary At this point we summarize the state Jnctions that we have developed U internal energy U PV enthalpy S entropy A U TS Helmholtz 39ee ener G U PV TS H TS Gibbs free energy Please note that we can express each of these in a differential form This simply refers to the possible changes in each function expressed in terms of its dependent variables dH dU PdVVdP d U TdS SdT dG dH TdS SdT 9232008 Question We have shovm the dU TdS PdV This means that the natural variables of internal energy are entropy and volume What are the natural variables of the enthalpy A dH TdS VdP entropy and pressure B dH TdS PdV entropy and volume 0 dH SdT VdP tem erature and ressure D dH SdT PdV temperature and volume Q uestion We have shovm the dU TdS PdV This means that the natural variables ofinternal energy are entropy and volume What are the natural variables of the enthalpy A dH TdS VdP entropy and pressure B dH TdS PdV entropy and volume 0 dH SdT VdP tem erature and ressure D dH SdT PdV temperature and volume Question We have shovm the dU TdS PdV This means that the natural variables of internal energy are entropy and volume What are the natural variables of the Gibbs 39ee energy A dG TdS VdP entropy and pressure B dG TdS PdV entropy and volume e and ressure D dG SdT PdV temperature and volume Q uestion We have shovm the dU TdS PdV This means that the natural variables ofinternal energy are entropy and volume What are the natural variables of the Gibbs free energy A dG TdS VdP entropy and pressure m a C II a n 39U a lt 8 EL 3 1 lt m 1 a lt me n ressure o u C dG SdT VdP tem erature a d D dG SdT PdV temperature and volume Chemistry 433 Lecture 12 The Third Law NC State University The Third Law of Thermodynamics The third law ofthermodynamics states that every substance has a positive entropy but at zero Kelvin the entropy is zero for a perfectly crystalline substance The third law introduces a numerical scale forthe entropy Stated succinctly 80 0 for all perfectly ordered crystalline materials It is commonly said that all motion ceases at absolute zero Certainly all translation and rotation have ceased at absolute zero However the nuclei still vibrate about their equilibrium positions in socalled zero point motion r a olecules are in their lowest vibrational state and entropy is zer in any case provided there are no imperfections in the crystal Residual Entropy The rst statistical picture of entropy was that of Boltzmann The Jnction W that represents the number ofways we can distribute N particles into a number of states Using the function W the entropy can be expressed as S k In At zero Kelvin the s stem is in its lowest ener state a perfect crystal there is only one way to distribute the energy and W 1 therefore However the entropy not equal to zero at T 0 K if the substance is not a perfect crystal Although the residual entropy in su cases is a s all c rection to the entropy calculated for chemical reactions it still leads to an important concept CO an Imperfect Crystal The molecule 00 has a very small dipole moment and there is a nite chance that 00 will crystallize as OCOCO instead ofCOOCCO For each 00 molecule there are two possible orientations of the molecule therefore there are two ways each 00 can exist in the lattice The number ofwa s er molecule is w 2 for each 00 lfwe have N 00 molecules there are wN wa s or 2N wa s that all ofthe 00 can be distributed Therefore the entropy at zero Kelvin is SkanklnvWNkln wnR In 2 The entropy at zero Kelvin is known as residual entropy There are number of substances that show similar statistical variations in orientation that lead to a residual entropy Question Suppose there are four ways the the molecule CFClBrl can be oriented in a crystal lattice at zero Kelvin What is the molar residual entropy A4R ln2 B 2R In 4 c R ln2 D R ln4 Question Suppose there are four ways the the molecule CFClBrl can be oriented in a crystal lattice at zero Kelvin What is the molar residual entropy A4R ln2 B 2R In 4 c R ln2 D R ln4 Question Which phrase best summarizes residual entropy A the entropy ofimperfect crystals B the entropy ofthe lowest energy state C the entropy of rotation D the entropy of residues Question Which phrase best summarizes residual entropy A the entropy ofimperfect crystals B the entropy ofthe lowest energy state C the entropy of rotation D the entropy of residues The Temperature Dependence of Entropy We have seen calculations for the entropy change for processes However it is also possible to calculate the absolute entropy We can begin with the de nition dS dqmT The heat transferred during a process at constant volume is dqWEV CVdT Thus the entropy change at constant volume is AS smr 50 I WT We have kept the equation general by showing CVT as a function oftemperature This calculation ofthe entropy is valid only at constant V At constant P we nd an analogous expression Starting with the heat transferred dqp Ev deT T We have Assmis0f w n The temperature dependence of the heat capacity The third law of thermodynamics shows that It is necessary to treat CpT as a function oftemperature for this reason alone The Einstein and Debye theories of heat heat capacity of at these low temperatures The Debye law states that the heat capacity depends on T3 nearT 0 K At temperatures close to T 0 CW aT3 The constant a is an empirical constant For practical calculation ofthe entropy experimental values can be used and the integrals are evaluated numerically Question lfthere is no residual entropy then atT 0 Kwe can state that ASgt0andegt0 DS0ande0 Question lfthere is no residual entropy then atT 0 Kwe can state that ASgt0andegt0 DS0ande0 Question Cunsider tne statement cy e u as T u Wnten cunsequence ts valid A it ts harder tn neat matternearapsuiute zeru t a ts c it gets eastertet reaen apsetiute zeret peeause tt ts easter r t E D Eutn El and C are current Question Cunstuertne statement cy e u as T u Wnten cunsequence ts valid A it ts harder tn neat matternear apsuiute zeru tnat ts D Eutn El and C are eurrect Absolute Entropy Tne apsetiute Entropy ean be calculated rretm u tet any temperature T ustng tne tntegrai pr tne functtun Graphical representation cpmT irtnere ts a pnase transtttetn between D and temperature T M We can aisu eaieuiate tne euntrtputtun tn tne Entrupyfrum m tne transtttun E mcmur ASH wcmu A H cmur g sum t m t at t 3 mi t t t Jr ne neat eapaetty etr eaen pnase ts different and tne neat a at tie a 25a 2amp0 eapaetttes are alsu a runettetn at temperature Remember quotquot 39 quotquot quot tnattne neat eapaetty appruaenes zeru and tne temperature guestu zeret su tnattne cp ts a strung runettetn etrT at very 3m Im cTdr luvv temperature a r Graphical representation an wt 54 reluctante Graphical representation Liquid 1 Sulld a an inn 9 2m 2w unwittmtxt quot Omar Aw Wquot Ctmdr AWH r rm h r T Graphical representation ii an m isa 2i 2 Macmillan cTdT f Tm MW 17quot 3de Y r Graphical representation ui m Liguid iw scilid u N n in 2m int 59 2i renown ixi I M cilndr g Am a T rm w crdr AWH f Tim7 Graphical representation Gas Liguid scilid a an im i 200 mimiuuiei Omar AM 7quot Cimd T Tm T Graphical representation Gas Liguid a m in m 200 rtnipmiuiuixi m 7de AWN w crdr r TS m r Absolute Entropies can be used to calculate Reaction Entropies Entruples are tabulated ln urder tn facilitate the ealeulatlun ciftne entrupy cnange cif cnemical reacticins Furthe general reacticin aA tne standard entrupy cnange is glven by A a yS Y 25 r a5 A r b5 El Wnere tne absulute Entruples squot are mcilarguantities the tne significant difference eurnpared tci enthalpy There s e u ed tnat case tne enthalpy cif fcirmaticin Was set arbitrarily tci zerci since entruples are zerci and T El vae can use 5a as an abscilute guantity The standard reaction entropy The standard reacticin entrupy A5quot istne difference between tne standard mcilar entruples cf the reactants and pruduets Witn eacn term Welghted by the lelent stciicnicimetric ccie As Z vspfodumsr Z vsiueadants The standard state is fcir reactants and prciducts at l bar cifpressure The unit cif energy used lSJm ch lMF39ORTANT Dci ncit ccinfuse entrupy and enthalp v One ccimmcin mistake is tci set tne Entruples cifelements ti egual tci zerci as cine dcie Elements have an entrupythat is net zerci temperature is T El K sfcirentnalpies ciffcirma cin unless the An example formation of H20 We apply the absolute entropies osz 02 and H20 to the n for calculation ofthe entropy of reactio 2H29 029 gt 2 H200 The entropy change is AS 2 S Hzoi 3102i 9 239H2i 9 200 2131 205 JmolK 327 JmolK This result is not surprising when you consider that 3 moles as are being consumed A gas has a greater number oftranslational degrees of 39eedom than a li 39 On the other hand it is dif cult at rst to understand how a spontaneous reaction can have such a large negative entropy Question Consider the formation ofCOz 05 029 gt 0029 Which is correct A AS 0 equilibrium B AS AHT phase transition 0 A0 pug g 802 g S 0 for solids D AS 8002 g 802 g SC s reaction Question Consider the formation ofCOz 05 029 gt 0029 Which is correct A AS39 o equilibrium B AS39 AHT phase transition 0 o ouvz g 802 g S 0 for solids D AS 8002 g 802 g SC s reaction The spontaneity of chemical reactions The spontaneity of chemical reactions must be understood by consideration of both the system and the surroundings This is one of the most subtle points ofthermodynamics The heat dissipated by the negative enthalpy change in the water proceeds with explosive force The heat dissipated in the surroundings AH is also an entropy term The entropy change in the surroundings is A H6 r36 The entropy change ofthe surroundihgs is As 572 kJmol1298 K 192 X 103 JmolK which is much larger than the negative entropy change ofthe system 327 JmolK Question An endothermic reaction can be spontaneous if AASgt0 BASlt0 C AS D none ofthe above Question An endothermic reaction can be spontaneous if AASgt0 BASlt0 C AS D none ofthe above
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