Introductory Physical Chemistry
Introductory Physical Chemistry CH 331
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This 2 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 331 at North Carolina State University taught by Stefan Franzen in Fall. Since its upload, it has received 16 views. For similar materials see /class/223995/ch-331-north-carolina-state-university in Chemistry at North Carolina State University.
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Date Created: 10/15/15
Chemistry 331 Lecture 12 Free Energy Functions NC State Un39vers39ty System and surroundings both play in role in the entropy In an isolated system the criterionds gt 0 indicates that a process is spontaneous In general we must consider dSsvs for the system and dSSM for surroundings Since we can think ofthe entire universe as an isolated systemdsm gt 0 The entropy tends to increase for the universe as a whole lfwe decompose dsm into the entropy change for the system and that for the surroundings we have a criterion for spontaneity for the system that also requires consideration ofthe entropy change in the surroundings The free energy functions will allow us to eliminate consideration ofthe surroundings and to express a criterion for spontaneity solely in terms of parameters that depend on the system Free Energy at Constant T and V Starting wi h the First Law dU 5 At constant temperature and volume we have SW 0 and M Recall that d3 2 6qT so we have dU STd which leads to dS S 0 Since T and V are constant we can write this as u Ts lt o The quantity in parentheses is a measure ofthe spontaneity ofthe system that depends on known state functions Definition of Helmholtz Free Energy We de ne a new state function A U TS such that dA S0 We call Athe Helmholtz free ener y At constant T and V the Helmholtz 39ee energy will decrease until all possible spontaneous processes have occurred At that point the system will be in equilibrium The condition for equilibrium is dA 0 time Definition of Helmholtz Free Energy Expressing the change in the Helmholtz free energy we have for an isothermal change from 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 e write AA AU TAS S 0 at constant T and lfAA is greater than zero a process is not spontaneous It can occur ifwork is done on the system however The L l L ILIA Noting theqEV TAS we have A qvev According to the rst law AU qEV wEV so wEV reversible isothermal A represents the maximum amount of reversible work that can be extracted 39om the system De nition of Gibbs Free Energy Most reactions occur at constant pressure rather than constant volume Using the facts that qw TdS andwEV PdV we have dU S Td P which can be written dU TdS PdV S 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 constant T and P We de ne a state function G U PV T3 H TS Thus dG S 0 at constant T and P The quantity G is called the Gibb39s 39ee energy I a system at constant T and P the Gibb39s energy will decrease as the result of spontaneous processes until the system reaches equilibrium where d6 0 Comparing Gibbs and Helmholtz The quantity G is called the Gibb39s 39ee 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 0 Comparing the Helmholtz and Gibb39s 39ee energies we see that AVT and GPT are completely analogous except that A is valid at constantV and G is valid at constant P We can see that G A PV which is exactly analogous to U the relationship between enthalpy and internal energy For chemical processes we see AG AH TAS g 0 at constant T and 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 ofthe Gibbs energy However from this point we will consider the implications ofthe Gibbs energy for physical and chemical processes There are four possible combinations ofthe sign of AH and AS in the Gibbs free energy change gt 0 u A A Exotherrnic sontaneous forT ltAHAS lt0 Exotherrnic sontaneous forallT muim 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 wr39te 4G 6312qu Gszowj where we have expressed G as a molar 39ee energy From the de nition of free energy we have 49 AM 7 TAWS The magnitude ofthe molar enthalpy ofvaporization is 407 kJmol and that ofthe entropy is 1089 JlmolK Thus AG4u 651d maf r NKISKKlEIBB J K mar Gibbs energy for a phase change However ifwe were to calculate the free energy of vaporization at 36315 Kwe would nd that it is 11 kanol so vaporization is not spontaneous at that temperature lfwe consider the free energy ofvaporization at 38315 K it is 108 kJmol and so the process is spontaneous DG lt 0 State Function Summary developed U internal energy U PV enthalpy S entropy A U T3 Helmholtz free energ G U PV Ts H Ts Gibbs 39ee energy Please note that we can express each ofthese in a differential rm This simply refers to the possible changes in each function expressed in terms ofits dependent variables dH dUPdVVdP dAdUTdSSdT dG dH TdSSdT
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