Physics 301 16-Nov-2005 21-1The Direction of a ReactionSuppose we have a reaction such asA + B ↔ C ,which has come to equilibrium at some temperature τ. Now we raise the temperature.Does the equilibrium shift to the left (more A and B) or to the right (more C)?The heat of reaction at constant pressure, Qp, is the heat that must be supplied to thesystem if the reaction goes from left to right. If Qp> 0, heat is absorbed and the reactionis call ed endothermic. If Qp< 0, heat is released and the reaction is call ed exothermic.For a reaction at constant pressure, the heat is the change in the enthalpy of t hesystem, Qp= ∆H. We haveH = G + τσ ,andσ = −∂G∂τp,Ni,soH = G − τ∂G∂τp,Ni= −τ2∂∂τGτp,Ni.What we actually want to do is to change t he temperature slightly. Then t he system i sno longer in equilibrium and the reaction (in the forward or reverse direction) w ill haveto occur in order to restore equilibrium. When the reaction occurs from left to right, thechange i n particle number is ∆Ni= −νiand the change in G is∆G = −Xiµiνi.If this is 0, we have the equilibrium condition (but we’ve taken it out of equilibrium bychanging the temperature). The change in H isQp= ∆H = −τ2∂∂τ∆Gτp,Ni= +τ2 ∂∂τ Xiµiνiτ!!p,Ni.The chemical potential isµi= τ logxip/τni,QZi,int.Copyrightc 2005, Princeton University Physics Department, Edward J. GrothPhysics 301 16-Nov-2005 21-2We substitute into our expression for Qpand obtain,Qp= τ2∂∂τXi(νilog (xip) − νilog (τ ni,QZi,int)) ,= −τ2∂∂τXi(νilog (τ ni,QZi,int)) ,= −τ2∂∂τXi(log (τ ni,QZi,int)νi) ,= −τ2∂∂τlog Yi(τ ni,QZi,int)νi!,= −τ2∂∂τlog Kp(τ) .We’ve related the heat of reaction to the equilibrium constant! This i s called van’t Hoff’sequation. A note on signs: I’ve assumed that the νiare positive on the left hand side ofthe reacti on and negative for the right hand side o f the reaction. Mandl (who provides thebasis for this section) assumes the opposite, so we wind up with our equilibrium constantsbeing inverses of each other and opposite signs in the van’t Hoff equation.In any case, our law of mass action has the concentrations of the left hand sidereactants in the numerator and the right hand side reactants in the denominator. So anincrease in the equilibrium constant means the reaction goes to the left and a decreasemeans the reaction goes to the right. We see that if Qpis positive (we have to add heatto go from left to ri ght, an endothermic reaction), then our equilibrium constant decreaseswith temperature. This means i ncreasing the temperature moves the reaction to the right.Rule of thumb: increasing the temperature causes the reaction t o go towards w hat everdirection it can absorb energy. We’ve just shown that increasing the temperature drivesan endothermic reaction to the right. It will drive an exothermic reaction to the left.Copyrightc 2005, Princeton University Physics Department, Edward J. GrothPhysics 301 16-Nov-2005 21-3Application: the Saha Eq uat ionThis section is related to K&K, chapter 9, problem 2. Consider the ionization ofatomic hydrogen,p++ e−↔ H .Ionizing hydrogen from its ground state requires an energy of 13.6 eV, and as the abovereaction is wri tten, it’s exothermic from left to right. If we are considering low densitygases, we can treat them as classical ideal gases and apply our law of mass action:[p+][e−][H]=(np,QZp,int)(ne,QZe,int)nH,QZH,intexp(I/τ),where t he partition function for the hydrogen atom i s to be computed with the ground stateat the zero of energy, as we’ve taken explicit account of the binding energy I = 13.6 eV.This (or more properly, some of the forms we will derive below) is called the Saha equati on.Some of the factors in the equilibrium constant are easy to calculate and others arehard to calculate! Let’s do the easy o nes. First of all, the mass of a proton and the mass ofa hydrogen atom are almost the same, so the quantum concentrations of the proton and thehydrogen are almost the same and we can cancel them out. The quantum concentrationof the electron isne,Q=meτ2π¯h23/2.The internal partiti on functions for the electron and proton are both just 2, since each hasspin 1/2 . This leaves us with the internal partition function of the hydrogen atom. This iscomplicated. First of all, the electron and proton each have two spin states, so whateverelse is going on there is a factor of four due to the spins.Aside: i n fact the spins can combine with the orbital angular momentum to give atotal angular momentum. In the gro und state, the orbital angular momentum is zeroand the spins can be parallel to give a to tal angular momentum of 1¯h with 3 states oranti-parallel to give a total angular momentum of 0 with 1 state. The parallel states areslightly higher in energy than the anti-parallel state. Transitions between these states arecalled hyperfine transitions and result in the 21 cm line which is radiated and a bsorbedby neutral hydrogen throughout our galaxy and others. In any case, the energy differencebetween these states is small enough to be ignored in computing the internal partitionfunction for the purposes of the Saha equation.When all is said and done, we have[p+][e−][H]= 4meτ2π¯h23/2e−I/τ1ZH,int,Copyrightc 2005, Princeton University Physics Department, Edward J. GrothPhysics 301 16-Nov-2005 21-4where t he factor of four accounts for the two spin states of the proton and the two spinstates of the electron (there is a factor of four in the hydrogen partiti on function as well).If the temperature is small compared to the binding energy of hydrogen ( w hich means it’ssmall compared to the difference between the first excit ed state and the ground state),then we might as well approximate the partition function as 4. This gives,[p+][e−][H]≈meτ2π¯h23/2e−I/τ.If we have only hydrogen and ionized hydrogen, [p+] = [e−] and[e−] ≈p[H]meτ2π¯h23/4e−I/2τ.Some points to note: the fact t hat the exponential has −I/2τ indicates that this is amass action effect, not a Boltzmann factor effect. If there is another source of electrons (forexample, heavier element s whose outer electrons are loosely bound), the reaction wouldshift to favor more hydrogen and fewer protons. The Saha equation applies to gases inspace or stars as well as donor atoms in semi-conductors