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Question 16.CSGP.40: Assume the equilibrium mole fractions of oxygen and nitrogen......

Assume the equilibrium mole fractions of oxygen and nitrogen are close to those in air, find the equilibrium mole fraction for NO at 3000 K, 500 kPa disregarding dissociations.

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Assume the simple reaction to make NO as:

\begin{array}{cccl|l}N _2 +&O_2 \Leftrightarrow &2 NO & Ar & n _{ N 2}=0.78- x \\0.78 & 0.21& 0 & 0.01 & n _{ O 2}=0.21- x \\- x & – x & +2x && n _{ NO }=2 x , \quad n _{ ar }=0.01 \\\hline 0.78- x & 0.21- x & 2x & 0.01 & n _{ tot }=1.0\end{array}

From A.11 at 3000 K, ln K = -4.205, K = 0.014921

\begin{aligned}& K =\frac{4 x ^2}{(0.78- x )(0.21- x )}\left(\frac{ P }{ P ^0}\right)^0 \quad \Rightarrow \\& \frac{ x ^2}{(0.78- x )(0.21- x )}=\frac{0.014921}{4}=0.00373 \text { and } 0< x <0.21\end{aligned}

Solve for x: x = 0.0230 ⇒    y _{ NO }=\frac{2 x }{1.0}= 0 . 0 4 6

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