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

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.

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

A practical example: the Haber process ½ N2(g) + 3/2 H2(g) = NH3(g) ΔHºr(298 K) = -46.21 kJ/mol ΔGºr(298 K) = -16.74 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….

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!!

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:

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!

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)

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 = 20.92 kJ, then:

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 = 0.1866; this means that we have an approximate yield of 18.7% of conversion of ethylene in ethanol.