Question

For the reaction:
$ CH _{4}( g )+2 O _{2}( g ) \rightleftharpoons CO _{2}( g )+2 H _{2} O ( l ) $
$\Delta H _{ r }=-170.8 \,kJ\,mol ^{-1}$
Which of the following statements is not true?

• A. At equilibrium, the concentrations of $C O_2(g)$ and $H_2O (l)$ are not equal
• B. The equilibrium constant for the reaction is given by $ K_p = \frac{ [ CO_2 ] }{ [ CH_4 ] [ O_2 ] } $
• C. Addition of $CH _{4}(g)$ or $O _{2}(g)$ at equilibrium will cause a shift to the right
• D. The reaction is exothermic

Answer 1

Answer: (B)

The given reaction is:-
$CH _{4}( g )+2 O _{2}( g ) \rightleftharpoons CO _{2}( g )+2 H _{2} O ( g )$
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, $\left[ H _{2} O \right]=1$
$\Rightarrow K _{ C }=\frac{\left[ CO _{2}\right]}{\left[ CH _{4}\right]\left[ O _{2}\right]^{2}}$
$\because \Delta Hr =-170.8 \,KJ / mol$ 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.