Equilibrium for a general reaction

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Expressions of the equilibrium constant K The equilibrium constant, K, (a dimensionless quantity) can be expressed in terms of fugacities for gas phase reactions or activities for aqueous phase reactions. Fugacity ( a dimensionless quantity) is equal to the numerical value of partial pressure, i.e. p j/p where p = 1 bar). The activity, a, is equal to the numerical value of the molality, i.e. bj/b where b = 1 mol kg-1. For reactions occurring in electrolyte solutions Effects of the interactions of ions on the reaction process should be considered. Such a factor can be expressed with the activity coefficient, , which denotes distance from the ideal system where there is no ion-interaction. The activity shall now be calculated as j = j*bj/b

For a reaction A + B C + D K = aC a D C D bC bD a Aa B A B b AbB K K b The activities of solids are equal to 1 (solid) = 1 (!!!) Illustration: Express the equilibrium constant for the heterogeneous reaction NH4Cl(s) NH3(g) + HCl(g) Solution: In term of fugacity (i.e. partial pressure): Kp = In term of molar fraction: Kx =

Estimate reaction compositions at equilibrium Example 1: Given the standard Gibbs energy of reaction H2O(g) H2(g) + 1/2O2(g) at 2000K is + 135.2 kJ mol-1, suppose that steam at 200k pa is passed through a furnace tube at that temperature. Calculate the mole fraction of O2 present in the output gas stream. Solution: (details will be discussed in class) lnK = - (135.2 x 103 J mol-1)/(8.3145 JK-1mol-1 x 2000K) = - 8.13037 K = 2.9446x10-4 K= ( PO2 / P )1 / 2 ( PH 2 / P ) PH 2O / P

Ptotal = 200Kpa assuming the mole fraction of O2 equals x PO2 = x* Ptotal, PH2 = 2(x*Ptotal) PH2O = Ptotal PO2 PH2 = (1-3x)Ptotal Equilibria in biological systems Biological standard state: pH = 7. For a reaction: A + vH+(aq) P rG = rG + RT ln( bP v b A bH

) 1 = rG + RT ln([ H ]v ) RT ln bp bA the first two terms of the above eq. form rG rG = rG + 7vRTln10 Example: For a particular reaction of the form A B + 2H + in aqueous solution, it was found that rG = 20kJ mol-1 at 28oC. Estimate

the value of rG. Solution: rG = rG + 7vRTln10 here v = - 2 !!! rG = 20 kJ mol-1 + 7(-2)(8.3145x10-3 kJ K-1mol-1) x(273+ 28K)ln10 = 20 kJ mol-1 80.676 kJ mol-1 = -61 kJ mol-1 (Note that when measured with the biological standard, the standard reaction Gibbs energy becomes negative!) Molecular Interpretation of equilibrium The response of equilibria to reaction conditions Equilibria respond to changes in pressure,

temperature, and concentrations of reactants and products. The equilibrium constant is not affected by the presence of a catalyst. How equilibria respond to pressure Equilibrium constant K is a function of the standard reaction Gibbs energy, rG . Standard reaction Gibbs energy rG is defined at a single standard pressure and thus is independent of pressure. The equilibrium constant is therefore independent of pressure: ( K ) 0 p T K is independent of pressure does NOT mean that the equilibrium composition is independent

of the pressure!!! consider the reaction 2A(g) B(g) assuming that the mole fraction of A equals xA at quilibrium, then xB = 1.0 xA, K= (1.0 x A ) Ptotal / P ( x A Ptotal / P ) 2 (1.0 x A ) P 2 x A Ptotal because K does not change, xA must change in response to any variation in Ptotal!!! Le Chateliers Principle A system at equilibrium, when

subject to a disturbance, responds in a way that tends to minimize the effect of the disturbance. Example: Predict the effect of an increase in pressure on the Haber reaction, 3H2(g) + N2(g) 2NH3(g). Solution: According to Le Chateliers Principle, an increase in pressure will favor the product. prove: K= 2 p NH 3 3

p N pH 2 2 2 2 x NH p total 3 3 x N 2 ptotal x H 2 3

ptotal 2 x NH 3 3 xN2 xH p2 2 total Kx 2 ptotal Therefore, to keep K unchanged, the equilibrium mole fractions Kx will change by a factor of 4 if doubling the pressure ptotal.

The response of equilibria to temperature According to Le Chateliers Principle: Exothermic reactions: increased temperature favors the reactants. Endothermic reactions: increased temperature favors the products. The vant Hoff equation: (a) (b) d ln K dT d ln K 1 d( ) T

r H RT 2 r H R (7.23a) (7.23b) Derivation of the vant Hoff equation: r G ln K RT Differentiate lnK with respect to temperature

d ln K 1 d ( r G / T ) dT R dT Using Gibbs-Helmholtz equation (eqn 3.53 8th edition) d ( r G / T ) r H dT T2 thus

d ln K dT r H RT 2 Because d(1/T)/dT = -1/T2: d ln K 1 d( ) T r H R

d ln K 0 , suggesting For an exothermic reaction, rH < 0, thus dT that increasing the reaction temperature will reduce the equilibrium constant.

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