Chapter 6, Problem 6.85P

Consider the endothermic conversion of oxygen to ozone: ${\text{3 O}}_{\text{2}}\left(g\right)\rightleftarrows {\text{2 O}}_{\text{3}}\left(g\right)$ . What effect does each of the following changes have on the direction of equilibrium?
(a) decrease ${\text{[O}}_{\text{3}}\text{]}$
(b) decrease ${\text{[O}}_{\text{2}}\text{]}$
(c) increase ${\text{[O}}_{\text{3}}\text{]}$
(d) decrease temperature
(e) add a catalyst
(f)Increase pressure

Interpretation Introduction

(a)

Interpretation:

The effect of the decrease in the concentration of O₃ to the direction of equilibrium should be explained.

Concept Introduction:

The net concentrations of the reactants and products do not change at equilibrium. They are used to define an expression and the equilibrium constant (K) which has a characteristic value.

For a reaction at a given temperature,

  $\text{K=}\frac{\left[\text{Products}\right]}{\left[\text{reactants}\right]}$

Le Chatelier's principle is a general rule used to explain the effect of a change in reaction conditions on equilibrium.

Le Chatelier's principle states:

If a chemical system at equilibrium is disturbed, equilibrium will shift in the direction to counteract the change.

Answer to Problem 6.85P

Equilibrium favor shift to right.

Explanation of Solution

Given information:

  ${\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

The equilibrium reaction is represented as follows:

  $\begin{array}{l}{\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}\\ \\ \text{K = }\frac{{[O_3]}^2}{{[O_2]}^3}\end{array}$

Reactants and products are in equilibrium. With decrease in the O₃ concentration, equilibrium is disturbed. Then rate of the forward reaction increases.

Equilibrium favors shift to right.

Interpretation Introduction

(b)

Interpretation:

The effect of the decrease in the concentration of O₂ to the direction of equilibrium should be explained.

Concept Introduction:

The net concentrations of the reactants and products do not change at equilibrium. They are used to define an expression and the equilibrium constant (K) which has a characteristic value.

For a reaction at a given temperature,

  $\text{K=}\frac{\left[\text{Products}\right]}{\left[\text{reactants}\right]}$

Le Chatelier's principle is a general rule used to explain the effect of a change in reaction conditions on equilibrium.

Le Chatelier's principle states:

If a chemical system at equilibrium is disturbed, equilibrium will shift in the direction to counteract the change.

Answer to Problem 6.85P

Equilibrium favors shift to left.

Explanation of Solution

Given information:

  ${\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

The given equilibrium reaction is as follows:

  $\begin{array}{l}{\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}\\ \\ \text{K = }\frac{{[O_3]}^2}{{[O_2]}^3}\end{array}$

Reactants and products are in equilibrium. With decrease in the O₂ concentration, equilibrium is disturbed. Then rate of the reverse reaction increases.

Equilibrium favors shift to left.

Interpretation Introduction

(c)

Interpretation:

The effect of the decrease in the concentration of O₃ to the direction of equilibrium should be explained.

Concept Introduction:

The net concentrations of the reactants and products do not change at equilibrium. They are used to define an expression and the equilibrium constant (K) which has a characteristic value.

For a reaction at a given temperature,

  $\text{K=}\frac{\left[\text{Products}\right]}{\left[\text{reactants}\right]}$

Le Chatelier's principle is a general rule used to explain the effect of a change in reaction conditions on equilibrium.

Le Chatelier's principle states:

If a chemical system at equilibrium is disturbed, equilibrium will shift in the direction to counteract the change.

Answer to Problem 6.85P

Equilibrium favors shift to left.

Explanation of Solution

Given information:

  ${\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

The given equilibrium reaction is represented as follows:

  $\begin{array}{l}{\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}\\ \\ \text{K = }\frac{{[O_3]}^2}{{[O_2]}^3}\end{array}$

Reactants and products are in equilibrium. With decrease in the O₃ concentration, equilibrium is disturbed. Then rate of the reverse reaction increases.

Equilibrium favors shift to left.

Interpretation Introduction

(d)

Interpretation:

The effect of the decrease in temperature to the direction of equilibrium should be explained.

Concept Introduction:

The net concentrations of the reactants and products do not change at equilibrium. They are used to define an expression and the equilibrium constant (K) which has a characteristic value.

For a reaction at a given temperature,

  $\text{K=}\frac{\left[\text{Products}\right]}{\left[\text{reactants}\right]}$

Le Chatelier's principle is a general rule used to explain the effect of a change in reaction conditions on equilibrium.

Le Chatelier's principle states:

If a chemical system at equilibrium is disturbed, equilibrium will shift in the direction to counteract the change.

Answer to Problem 6.85P

Equilibrium favors shift to left.

Explanation of Solution

Given information:

  ${\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

The given equilibrium reaction is represented as follows:

  $\begin{array}{l}{\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}\\ \\ \text{K = }\frac{{[O_3]}^2}{{[O_2]}^3}\end{array}$

Reactants and products are in equilibrium. This is endothermic reaction.

Endothermic reaction absorbs energy to drive the forward reaction.

  ${\text{Temperature + 3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

With decrease in temperature, equilibrium is disturbed. Then rate of the reverse reaction increases.

Equilibrium favors shift to left.

Interpretation Introduction

(e)

Interpretation:

The effect of the addition of a catalyst to the direction of equilibrium should be explained.

Concept Introduction:

The net concentrations of the reactants and products do not change at equilibrium. They are used to define an expression and the equilibrium constant (K) which has a characteristic value.

For a reaction at a given temperature,

  $\text{K=}\frac{\left[\text{Products}\right]}{\left[\text{reactants}\right]}$

Le Chatelier's principle is a general rule used to explain the effect of a change in reaction conditions on equilibrium.

Le Chatelier's principle states:

If a chemical system at equilibrium is disturbed, equilibrium will shift in the direction to counteract the change.

Answer to Problem 6.85P

No change for the equilibrium.

Explanation of Solution

Given information:

  ${\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

  $\begin{array}{l}{\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}\\ \\ \text{K = }\frac{{[O_3]}^2}{{[O_2]}^3}\end{array}$

Reactants and products are in equilibrium. This is endothermic reaction.

Endothermic reaction absorbs energy to drive the forward reaction.

  ${\text{Catalyst + 3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

Catalyst can increase the reaction rate by decreasing activation energy.

With the addition of a catalyst, equilibrium is not change.

Equilibrium does not change.

Interpretation Introduction

(f)

Interpretation:

The effect of the increase in pressure to the direction of equilibrium should be explained.

Concept Introduction:

The net concentrations of the reactants and products do not change at equilibrium. They are used to define an expression and the equilibrium constant (K) which has a characteristic value.

For a reaction at a given temperature,

  $\text{K=}\frac{\left[\text{Products}\right]}{\left[\text{reactants}\right]}$

Le Chatelier's principle is a general rule used to explain the effect of a change in reaction conditions on equilibrium.

Le Chatelier's principle states:

If a chemical system at equilibrium is disturbed, equilibrium will shift in the direction to counteract the change.

Answer to Problem 6.85P

Equilibrium favors shift to right.

Explanation of Solution

Given information:

  ${\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

  $\begin{array}{l}{\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}\\ \\ \text{K = }\frac{{[O_3]}^2}{{[O_2]}^3}\end{array}$

Reactants and products are in equilibrium. Both reactant and products are gaseous molecules.

Therefore, with the increase in the pressure, the equilibrium will shift towards the side of the reaction with fewer gas molecules.

  ${\text{3O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2O}}_{\text{3}}\text{(g)}$

Equilibrium shift to the right side.