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In this case, according to the described chemical reaction, which takes place between carbon monoxide and hydrogen to produce methanol at 260 °C and 40 atm:
[tex]CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)[/tex]
It is possible to calculate the pressure-based equilibrium constant via:
[tex]Kp=exp(-\frac{\Delta G\°}{RT} )[/tex]
Whereas the change in the Gibbs free energy for the reaction is calculated with the following, assuming these changes can be assumed constant for the temperature range (25°C to 260°C):
[tex]\Delta G\°=\Delta H\°-T\Delta S\°[/tex]
Whereas the change in both enthalpy and entropy are based on enthalpies of formation and standard entropies of both carbon monoxide and methanol respectively (exclude hydrogen as it is a single molecule of the same atom):
[tex]\Delta H\°=-166.3-(-137.3)=-29kJ/mol\\\\\Delta S\°=0.1268-0.1979=-0.0711 kJ/mol-K[/tex]
Thus:
[tex]\Delta G\°=-29\frac{kJ}{mol} -(260+273.15)K*(-0.0711\frac{kJ}{mol*K} )=-66.91 kJ/mol[/tex]
Hence, the pressure-based equilibrium constant will be:
[tex]Kp=exp(-\frac{-66,091\frac{J}{mol}}{8.314\frac{J}{mol*K} *(260+273.15)K} )=3.592x10^6[/tex]
Next, we calculate the concentration-based equilibrium constant:
[tex]Kc=\frac{Kp}{RT^{\Delta \nu}} =\frac{3.592x10^6}{((8.314\frac{J}{mol*K})(260+273.15)K)^{(1-2-1)}}= 7.058x10^{13}[/tex]
After that, we calculate the volume for us to get concentrations for the involved species at equilibrium:
[tex]V=\frac{(200+350)mol*0.08206\frac{atm*L}{mol*K}(260+273.15)K}{40atm} =601.6L[/tex]
[tex][CO]_0=\frac{200mol}{601.6L}=0.332M[/tex]
[tex][H_2]_0=\frac{350mol}{601.6L}=0.582M[/tex]
Then, the equilibrium expression and solution according to the ICE chart:
[tex]7.058x10^{13}=\frac{x}{(0.332-x)(0.582-2x)^2}[/tex]
Whose physically-consistent solution would be x = 0.29 M, it means that the equilibrium conversions are:
[tex]X_{CO}=\frac{0.29M}{0.332M}=0.873(87.3\%) \\\\X_{H_2}=\frac{0.29M*2}{0.582m}=0.997(99.7\%)[/tex]
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