Question

An equilibrium mixture of 3 kmol of CO, 2.5 kmol of O2, and 8 kmol of N2 is heated to 2600 K at a pressure of 5 atm. Determine the equilibrium composition of the mixture for these conditions.

Answer 1

Answer:

$x_{CO}=0.0203\\x_{O_2}=0.0926\\x_{CO_2}=0.227\\x_{N_2}=0.660$

Explanation:

Hello,

In this case, we consider the reaction:

$CO(g)+\frac{1}{2} O_2(g)\rightleftharpoons CO_2$

For which the law of mass action is expressed as:

$Kp=\frac{n_{CO_2}}{n_{CO}*n_{O_2}^{1/2}} (\frac{P}{n_{Tot}} )^{1-1-1/2}$

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:

$K=exp(2.801)=16.46$

In such a way, in terms of the change $x$ the equilibrium goes:

$16.46=\frac{x}{(3kmol-x)*(2.5kmol-0.5x)^{0.5}} (\frac{5}{13.5kmol-0.5x} )^{-0.5}$

Hence, solving for $x$:

$x=2.754kmol$

Thus, the moles at equilibrium:

$n_{CO}=3-2.754=0.246kmol\\n_{O_2}=2.5-0.5(2.754)=1.123kmol\\n_{CO_2}=x=2.754kmol\\n_{N_2}=8kmol$

Finally the compositions:

$x_{CO}=\frac{0.246}{0.246+1.123+2.754+8} =0.0203\\\\x_{O_2}=\frac{1.123}{0.246+1.123+2.754+8} =0.0926\\\\x_{CO_2}=\frac{2.754}{0.246+1.123+2.754+8} =0.227\\\\x_{N_2}=\frac{8}{0.246+1.123+2.754+8} =0.660$

Best regards.