11.9 Consider the combustion of methane: CH4(g) + 202(g) CO2(g) + 2H20(g) Suppos

Question

11.9 Consider the combustion of methane: CH4(g) + 202(g) CO2(g) + 2H20(g) Suppose the reaction occurs in a flowing gas mixture of methane and air. Assume that the pressure is constant at bar, the reactant mixture is at a temperature of 298.15 K and has stoichiometric proportions of methane and oxygen, and the reaction goes to completion with no dissociation. For the quantity of gaseous product mixture containing mol CO2, 2 mol H20, and the nitrogen and other substances remaining from the air, you may use the approximate formula Cp (P) = a + bT, where the coefficients have the values a = 297.0 J K-1 and b = 8.520 × 10-2 J K-2, Solve Eq. 11.61 for T2 to estimate the flame temperature to the nearest kelvin,

Explanation / Answer

Lets take 1 mole CH4 and 2 mol O2

moles of CO2 and H2O produced = 1 mol and 2 mol respectively

now with O2 and when air is used ,nitrogen will also will be coming

for 21 mol O2 we have 79 mol N2

for 2 mol O2 we have 79/21*2 = 7.52 mol N2 won't react and come in product

total moles of product = 7.52 + 1 + 2 = 10.52 mol

heat of combustion of methane = - 802.3 kJ/mol

heat released = 802.3 kJ/mol x 1mol = 802.3 kJ = 802300 J

heat absorbed by products = nCpdT = 10.52 mol x (aT + bT2/2) from T to 298.15 K

802300 J = 10.52 x {297 j/K x(T-298.15) + 8.52 x 10-2 J/(K2.2) (T2-298.152)}

76264.258 = 297 j/K x(T-298.15) + 8.52 x 10-2 J/(K2.2) (T2-298.152)

76264.258 = 297T - 88550.55 + 8.52 x 10-2T2 - 7573.719

0 = 8.52 x 10-2T2 + 297T -172387.7

T = 571 K

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