Chemical equilirium, the law of mass action, and the reaction heat

Transcription

1 Chemical equilirium, the law of mass action, and the reaction heat Supplementary notes for the astrochemistry course, spring 2013 Jorma Harju March 21, Equilibrium constant and the law of mass action Most reactions we are dealing with in astrochemistry are of the type A+B k + C+D k The equilibrium constant, K(T, is defined by K(T k + k = N C,eqN D,eq N A,eq N B,eq, where N A,eq, N B,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: K(T = Z CZ D Z A Z B The partition function Z depends on energy spectrum of the molecule For a particle obeying the Boltzmann statistics Z(T = Σ i g i exp(e i /, where g i is the statistical weight (degeneracy, and E i is the energy of the state i The total partition funtion is a product of the electronic, translational, vibrational, and rotational partition functions, Z = Z el Z tr Z vib Z rot Normally only the ground electronic state is thermally accessible, so Z el = g el, where g el 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 E 0 The partition function in terms of E 0 is Z = Σg i e (E i E 0 / = e E0/ Z, implying Z = Z e E0/ Omitting the primes, K(T can be written as K(T = Z ( CZ D E0 exp, Z A Z B 1

2 where E 0 is the difference between the ZPVEs (products reactants: E 0 = E C 0 +E D 0 E A 0 E B 0 Sometimes E 0 is called the enthalpy of the reaction (eg, Ramanlal & Tennyson Why this assignment can be made is discussed below The translational partition function, Z tr, contains the term m 3/2, where m is the molecular mass In our example reaction other terms of Z tr are reduced in the expression of K(T, and therefore the equilibrium constant can be written as K(T N C,eqN D,eq N A,eq N B,eq = ( mc m D m A m B 3/2 Z C,int Z D,int Z A,int Z B,int exp ( E0, where Z int = Z rot Z vib 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: H + 3 +HD k + H 2 D + +H 2 k The equilibrium constant is K(T = N(H 2D + eq N(H 2 eq N(H + 3 eq N(HD eq = ( 3/2 4 2 Z(H 2 D +,intz(h 2,int 3 3 Z(H + 3,intZ(HD,int ( E0 exp The difference between the ZPVEs is E 0 /k = E 0 (H 2 D + /k + E 0 (H 2 /k E 0 (H + 3/k + E 0 (HD/k = 1395 K 1 A negative H E 0 means that the reaction from left to right is exothermic InFig1weshowthepartitionfunctionsZ(H 2 D +,int,z(h 2,int,Z(H + 3,int,andZ(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 or equivalently T 2 T lnk(t = H k, lnk(t 1 = H k T 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 Atlowtemperatures(1/T > 01K 1 theslopeindicatesthat H/k 232K ( ZPVE minus the rotational ZPE of H + 3, whereas at elevated temperatures (1/T < 002K 1 where the rotational levels of H + 3 are populated, H/k 140 K (= ZPVE In what follows we derive the equations used in this example 1 The ground rotational state of H + 3 is forbidden by the Pauli exclusion principle, and the lowest allowed state J,K = 1,1 lies 923 K above the ground state Often this rotational zero-point energy is added to the zero-point energy of H + 3, and E 0/k becomes K When using this value, we should also give the energies of Z(H + 3,int with respect to the state J,K = 1,1 2

3 Figure 1: Internalpartitionfunctions, Z int = Z vib Z rot, ofh 2 D +, H 2, H + 3,andHDasfunctions oft NotethatforH + 3 theenergiesarecalculatedfromtheforbiddenrotationalgroundstate, and therefore Z(H + 3,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, lnk, of reaction H + 3 +HD H 2 D + + H 2 as functions of T (left and 1/T (right Also shown are two approximations ln K(T H/ from the van t Hoff equation At elevated temperatures, the assumption H/k = ZPE 1 /k = 1395 K (the difference of zero-point vibrational energies seems to work well, whereas at low temperatures H/k = ZPE 2 /k = 2318 K ( ZPVE combined with the rotational zero-point energy of H + 3 gives a better agreement 3

4 2 Thermodynamic potentials The thermodynamic properties of a gas, eg, 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 + VdP + µdn, df = SdT PdV +µdn and dg = SdT +VdP +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 +Vdp+Σ i µ i dn i (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, = µ i (2 T P,N i P T,N i N i 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 i = = = (3 N i N i N i S,V 3 Chemical equilibrium Consider the chemical reaction S,P ν A A+ν B B k + ν C C+ν D D, k occurring in a mixture of reacting substances The coefficients ν A etc are the stoichiometric coefficients For example, in the reaction N 2 + 3H 2 2NH 3 (+ E, the coefficients are 1, 3, and 2, and in the reaction NH 3 + H 2 O NH OH all coefficients have the value 1 The numbers of particles A,,D, denoted by N A,,N D, in a volume element V change until 4 T,V

5 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 ν A A ν B B+ν C C+ν D D = 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, eg, 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 dn A = G N A + G N B dn B dn A + G N C dn C dn A + G N D dn D dn A = 0 (6 The relative changes of the numbers N A,N D are determined by the reaction equation (4: dn B dn A = ν B ν A, dn C dn A = ν C ν A, dn D dn A = ν D ν A Multiplying Equation (6 by ν A and using the definition of the chemical potential: µ A = G N A, µ B = G N B, we obtain the condition of chemical equilibrium: ν A µ A ν B µ B +ν C µ C +ν D µ D = 0 In other words, the reaction is in equilibrium when the chemical potentials of the reactants and products are equal 5

6 We express this in a more compact form by denoting the substances by A 1,A 2, instead of A,B Then the reaction equation becomes Σν i A i = 0 (7 Note that now the sign + or is incorporated in the stoichiometric coefficient ν i Using this notation the condition for chemical equilibrium can be written as Σ i ν i µ i = 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, eg K(T N C,eqN D,eq N A,eq N B,eq (see below 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, P = N V (9 the partial pressure P i of the component i can be written as P i = N i V = N i N N V = N i N P, where N i 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 = ΣP i Furthermore, the free energy and the Gibbs energy of the mixture can be written as sums: F(N,T,V = ΣF i (N i,t,v (10 G(N,T,P = ΣG i (N i,t,p i (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 = ln N! Z(T,VN N ln N Z(T,V, (12 6

7 where Z(T,V is the partition function (or sum over states of an individual molecule, and e = 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 = ln{z(t,v} The distribution of molecules of an ideal gas among the various states is called the Boltzmann distribution Accordingtothisdistributiontheprobabilityp j thatamoleculeisintheenergy state E j can be obtained from p j = 1 Z(T e E j/k B T, where k B is the Boltzmann constant ( JK 1, and T is the gas temperature From the normalization condition Σp j = 1 we obtain for the partition function the expression Z = Σe E j/ In a mixture of gases, the free energy of component i is { } e F i (N,T,V = N i ln Z i (T,V N i, where N i is the number of particles i in volume V and Z i (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 i (N,T,V = ΣN i ln Z i (T,V (13 N i 6 Partition function The partition function Z(T, V can be divided into translational or kinetic and internal parts The translational partition function, Z tr (T,V, takes into account the energy quantization of a particle in a vessel of confined volume The internal partition function, Z int (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 = Z tr (T,VZ int (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, eg Mandl Ch 72; Landau & Lifshitz 38, resulting in the expression Z tr (T,V = V ( 3/2 2πm (15 h 2 The internal partition function can be divided into the following factors (which are independent in the first approximation: Z int (T = e ǫe/ Z rot (TZ vib (T, 7

8 where ǫ e is the electronic energy of the molecule, Z rot is the rotational partition function: and Z vib is the vibrational partition function: Z rot (T Σ J=0(2J +1e BJ(J+1/, Z vib (T Σ v=0e hν 0(v+ 1 2 / (Above we have use the approximations of a rigid linear rotor and a harmonic oscillator More accurate formula for Z rot and Z vib 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, F tr, rotational free energy, F rot = N ln(z rot, the vibrational free energy, F vib = N ln(z vib, and the electronic free energy, Nǫ e According to the Boltzman distribution, at low temperatures almost all molecules are at their ground states: p j = e E j/ 0, when E j > 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, ǫ e,0 = 0, implying e ǫ e,0/ = 1 At low temperatures, the rotational partition function approaches the statistical weight of the ground rotational state, Z rot (T g 0 (1 for a linear rotor, when T 0, because the rotational energy levels (normally start from zero In contrast, the vibrational partition function Z vib (T e 1 2 hν 0/ The lowest possible vibrational energy of a molecule, 1 2 hν 0, 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 G i = F i +P i V Because the natural variables of G are N,T, and P, we replace V using the ideal gas equation of state: V = N i P i We have to make this substitution also in the translational partition function: Z i,tr (T,P i = N ( 3/2 i 2πmi P i h 2 After these substitutions we obtain G i (N i,t,p i = N i ln { e P i ( 3/2 2πmi Z i,int(t} +N i This expression can be simplified by separating the term lne(= 1 from the logarithm, because N i lne+n i = 0: { 3/2 } Zi,int (T h 2 G i (N i,t,p i = N i ln { = N i lnp i N i ln ( 2πmi P i h 2 8 ( 2πm i h 2 3/2 } Zi,int (T

9 We can spare some pain by collecting the terms that depend on T only on the right hand side under a function f i (T, defined as { ( 3/2 2πmi f i (T ln Z i,int(t} (16 h 2 The Gibbs energy of the mixture is then G(N,T,P = ΣG i (N i,t,p i = ΣN i lnp i +ΣN i f i (T (17 If we wish to use the total pressure P instead of the partial pressure P i, we can make the substitution P i = N i /N P: G(N,T,P = ΣN i lnp +ΣN i ln N i N +ΣN if i (T (18 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 ( { } Fi e µ i = = ln Z i (T,V N i ( 1 { } Zi (T,V = ln (19 N i N i N i N i T,V Here we again utilized algebraic rules: { } e ln Z i (T,V = lne lnn i +lnz i (T,V N i The case of the Gibbs energy is very simple: ( Gi µ i = = lnp i +f i (T (20 N i T,P i By substituting P i = N i /V and the definition of the function f i (T one can see that this expression is identical with the one obtained from F i On grounds of the relationship µ i = G i /N i we can interpret µ as the Gibbs energy per molecule (eg, Mandl Ch 81; Landau & Lifshitz 24 In Atkins (Ch 6 the chemical potential of a species is defined as the molar Gibbs function, G m, ie 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 +Nln(P/atm For the molar Gibbs function, ie the chemical potential as defined by Atkins, one gets G m (P = µ(p = µ +RTln(P/atm, 9

10 where R is the gas constant (R = N 0 k, N 0 is the Avogadro number Using the expression for G we obtained previously, and the same reference pressure as Atkins, P ref = 1 atm, the Gibbs energy of one mole of ideal gas can be written as We can thus make the following identifications: G m (P,T = N 0 ln(p/atm+n 0 ln(p ref +N 0 f(t G (T (Atkins = Nln(P ref +Nf(T µ (T (Atkins = N 0 ln(p ref +N 0 f(t (21 Similarly one could have defined a standard chemical potential per molecule as µ i = ln(p ref +f(t (22 We note that the reference pressure P ref = 1 atm 10 5 Pa is extremely high compared with interstellar conditions In molecular clouds, the pressure is typically P Pa, and a perhaps a few orders of magnitude higher in the nuclei of dense cores 9 The equilibrium constant in terms of numbers N i Here we use the expression for the chemical potential otained from the free energy Substituing Eq 19, { } Zi (T,V µ i = ln, to the condition for chemical equilibrium we get { } Zi (T,V Σν i µ i = Σν i ln = 0 The number of molecules i in chemical equilibrium are denoted by N i,eq Dividing by and using some algebra the equation can be written as { } Zi (T,V Σν i ln = Σln{Z i (T,V} ν i Σln{N i,0 } ν i = 0 N i,0 We move the second term to the right and take both sides as arguments of an exponential function: e Σln{Z i(t,v} ν i = e Σln{N i,0} ν i, which is equivalent with N i N i,eq Z 1 (T,V ν 1 Z 2 (T,V ν 2 Z 3 (T,V ν 3 = N ν 1 1,eqN ν 2 2,eqN ν 3 3,eq or Π ( N ν i i,eq = Π(Zi (T,V ν i 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 K(T,V Π ( N ν i i,eq = Π(Zi (T,V ν i (23 As an example we consider the reaction A+B k + C+D k 10

11 For this reaction the total number of particles is constant, Σν i = 0, and ν i = 1 for all i Eq 23 means that the relative numbers in equilibrium can be calculated from the following formula: N C,eq N D,eq = k + K(T = Z CZ D N A,eq N B,eq k Z A Z B K does not depend on the volume because Vs is reduced: V Σ iν i = 1 Here the numbers N i can be replaced by number densities n i 10 The equilibrium constant in terms of partial pressures The equilibrium constant in terms of partial pressures, K p (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 Σ i ν i µ i = Σ i ν i [lnp i +f i (T] = 0 Dividing by we obtain Σ i ν i lnp i = Σ i ν i f i (T The function f i (T/( 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 e Σ iν i lnp i = e Σ ilnp i ν i = Π(P ν i i = e Σ iν i f i (T/(, where Π(P ν i i = P ν 1 1 P ν 2 2 P ν 3 3 Here we have used the relations lna p = plna and e lna = a The product Π(P ν i i is defined as the equilibrium constant for partial pressures, K P (T, and it can be thus written as K P (T Π(P ν i i = e Σ iν i f i (T/( (24 The same result can be obtained from Eq 23 by using the equation of state of an ideal gas, PV = N, and the relationship between the numbers N i, partial pressures, P i, and concentrations [i]: [i] N i N = P i P Substituting N i,0 = NP i,0 /P to Eq 23, Π ( N ν i i,0 = Π(Zi (T,V ν i we obtain K in terms of partial pressures: Π ( P ν i i,0 KP (T = Π ( P ν i ( N i K(T,V ν ( PV = Π 2πmi 3/2 νi N h 2 Zi,int (T ( = Π ( 2πm i 3/2 νi h 2 Zi,int (T ( { } = Π exp f νi i(t { } = exp, 11 Σ ν if i (T

12 Likewise, we one can derive the equilibrium constant in terms of concentrations substituting N i = N[i]: Π([i] ν i 0 K C (T,P = N Σν i K(T,V (25 Eq 24 is equivalent with the one presented in textbooks of chemistry (eg Atkins Ch 9, relating K P to the standard molar Gibbs function, G m (= G m(products G m(reactants of the reaction : RTlnK P (Atkins = G m (26 HerethepartialpressuresinK P arereplacedbyratiosp i /P ref, wherethereferencepressureis1atm Making the substitution P i (P i /P ref P ref in Eq 24 we obtain: K P (T = Π(P νi i = Π((P i /P ref νi Π(P νi ref = e Σiνifi(T/( Taking the natural logarithm of the equation on the right we obtain lnπ((p i /P ref νi = lnπ(p νi ref Σ iν i f i (T/( = Σ i ν i [lnp ref +f i (T/(] = Σ i ν i µ i /( In the last equation we have used the standard chemical potential per molecule defined in Eq 22 The left hand side the equilibrium constant K P as defined by Atkins Multiplying by N 0 (= RT we obtain the desired formula, because N 0 µ i = G m,i and Σ iν i G m,i = G m, ie the change of molar Gibbs energy in the reaction On grounds of this equation one can estimate G m(t of a reaction by determining the equilibrium constant (ie 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 K P (T (Eqs 23 and 24, simply by writing the expressions of the translational partition function explicitly on the right hand side So we get K(T,V Π ( N ν i i,eq = Π(Zi (T,V ν i = Π([Z i,tr (T,VZ i,int (T] ν i ([ = Π V ( ] 2πm 3/2Zi,int νi (T h 2 [2πV ] 3/2νi = Π( m 3/2ν i h 2 i Z i,int (T ν i In the situation where the total number of particles does not change, Σ i ν i = 0, the constants on the right, except the masses m i are reduced (note that also V and T are constants here, and the equation above can be simplified: K(T = Π ( N ν i i,eq ( = Π m 3/2ν i i Z i,int (T ν i We now manipulate the right hand side of Eq 24: { } { } f exp Σ i ν i (T i = exp Σ i ν i ln f i(t { [ = exp Σ i ν i ln ( 2πm i 3/2 ]} h 2 Zi,int (T ([ = Π ( 2πm i 3/2 ] νi h 2 Zi,int (T (27 In case Σ i ν i = 0 all the constants except the masses m i cancel each other in the product, and we obtain ( K P (T Π(P ν i i = Π m 3/2ν i i Z i,int (T ν i (28 12

13 12 Reaction heat The heat absorbed by a reaction in constant pressure corresponds to the change of the enthalpy, H, of the system, δq P = δh = δ(e +PV, whereas the heat absorbed in constant volume equals to the change of internal energy, E: δq V = δe The enthalpy can be derived from the Gibbs energy using the previously mentioned relationships between H, G, and S: ( [ G H = G+TS = G T = T 2 G T N,P T + 1 ( ] ( G = T 2 G 2 T T N,P T T 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 E = F +TS = F T = T 2 F T N,V T T ( ] ( F = T 2 F T N,V T T 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 = Σµ i N i, and then δg = Σµ i δn i According to the reaction equation, the number of molecules i change in δn elementary reactions by δn i = ν i δn Note that here we use the convention that the coefficients ν i of the products (on the right are positive, and the coefficients of the reactants are negative Substituting δn i to the expression of δg we get δg = δnσν i µ i We can proceed by writing the chemical potential in either of the forms derived above Choosing the expression µ i = lnp i +f i (T which is more natural in connection with the Gibbs energy we obtain δg = δnσν i [ lnp i +f i (T] = δn[σν i lnp i +Σ ν if i (T ] = δn[σν i lnp i lnk P (T], In the last form we have used the formula of K P (T derived above The substitution of µ i in terms of numbers yields δg = δn[σν i lnn i lnk(t,v] = δn[lnπ(n ν i i lnπ ( N ν i i,0 ] 13

14 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: : δg < 0 Σν i lnp i < lnk P (T Π(P ν i i < Π ( P ν i i,eq Π(N ν i i < Π ( N ν i i,eq : δg > 0 Σν i lnp i > lnk P (T Π(P ν i i > Π ( P ν i i,eq Π(N ν i i > Π ( N ν i i,eq The reaction heat in constant pressure can be calculated using the relationship between H and G: δq P = δh = T 2( T = δn 2 δg T T,P [Σν T ilnp i lnk P (T] = δn 2 lnk T P(T If we choose δn = N 0 (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: T lnk P(T = H RT 2, (29 where R N 0 k is the so called gas constant From this equation we see that the heat absorbed by the reaction is positive, δq p > 0, when lnk T P(T > 0, ie when the equilibrium constant increases as a function of temperature The reaction is then endothermic Similarly, when T lnk P(T < 0, also δq p < 0, and the reaction is exothermic Eq 29 is equivalent with lnk P (T 1 T = H R (30 The temperature parameter, β 1, is useful in calculations involving the partition function (Z(β = Σg i e βe i, = 1 Using β we get for the enthalpy of a single elementary T 2 β reaction in constant pressure: δh = lnk P(β β The reaction heat in constant volume, δq V = δe, can be derived starting from the change of free energy: δf = Σ F i N i δn i = Σµ i ν i δn { } = δnσν i ln Zi (T,V N i = δn[lnk(t,v Σν i lnn i ], which implies ( δe = T 2 δf T T N,V = δn 2 T lnk(t,v Using the temperature parameter β we get for a single reaction in constant volume: δe = lnk(β,v β 14

15 References [1] Atkins, PW, Physical Chemistry, Oxford University Press [2] Mandl, F, Statistical Physics, John Wiley & Sons, Ltd [3] Landau, LD & Lifshitz, EM, 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,