Fuel rich combustion and chemical equilibrium Combustion products from the reactant mixture of 1,000 mol of CO2, 500 mol of O2 and 500 mol of N2 consist of CO2, CO, O2, N2 and NO at 3,000 K and 1 atm. Determine the equilibrium composition of the combustion product.

Question

Fuel rich combustion and chemical equilibrium
Combustion products from the reactant mixture of 1,000 mol of [latex]CO_{2}[/latex], 500 mol of [latex]O_{2}[/latex] and 500 mol of [latex]N_{2}[/latex] consist of [latex]CO_{2}, CO, O_{2}, N_{2}[/latex] and NO at 3,000 K and 1 atm. Determine the equilibrium composition of the combustion product.

Accepted Answer

Since the reactant mixture and the product compounds are already known, the combustion stoichiometry can be described as

[latex]CO_{2}+\frac{1}{2}O_{2}+\frac{1}{2}N_{2} \rightarrow aCO+bNO+cCO_{2}+dO_{2}+eN_{2}[/latex]

with a, b, c, d, e as coefficients to be determined.
From the mass balances for C, O, and N we can set up three equations:
Carbon balance : a + c + 1 (1)
Oxygen balance : a + b + 2c + 2d = 3 (2)
Nitrogen balance : b + 2e = 1 (3)

We need two more equations, which can be obtained from two equilibrium reactions:
[latex]CO_{2}\leftrightarrow CO+½O_{2}[/latex] [latex](R_{1})[/latex]
[latex]½O_{2} + ½N_{2} ↔ NO[/latex] [latex](R_{2})[/latex]

The corresponding chemical equilibrium constants can be found in Table 3.2, which gives,
[latex]K_{P1}[/latex] (3,000 K) = exp(-1.111) = 0.3273
[latex]K_{P2}[/latex] (3,000 K) = exp(-2.114) = 0.1222

In the combustion products, for a mole of CO, there are b mole of NO, c mole of [latex]CO_{2}[/latex], d mole of [latex]O_{2}[/latex] and e mole of [latex]N_{2}[/latex], therefore, the mole fraction of the species in the combustion products can be determined as
[latex]y_{CO}=\frac{a}{a + b + c + d + e}, y_{NO}=\frac{b}{a + b + c + d + e}, y_{CO_{2}}=\frac{c}{a + b + c + d + e}[/latex]

[latex]y_{O_{2}}=\frac{d}{a + b + c + d + e}, y_{N_{2}}=\frac{e}{a + b + c + d + e}[/latex]

From the equilibrium chemical reaction [latex]CO_{2} ↔ CO + ½O_{2}[/latex], the chemical equilibrium constant at 1 atm is
[latex]K_{P1}=\frac{y_{co} \cdot y^{1/2}_{O_{2}} }{y_{CO_{2}}}=\frac{ad^{1/2}}{c}(a+b+c+d+e)^{-1/2}=0.3273[/latex] (4)

Similarly, for the second equilibrium reaction [latex]½ O_{2} + ½ N_{2}[/latex] ↔ NO, its equilibrium constant
[latex]K_{P2}=\frac{y_{NO}}{y^{1/2}_{O_{2}}y^{1/2}_{N_{2}}}=\frac{b}{d^{1/2}e^{1/2}}=0.1222[/latex] (5)

From Eqs. (1)–(3), we can get
c = 1 - a
d = 1/2(1 + a - b)
e = 1/2(1 - b)

Put them into (4) and (5), we can get
[latex]\frac{a}{1 - a}\left(\frac{1 + a - b}{4 + a} \right)^{1/2}=0.3273[/latex] (4′)
[latex]\frac{2b}{\left[(1 + a - b)(1 - b)\right]^{1/2}}=0.1222[/latex] (5′)

Solving Eqs. (4′) and (5′) we get
a = 0.3745 b = 0.0675 c = 0.6255 d = 0.6535 e = 0.4663

Then the mole fraction of each species is determined as follows:
[latex]y_{CO} = 0.17, y_{NO} = 0.03, y_{CO_{2}} = 0.29, y_{O_{2}} = 0.30, y_{N_{2}} = 0.21.[/latex]

Table 3.2 Equilibrium constants based on partial pressure for chemical reactions

T (K)	[latex]ln(K_{P})[/latex]
T (K)	[latex]½O_{2} + ½N_{2} ↔ NO[/latex]	[latex]CO_{2} + H_{2} ↔ CO + H_{2}O[/latex]	[latex]CO_{2} ↔ CO + ½O_{2}[/latex]	[latex]H_{2}O ↔ H_{2} + ½ O_{2}[/latex]
298	−35.052	−11.554	−103.762	−92.208
500	−20.295	−4.9252	−57.616	−52.691
1,000	−9.388	−0.366	−23.529	−23.163
1,200	−7.569	0.3108	−17.871	−18.182
1,400	−6.27	0.767	−13.842	−14.609
1,600	−5.294	1.091	−10.83	−11.921
1,800	−4.536	1.328	−8.497	−9.826
2,000	−3.931	1.51	−6.635	−8.145
2,000	−3.433	1.648	−5.12	−6.768
2,400	−3.019	1.759	−3.86	−5.619
2,600	−2.671	1.847	−2.801	−4.648
2,800	−2.372	1.918	−1.894	−3.812
3,000	−2.114	1.976	−1.111	−3.086
3,200	−1.888	2.022	−0.429	−2.451
3,400	−1.69	2.061	0.169	−1.891
3,600	−1.513		0.701	−1.392
3,800	−1.356		1.176	−0.945
4,000	−1.216		1.599	−0.542

Question 3.8: Fuel rich combustion and chemical equilibrium Combustion pro...

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