$CH _4( g )+2 O _2( g ) \rightleftharpoons CO _2( g )+2 H _2 O ( g )$
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.
$(I)\, CO_{2(g)} + H_2O_{(g)} $ $\rightleftharpoons$ $ CO_{2(g)} + H_{2(g)} ;\, k_1$
$(II) \,CH_{4(g)} + H_2O_{(g)} $ $\rightleftharpoons$ $ CO_{(g)} + 3H_{2(g)} ;\, k_2$
$(III) \,CH_{4(g)} + 2H_2O_{(g)} $ $\rightleftharpoons$ $ CO_{2(g)} + 4H_{2(g)} ; \,k_3$ તો તેમના સંતુલન અચળાંકો વચ્ચે સાચો સંબંધ........ છે.
$NO(g) \rightarrow \frac{1}{2} N_2(g)+ \frac{1}{2} O_2(g)$ સમાન તાપમાને શું થશે? :
તાપમાન કે જે $K _{ C }=20.4$ અને $K _{ P }=600.1,$ ....... $K$
[ધારો કે બધા વાયુઓ આદર્શ છે અને $R =0.0831\, L\,bar \, K ^{-1} mol ^{-1}]$