Question 16.CSGP.94: Redo the previous Problem but include the dissociation of ox......

$\text { (1) } O _2⇔2 O \quad \ln K =-7.52, \quad K _1=0.000542=\left( y _{ O }^2 / y _{ O 2}\right) P / P _{ o }$

$\text { (2) } N _2 \Leftrightarrow>2 N \quad \operatorname{lnK}=-28.313, \quad K _2=5.056 \times 10^{-13}=\left( y _{ N }^2 / y _{ N 2}\right) P / P _{ o }$

$\text { (3) } N _2+ O _2 \Leftrightarrow 2 NO \quad \operatorname{lnK}=-5.316, \quad K _3=0.00491=\left( y _{ NO }^2 / y _{ N 2} y _{ O 2}\right)$

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 = a = 0 solve for c from the last eq. then for a from the first and finally given the (a,c) solve for b from the second equation. The order chosen according to expected magnitude $K _3> K _1> K _2$

$a =0.001626, b =0.99 \times 10^{-7}, c =0.01355 \Rightarrow$

Indeed it is a very small effect to include the additional dissociations.