Answer:
[tex]x_{CO}=0.0203\\x_{O_2}=0.0926\\x_{CO_2}=0.227\\x_{N_2}=0.660[/tex]
Explanation:
Hello,
In this case, we consider the reaction:
[tex]CO(g)+\frac{1}{2} O_2(g)\rightleftharpoons CO_2[/tex]
For which the law of mass action is expressed as:
[tex]Kp=\frac{n_{CO_2}}{n_{CO}*n_{O_2}^{1/2}} (\frac{P}{n_{Tot}} )^{1-1-1/2}[/tex]
Whereas the exponents are referred to the stoichiometric coefficients in the chemical reaction. Moreover, in table A-28 (Cengel's thermodynamics) the natural logarithm of the undergoing reaction at 2600 K is 2.801, thus:
[tex]K=exp(2.801)=16.46[/tex]
In such a way, in terms of the change [tex]x[/tex] the equilibrium goes:
[tex]16.46=\frac{x}{(3kmol-x)*(2.5kmol-0.5x)^{0.5}} (\frac{5}{13.5kmol-0.5x} )^{-0.5}[/tex]
Hence, solving for [tex]x[/tex]:
[tex]x=2.754kmol[/tex]
Thus, the moles at equilibrium:
[tex]n_{CO}=3-2.754=0.246kmol\\n_{O_2}=2.5-0.5(2.754)=1.123kmol\\n_{CO_2}=x=2.754kmol\\n_{N_2}=8kmol[/tex]
Finally the compositions:
[tex]x_{CO}=\frac{0.246}{0.246+1.123+2.754+8} =0.0203\\\\x_{O_2}=\frac{1.123}{0.246+1.123+2.754+8} =0.0926\\\\x_{CO_2}=\frac{2.754}{0.246+1.123+2.754+8} =0.227\\\\x_{N_2}=\frac{8}{0.246+1.123+2.754+8} =0.660[/tex]
Best regards.
