Chapter 9, Problem 9.89EP

(a)

Interpretation Introduction

Interpretation:

For the reaction ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\overset{}{\rightleftharpoons }}{\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}\text{(g) + heat}$ the change in equilibrium shift has to be determined when the concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ is increased.

Concept Introduction:

Le Chatelier’s principle:

If some forces applied, the system at equilibrium will get disrupted.  This change in equilibrium can be due to the change in pressure or temperature.  The change in reactant concentration can also disrupt the equilibrium.  Over time, the forward and backward reaction become equal and will attain a new equilibrium.  The equilibrium will shifts to right, if more products are formed and the system will shifts to left, if more reactants are formed.

The principle states that if some stress is applied to the system at equilibrium, the system will adjust itself in a direction which reduces the stress.

Concentration Changes:

Addition of reactant or product or removal of reactant or product from a system at equilibrium will affects the equilibrium.  If some reactant is added to a system at equilibrium, then the equilibrium will shifts to the product side, so that the added reactant get consumed.  If product is added then the equilibrium will shift towards left side.

Example:

  ${\text{N}}_{\text{2}}{\text{(g)+3H}}_{\text{2}}\text{(g)}\stackrel{}{\rightleftharpoons }{\text{2NH}}_{\text{3}}\text{(g)}$

If ${\text{N}}_{\text{2}}$ is added then the equilibrium will shift towards right.

If ${\text{NH}}_{\text{3}}$ is added then the equilibrium will shift towards left.

If ${\text{N}}_{\text{2}}$ is removed then the equilibrium will shift towards left.

Answer to Problem 9.89EP

The equilibrium will shift towards left on increasing the concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$.

Explanation of Solution

The given reaction is:

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\overset{}{\rightleftharpoons }}{\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}\text{(g) + heat}$

On increasing the concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}$ will shift the equilibrium towards left.  Increasing the concentration of product will shift the equilibrium towards left, thus using up some of the added product.

Hence the equilibrium will shifts to the left.

(b)

Interpretation Introduction

Interpretation:

For the reaction ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\overset{}{\rightleftharpoons }}{\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}\text{(g) + heat}$ the change in equilibrium shift has to be determined when the concentration of ${\text{C}}_{\text{6}}{\text{H}}_6$ is decreased.

Concept Introduction:

Le Chatelier’s principle:

The principle states that if some stress is applied to the system at equilibrium, the system will adjust itself in a direction which reduces the stress.

Concentration changes:

If the concentration of the reactant is increased, then the equilibrium will shift towards the right (product side).  If the concentration of product is increased then the equilibrium will shift towards reactant side.  Finally a new equilibrium will attain.

Answer to Problem 9.89EP

The equilibrium shifts towards left, on decreasing the concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$.

Explanation of Solution

The given reaction is:

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\overset{}{\rightleftharpoons }}{\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}\text{(g) + heat}$

Decreasing the concentration of reactant will shift the equilibrium towards the reactant side, thus producing more of the substance that was removed.  On decreasing the concentration of ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}$ the equilibrium will shift towards left.

(c)

Interpretation Introduction

Interpretation:

For the reaction ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\overset{}{\rightleftharpoons }}{\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}\text{(g) + heat}$ the change in equilibrium shift has to be determined when the temperature is increased.

Concept Introduction:

Le Chatelier’s principle:

If some forces applied, the system at equilibrium will get disrupted.  This change in equilibrium can be due to the change in pressure or temperature.  The change in reactant concentration can also disrupt the equilibrium.  Over time, the forward and backward reaction become equal and will attain a new equilibrium.  The equilibrium will shifts to right, if more products are formed and the system will shifts to left, if more reactants are formed.

The principle states that if some stress is applied to the system at equilibrium, the system will adjust itself in a direction which reduces the stress.

Temperature Changes

Heat is one of the products in exothermic reaction and heat is used up in endothermic reaction.

Consider an exothermic reaction;

    ${\text{H}}_{\text{2}}{\text{(g)+F}}_{\text{2}}\text{(g)}\underset{}{\rightleftharpoons }\text{2HF(g)+heat}$

If heat is added up, then the reaction will shift to left so that the amount of heat will decrease.

Lowering the temperature will make the reaction to shift towards right.

Consider an endothermic reaction;

    ${\text{Heat+2CO}}_{\text{2}}\text{(g)}\underset{}{\rightleftharpoons }{\text{2CO(g)+O}}_{\text{2}}\text{(g)}$

Increase in temperature will shift the reaction towards right.

If heat is added up, then the reaction will shift towards right.

Answer to Problem 9.89EP

The equilibrium shifts to left on increasing the temperature.

Explanation of Solution

The given reaction is:

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\overset{}{\rightleftharpoons }}{\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}\text{(g) + heat}$

Increasing the temperature of the reaction promotes endothermic reaction, thus the equilibrium will shift towards left.  Hence, on increasing temperature, the equilibrium moves towards left.

(d)

Interpretation Introduction

Interpretation:

For the reaction ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\overset{}{\rightleftharpoons }}{\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}\text{(g) + heat}$ the change in equilibrium shift has to be determined when the pressure is decreased.by increasing the volume of the container.

Concept Introduction:

Le Chatelier’s principle:

The principle states that if some stress is applied to the system at equilibrium, the system will adjust itself in a direction which reduces the stress.

Pressure Changes:

Only the gaseous reactants and products get affected by the pressure change.

Consider the reaction:

  ${\text{2H}}_{\text{2}}{\text{(g)+O}}_{\text{2}}\text{(g)}\underset{}{\to }{\text{2H}}_{\text{2}}\text{O(g)}$

3 moles of reactant gives 2 moles of product.  Increase in pressure will shift the reaction towards the side which have fewer molecules.

Answer to Problem 9.89EP

The equilibrium shifts to left, on decreasing pressure by increasing the volume.

Explanation of Solution

The given reaction is:

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{6}}{\text{(g) + 3H}}_{\text{2}}\text{(g) }\underset{}{\overset{}{\rightleftharpoons }}{\text{ C}}_{\text{6}}{\text{H}}_{\text{12}}\text{(g) + heat}$

Decreasing the pressure by increasing the volume of the container will decrease the concentration of the gases.  Decrease in pressure will shift the equilibrium towards the side which contains more number of moles.  So here the reactant side is having more number of moles, and then the equilibrium will shift towards the reactant side.

Hence the equilibrium will shift towards left side.