SOx chemical equilibrium Consider a mixture of 64 kg of SO2 and 32 kg of O2 mixed with 56 kg of inert N2 in a sealed reactor. After ignition the system eventually reaches chemical equilibrium at temperature T = 1,000 K and pressure P = 2 atm. Assume nitrogen does not participate in the chemical

Question

SOx chemical equilibrium
Consider a mixture of 64 kg of [latex]SO_{2}[/latex] and 32 kg of [latex]O_{2}[/latex] mixed with 56 kg of inert [latex]N_{2}[/latex] in a sealed reactor. After ignition the system eventually reaches chemical equilibrium at temperature T = 1,000 K and pressure P = 2 atm. Assume nitrogen does not participate in the chemical reactions. If the equilibrium constant (KP) at 1,000 K is 1.8, determine the corresponding chemical mole fractions in the equilibrium products.

Accepted Answer

The mole amount of the mixture before reaction can be determined as follows:
[latex]n_{SO_{2}}=\frac{64,000 g}{64 g/mole}[/latex] = 1,000 mole

[latex]n_{O_{2}}=\frac{32,000 g}{32 g/mole}[/latex] = 1,000 mole

[latex]n_{N_{2}}=\frac{56,000 g}{28 g/mole}[/latex] = 2,000 mole

After ignition the chemical reaction most likely proceed as follows:
[latex]SO_{2}+\frac{1}{2}O_{2}\leftrightarrow SO_{3}[/latex]

Since [latex]N_{2}[/latex] does not participate in the reaction as an inert gas, it works only as a dilution gas throughout the process. When the final products reach equilibrium, we assume there is s mole of [latex]SO_{3}[/latex]. Then we can determine the mole amount of other gases before combustion and after reaching equilibrium by mass balance as follows.

Mole amount	[latex]n_{SO_{2}}[/latex]	[latex]n_{O_{2}}[/latex]	[latex]n_{SO_{3}}[/latex]	[latex]n_{N_{2}}[/latex]	Total n
Prior to combustion	1,000	1,000	0	2,000	4,000
At equilibrium after reaction	1,000 − s	1,000 − 0.5 s	s	2,000	4,000 − 0.5 s

The partial pressure-based equilibrium constant can be calculated using Eq. (3.31)

[latex]K_{P}=\frac{n^{c}_{C}n^{d}_{D}}{n^{a}_{A}n^{b}_{B}}\left(\frac{P}{n} \right)^{c+d-a-b}[/latex] (3.31)

[latex]K_{P}=\frac{n_{SO_{3}}}{n_{SO_{2}}n^{0.5}_{O_{2}}} \left(\frac{P}{n} \right)^{1-1-0.5}=\frac{s}{(1,000-s)(1,000-0.5s)^{0.5}}\left(\frac{2}{4,000-0.5s} \right)^{-0.5}[/latex]

At 1,000 K, [latex]K_{P}[/latex] = 1.8, by iteration we can solve this equation and get

s = 515

So the final mole amount of the species are determined as follows.

	[latex]SO_{2}[/latex]	[latex]O_{2}[/latex]	[latex]SO_{3}[/latex]	[latex]N_{2}[/latex]	Total
[latex]n_{i}(mole)[/latex]	485	742.5	515	2,000	3742.5

Question 3.7: SOx chemical equilibrium Consider a mixture of 64 kg of SO2 ...

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