\(CH _4( g )+2 O _2( g ) \rightleftharpoons CO _2( g )+2 H _2 O ( g )\)
Now, \(K _{ C }=\frac{\left[ CO _2\right]\left[ H _2 O \right]^2}{\left[ CH _4\right]\left[ O _2\right]^2}\)
Now, \(H _2 O\) is pure liquid, so, \(\left[ H _2 O \right]=1\)
\(\Rightarrow \quad K _{ C }=\frac{\left[ CO _2\right]}{\left[ CH _4\right]\left[ O _2\right]^2}\)
\(\because \Delta Hr =-170.8 \,KJ / mol\) is negative, so reaction is exothermic by adding \(O _2\) (g) or \(CH _4( g )\) at equilibrium, by Le Chatelier's principle, the equilibrium shift towards right side.
$CO\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to C{O_2}\left( g \right)$