Chemical Reaction Equilibria Chapter 13-Part II

Equilibrium constant K Standard Gibbs free energy change of reaction where K is a function of temperature

what is the effect of T on the equilibrium constant K? if the reaction is endothermic, DH0 >0, K increases when T increases if the reaction is exothermic, DH0 <0, K decreases when T increases

if DH0 is independent of T T’ is a reference temperature so, ln K vs 1/T is a straight line (Fig. 13.2)

However, in general DH0 is a function of T In Chapter 4 you learned how to calculate standard heats of reaction However, in general DH0 is a function of T note that we indicate DM0 as a reaction property at T and DM0o is that property evaluated at To

DG0=DH0-TDS0 the integral involves the DCp of reaction, i.e, the temperature dependence of the heat capacity of each species Cp/R = A+BT+CT2+DT-2 the integral is given by equation 4.19 and includes terms like: and the integral also includes similar terms, see eqn. (13.19)

We can rewrite K K=K0 K1 K2 K0 is K at the reference temperature To K1 is given by K2 is given by

DGo data Tables, for example TRC Tables (TAMU) Some values of DGo and DHo at 298 K Table C.4, appendix C.

an example the following reaction reaches equilibrium at 500oC and 2 bar: 4HCl (g) + O2 (g)  2H2O(g) + 2Cl2(g) If the system initially contains 5 mol HCl for each mole of oxygen, what is the composition of the system at equilibrium? Assume ideal gases. First calculate n and no DHo(298 K) DGo (298 K)

DHo(298 K)= -114408 J/mol; DGo(298 K)=-75948 J/mol (from Table C.4) to calculate DG at 500oC=773 K get parameters A, B, C, D for all the species from Table C.1 then calculate DA, DB, DD and then calculate DG (773K)=-1.267x104 J/mol using (13.18) K=exp(-DG/RT)=7.18

mole fractions as a function of e yHCl=(5-4e)/(6-e) yO2=(1-e)/(6-e) yH2O=2e/(6-e) yCl2=2e/(6-e) solve for e, and calculate the equilibrium mole fractions