Chemical Equilirium, the Law of Mass Action, and the Reaction Heat Supplementary Notes for the Astrochemistry Course

Chemical equilirium, the law of mass action, and the reaction heat Supplementary notes for the astrochemistry course

Jorma Harju

February 19, 2019

1 Equilibrium constant and the law of mass action

Most reactions we are dealing with in astrochemistry are of the type

k+ A+B ↽⇀ C+D . k−

The equilibrium constant, K(T ), is deﬁned by

k N N K(T ) ≡ + = C,eq D,eq , k− NA,eqNB,eq where NA,eq, NB,eq, etc. are the numbers of the substances A, B, ... in given volume and temperature when chemical equilibrium is established, that is, when the number densities do not change. The equilibrium constant can be calculated from the partition functions (sums over states), Z(T ), of the species. For the reaction given above we obtain the formula:

Z Z K(T )= C D . ZAZB

The partition function Z depends on energy spectrum of the molecule. For a particle obeying the Boltzmann statistics Z(T ) = Σigi exp(−Ei/kT ), where gi is the statistical weight (degeneracy), and Ei is the energy of the state i. The total partition funtion is a product of the electronic, translational, vibrational, and rotational partition functions, Z = ZelZtrZvibZrot. Normally only the ground electronic state is thermally accessible, so Zel = gel, where gel is the degeneracy of the ground state (usually = 1). The energies in the partition functions are usually given relative to the zero-point vibrational ′ energy, ZPVE, which we denote by E0. The partition function in terms of E0 is Z = −(Ei−E0)/kT E0/kT ′ −E0/kT Σgie = e Z, implying Z = Z e . Omitting the primes, K(T ) can be written as Z Z −∆E K(T )= C D exp 0 , Z Z kT A B

1 where ∆E0 is the diﬀerence between the ZPVEs (products − reactants): C D A B ∆E0 = E0 + E0 − E0 − E0 .

Sometimes ∆E0 is called the enthalpy of the reaction (e.g., Ramanlal & Tennyson). Why this assignment can be made is discussed below.

3/2 The translational partition function, Ztr, contains the term m , where m is the molecular mass. In our example reaction other terms of Ztr are reduced in the expression of K(T ), and therefore the equilibrium constant can be written as N N m m 3/2 Z Z −∆E K(T ) ≡ C,eq D,eq = C D C,int D,int exp 0 , N N m m Z Z kT A,eq B,eq A B A,int B,int where Zint = ZrotZvib is the partition function related to the internal energy states of the molecule. This form is known as the law of mass action.

As an example we consider the most important deuteration reaction in interstellar molecular clouds: k + + + H3 + HD ↽⇀ H2D + H2 k− The equilibrium constant is N(H D+) N(H ) 4 · 2 3/2 Z(H D+, int)Z(H , int) −∆E K(T )= 2 eq 2 eq = 2 2 exp 0 . N(H+) N(HD) 3 · 3 Z(H+, int)Z(HD, int) kT 3 eq eq 3 + + The diﬀerence between the ZPVEs is ∆E0/k = E0(H2D )/k + E0(H2)/k − E0(H3 )/k + 1 E0(HD)/k = −139.5 K. A negative ∆H ∼ ∆E0 means that the reaction from left to right is exothermic. + + In Fig. 1 we show the partition functions Z(H2D , int), Z(H2, int), Z(H3 , int), and Z(HD, int) as functions of T . The equilibrium constant calculated from these partition functions is shown in Fig. 1.

The van ’t Hoﬀ equation relates the temperature dependence of the equilibrium constant to the enthalpy change, ∆H, of the reaction (reaction heat in constant pressure). For a single elementary reaction this equation can be written as

∂ ∆H T 2 ln K(T )= , ∂T k or equivalently ∂ ln K(T ) ∆H 1 = − . ∂ T k The latter form shows that the reaction enthalpy can be obtained from the slope of the lnK vs. 1/T diagram. It is evident from Fig. 1 that the reaction enthalpy depends on the temperature. At low temperatures (1/T > 0.1K−1) the slope indicates that ∆H/k ∼−232K + (∆ZPVE minus the “rotational ZPE” of H3 ), whereas at elevated temperatures (1/T < −1 + 0.02K ) where the rotational levels of H3 are populated, ∆H/k ∼−140 K (= ∆ZPVE). In what follows we derive the equations used in this example.

1 + The ground rotational state of H3 is forbidden by the Pauli exclusion principle, and the lowest allowed state J, K = 1, 1 lies 92.3 K above the ground state. Often this “rotational zero-point energy” is added to + the zero-point energy of H3 , and ∆E0/k becomes -231.8 K. When using this value, we should also give the + energies of Z(H3 , int) with respect to the state J, K = 1, 1.

2 + + Figure 1: Internal partition functions, Zint = ZvibZrot, of H2D ,H2,H3 , and HD as functions + of T . Note that for H3 the energies are calculated from the forbidden rotational ground state, + and therefore Z(H3 , int) → 0 when T → 0. For other molecules, Z approaches the statistical weight of the ground state. The contributions of diﬀerent nuclear spin variants are indicated when applicable. The partition functions are adopted from Hugo et al.

+ Figure 2: The natural logarithm of the equilibrium constant, ln K, of reaction H3 + HD ↔ + H2D + H2 as functions of T (left) and 1/T (right). Also shown are two approximations ln K(T ) ∼ −∆H/kT from the van ’t Hoﬀ equation. At elevated temperatures, the as- sumption ∆H/k = ∆ZPE1/k = −139.5 K (the diﬀerence of zero-point vibrational energies) seems to work well, whereas at low temperatures ∆H/k =∆ZPE2/k = −231.8 K (∆ZPVE + combined with the “rotational zero-point energy” of H3 ) gives a better agreement.

3 2 Thermodynamic potentials

The thermodynamic properties of a gas, e.g., the equation of state, can be derived from one of the following thermodynamic potentials: the internal energy , E(S,V,N) (the total energy contained by a thermodynamic system), the enthalpy H(S,P,N) (the total energy of a thermodynamic system including the internal energy and the energy needed to establish the space and pressure of the system), the free energy F = F (T,V,N) (also called the Helmholtz function, the energy that can be converted into work at a constant temperature and volume), or the Gibbs energy G = G(T,P,N) (also called the Gibbs function, the energy that can be converted into work at a uniform temperature and pressure). The choice of the most appropriate potential for our purpose depends on which of the quantities S (entropy), V (volume), N (number of particles), T (temperature), and P (pressure) we wish to use as variables. If one of potantials is known, the others can be derived using the the following relations: H = E +PV , F = E −TS, and G = H −TS = F +PV . In chemical applications the Gibbs energy, G(T,P,N), is particularly useful because the pressure and the temperature are usually under control, when the number of particles and the composition changes. If the number of the gas particles, N, is allowed to change the first law of thermodynamics (conservation of the energy) is expressed by the equation dE = TdS − PdV + µdN, where µ is the chemical potential of the gas. Correspondingly, the total differentials of the enthalpy, the free energy, and the Gibbs energy can be written as dH = TdS + V dP + µdN, dF = −SdT − PdV + µdN and dG = −SdT + V dP + µdN. In a mixture of gases the chemical potential µ is defined separately for each component i, and a general change of the Gibbs energy can be written as dG = −SdT + V dp + ΣiµidNi . (1) This formula is sometimes called the master equation of chemical thermodynamics (Atkins, Ch. 8). From this we can make the identifications

∂G ∂G ∂G = −S , = V , = µi . (2) ∂T ∂P ∂Ni P,Ni T,Ni P,T The variables kept constant in the derivates are indicated with subscripts. The chemical potential can be obtained also from the other thermodynamic potentials in the following way: ∂E ∂H ∂F µ = = = . (3) i ∂N ∂N ∂N i S,V i S,P i T,V

3 Chemical equilibrium

Consider the chemical reaction

k+ νAA+ νBB ↽⇀ νCC+ νDD , k− occurring in a mixture of reacting substances. The coefficients νA etc. are the stoichiometric coefficients. For example, in the reaction N2 + 3H2 → 2NH3(+∆E), the coefficients are 1, + − 3, and 2, and in the reaction NH3 + H2O ↽⇀ NH4 + OH all coefficients have the value 1. The numbers of particles A,...,D, denoted by NA,...,ND, in a volume element V change until

4 an equilibrium state is established in which the numbers no more change. In this state the reaction proceeds in both direction at equal rates. In what follows we derive the condition for chemical equilibrium. The reaction equation can be also written as

−νAA − νBB+ νCC+ νDD=0 . (4) The stoichiometric coeﬃcients of the products of a reaction proceeding from left to right are chosen here to be positive, and the coeﬃcients of the reactants are negative. Assume that the reaction occurs at constant pressure, P , and temperature, T . In such processes the Gibbs energy, G, tends to a minimum. This is an implication of the second law of thermodynamics which can be formulated, e.g., in the following way: In natural processes the entropy of a closed system increases, and reaches a maximum in the state of equilibrium. When the system can exchange heat with its surroundings, the “closed system” where the entropy must increase is the universe. For a reaction occurring in constant temperature the change in the Gibbs energy can be written as ∆G =∆H − T ∆S (T = const.) , (5) where ∆H and ∆S are the enthalpy and entropy changes of the system repectively. For spontaneuous reactions ∆G< 0. At low temperatures T ∆S is often small, and so ∆G ∼ ∆H, whereas at higher temperatures T ∆S may dominate. For an exothermic reaction the enthalpy change is negative (∆H < 0). Also endothermic reactions (∆H > 0) can occur spontaneously provided that the overall entropy in the universe (system plus surroundings) increase. The entropy change of the surroundings is −∆H/T (heat transferred to the surroundings at constant pressure is −∆H). This change must be compensated by the entropy change ∆S of the system. The natural variables of the Gibbs energy are the temperature, the pressure, and the number of particles: G = G(T,P,N). Therefore the minimum Gibbs energy in constant pressure and temperature corresponds to the zero point of its derivative with respect to the number of one of the substances participating in the reaction. We choose the substance A:

dG ∂G ∂G dN ∂G dN ∂G dN = + B + C + D = 0 . (6) dNA ∂NA ∂NB dNA ∂NC dNA ∂ND dNA

The relative changes of the numbers NA,...ND are determined by the reaction equation (4):

dN ν dN ν dN ν B = B , C = − C , D = − D . dNA νA dNA νA dNA νA

Multiplying Equation (6) by −νA and using the deﬁnition of the chemical potential: ∂G ∂G µA = , µB = ..., ∂NA ∂NB we obtain the condition of chemical equilibrium:

−νAµA − νBµB + νCµC + νDµD =0 In other words, the reaction is in equilibrium when the chemical potentials of the reactants and products are equal.

5 We express this in a more compact form by denoting the substances by A1, A2,... instead of A,B.... Then the reaction equation becomes

ΣνiAi =0 . (7)

Note that now the sign + or − is incorporated in the stoichiometric coeﬃcient νi. Using this notation the condition for chemical equilibrium can be written as

Σiνiµi =0 . (8)

The choice that the stoichiometric coeﬃcients of reactants are negative depends on the fact that we will deﬁne the equilibrium constant, K(T ), as the abundance ratio of the products over the reactants in equilibrium, e.g. K(T ) ≡ NC,eqND,eq (see below). NA,eqNB,eq

4 Mixtures of ideal gases

Tenuous gas consisting of various substances can be approximated by a mixture of ideal gases. In an ideal gas, the interaction between molecules are negligible from the thermodynanamic point of view. In this mixture thermodynamic quantities like the internal energy, E, and the entropy, S, are sums of those of the component gases. Using the ideal gas equation of state, NkT P = (9) V the partial pressure Pi of the component i can be written as N kT N NkT N P = i = i = i P , i V N V N where Ni is the number of molecules i, P is the total pressure of the gas, T is the kinetic temperature of the gas, and V is the volume ﬁlled by the gas. The total pressure is the sum of the partial pressures: P = ΣPi. Furthermore, the free energy and the Gibbs energy of the mixture can be written as sums:

F (N,T,V ) = ΣFi(Ni,T,V ) (10)

G(N,T,P ) = ΣGi(Ni,T,Pi) (11)

In the following we derive the Gibbs energy and the chemical potential for a mixture of ideal gases.

5 Free energy of an ideal gas

In statistical physics the following expression is derived for the free energy of an ideal gas:

1 e F (T,V,N)= −kT ln Z(T,V )N ≈−NkT ln Z(T,V ) , (12) N! N n o

6 where Z(T,V ) is the partition function (or sum over states) of an individual molecule, and e = 2.71828... is the base of natural logarithms. Stirling’s approximation in Eq. 12 is valid for large numbers N. The free energy for a single molecule in volume V is

F (T,V )= −kT ln {Z(T,V )} .

The distribution of molecules of an ideal gas among the various states is called the Boltzmann distribution. According to this distribution the probability pj that a molecule is in the energy state Ej can be obtained from 1 p = e−Ej /kBT , j Z(T ) −23 −1 where kB is the Boltzmann constant (1.3810 JK ), and T is the gas temperature. From the normalization condition Σpj = 1 we obtain for the partition function the expression

Z = Σe−Ej /kT .

In a mixture of gases, the free energy of component i is e F (N,T,V )= −N kT ln Z (T,V ) , i i N i i where Ni is the number of particles i in volume V and Zi(T,V ) is the partition function for a particle of type i. The temperature of the mixture is T . The free energy of the whole mixture is then e F (N,T,V ) = ΣF (N,T,V )= −ΣN kT ln Z (T,V ) . (13) i i N i i 6 Partition function

The partition function Z(T,V ) can be divided into translational or kinetic and internal parts. The translational partition function, Ztr(T,V ), takes into account the energy quantization of a particle in a vessel of conﬁned volume. The internal partition function, Zint(T ), depends on the particle’s internal degrees of freedom (electronic states, vibration, rotation, etc.). So the total partition function can be written as

Z(T,V )= Ztr(T,V ) Zint(T ) . (14)

The translational partition function of an ideal gas can derived quasi-classically by analyzing the distribution of molecules in the phase space extended by the moments p and coordinates q (see, e.g. Mandl Ch. 7.2; Landau & Lifshitz § 38), resulting in the expression

2πmkT 3/2 Z (T,V )= V . (15) tr h2 The internal partition function can be divided into the following factors (which are indepen- dent in the ﬁrst approximation):

−ǫe/kT Zint(T )= e Zrot(T )Zvib(T ) ,

7 where ǫe is the electronic energy of the molecule, Zrot is the rotational partition function:

∞ −BJ(J+1)/kT Zrot(T ) ≈ ΣJ=0(2J + 1)e , and Zvib is the vibrational partition function:

1 ∞ −hν0(v+ 2 )/kT Zvib(T ) ≈ Σv=0e . (Above we have use the approximations of a rigid linear rotor and a harmonic oscillator. More accurate formula for Zrot and Zvib can be found in the literature.) The logarithm of a product can be written as the sum of logarithms, and therefore also F can be presented as the sum of the ’kinetic free energy’, Ftr, ’rotational free energy’, Frot = −NkT ln(Zrot), the ’vibrational free energy’, Fvib = −NkT ln(Zvib), and the ’electronic free energy’, Nǫe. According to the Boltzman distribution, at low temperatures almost all molecules are at −Ej /kT their ground states: pj = e → 0, when Ej > 0. The behaviour of the partition function at a low temperature depends on the ground state energy. We can choose the electronic ground state as the zero point, ǫe,0 = 0, implying e−ǫe,0/kT = 1. At low temperatures, the rotational partition function approaches the statistical weight of the ground rotational state, Zrot(T ) → g0 (1 for a linear rotor), when T → 0, because the rotational energy levels (normally) start from zero. In contrast, the 1 − hν0/kT vibrational partition function Zvib(T ) → e 2 . The lowest possible vibrational energy 1 of a molecule, 2 hν0, is called the zero-point energy (ZPVE). This energy is important for chemical fractionation at low temperatures.

7 The Gibbs energy of a mixture of ideal gases

According to the relationship between the Gibbs and free energies, we can write the Gibbs energy of the component i as Gi = Fi + PiV . Because the natural variables of G are N,T , and P , we replace V using the ideal gas equation of state: N kT V = i . Pi We have to make this substitution also in the translational partition function:

N kT 2πm kT 3/2 Z (T, P )= i i . i,tr i P h2 i After these substitutions we obtain ekT 2πm kT 3/2 G (N ,T,P )= −N kT ln i Z (T ) + N kT . i i i i P h2 i,int i ( i )

This expression can be simpliﬁed by separating the term ln e(= 1) from the logarithm, because −NikT ln e + NikT = 0:

kT 2πmikT 3/2 G (N ,T,P ) = −N kT ln 2 Z (T ) i i i i Pi h i,int 3/2 n 2πmikT o = NikT ln Pi − NikT ln kT h2 Zi,int(T ) . n o 8 We can spare some pain by collecting the terms that depend on T only on the right hand side under a function fi(T ), deﬁned as

2πm kT 3/2 f (T ) ≡−kT ln kT i Z (T ) . (16) i h2 i,int ( )

The Gibbs energy of the mixture is then

G(N,T,P ) = ΣGi(Ni,T,Pi) = ΣNikT ln Pi + ΣNifi(T ) . (17)

If we wish to use the total pressure P instead of the partial pressure Pi, we can make the substitution Pi = Ni/N P : N G(N,T,P ) = ΣN kT ln P + ΣN kT ln i + ΣN f (T ) . (18) i i N i i

8 Chemical potential of an ideal gas

It is straightforward to derive the chemical potential of a component i in a mixture of ideal gases starting from one of the thermodynamical potentials (see Eqs. (2) and (3)). Using the free energy we obtain

∂F e 1 Z (T,V ) µ = i = −kT ln Z (T,V ) − N kT (− )= −kT ln i (19) i ∂N N i i N N i T,V i i i Here we again utilized algebraic rules:

e ln Z (T,V ) = ln e − ln N + ln Z (T,V ) . N i i i i

The case of the Gibbs energy is very simple:

∂Gi µi = = kT ln Pi + fi(T ) . (20) ∂Ni T,Pi

By substituting Pi = NikT/V and the deﬁnition of the function fi(T ) one can see that this expression is identical with the one obtained from Fi. On grounds of the relationship µi = Gi/Ni we can interpret µ as the Gibbs energy per molecule (e.g., Mandl Ch. 8.1.; Landau & Lifshitz § 24).

In Atkins (Ch. 6) the chemical potential of a species is deﬁned as the molar Gibbs function, Gm, i.e. the Gibbs energy per mole. Atkins deﬁnes a“standard Gibbs function”, G⊖, as the Gibbs energy of an ideal gas at a pressure of 1 atm, and writes the pressure dependence of the G as

G(P )= G⊖ + NkTln(P/atm) .

For the molar Gibbs function, i.e. the chemical potential as deﬁned by Atkins, one gets

⊖ Gm(P )= µ(P )= µ + RTln(P/atm) ,

9 where R is the gas constant (R = N0k, N0 is the Avogadro number). Using the expression for G we obtained previously, and the same reference pressure as Atkins, Pref = 1 atm, the Gibbs energy of one mole of ideal gas can be written as

Gm(P,T )= N0kTln(P/atm)+ N0kTln(Pref )+ N0f(T ) . We can thus make the following identiﬁcations:

⊖ G (T )(Atkins) = NkTln(Pref )+ Nf(T ) ⊖ (21) µ (T )(Atkins) = N0kTln(Pref )+ N0f(T ) . Similarly one could have deﬁned a “standard” chemical potential per molecule as

⊖ µi = kTln(Pref )+ f(T ) . (22) 5 We note that the reference pressure Pref = 1 atm ∼ 10 Pa is extremely high compared with interstellar conditions. In molecular clouds, the pressure is typically P ∼ 10−12 Pa, and a perhaps a few orders of magnitude higher in the nuclei of dense cores.

9 The equilibrium constant in terms of numbers Ni

Here we use the expression for the chemical potential otained from the free energy. Substi- tuing Eq. 19, Z (T,V ) µ = −kT ln i , i N i to the condition for chemical equilibrium we get Z (T,V ) Σν µ = −kT Σν ln i =0 . i i i N i,eq The number of molecules i in chemical equilibrium are denoted by Ni,eq. Dividing by −kT and using some algebra the equation can be written as Z (T,V ) Σν ln i = Σln {Z (T,V )}νi − Σln {N }νi =0 . i N i i,0 i,0 We move the second term to the right and take both sides as arguments of an exponential function: ν ν ⇔ eΣ ln{Zi(T,V )} i = eΣ ln{Ni,0} i , which is equivalent with

ν1 ν2 ν3 ν1 ν2 ν3 Z1(T,V ) Z2(T,V ) Z3(T,V ) ... = N1,eq N2,eq N3,eq... or νi νi Π Ni,eq =Π(Zi(T,V ) ) . The left hand side is called the equilibrium constant, K. Here it is expressed in terms of the numbers of molecules in constant volume V and temperature T . We therefore write

νi νi K(T,V ) ≡ Π Ni,eq =Π(Zi(T,V ) ) . (23) As an example we consider the reaction

k+ A+B ↽⇀ C+D . k−

10 For this reaction the total number of particles is constant, Σνi = 0, and |νi| = 1 for all i. Eq. 23 means that the relative numbers in equilibrium can be calculated from the following formula: N N k Z Z C,eq D,eq = + ≡ K(T )= C D . NA,eqNB,eq k− ZAZB

Σiνi K does not depend on the volume because V s is reduced: V = 1. Here the numbers Ni can be replaced by number densities ni.

10 The equilibrium constant in terms of partial pressures

The equilibrium constant in terms of partial pressures, Kp(T ), for an ideal gas can be derived either from the expression for K(T,V ) above (Eq. 23), or by substituting the chemical potential from Eq. 20 to the condition for chemical equilibrium (Eq. 8). The latter method gives Σiνiµi = Σiνi[kTlnPi + fi(T )]=0 . Dividing by −kT we obtain f (T ) Σ ν lnP = −Σ ν i . i i i i i kT

The function fi(T )/(−kT ) on the right is a logarithm as can be seen in the deﬁnition of f(T ) (Eq. 16), and both sides can be written as products. First we, however, leave the right hand side unaltered, and note that the equation above is equivalent with

ν ΣiνilnPi ΣilnPi i νi −Σiνifi(T )/(kT ) e = e =Π(Pi )= e ,

νi ν1 ν2 ν3 p ln a where Π(Pi )= P1 P2 P3 .... Here we have used the relations lna = p ln a and e = a.

νi The product Π (Pi ) is deﬁned as the equilibrium constant for partial pressures, KP (T ), and it can be thus written as

νi −Σiνifi(T )/(kT ) KP (T ) ≡ Π(Pi )= e . (24)

The same result can be obtained from Eq. 23 by using the equation of state of an ideal gas, PV = NkT , and the relationship between the numbers Ni, partial pressures, Pi, and concentrations [i]: N P [i] ≡ i = i . N P Substituting Ni,0 = NPi,0/P to Eq. 23,

νi νi Π Ni,0 =Π(Zi(T,V ) ) . we obtain K in terms of partial pressures: ν νi P i Π Pi,0 ≡ KP (T ) = Π N νi K(T,V ) νi PV 2πmikT 3/2 = Π N h2 Zi,int(T ) 3/2 νi 2πmikT = Π kT h2 Zi,int(T ) νi fi(T ) = Π exp − kT nνifi(T ) o = exp −Σ kT , n o 11 Likewise, we one can derive the equilibrium constant in terms of concentrations substituting Ni = N[i]: νi −Σνi Π([i]0 ) ≡ KC (T, P )= N K(T,V ) . (25)

Eq. 24 is equivalent with the one presented in textbooks of chemistry (e.g. Atkins Ch. 9), relating KP to ⊖ ⊖ ⊖ the “standard molar Gibbs function”, ∆Gm (= Gm(products) − Gm(reactants)) of the reaction : ⊖ RTlnKP (Atkins) = −∆Gm . (26)

Here the partial pressures in KP are replaced by ratios Pi/Pref , where the reference pressure is 1 atm. Making the substitution Pi → (Pi/Pref )Pref in Eq. 24 we obtain:

νi νi νi −Σiνifi(T )/(kT ) KP (T )=Π(Pi )=Π((Pi/Pref ) )Π(Pref )= e . Taking the natural logarithm of the equation on the right we obtain

νi νi lnΠ((Pi/Pref ) ) = −lnΠ(Pref ) − Σiνifi(T )/(kT ) = −Σiνi[lnPref + fi(T )/(kT )] ⊖ = −Σiνiµi /(kT ) . In the last equation we have used the “standard chemical potential per molecule” deﬁned in Eq. 22. The left hand side the equilibrium constant KP as deﬁned by Atkins. Multiplying by N0kT (= RT ) we obtain ⊖ ⊖ ⊖ ⊖ the desired formula, because N0µi = Gm,i and ΣiνiGm,i = ∆Gm, i.e. the change of molar Gibbs energy ⊖ in the reaction. On grounds of this equation one can estimate ∆Gm(T ) of a reaction by determining the equilibrium constant (i.e. equilibrium partial pressures) as a function of temperature.

11 The law of mass action

The law of mass action can be obtained from the formula for the equilibrium constants, K(T,V ) and KP (T ) (Eqs. 23 and 24), simply by writing the expressions of the translational partition function explicitly on the right hand side. So we get

νi νi K(T,V ) ≡ Π Ni,eq = Π(Zi(T,V ) ) νi = Π([Zi,tr(T,V )Zi,int(T )] ) νi 2πmkT 3/2 = Π V h2 Zi,int(T ) 3/2νi 3/2ν h2πV kT i i νi = Π h2 mi Zi,int(T ) In the situation where the total number of particles does not change, Σiνi = 0, the constants on the right, except the masses mi are reduced (note that also V and T are constants here), and the equation above can be simpliﬁed:

νi 3/2νi νi K(T ) = Π Ni,eq = Π mi Zi,int(T ) . (27) We now manipulate the right hand side of Eq. 24:.

fi(T ) fi(T ) exp −Σiνi kT = exp Σiνiln kT 3/2 n o n o 2πmikT = exp Σiνi ln kT h2 Zi,int(T ) 3/2 νi n 2πmhikT io = Π kT h2 Zi,int(T ) . h i In case Σiνi = 0 all the constants except the masses mi cancel each other in the product, and we obtain

νi 3/2νi νi KP (T ) ≡ Π(Pi ) = Π mi Zi,int(T ) . (28) 12 12 Reaction heat

The heat absorbed by a reaction in constant pressure corresponds to the change of the enthalpy, H, of the system, δQP = δH = δ(E + PV ) , whereas the heat absorbed in constant volume equals to the change of internal energy, E:

δQV = δE .

The enthalpy can be derived from the Gibbs energy using the previously mentioned relation- ships between H, G, and S:

∂G G 1 ∂G ∂ G H = G + TS = G − T = −T 2 − + = −T 2 . ∂T T 2 T ∂T ∂T T N,P " N,P # N,P

Similarly, we see from the diﬀerential of the free energy, F , that ∂F S = − , ∂T N,V and the internal energy can be obtained from

∂F F 1 ∂F ∂ F E = F + TS = F − T = −T 2 − + = −T 2 . ∂T T 2 T ∂T ∂T T N,V " N,V # N,V

Consider the change of the Gibbs energy in δn elementary reactions. For the mixture of gases the Gibbs energy can be written as G = ΣµiNi, and then

δG = ΣµiδNi .

According to the reaction equation, the number of molecules i change in δn elementary reactions by δNi = νiδn. Note that here we use the convention that the coeﬃcients νi of the products (on the right) are positive, and the coeﬃcients of the reactants are negative. Substituting δNi to the expression of δG we get

δG = δnΣνiµi .

We can proceed by writing the chemical potential in either of the forms derived above. Choosing the expression µi = kT ln Pi + fi(T ) which is more natural in connection with the Gibbs energy we obtain

δG = δnΣνi[kT ln Pi + fi(T )] νifi(T ) = δnkT [Σνi ln Pi + Σ kT ] = δnkT [Σνi ln Pi − ln KP (T )] ,

In the last form we have used the formula of KP (T ) derived above. The substitution of µi in terms of numbers yields

νi νi δG = δnkT [Σνi ln Ni − ln K(T,V )] = δnkT [lnΠ(Ni ) − ln Π Ni,0 ] 13 The general direction of the reaction can be seen in the sign of δG. In natural processes the Gibbs energy tends to a minimum. The reaction proceeds from the left to the right if δG < 0, and from the right to the left if δG > 0:

νi νi νi νi → : δG < 0 ⇔ Σνi ln Pi < ln KP (T ) ⇔ Π(Pi ) < Π Pi,eq ⇔ Π(Ni ) < Π Ni,eq

νi νi νi νi ← : δG > 0 ⇔ Σνi ln Pi > ln KP (T ) ⇔ Π(Pi ) > Π Pi,eq ⇔ Π(Ni ) > Π Ni,eq The reaction heat in constant pressure can be calculated using the relationship between H and G:

2 ∂ δG δQP = δH = −T ∂T T T,P 2 ∂ = −δnkT ∂T [Σνi ln Pi − ln KP (T )] 2 ∂ = δnkT ∂T ln KP (T )

If we choose δn = N0 (the Avogadro number) the number of reactions corresponds to one mole, and we get the equation to a more familiar form (van ’t Hoﬀ’s equation): ∂ ∆H ln K (T )= , (29) ∂T P RT 2 where R ≡ N0k is the so called gas constant. From this equation we see that the heat ab- ∂ sorbed by the reaction is positive, δQp > 0, when ∂T ln KP (T ) > 0, i.e. when the equilibrium constant increases as a function of temperature. The reaction is then endothermic. Similarly, ∂ when ∂T ln KP (T ) < 0, also δQp < 0, and the reaction is exothermic. Eq. 29 is equivalent with ∂ ln KP (T ) ∆H 1 = − . (30) ∂ T R

1 The temperature parameter, β ≡ kT , is useful in calculations involving the partition function −βEi ∂ 1 ∂ (Z(β) = Σgie , ∂T = − kT 2 ∂β ). Using β we get for the enthalpy of a single elementary reaction in constant pressure: ∂ ln K (β) δH = − P . ∂β

The reaction heat in constant volume, δQV = δE, can be derived starting from the change of free energy:

δF = Σ ∂Fi δN = Σµ ν δn ∂Ni i i i = −δnkT Σν ln Zi(T,V ) i Ni = −δnkT [ln K(T,Vn ) − oΣνi ln Ni] , which implies ∂ δF ∂ δE = −T 2 = δnkT 2 ln K(T,V ) . ∂T T ∂T N,V Using the temperature parameter β we get for a single reaction in constant volume: ∂ ln K(β,V ) δE = − . ∂β

14 References

[1] Atkins, P.W., Physical Chemistry, Oxford University Press

[2] Mandl, F., Statistical Physics, John Wiley & Sons, Ltd

[3] Landau, L.D. & Lifshitz, E.M., Statistical Physics, Pergamon Press Ltd

[4] Ramanlal, J., Tennyson, J. 2004, Mon. Not. R. Astron. Soc. 354, 161

[5] Hugo, E., Asvany O., Schlemmer, S. 2009, Journal of Chemical Physics 130, 164302