General Physical Chemistry I
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1 General Physical Cheistry I Lecture 12 Aleksey Kocherzhenko Aril 2, 2015"
2 Last tie "
3 Gibbs free energy" In order to analyze the sontaneity of cheical reactions, we need to calculate the entroy changes in both the syste and the surroundings" Notice, however, that the change in the entroy of the surroundings is exressed in ters of a roerty of the syste the reaction enthaly:" rs sur = H r hus, the condition for sontaneity of a cheical reaction," ds universe =ds sys +ds sur > 0 can be exressed in ters of the roerties of the syste only:" ds sys, dh sys ds sys < 0 J. Willard Gibbs" dh sys > 0 Introduce the Gibbs free energy:" G = H S At constant teerature:" dg =dh ds he Gibbs free energy only deends on state functions, so it is a state function itself! Criterion for sontaneity of cheical reactions (or other rocesses):" dg <0
4 Sontaneity and equilibriu conditions" ds universe =ds sys +ds sur ) ds sys =ds universe ds sur he change in the Gibbs free energy can then be written as follows:" dg =dh sys ds sys = dh sur =dh sys ds universe {z } =ds sys =dh sys ds universe + dh sys {z }!! because"dh sys = dh sur = = ds universe At constant teerature and ressure, the criterion for sontaneity of a rocess is:" ds universe > 0, dg = ds universe < 0 Equilibriu is achieved when"dg =0
5 Phase transitions"
6 Variation of the Gibbs free energy" Gibbs free energy:" G = H S he change in the Gibbs free energy is:" dg =dh ds Sd =du + dv + V d ds Sd H = U + V he first law of therodynaics:" dq =du +da If the transition fro the initial to the final state is reversible:! ) du =dq da {z} =dv ds = dq = ds dv or" dq = ds for a reversible rocess! ) the change of the Gibbs free energy in a reversible rocess is:! dg = ds dv + dv + V d ds Sd = V d Sd But the Gibbs free energy is a state function: does not deend on the rocess by which the syste transfers fro the initial to the final state!
7 Gibbs free energy variation at" = const Gibbs free energy variation:"dg = V d If" = const Sd ) d =0 ) dg = Sd An increase in the teerature at constant ressure leads to a decrease" in the Gibbs free energy" he Gibbs free energy is an extensive quantity (deends on the aount of substance in syste)" ) use olar Gibbs free = S Ø Largest value for gas (ost disordered)" Ø Saller for liquid (ore ordered)" Ø Sallest for solid (ost = S By the statistical definition, entroy = natural logarith of the # of icrostates that corresond to a articular acrostate ositive const" # of icrostates 1, thus the logarith is 0, and the entroy 0" he derivative of is the sloe of " G G ( ) with If the sloe of a function is negative, then the function decreases as the variable increases"
8 Gibbs free energy variation at" = S Ø Saller for liquid (ore ordered)" Ø Largest value for gas (ost disordered)" Ø Sallest for solid (ost ordered)" If the olar entroy is assued to be constant over a range of teeratures:" Largest sloe for gas" Phase transition" Sallest sloe for solid" Phase transition" Sontaneous rocesses tend towards lowest Gibbs free energy à ost stable hase has lowest Gibbs free energy! If at soe ressure liquid is never the hase with lowest Gibbs free energy à subliation occurs!
9 Gibbs free energy variation at" = const Gibbs free energy variation:"dg = V d Sd If" = const ) d =0 ) dg = V d or:" An increase in the ressure at constant teerature leads to an increase in the Gibbs free energy" () is not a straight line" G (g) For = V {z} >0 = V G (s) () G (l) and" are straight lines " () Ø Solids and liquids are (alost) incoressible à " V (s) const < V (l) const with a few excetions, ost notably H 2 O" (solids are usually denser than liquids)" Ø Gasses are coressible, for erfect gasses: "V = R ) V = R G (g) () Sloe for is not constant à not a straight line"
10 Gibbs free energy variation at" = const Ø For erfect gasses: "V = R Variation of the Gibbs free energy with ressure for a erfect gas:" Z f Z f = i V ()d () is not a straight line" G (g) G (s) () = R G (l) and" are straight lines " i () d ) V = R G (g) = G (g) ( f ) G (g) ( i ) = R (ln f ln i ) or:" G (g) Standard olar" Gibbs free energy" = = R ln f i () =G + R ln 1 bar" he stable hase of a substance is deterined by both the teerature and the ressure"
11 Phase diagras" Mas showing the therodynaically stable hases for each and," hase regions in a hase diagra are searated by hase boundaries:! Solid and liquid hases in equilibriu,"g (l) = G (s) Liquid and gas hases in equilibriu,"g (g) = G (l) Points:" Solid and gas hases in equilibriu," G (g) = G (s) A gas hase stable" B gas and liquid in equilibriu" C liquid hase stable" D liquid and solid in equilibriu" E solid hase stable"
12 Characteristic oints" Noral vs. anoalous liquid! Highest teerature at which a liquid can exist" Lowest ressure at which a liquid can exist" (for a noral liquid, also the lowest teerature at which liquid can exist)" Equilibriu between hases is dynaic: there are as any olecules leaving a hase er second as there are olecules coing back to it!
13 he hase rule" Four hases of a single substance can t coexist at utual equilibriu why?! If they did, we would need the following indeendent conditions to be satisfied:! G (I) (, )=G (II) (, ) G (II) G (III) Phase indices" (, )=G (III) (, ) (, )=G (IV) (, ) hree equations, but only two variables" he hase rule! F = C P +2 Generally, these conditions cannot all be satisfied! # of coexisting hases" # of coonents (different substances)" # of degrees of freedo (indeendent variables: ressure, teerature, ole fractions)"
14 Phase diagras of H 2 O and CO 2" Water" Carbon dioxide" 1 bar" 273 K" 373 K" At 1 bar, water freezes at 0 C (273 K) and boils at 100 C (373 K)" At low ressure, the structure of ice is deterined by H-bonds" At high ressures, there are ultile hases of ice with higher densities" At 1 bar, carbon dioxide does not have a stable liquid hase for any " he trile oint is at 5 bar and 217 K, liquid CO 2 exists at higher ressures"
15 Calculating hase boundaries " wo hases are in equilibriu if" V (II) V (I) {z } transitionv G (I) (, )=G (II) (, ) So long as we are on a hase boundary, this condition ust reain satisfied as we change the ressure and the teerature" he variation in the olar Gibbs free energy is: "dg = V d S d As we change and, the change in the olar Gibbs free energy for two hases needs to be the sae in order to reain on the hase boundary:" dg (I) =dg (II) Rearrange the ters:" ) V (I) d S (I) d = V (II) d S (II) d = ) d = transition S transitionv d S (II) S (I) {z } transitions d d
16 We found that:" he Claeyron equation " d constant and :" transition S transitionv d transitions = transition H ) d = transition H transition V d Claeyron equation" Éile Claeyron" As we ove along the hase boundary, a sall change in teerature leads to a corresonding sall change in ressure given by the Claeyron equation" ) d d = transition H transition V for elting, enthaly of fusion is > 0 (elting is always endotheric)" density of noral liquids is saller than for solids, density of anoalous liquids is larger" Sloe of the hase boundary" à Sloe of solid-liquid hase boundary is ositive for noral and negative for anoalous liquids"
17 he Clausius-Claeyron equation " he Claeyron equation:" d = transition H transition V d For the liquid-solid transition, the change in volue for the transition is alost constant across a range of teeratures" à for gasses, the volue deends on teerature" V (g) V (l) ) transition V = V (g) à if the vaor is assued to be a erfect gas:" ) d transitionh R 2 d V (l) V (g) V (g) = R Rudolh Clausius" or:" d {z} d(ln ) transition H R 2 d Éile Claeyron"
18 d(ln) he Clausius-Claeyron equation " transition H R 2 d Clausius-Claeyron equation in differential and integrated for" Exercise:! Estiate the change in ressure that is required to lower the elting oint of ice fro 0 to 3 ºC. " he densities of ice and water at 0 ºC are and g c 3, resectively. he enthaly of fusion for H 2 O at 0 ºC is kj ol 1." ) Z ln f ln i ) ln f i d(ln) Z f i transition H R transitionh R 2 d 1 i 1 f
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