Chemical Thermodynamics 2018/2019

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Chemical Thermodynamics2018/2019
9th Lecture Calculating Equilibrium
Quantities Valentim M B Nunes, UD de Engenharia
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Introduction
The calculation of equilibrium quantities its 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.

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Le Chateliers 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
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A practical example the Haber process
½ N2(g) 3/2 H2(g) NH3(g)
?Hºr(298 K) -46.21 kJ/mol ?Gºr(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.
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Haber process
Raising the temperature to 800 K speed it up.
But, since ?Hºlt0, the reaction is exothermic, Le
Chatelier tell us that the equilibrium will shift
towards the reactants.
What to do? Recall that
In this case, ?n -1, then if we increase the
pressure the equilibrium shifts toward the
products. For instance at 100 bar
much better!!
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Effect of total pressure example
N2O4(g) 2 NO2(g)
N2O4 NO2 Total
Initial of mol 1 0 1
at equilibrium 1-? 2? 1?
Molar fraction 1
Partial pressure p
Equilibrium constant comes
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Effect of total pressure example
Kp doesnt depend on pressure, but ? 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, ? 0.45, that means 45 of dissociation
At 10-5 bar, ? 0.999, that means approximately
100 of dissociation!
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Another example
CO2(g) CO(g) ½ O2(g)
CO2 CO O2 total
Initial of mol 1 0 0 1
at equilibrium 1-? ? ½ ? 1 ½ ?
Molar fraction 1
Partial pressure p
? ? (T,p½)
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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, ?Gº 20.92 kJ, then
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Synthesis of ethanol
C2H4 H2O C2H5OH total
Initial of mol 1 5 0 6
at equilibrium 1-? 5-? ? 6-?
Molar fraction 1
Partial pressure p
? 0.1866 this means that we have an
approximate yield of 18.7 of conversion of
ethylene in ethanol.