Applying the condition for equilibrium to this equilibrium, we get (1) n i i =, r G and 5 i

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CHEMICAL EQUILIBRIA The Thermodyamc Equlbrum Costat Cosder a reversble reacto of the type 1 A 1 + 2 A 2 + W m A m + m+1 A m+1 + Assgg postve values to the stochometrc coeffcets o the rght had sde ad egatve values to the oes o the left, such a reacto ca be wrtte the geeral form 5 A = 0 Applyg the codto for equlbrum to ths equlbrum, we get 5 = 0 (1) However, Substtutg above, we get = + RT l a 5 = 5 + RT 5 l a = 5 + RT l 2 a (2) (3) Usg Eq (1), ad the deftos 5 =, r G ad 5 =, r G, we get, r G = RT l 2 a Sce the stadard free eergy s oly a fucto of temperature (t s defed wth respect to a stadard pressure, whch s a costat), the quatty the logarthm wll be a costat at costat temperature as log as Eq (1) s satsfed, e, the system s equlbrum Therefore, we wrte K = 2 a, ad, r G = RT l K (4) Eq (4) s the defto of the thermodyamc equlbrum costat 1

Chemcal Equlbrum Ivolvg Gases Cosder a reacto of the type A A(g) + B B(g) W C C(g) + D D(g) Durg the reacto, the partal fugacty of each gas chages The resultg free eergy chage for each gas ca be expressed as,g A = A RT l f A,2 f, A,1,G B = B RT l f B,2 f, ad so o B,1 If we assume that the reacto starts wth each gas at the stadard pressure, deoted as p, we may wrte G A = G A + A RT l G B = G B + B RT l f A p, f B p, ad so o It s mportat to remember that the free eerges the expressos above are NOT molar quattes, e, we eed to keep md that G A = A G m, A, G A = A G m, A, etc Now, the free eergy chage for the reacto may be wrtte as, r G =, r G + RT l f C/p C f D /p D f A/p A f B /p B, where ad r G s smlarly defed r G = (G C + G D ) (G A + G B ) = ( C G m, C + D G m, D ) ( A G m, A + B G m, B ), 2

Now, at equlbrum, r G = 0 ad so, we get, r G = RT l f C/p C f D /p D f A/p A f B /p B eq The partal pressures that eter to ths expresso are the values measured at equlbrum ad, therefore, s a costat at a gve temperature Note that, because of the dvso of each pressure term by the stadard pressure, the quatty wth the square brackets s dmesoless I other words,, r G = RT l K p,or K p = exp, r G /(RT) For deal gases, the partal fugactes are replaced by the partal pressures The equlbrum costat ca also be expressed terms of cocetratos For example, assumg deal behavor, we substtute p A = A RT/V, etc, ad recogze that A /V = [A], the cocetrato of A etc, to get K p = K C RT C + D A B, p where K C = [ C] C [D] D [A] A [B] B Also, by recallg that p A = x A p tot, where x A s the mole fracto of A ad p s the total pressure at equlbrum, we ca express the equlbrum costat terms of mole-fractos K x : K P = K x p C + D A B, p where K x = x C C x D D x A A x B B 3

Chemcal Equlbra Soluto: Cosder a reacto of the type P P(aq) + Q Q(aq) W R R(aq) + S S(aq) If the stadard state s take to be 1 mol of solute per klogram of solvet (e, a 1 molal soluto), a defto s possble wth referece to a Hery s Law stadard state, so that a = γ m K a = a R R a S S a P P a Q Q =? R R? S S? P P? Q m R R m S S Q m P P m Q Q Heterogeeous equlbrum: The actvty of pure solds ad lquds s accepted to be exactly equal to 1 So, the equlbrum costat for the reacto s H 2 O(l) W H + (aq) + OH (aq) K = a H +a OH a H2 O = a H +a OH The equlbrum costat for CaCO 3 (s) W CaO(s) + CO 2 (g) s K = a CaO(s)a CO2 (g) a CaCO3 (s) = a CO2 (g) = f CO 2 p 4

Le Chateler Prcple: If a system dyamc equlbrum s dsturbed, the system wll adjust tself such a way as to couteract, as far as possble, the effect of that chage Types of dsturbaces to chemcal equlbra we wll cosder are: (a) chage volume, (b) chage pressure, (c) chage temperature, (d) chage cocetrato or partal pressure (a) chage volume: Cosder the reacto AB(aq) W A + (aq) + B (aq) The equlbrum costat K C (gorg the stadard cocetrato terms for smplcty) s gve by: K C = c A + c B c AB = ( A +/V) ( B /V) ( AB /V) = A + B AB 1 V So, f the umber of moles of each speces remaed the same, as the system s dluted, the equlbrum costat decreases Sce ths caot happe (K C s a costat at a gve temperature), the umber of moles of A + (aq) ad B (aq) must crease wth creasg volume (b) chage pressure: Cosder the reacto N 2 (g) + 3H 2 (g) W 2NH 3 (g) Smlar argumets as above ca be made for K p as pressure s chaged However, t s suffcet to ote that there are 4 moles of gases o the reactat sde ad oly two moles of gases o the product sde Whe pressure s creased holdg temperature costat, the equlbrum shfts the drecto of the products so that the volume shrks Whe pressure s decreased, the equlbrum shfts the drecto of the reactats so that volume s creased (c) chage temperature: If a reacto s exothermc the forward drecto, a decrease temperature drves the reacto forward to couteract the decrease temperature A crease temperature wll drve the reacto the reverse drecto sce the reverse reacto wll be edothermc (d) chage cocetrato or partal pressure: Addg reactats or removg products drves the equlbrum towards products, removg reactats or addg products drves the equlbrum towards reactats 5

Temperature Depedece of Equlbrum Costats Applyg Eq (347) to a reversble reacto, we may wrte T, r G T =, rh P T 2 Sce, r G = RT l K p, substtutg for r G, we get T l K p = d p dt l K p =, rh RT 2 O rearragg ad tegratg, we get l K p =, rh RT + cost (5) whch mples that whe l(k p ) s plotted as a fucto of 1/T, the slope wll be equal to H /R However, we also kow that G = H T S Therefore, we get l K p =, rg RT =, rh RT +, rs R Comparg Eqs (5) ad (6), we see that f r H s depedet of temperature, the tercept of the le wll be equal to the etropy chage for the process A more sophstcated treatmet could clude the temperature depedece of the reacto ethalpy so that we wrte, r H =,H I +,at +,b 2 T2,c T +,d 3 T3 ad a smlar expresso for the etropy of reacto It s also possble to show that T l K C d p dt l K C =, ru RT 2, where the umercal value of the equlbrum costat (wthout uts) s to be used the logarthm (6) 6