Chemical equilibrium: the principles 자연과학대학 화학과 박영동 교수

Chemical equilibrium: the principles 7.1 Thermodynamic background The reaction Gibbs energy The variation of Δ r G with composition Reactions at equilibrium The standard reaction Gibbs energy The equilibrium composition The equilibrium constant in terms of concentration 7.2 The response of equilibria to the conditions The presence of a catalyst The effect of temperature The effect of compression

Chapter 7. Chemical equilibrium: the principles Be familiar with the following terminologies or ideas: - The reaction Gibbs energy ΔG R - The variation of ΔG R with composition - Determination of equilibrium constants and the equilibrium concentration - The effect of temperature on the equilibrium constant - The effect of pressure

Derivation of the general equilibrium expression dG R = VdP - SdT +  (i=from 1 to N s ) μ i dn i Let ξ be the extent of chemical reaction and if the sign of (  G/  ξ) T,P at constant T and P determines whether the reaction has the spontaneity. (  G/  ξ) T,P ≤ 0 where “<” implies that it has spontaneity and “=” means the equilibrium. It can be shown that ΔG R = (  G/  ξ) T,P. For any reaction aA + bB → cC + dD the reaction Gibbs energy at const T and P can be expressed as in ΔG R =  (i= 1 to N s ) ν i μ i (1) where ν i denotes the stoichiometric coefficients like a, b, c, d. Remember that μ i = μ i o + RT ln a i and substitute this for eq (1), then ΔG R =  ν i μ i =  ν i μ i o + RT  ν i ln a i =ΔG R o + RT ln Π a i νi =ΔG R o + RT ln Q (2) where Q ≡ Π a i νi = (a C c a D d )/(a A a a B b )(3) At equilibrium ΔG R = 0 and ΔG R o = - RT ln K(4) where K ≡ (a C,eq c a D,eq d )/(a A,eq a a B,eq b ) The standard reaction Gibbs energy, ΔG R o, can be obtained by two different methods. (method 1) ΔG R o = ΔH R o - T ΔS R o (method 2) ΔG R o =  ν i ΔG f o (product,i) -  ν i ΔG f o (reactant,i)

Determination of equilibrium composition (Example) Find the equilibrium constant K of N 2 O 4 (g) ⇄ 2 NO 2 (g) at a given temperature T and the composition at equilibrium (Answer) (1) K can be obtained from eq (4) if ΔG R o of the reaction is known. (2) N 2 O 4 (g) ⇄ 2 NO 2 (g) Initial amount10 Equilibrium amount1-ξ 2ξTotal amount = 1 + ξ Equilibrium mole fraction (1-ξ)/ (1+ξ) 2ξ/ (1+ξ) K = [P(NO 2 )/P o ) 2 / [P(N 2 O 4 )/P o ) = {[2ξ/(1+ξ)](P/P o )} 2 / {[(1-ξ)/(1+ξ)](P/P o )} = 4ξ 2 (P/P o ) / (1 -ξ 2 ) Then the extent of the reaction ξ is given by ξ = 1 / [ 1 + (4/K)(P/P o )] 1/2

Temperature effect on K Let us first derive the Gibbs-Helmholtz equation. Remember that G=H - TS=H + T(  G/  T) P,{ni} (see eq of the special lecture)(6) [(  (G/T)/  T] P,{ni} = -G/T 2 + (1/T) (  G/  T) P,{ni} (7) From eqs (6) and (7) we obtain H = -T 2 [(  (G/T)/  T] P,{ni} (8) Equation (8) is referred to as the Gibbs-Helmholtz equation. For a change between state 1 and state 2, this equation can be written ΔH = -T 2 [(  (ΔG/T)/  T] P,{ni} For a chemical reaction occurring at standard state (constant pressure) with a fixed compositionΔH R o = -T 2 [(d(ΔG R o /T)/ dT] = RT 2 (d lnK/dT) Or (d lnK/dT) = ΔH R o / RT 2 (9) Eq. (9) is known as the van’t Hoff equation. The integral of eq (9) from T 1 to T 2 yields ln(K 2 /K 1 ) = (ΔH R o /R) (1/T 1 - 1/T 2 ) Also remember that ln K = -ΔG R o / RT = - ΔH R o / RT + ΔS R o /R

Reaction Gibbs Energy

The variation of Δ r G with composition

The standard reaction Gibbs energy

The endothermic reaction and temperature

The standard Gibbs energy of formation

The effect of temperature