Consider a system that consists initially of 1mol of CO and 3 mol of H2 at 1000K. The system pressure is 25 bar.The following reactions are to be considered: 2CO+2H2⇔CH4+CO2 (A)CO+3H2⇔CH4+H2O (B)CO2+H2⇔H2O+CO (C)When the equilibrium constants for reactions (A) and (B) are expressed in terms of the

Question

Consider a system that consists initially of 1mol of CO and 3 mol of [latex]H_{2}[/latex] at 1000K. The system pressure is 25 bar.The following reactions are to be considered:

[latex]2CO+2H_{2}\leftrightarrow CH_{4}+CO_{2}[/latex] (A)
[latex]CO+3H_{2}\leftrightarrow CH_{4}+H_{2}O[/latex] (B)
[latex]CO_{2}+H_{2}\leftrightarrow H_{2}O+CO[/latex] (C)

When the equilibrium constants for reactions (A) and (B) are expressed in terms of the partial pressures of the various species (in bar), the equilibrium constants for these reactions have the following values:

[latex]K_{P,A}=0.046[/latex] [latex]K_{P,B}=0.034[/latex] [latex]K_{P,C}=?[/latex]

Determine the equilibrium composition of the mixture.

Accepted Answer

The first step in the analysis is to determine if the chemical equations (A) to (C) are independent by applying the test described above. When one does this, one finds that only two of the reactions are independent. We will choose the first two for use in subsequent calculations. Let the variables [latex]\xi _{A}[/latex] and [latex]\xi _{B}[/latex] represent the equilibrium extents of reactions A and B, respectively. A mole table indicating the mole numbers of the various species present at equilibrium may be prepared using the following form of equation (1.1.6):

[latex]n_{1}=n_{i0}+ u _{Ai}\xi _{A}+ u _{Bi}\xi _{B}[/latex]

Species	Initial mole number	Equilibrium mole number
CO	1	[latex]1-2\xi _{A}-\xi _{B}[/latex]
[latex]H_{2}[/latex]	3	[latex]3-2\xi _{A}-3\xi _{B}[/latex]
[latex]CH_{4}[/latex]	0	[latex]\xi _{A+\xi _{B}}[/latex]
[latex]CO_{2}[/latex]	0	[latex]\xi _{A}[/latex]
[latex]H_{2}O[/latex]	0	[latex]\xi _{B}[/latex]
Total	4	[latex]4-2\xi _{A}-2\xi _{B}[/latex]

The fact that none of the mole numbers can ever be negative places maximum values of [latex]\frac{1}{2}[/latex] on [latex]\xi _{A}[/latex] and 1 on [latex]\xi _{B}[/latex].
The values of [latex]K_{P}[/latex] for reactions A and B are given by

[latex]K_{P,A}=\frac{P_{CH_{4}}P_{CO_{2}}}{P_{CO}^{2}P_{H_{2}}^{2}}=\frac{Y_{CH_{4}}Y_{CO_{2}}}{Y_{CO}^{2}Y_{H_{2}}^{2}P^{2}}[/latex]

[latex]K_{P,B}=\frac{P_{CH_{4}}P_{H_{2}O}}{P_{CO}P_{H_{2}}^{3}}=\frac{Y_{CH_{4}}Y_{H_{2}O}}{Y_{CO}Y_{H_{2}}^{3}P^{2}}[/latex]

The mole fractions of the various species may be expressed in terms of [latex]\xi _{A}[/latex] and [latex]\xi _{B}[/latex], so that the above relations for [latex]K_{P}[/latex] become

[latex]K_{P,A}=\frac{\left ( \frac{\xi _{A}+\xi _{B}}{4-2\xi _{A}-2\xi _{B}} ight )\left ( \frac{\xi _{A}}{4-2\xi _{A}-2\xi _{B}} ight )}{\left ( \frac{1-2\xi _{A}-\xi _{B}}{4-2\xi _{A}-2\xi _{B}} ight )^{2}\left ( \frac{3-2\xi _{A}-3\xi _{B}}{4-2\xi _{A}-2\xi _{B}} ight )^{2}P^{2}}[/latex]

[latex]=\frac{\left ( \xi _{A}+\xi _{B} ight )\xi _{A}\left ( 4-2\xi _{A}-2\xi _{B} ight )^{2}}{\left ( 1-2\xi _{B}-\xi _{B} ight )^{2}\left ( 3-2\xi _{A}-3\xi _{B} ight )^{2}P^{2}}[/latex]

[latex]K_{P,B}=\frac{\left ( \frac{\xi _{A}+\xi _{B}}{4-2\xi _{A}-2\xi _{B}} ight )\left ( \frac{\xi _{B}}{4-2\xi _{A}-2\xi _{B}} ight )}{ \frac{1-2\xi _{A}-\xi _{B}}{4-2\xi _{A}-2\xi _{B}}\left ( \frac{3-2\xi _{A}-3\xi _{B}}{4-2\xi _{A}-2\xi _{B}} ight )^{3}P^{2}}[/latex]

[latex]=\frac{\left ( \xi _{A}+\xi _{B} ight )\xi _{B}\left ( 4-2\xi _{A}-2\xi _{B} ight )^{2}}{\left ( 1-2\xi _{A}-\xi _{B} ight )\left ( 3-2\xi _{A}-3\xi _{B} ight )^{3}P^{2}}[/latex]

Substitution of numerical values for P,[latex]K_{P,A}[/latex], and [latex]K_{P,B}[/latex] gives two equations in two unknowns. The resulting equations can be solved only by numerical techniques. In this case, a simple graphical approach can be employed in which one plots [latex]\xi _{A}[/latex] versus [latex]\xi _{B}[/latex] for each equation and notes the point of intersection. Values of [latex]\xi _{A}=0.128[/latex] and [latex]\xi _{B}=0.593[/latex] are consistent with these equations. Thus, at equilibrium,

Species	Mole number	Mole fraction
CO	0.151	0.059
[latex]H_{2}[/latex]	0.965	0.377
[latex]CH_{4}[/latex]	0.721	0.282
[latex]CO_{2}[/latex]	0.128	0.05
[latex]H_{2}O[/latex]	0.593	0.232
Total	2.558	1.000

Species	Initial mole number	Equilibrium mole number
CO	1	$1-2\xi _{A}-\xi _{B}$
$H_{2}$	3	$3-2\xi _{A}-3\xi _{B}$
$CH_{4}$	0	$\xi _{A+\xi _{B}}$
$CO_{2}$	0	$\xi _{A}$
$H_{2}O$	0	$\xi _{B}$
Total	4	$4-2\xi _{A}-2\xi _{B}$

Question 2.2: Consider a system that consists initially of 1mol of CO and ...

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