Question

Write equilibrium expressions for the following gas-phase reactions.

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftarrows }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{Xe}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathbf{}{\mathbf{F}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftarrows }{\mathbf{XeF}}_{\mathbf{6}}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{2}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{15}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftarrows }\mathbf{12}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathit{g}\mathbf{\right)} \end{gathered}$$

Short answer

The equilibrium expressions are:

$\begin{array}{c}\left(\mathrm{a}\right)\mathrm{K}={\left[\frac{{\left({\mathrm{P}}_{{\mathrm{CO}}_2}\right)}^{12}{\left({\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}\right)}^6}{{\left({\mathrm{P}}_{{\mathrm{C}}_6{\mathrm{H}}_6}\right)}^2{\left({\mathrm{P}}_{{\mathrm{O}}_2}\right)}^{15}}\right]}_{\mathrm{eq}}\\ \left(\mathrm{b}\right)\mathrm{K}={\left[\frac{{\left({\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}\right)}^2}{{\left({\mathrm{P}}_{{\mathrm{H}}_2}\right)}^2\left({\mathrm{P}}_{{\mathrm{O}}_2}\right)}\right]}_{\mathrm{eq}}\\ \left(\mathrm{c}\right)\mathrm{K}={\left[\frac{{\left({\mathrm{P}}_{{\mathrm{CO}}_2}\right)}^{12}{\left({\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}\right)}^6}{{\left({\mathrm{P}}_{{\mathrm{C}}_6{\mathrm{H}}_6}\right)}^2{\left({\mathrm{P}}_{{\mathrm{O}}_2}\right)}^{15}}\right]}_{\mathrm{eq}}\\ \end{array}$

Step-by-step solution

Concept of the law of mass action 

The rate of a chemical reaction at a particular temperature and moment is directly proportional to the product of the active masses of the reactants, according to the law.

For a gas phase reaction

$\mathbf{aA}\mathbf{+}\mathbf{bB}\mathbf{\rightleftharpoons }\mathbf{cC}\mathbf{+}\mathbf{dD}$

The equilibrium constant is written as:

$\mathbf{K}\mathbf{=}{\left[\frac{{\left(P_C\right)}^c{\left(P_D\right)}^d}{{\left(P_A\right)}^a{\left(P_B\right)}^b}\right]}_{\mathbf{e}\mathbf{q}}$

Equilibrium reaction  

The reactions are as follows:

$\left(\mathrm{a}\right)2{\mathrm{H}}_2\left(g\right)+{\mathrm{O}}_2\left(g\right)\rightleftarrows 2{\mathrm{H}}_2\mathrm{O}\left(g\right)$

$\mathrm{K}={\left[\frac{{\left({\mathrm{P}}_{\mathrm{C}}\right)}^{\mathrm{c}}{\left({\mathrm{P}}_{\mathrm{D}}\right)}^{\mathrm{d}}}{{\left({\mathrm{P}}_{\mathrm{A}}\right)}^{\mathrm{a}}{\left({\mathrm{P}}_{\mathrm{B}}\right)}^{\mathrm{b}}}\right]}_{\mathrm{eq}}$

$\left(\mathrm{b}\right)\mathrm{Xe}\left(g\right)+3{\mathrm{F}}_2\left(g\right)\rightleftarrows {\mathrm{XeF}}_6\left(g\right)$

$\mathrm{K}={\left[\frac{{\left({\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}\right)}^2}{{\left({\mathrm{P}}_{{\mathrm{H}}_2}\right)}^2\left({\mathrm{P}}_{{\mathrm{O}}_2}\right)}\right]}_{eq}$

Simplify the further 3^rd reaction:

$\left(\mathrm{c}\right)2{\mathrm{C}}_6{\mathrm{H}}_6\left(g\right)+15{\mathrm{O}}_2\left(g\right)\rightleftarrows 12{\mathrm{CO}}_2\left(g\right)+6{\mathrm{H}}_2\mathrm{O}\left(g\right)$

$\mathrm{K}={\left[\frac{{\left({\mathrm{P}}_{{\mathrm{CO}}_2}\right)}^{12}{\left({\mathrm{P}}_{{\mathrm{H}}_2\mathrm{O}}\right)}^6}{{\left({\mathrm{P}}_{{\mathrm{C}}_6{\mathrm{H}}_6}\right)}^2{\left({\mathrm{P}}_{{\mathrm{O}}_2}\right)}^{15}}\right]}_{\mathrm{eq}}$