Question 16.CSGP.107: Dry air is heated from 25°C to 4000 K in a 100-kPa constant-......

Air assumed to be 21% oxygen and 79% nitrogen by volume.
From the elementary reactions we have at 4000 K (A.11)

(1)  $O _2 \Leftrightarrow 2 O \quad K _1=2.221= y _{ O }^2 / y _{ O 2}$

(2)  $N _2 \Leftrightarrow 2 N \quad K _2=3.141 \times 10^{-6}= y _{ N }^2 / y _{ N 2}$

(3)  $N _2+ O _2 \Leftrightarrow>2 NO \quad K _3=0.08955= y _{ NO }^2 / y _{ N 2} y _{ O 2}$

Call the shifts a,b,c respectively so we get

From which the mole fractions are formed and substituted into the three equilibrium equations. The result is

which give 3 eqs. for the unknowns (a,b,c). Trial and error assume b = c = 0 solve for a from $K_1$ then for c from $K_3$ and finally given the (a,c) solve for b from $K_2$.
The order chosen according to expected magnitude $K _1> K _3> K _2$

[ If no reaction    $Q = n _{ O 2} \Delta \overline{ h }_{ O 2}+ n _{ N 2} \Delta \overline{ h }_{ N 2}=131849 \,kJ / kmol \text { air }$]