Chapter 7, Problem 7.25P

Interpretation Introduction

(a)

Interpretation:

Equilibrium expression for the reaction ${\text{2H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(g)}◰\; \hspace{1em}{\text{2H}}_{\text{2}}{\text{O(g)+O}}_{\text{2}}\text{(g)}$ must be written.

Concept Introduction:

These equilibrium concentrations are used for the calculation of equilibrium constant (${\text{K}}_{\text{eq}}$) which is defined as.

${\text{K}}_{\text{eq}}\text{=}\frac{\text{[Products]}}{\text{[Reactants]}}$.

Answer to Problem 7.25P

The equilibrium expression for the reaction ${\text{2H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(g)}◰\; \hspace{1em}{\text{2H}}_{\text{2}}{\text{O(g)+O}}_{\text{2}}\text{(g)}$ can be written as.

${\text{K}}_{\text{eq}}\text{=}\frac{{{\text{[H}}_{\text{2}}\text{O(g)]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{(g)]}}{{{\text{[H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(g)]}}^{\text{2}}}$.

Explanation of Solution

${\text{2H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(g)}◰\; \hspace{1em}{\text{2H}}_{\text{2}}{\text{O(g)+O}}_{\text{2}}\text{(g)}$

In this reversible reaction products are ${\mathrm{H}}_2\text{O}(g)$ and ${\text{O}}_{\text{2}}\text{(g)}$. Reactant is ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(g)}$

Now we know that ${\text{K}}_{\text{eq}}\text{=}\frac{\text{[Products]}}{\text{[Reactants]}}$

Coefficients of reactants and products are raised to power of the concentrations of reactants and products.

Thus, ${\text{K}}_{\text{eq}}\text{=}\frac{{{\text{[H}}_{\text{2}}\text{O(g)]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{(g)]}}{{{\text{[H}}_{\text{2}}{\text{O}}_{\text{2}}\text{(g)]}}^{\text{2}}}$.

Interpretation Introduction

(b)

Interpretation:

Equilibrium expression for the reaction ${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(g)}◰\; \hspace{1em}{\text{2NO}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}$ must be written.

Concept Introduction:

These equilibrium concentrations are used for the calculation of equilibrium constant (${\text{K}}_{\text{eq}}$) which is defined as.

${\text{K}}_{\text{eq}}\text{=}\frac{\text{[Products]}}{\text{[Reactants]}}$.

Answer to Problem 7.25P

The equilibrium expression for the reaction ${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(g)}◰\; \hspace{1em}{\text{2NO}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}$ can be written as.

${\text{K}}_{\text{eq}}\text{=}\frac{{{\text{[NO}}_{\text{2}}\text{(g)]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{(g)]}}{{{\text{[N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(g)]}}^{\text{2}}}$.

Explanation of Solution

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(g)}◰\; \hspace{1em}{\text{2NO}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}$

In this reversible reaction products are ${\mathrm{NO}}_2(g)$ and ${\text{O}}_{\text{2}}\text{(g)}$. Reactant is ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(g)}$

Now we know that ${\text{K}}_{\text{eq}}\text{=}\frac{\text{[Products]}}{\text{[Reactants]}}$

Coefficients of reactants and products are raised to power of the concentrations of reactants and products.

Thus, ${\text{K}}_{\text{eq}}\text{=}\frac{{{\text{[NO}}_{\text{2}}\text{(g)]}}^{\text{2}}{\text{[O}}_{\text{2}}\text{(g)]}}{{{\text{[N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(g)]}}^{\text{2}}}$.

Interpretation Introduction

(c)

Interpretation:

Equilibrium expression for the reaction ${\text{6H}}_{\text{2}}{\text{O(g)+6CO}}_{\text{2}}\text{(g)}◰\; \hspace{1em}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{(s)+6O}}_{\text{2}}\text{(g)}$ must be written.

Concept Introduction:

These equilibrium concentrations are used for the calculation of equilibrium constant (${\text{K}}_{\text{eq}}$) which is defined as.

${\text{K}}_{\text{eq}}\text{=}\frac{\text{[Products]}}{\text{[Reactants]}}$

Active mass of solid substances is taken as unity and thus not used in any equilibrium expression.

Answer to Problem 7.25P

The equilibrium expression for the reaction ${\text{6H}}_{\text{2}}{\text{O(g)+6CO}}_{\text{2}}\text{(g)}◰\; \hspace{1em}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{(s)+6O}}_{\text{2}}\text{(g)}$ can be written as.

${\text{K}}_{\text{eq}}\text{=}\frac{{{\text{[O}}_{\text{2}}\text{(g)]}}^{\text{6}}}{{{\text{[H}}_{\text{2}}\text{O(g)]}}^{\text{6}}{{\text{[CO}}_{\text{2}}\text{(g)]}}^{\text{6}}}$.

Explanation of Solution

${\text{6H}}_{\text{2}}{\text{O(g)+6CO}}_{\text{2}}\text{(g)}◰\; \hspace{1em}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{(s)+6O}}_{\text{2}}\text{(g)}$

In this reversible reaction products are ${\mathrm{C}}_6{\text{H}}_{12}{\text{O}}_6(s)$ and ${\text{O}}_{\text{2}}\text{(g)}$.Reactant are ${\text{H}}_{\text{2}}\text{O(g)}$and ${\text{CO}}_{\text{2}}\text{(g)}$.But ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is solid and so must be excluded from equilibrium expression.

Now we know that ${\text{K}}_{\text{eq}}\text{=}\frac{\text{[Products]}}{\text{[Reactants]}}$

Coefficients of reactants and products are raised to power of the concentrations of reactants and products.

Thus, ${\text{K}}_{\text{eq}}\text{=}\frac{{{\text{[O}}_{\text{2}}\text{(g)]}}^{\text{6}}}{{{\text{[H}}_{\text{2}}\text{O(g)]}}^{\text{6}}{{\text{[CO}}_{\text{2}}\text{(g)]}}^{\text{6}}}$.