$CH _4( g )+2 O _2( g ) \rightleftharpoons CO _2( g )+2 H _2 O ( g )$

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.