Dsrder and Suppse I have 10 partcles that can be n ne f tw states ether the blue state r the red state. Hw many dfferent ways can we arrange thse partcles amng the states? All partcles n the blue state: 1 cnfguratn _J J J J J J J J J J _ Dsrder and 9 partcles n the blue state; 1 partcle n the red state: 10 cnfguratns _J J J J J J J J J J _J J _J J J J J J J J _ Dsrder and 8 partcles n the blue state; 2 partcles n the red state: 90 cnfguratns _J J Dsrder and 7 partcles n the blue state; 3 partcles n the red state: 720 cnfguratns _J J J _J J J J J J J J _J J J J J J J _ Dsrder and 6 partcles n the blue state; 4 partcles n the red state: 4320 cnfguratns _J J J J Dsrder and 5 partcles n the blue state; 5 partcles n the red state: 21,600 cnfguratns _J J J J J _J J J J J J _J J J J J _ 1
Dsrder and 4 partcles n the blue state; 6 partcles n the red state: 4320 cnfguratns 3 partcles n the blue state; 7 partcles n the red state: 720 cnfguratns 2 partcles n the blue state; 8 partcles n the red state: 90 cnfguratns 1 partcle n the blue state; 9 partcles n the red state: 10 cnfguratns 0 partcles n the blue state; 10 partcles n the red state: 1 cnfguratn Dsrder and Whch cnfguratn f partcles wll have the hghest prbablty f beng bserved? Number f cnfguratns 20000 15000 10000 5000 0 0 1 2 3 4 5 6 7 8 9 10 Number f partcles n red state Dsrder and The mst prbable cnfguratn s that wth the hghest level f dsrder The partcles are spread between as many dfferent states as pssble s a measure f the amunt f dsrder n a system The 2 nd Law f Thermdynamcs states that entrpy shuld be maxmzed system cntans the hghest degree f dsrder pssble Dsrder and Ths prncple apples t a number f cmmn systems: Puttng ce n a warm drnk: Energetcally, there s n reasn why heat cannt flw frm the sld nt the lqud makng the drnk warmer and the ce clder Ths wuld cncentrate energy n the system, and nt spread t ut as much as pssble and the 2 nd Law mandates that heat flw frm ht t cld Dsrder and Ths prncple apples t a number f cmmn systems: Mxng f tw gases at the same pressure: When the valve separatng the gases s pened, the gases mx untl they are equally dspersed thrughut valve clsed the cntaner If there s n attractn between gases, there s n energetc reasn valve pened fr mxng Mxng s cnsequence f 2 nd Law and maxmum entrpy The change n entrpy f a system, ΔS, s related t ts heat cntent q ΔS = S S rev fnal ntal = T q rev s the heat exchanged between the system and ts surrundng durng a reversble prcess Reversble prcesses are thse n whch changes ccur n nfntesmally small steps durng an nfntely lng tme 2
s the nly thermdynamc state functn that can be measured n an abslute scale: Remember entrpy s a measure f dsrder; a pure sld at a temperature f abslute zer has n knetc energy (E T = 3/2 RT), s the mlecules d nt mve frm ther equlbrum pstns Ths system has n dsrder, s S = 0 The entrpy f a sld s lwer than that f a lqud whch s lwer than that f a gas S H2O(s) = 49.5 J/ml K S H2O(l) = 70.0 J/ml K S H2O(g) = 188.8 J/ml K Slds are mre rdered than lquds Lquds are mre rdered than gases usually ncreases upn dsslutn f a slute n a slvent The slute s mre dsrdered n slutn than n the sld phase Slutns f sme salts have decreased entrpy upn dsslutn because the ns are strngly hydrated the water slvent mlecules becmes mre rdered n the regn surrundng the n We can calculate entrpy usng Hess s Law aa + bb cc + dd ΔS rxn = (c S C + d S D ) (a S A + b S B ) In general: ΔS rxn = n S n S prds reacts where S s the entrpy f the cmpund and n s ts stchmetrc ceffcent Example: cmbustn f ethane C 2 H 6 (g) + 7/2 O 2 (g) 2 CO 2 (g) + 3 H 2 O(g) D yu thnk entrpy wll ncrease r decrease fr ths reactn? S (J/ml K) 206.6 205.1 213.7 188.8 ΔS rxn = [2(213.7) + 3(188.8)] [206.6 + 3.5(205.1)] = 69.4 J/ml K Is t pssble fr the entrpy f a system t decrease? H 2 O(g) H 2 O(l) ΔS cndensatn = 70.0 J/ml K 188.8 J/ml K = -118.8 J/ml K Is ths a vlatn f the 2 nd Law because the entrpy f the system s decreasng? 3
As water vapr cndenses, heat s transferred frm the gas phase mlecules t the surrundngs ΔS surr = q rev /T = q surr /T = ΔH surr /T ΔH surr s heat ncrease f surrundngs ΔH surr = -ΔH sys heat ganed by surrundngs was lst by system H 2 O(g) H 2 O(l) ΔH sys = -ΔH vap = -40.7 kj/ml ΔS surr = -ΔH sys /T = -(-40700 J/ml)/298 K = 136.6 J/ml K The entrpy f the surrundngs ncreases by 136.6 J/ml K If we nw examne the change n entrpy f the unverse: ΔS unverse = ΔS sys + ΔS surr = -118.8 J/ml K + 136.6 J/ml K = 17.8 J/ml K The entrpy f the unverse ncreases, mantanng the valdty f the 2 nd Law When a substance s nt n ts standard state (P 1 atm r [A] 1 M), the entrpy s gven by: S = S R lnp r S = S R ln[a] Ths allws calculatn under nn-standard state cndtns Fr the reactn aa + bb cc + dd c d [C] [D] ΔSrxn = ΔS - R ln rxn a b [A] [B] = ΔSrxn - R ln[q] c d [C] [D] Q = a b [A] [B] reactn qutent When a change takes place n a system, sme energy may be transferred n the frm f heat, and sme energy may be ccuped by changes n entrpy. The energy remanng may be used t perfrm ther types f wrk ths s called the free energy f the system Gbb s s defned as: ΔG = ΔH - TΔS 4
Example: Cmbustn f prpane C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) ΔG f(kj/ml) -23.49 0.00-394.36-228.57 ΔG rxn = [3(-394.36) + 4(-228.57)] [(-23.49) + 5(0.00)] = -2073.87 kj/ml The same rules apply t Gbb s as apply t enthalpy The ΔG rxn can be calculated frm the ΔG f (Gbb s f frmatn) f the reactants and prducts ΔGrxn = n ΔG () - n ΔG () f f prds reacts ΔG f f the elements n ther standard frm at any temperature s 0.00 kj/ml Gbb s s a ndcatr f the spntanety f a reactn: If ΔG s negatve, the reactn wll ccur spntaneusly as wrtten If ΔG s pstve, the reverse reactn wll ccur spntaneusly If ΔG = 0, the system s n equlbrum Example: Transtn f carbn frm damnd frm t graphte frm at standard pressure C damnd C graphte ΔG f(kj/ml) 2.9 0.0 ΔG rxn = 0.0 2.9 = -2.9 kj/ml Damnd spntaneusly transfrms t graphte at standard pressure and rm temperature Temperature can have an effect n the sgn f ΔG rxn ΔG rxn = ΔH rxn - TΔS rxn ΔH rxn ΔS rxn ΔG rxn - + - + - + - - - at lw T + at hgh T + + + at lw T - at hgh T ΔG rxn = ΔH rxn - TΔS rxn = ΔH rxn T(ΔS rxn R ln[q]) ΔH rxn ΔH rxn because pressure has very lttle effect n enthalpy ΔG rxn = ΔH rxn - TΔS rxn + RT ln[q] ΔH rxn - TΔS rxn = ΔG rxn ΔG rxn = ΔG rxn + RT ln[q] 5
At equlbrum, ΔG rxn = 0 and Q = K eq 0 = ΔG rxn + RT lnk eq ΔG rxn = -RT lnk eq relates equlbrum cnstant t thermdynamc functns f G, H, and S Example: Fnd K sp fr dsslutn f AgCl(s) - ΔG K = exp rxn eq RT AgCl(s) Ag + (aq) + Cl - (aq) ΔG f(kj/ml) -109.79 77.11-131.23 ΔG rxn = [77.11 131.23] [-109.79] = 55.67 kj/ml - 55670 J/ml -10 K eq = exp = 1.8 x 10 (8.314 J/ml K)(298 K) Same value as reprted K sp n tables f slublty prducts When ΔG rxn s negatve, the reactn prceeds spntaneusly The prducts are thermdynamcally favred Des ths mean that the reactn actually ccurs? Yes! But the tme frame fr reactn may be remarkably slw damnd graphte methane n ar CO 2 and H 2 O The knetcs f the reactn (actvatn barrer, rate ceffcent) may be s slw that we d nt bserve a chemcal change 6