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The equilibrium constant for the reaction N2 + O2<--> 2NO is 1.69 × 10^?3 at 2300...

The equilibrium constant for the reaction N2 + O2<--> 2NO is 1.69 × 10^?3 at 2300 K. A mixture consisting of 2 mol of nitrogen and 1.0 mol of oxygen is heated to 2300 K at 1 bar and allowed to come to equilibrium. Calculate the mole fraction of NO at equilibrium.

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Answer #1

given equilibrium constant for the reaction
N2 + O2 -> 2NO
k = 1.69*10^-3 at T = 2300 K

now, number of moles of N2 = 2 mol
number of moles of oxygen = 1 mol
mole fraction st t = 0
N2 -> 2/3
O2 -> 1/3

mole fraction at equilibrium
N2 -> (2 - x)/(3)
O2 -> (1 - x)/3
NO -> 2x/3

hence
k = 18x/(2 - x)(1 - x)
(x^2 - 3x + 2) = 18x/k = 10650.8875739644x
x^2 -10653.8875739 x + 2 = 0
x = 10653.88738620 or 1.87692656*10^-4

now, x < 1
hence
mole fraction of NO at equilibrium = 2x/3 = 1.251*10^-4

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