Physics Nov The Direction of a Reaction A + B C,

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1 Physcs Nov Suppose we have a reacton such as The Drecton of a Reacton A + B C whch has come to equlbrum at some temperature τ. Now we rase the temperature. Does the equlbrum shft to the left (more A and B) or to the rght (more C)? The heat of reacton at constant pressure Q p s the heat that must be suppled to the system f the reacton goes from left to rght. If Q p > 0 heat s absorbed and the reacton s called endothermc. If Q p < 0 heat s released and the reacton s called exothermc. For a reacton at constant pressure the heat s the change n the enthalpy of the system Q p = H. We have H = G + τσ and so H = G τ σ = ( ) G pn ( ) ( G = τ 2 pn ( )) G. τ pn What we actually want to do s to change the temperature slghtly. Then the system s no longer n equlbrum and the reacton (n the forward or reverse drecton) wll have to occur n order to restore equlbrum. When the reacton occurs from left to rght the change n partcle number s N = ν and the change n G s G = µ ν. If ths s 0 we have the equlbrum condton (but we ve taken t out of equlbrum by changng the temperature). The change n H s Q p = H = τ 2 ( ( )) ( ( G = +τ 2 τ pn µ ν τ )) pn. The chemcal potental s ( ) x p/τ µ = τ log n Q Z nt.

2 Physcs Nov We substtute nto our expresson for Q p and obtan Q p = τ 2 = τ 2 = τ 2 = τ 2 log (ν log (x p) ν log (τ n Q Z.nt )) (ν log (τ n Q Z.nt )) (log(τ n Q Z.nt ) ν ) = τ 2 log K p(τ). ( ) (τ n Q Z.nt ) ν We ve related the heat of reacton to the equlbrum constant! Ths s called van t Hoff s equaton. A note on sgns: I ve assumed that the ν are postve on the left hand sde of the reacton and negatve for the rght hand sde of the reacton. Mandl (who provdes the bass for ths secton) assumes the opposte so we wnd up wth our equlbrum constants beng nverses of each other and opposte sgns n the van t Hoff equaton. In any case our law of mass acton has the concentratons of the left hand sde reactants n the numerator and the rght hand sde reactants n the denomnator. So an ncrease n the equlbrum constant means the reacton goes to the left and a decrease means the reacton goes to the rght. We see that f Q p s postve (we have to add heat to go from left to rght an endothermc reacton) then our equlbrum constant decreases wth temperature. Ths means ncreasng the temperature moves the reacton to the rght. Rule of thumb: ncreasng the temperature causes the reacton to go towards whatever drecton t can absorb energy. We ve just shown that ncreasng the temperature drves an endothermc reacton to the rght. It wll drve an exothermc reacton to the left.

3 Physcs Nov Applcaton: the Saha Equaton Ths secton s related to K&K chapter 9 problem 2. Consder the onzaton of atomc hydrogen p + + e H. Ionzng hydrogen from ts ground state requres an energy of 13.6 ev and as the above reacton s wrtten t s exothermc from left to rght. If we are consderng low densty gases we can treat them as classcal deal gases and apply our law of mass acton: [p + ][e ] [H] = (n pq Z pnt )(n eq Z ent ) n HQ Z Hnt exp(i/τ) where the partton functon for the hydrogen atom s to be computed wth the ground state at the zero of energy as we ve taken explct account of the bndng energy I = 13.6 ev. Ths (or more properly some of the forms we wll derve below) s called the Saha equaton. Some of the factors n the equlbrum constant are easy to calculate and others are hard to calculate! Let s do the easy ones. Frst of all the mass of a proton and the mass of a hydrogen atom are almost the same so the quantum concentratons of the proton and the hydrogen are almost the same and we can cancel them out. The quantum concentraton of the electron s ( ) 3/2 me τ n eq = 2π h 2. The nternal partton functons for the electron and proton are both just 2 snce each has spn 1/2. Ths leaves us wth the nternal partton functon of the hydrogen atom. Ths s complcated. Frst of all the electron and proton each have two spn states so whatever else s gong on there s a factor of four due to the spns. Asde: n fact the spns can combne wth the orbtal angular momentum to gve a total angular momentum. In the ground state the orbtal angular momentum s zero and the spns can be parallel to gve a total angular momentum of 1 h wth 3 states or ant-parallel to gve a total angular momentum of 0 wth 1 state. The parallel states are slghtly hgher n energy than the ant-parallel state. Transtons between these states are called hyperfne transtons and result n the 21 cm lne whch s radated and absorbed by neutral hydrogen throughout our galaxy and others. In any case the energy dfference between these states s small enough to be gnored n computng the nternal partton functon for the purposes of the Saha equaton. When all s sad and done we have [p + ][e ] [H] ( ) 3/2 me τ = 4 2π h 2 e I/τ 1 Z Hnt

4 Physcs Nov where the factor of four accounts for the two spn states of the proton and the two spn states of the electron (there s a factor of four n the hydrogen partton functon as well). If the temperature s small compared to the bndng energy of hydrogen (whch means t s small compared to the dfference between the frst excted state and the ground state) then we mght as well approxmate the partton functon as 4. Ths gves [p + ][e ] [H] ( ) 3/2 me τ 2π h 2 e I/τ. If we have only hydrogen and onzed hydrogen [p + ] = [e ] and [e ] ( ) 3/4 me τ [H] 2π h 2 e I/2τ. Some ponts to note: the fact that the exponental has I/2τ ndcates that ths s a mass acton effect not a Boltzmann factor effect. If there s another source of electrons (for example heaver elements whose outer electrons are loosely bound) the reacton would shft to favor more hydrogen and fewer protons. The Saha equaton apples to gases n space or stars as well as donor atoms n sem-conductors (modfed for the approprate physcal characterstcs of the atom and the medum). In fact we can do a lttle more wth the Saha equaton. Let s consder an atom whch has several electrons and ask about the onzaton equlbrum between the ons that have been onzed tmes and those that have been onzed + 1 tmes n +1 [e ] = (n +1Q Z +1nt )(n eq Z ent ) n n Q Z nt exp(i +1 /τ) where n +1 and n are the concentratons of the two ons n +1Q and n Q are the quantum concentratons of the two ons whch are essentally the same so we cancel them out Z +1nt and Z nt are the nternal partton functons of the two ons and I +1 s the dfference n bndng energy between the two ons. That s I +1 s the energy requred to remove an electron from on and produce on +1. Now each on wll have some nternal structure and energy levels. We let ɛ +1j be the energy (relatve to 0 for the on ground state) of the j th state of on + 1. Ths state has multplcty g +1j. (If there s more than one state at a gven energy we say that energy s degenerate and the multplcty s the number of such states. Sometmes the multplcty s called the degeneracy or the statstcal weght.) Smlarly ɛ k and g k are the energy and multplcty of the k th state of on. The fracton of ons + 1 whch are n state j s gven by a Boltzmann factor n +1j n +1 = g +1j e ɛ +1j/τ Z +1nt.

5 Physcs Nov If we substtute ths expresson nto the Saha equaton and also substtute the quantum concentraton of the electrons and the nternal partton functon of the electrons (2) we get n +1j [e ] n k = 2g +1j g k ( ) 3/2 me τ 2π h 2 e (I +1 + ɛ +1j ɛ k )/τ. Ths form of the Saha equaton connects the concentraton of ons n varous energy levels to the electron concentraton and the temperature. Note that we managed to get rd of the nternal partton functons. Of course now we have a relaton connectng concentratons of states of a gven energy level rather than concentratons of ons of a gven onzaton. We can apply the above expresson to hydrogen (agan!). There are only two onzaton states. We let = 0 and k = 0 so n k s the concentraton of hydrogen atoms n the ground state (whch has multplcty g 00 = 4 and energy ɛ 00 = 0. The onzed state s just a proton whch has a multplcty of 2 and no excted states. So [p + ][e ( ) 3/2 ] me τ = n 00 2π h 2 e I/τ whch s essentally the same equaton we had before except that now t ncludes only hydrogen atoms n the ground state and t s exact. Phase Transtons Phase transtons occur throughout physcs. We are all famlar wth meltng ce and bolng water. But other knds of phase transtons occur as well. Some solds when heated through certan temperatures change ther crystal structure. For example sulfur can exst n monoclnc or rhombc forms. When ron s cooled below the Cure pont t spontaneously magnetzes. The Cure pont of ron s T c = 1043 K. A typcal chunk of ron has no net magnetzaton because t magnetzes n small domans wth the drecton of the magnetc feld orented at random. The magnetzaton even n the small domans dsappears above the Cure temperature. The transton between the normal and superflud states of 4 He s a phase transton as are the transtons between normal and superconductng states n superconductors. You ve probably heard about the symmetry breakng phase transtons that mght have occurred n the very early unverse as the unverse cooled from ts extremely hot ntal state. Such transtons broke the symmetry of the fundamental forces causng there to be dfferent couplngs for the strong weak electromagnetc and gravtatonal force. The latent heat released n such a transton mght have drven the unverse nto a state of very rapd expanson (nflaton).

6 Physcs Nov The spontaneous magnetzaton of ron as t s cooled below the Cure temperature s an example of a symmetry breakng transton. Above the Cure pont the atomc magnets (spns) are orented at random (by thermal fluctuatons). So any drecton s the same as any other drecton and there s rotatonal symmetry. Below the Cure pont (and wthn a sngle doman) all the atomc magnets are lned up so a sngle drecton s pcked out and the rotatonal symmetry s broken. Ths s not an exhaustve lst of phase transtons! Even so we wll not have tme to dscuss all these knds of phase transtons. We wll start wth somethng smple lke the lqud to gas transton. Phase Dagrams Suppose we do some very smple experments. We place pure water nsde a contaner whch keeps the amount of water constant and doesn t allow any other knds of molecules to enter. The contaner s n contact wth adjustable temperature and pressure reservors. We dal n a temperature and a pressure wat for equlbrum to be establshed and then see what we have. For most pressures and temperatures we wll fnd that the water s all sold (ce) all lqud or all vapor (steam). For some temperatures and pressures we wll fnd mxtures of sold and vapor or sold and lqud or lqud and vapor. The fgure shows a schematc plot of a phase dagram for water. I ddn t put any numbers on the axes whch s why t s schematc. (Also there are several knds of ce whch we re gnorng!) K&K gve a dagram but t doesn t have any resoluton at the trple pont. Note that the frst fgure (whch we ll talk about some more n a mnute) s somethng lke a map: t says here we have vapor there we have sold etc. The second fgure s a schematc of a pv dagram showng an sotherm. For an deal gas we would have a hyperbola. For the sotherm as shown we have pure lqud on the branch to the left of pont a pure vapor to the rght of pont b and along the segment from a to b we have a mxture of lqud and vapor. If we move along ths sotherm from left to rght we are essentally movng down a vertcal lne n the pτ dagram. To the left of pont a we are movng to lower pressures wth lqud water. from a to b we are stuck at the lne n the pτ dagram that dvdes the lqud from the vapor regon and to the rght of b we are movng down n the vapor regon. So the

7 Physcs Nov entre transton from all lqud to all vapor whch s a to b n the pv dagram happens n a sngle pont n the pτ dagram. At ths pont the water has a fxed temperature and pressure and what adjusts to match the volume s the relatve amounts of lqud and vapor. Now at each locaton n the pτ dagram we fx the temperature and pressure and let the system come to equlbrum. The equlbrum condton s that the Gbbs free energy s mnmzed. Ignorng for the moment the fact that the water can be a sold the Gbbs free energy s G(pτN l N v ) = N l µ l (pτ) + N v µ v (pτ) where the subscrpts l and v refer to the lqud and vapor and we ve made use of the fact that for a sngle component substance the chemcal potental can be wrtten as a functon of p and τ only. There are several ways we mght mnmze G. Frst of all f µ l (pτ) < µ v (pτ) then we mnmze G by settng N l = N and N v = 0 where N s the total number of water molecules. In other words the system s entrely lqud. If µ v (pτ) < µ l (pτ) we mnmze the free energy by makng the system entrely vapor. Fnally f µ l (pτ) = µ v (pτ) we can t change the free energy by changng the amount of vapor and lqud so we can have a mxture wth the exact amounts of lqud and vapor determned by other constrants (such as the volume to be occuped). So what we ve just shown s that where lqud and vapor coexst n equlbrum we must have µ l (pτ) = µ v (pτ) whch s exactly the same condton we would have come up wth had we consdered the reacton H 2 O lqud H 2 O vapor. Ths s a relaton between p and τ and t descrbes a curve on the pτ dagram. It s called the vapor pressure curve. by Wth smlar arguments we deduce that sold and vapor coexst along the curve defned µ s (pτ) = µ v (pτ) whch s called the sublmaton curve and sold and lqud coexst along the curve µ s (pτ) = µ l (pτ) whch s the meltng curve. If we have all three chemcal potentals equal smultaneously µ s (pτ) = µ l (pτ) = µ v (pτ)

8 Physcs Nov we have two condtons on p and τ and ths defnes a pont. Ths unque (for each substance) pont where sold lqud and vapor all coexst s called the trple pont. For water T t = K p t = 4.58 mm Hg. Actually ths s now used to defne the Kelvn scale. If a substance has more than three phases t can have more than one trple pont. For example the two crystallne phases of sulfur gve t four phases and t has three trple ponts. The vapor pressure curve eventually ends at a pont called the crtcal pont. At ths pont one can t tell the dfference between the lqud phase and the vapor phase. We ll say more about ths later but for now consder that as you go up n temperature you get suffcently volent motons that bndng to neghborng molecules (a lqud) becomes a neglgble contrbuton to the energy. As one goes up n temperature the heat of vaporzaton decreases. At the crtcal pont t s zero. The crtcal pont for water occurs at T c = K p c = atm. Another way to thnk of the phase dagram and the coexstence curves s to magne a 3D plot. Pressure and temperature are measured n a horzontal plane whle µ(pτ) s plotted as heght above the plane. Ths defnes a surface. In fact we have several surfaces one for µ s µ l and µ v. We take the overall surface to be the lowest of all the surfaces remember we re tryng to mnmze G. Where µ v s the lowest we have pure vapor etc. Where two surfaces ntersect we have a coexstence curve. Of course the phase dagram corresponds to equlbrum. It s possble to have lqud n the vapor regon (superheated) or sold regon (supercooled) etc. but these stuatons are unstable and the system wll try to get to equlbrum. Whether ths happens rapdly or slowly depends on the detals of the partcular stuaton.