1. Chemical Thermodynamics 2013/2014
9th Lecture: Calculating Equilibrium Quantities
Valentim M B Nunes, UD de Engenharia

2. Introduction
The calculation of equilibrium quantities it’s a very important subject both at lab scale and industrial scale. Thermodynamic principles give us the tools to calculate the yield of a given reaction, and to predict the effect of variables, like temperature or pressure, on reaction yields.

3. Le Chatelier’s Principle
Temperature; recall that,
Increase in temperature
 Endothermic reaction is favored
 Exothermic reaction is extinguished
Decrease in temperature
 Endothermic reaction is extinguished
 Exothermic reaction is favored
Pressure: recall that,
Increase of pressure
The system shifts to lower volume, then it will evolve in the direction corresponding to a lower number of gaseous molecules
Decrease of pressure
The system shifts to higher volume, then it will evolve in the direction corresponding to a higher number of molecules

4. A practical example: the Haber process
½ N2(g) + 3/2 H2(g) = NH3(g)
ΔHºr(298 K) = kJ/mol
ΔGºr(298 K) = kJ/mol
Increase of temperature
Decrease of temperature
Increase of pressure
Decrease of pressure OK
At room temperature the reaction is slow! This is kinetics, not thermodynamics….

5. Haber process
Raising the temperature to 800 K speed it up. But, since ΔHº<0, the reaction is exothermic, Le Chatelier tell us that the equilibrium will shift towards the reactants.
What to do? Recall that
In this case, Δn = -1, then if we increase the pressure the equilibrium shifts toward the products. For instance at 100 bar:
much better!!

6. Effect of total pressure: example
N2O4(g) = 2 NO2(g)
N2O4
NO2
Total
Initial # of mol 1
# at equilibrium
1-ξ 2ξ
1+ξ
Molar fraction
Partial pressure p
Equilibrium constant comes:

7. Effect of total pressure: example
Kp doesn’t depend on pressure, but ξ will shift with pressure! This is a very important quantity as it defines the amount of product that we can obtain from one reaction at a given temperature. At 298 K:
Rearranging the equation for Kp we obtain:
At 1 bar, ξ = 0.45, that means 45% of dissociation
At 10-5 bar, ξ = 0.999, that means approximately 100% of dissociation!

8. Another example
CO2(g) = CO(g) + ½ O2(g)
CO2 CO O2
total
Initial # of mol 1
# at equilibrium
1-ξ ξ
½ ξ
1+ ½ ξ
Molar fraction
Partial pressure p
ξ = ξ (T,p½)

9. Synthesis of ethanol
Let us suppose that we want to calculate the extent of the reaction of synthesis of ethanol, reacting ethylene and water, in gaseous phase, at 250 ºC and 34 atm (~34 bar). Assume that we mixture 5 moles of water vapor for each mole of ethylene. The reaction is:
C2H4(g) + H2O(g) = C2H5OH(g)
At this temperature, ΔGº = kJ,
then:

10. Synthesis of ethanol
C2H4
H2O
C2H5OH
total
Initial # of mol 1 5 6
# at equilibrium
1-ξ
5-ξ ξ
6-ξ
Molar fraction
Partial pressure p
ξ = ; this means that we have an approximate yield of 18.7% of conversion of ethylene in ethanol.