Question

For the reaction:

CH${}_4$(g) + 2O${}_2$(g) $\rightleftharpoons$ CO${}_2$(g) + 2H${}_2$O$\left(l\right)$

$\Delta$H${}_r$ = -170.8 kJmol${}^{-1}$

Which of the following statements is not true?

• A. At equilibrium, the concentrations of CO${}_2$(g) and H${}_2$O$\left(l\right)$ are not equal.
• B. The equilibrium constant for the reaction is given by $K_C$= $\frac{\left[C{O}_2\right]}{\left[C{H}_4\right]\left[O_2\right]}$.
• 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

The correct option is B The equilibrium constant for the reaction is given by $K_C$= $\frac{\left[C{O}_2\right]}{\left[C{H}_4\right]\left[O_2\right]}$.

The given reaction is:-

${CH}_4\left(g\right)+2O_2\left(g\right)\rightleftharpoons {CO}_2\left(g\right)+2H_{2}O\left(g\right)$

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, $\left[H_{2}O\right]=1$

$\Rightarrow K_C=\frac{\left[{CO}_2\right]}{\left[{CH}_4\right]{\left[O_2\right]}^2}$

$\because △Hr=-170.8KJ/mol$ is negative, so reaction is exothermic by adding $O_2\left(g\right)$ or ${CH}_4\left(g\right)$ at equilibrium, by Le Chatelier's principle, the equilibrium shift towards right side.