1. Chemical Reaction Equilibria
Chapter 13-Part II

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

5. 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

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

8. 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

9. 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)

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

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

12. 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)

13. 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

14. 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