Chapter 7, Problem 7.94QP

(a)

Interpretation Introduction

Interpretation:

Effect of addition of $\text{C}$ in the given equilibrium reaction has to be predicted.

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

Concept Introduction:

LeChatelier’s principle states that if a stress is applied on the system that is present at equilibrium, then the system will respond to the stress applied by altering the composition at equilibrium in a way to minimize the stress that is applied.

Explanation of Solution

Given equilibrium reaction is,

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

On addition of $\text{C}(s)$, there will not be any change in the equilibrium because the added reactant is a solid.  As the substance added is a solid, addition or removal of solid substance does not have any effect on the equilibrium.

(b)

Interpretation Introduction

Interpretation:

Effect of addition of ${\text{H}}_{\text{2}}$ in the given equilibrium reaction has to be predicted.

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given equilibrium reaction is,

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

Addition of ${\text{H}}_{\text{2}}$ results in increase of concentration of the reactant.  Therefore, the equilibrium will be shifted towards the right and more amount of product will be formed.

(c)

Interpretation Introduction

Interpretation:

Effect of removal of ${\text{CH}}_{\text{4}}$ in the given equilibrium reaction has to be predicted.

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given equilibrium reaction is,

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

Removal of ${\text{CH}}_{\text{4}}$ results in decrease in concentration of the product.  Therefore, the equilibrium will be shifted towards the right and more amount of product will be formed.

(d)

Interpretation Introduction

Interpretation:

Effect of increase in temperature in the given equilibrium reaction has to be predicted.

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given equilibrium reaction is,

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

When stress is applied on a system that is at equilibrium, the system adjusts itself to get relieved from the stress.  As in the given equilibrium reaction, heat energy is liberated, it is an exothermic reaction.  Increase in temperature shifts the equilibrium to the left.  Therefore, concentration of reactant gets increased.

(e)

Interpretation Introduction

Interpretation:

Effect of addition of catalyst to the given equilibrium reaction has to be predicted.

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given reaction is,

    $\text{C}(s)+2{\text{H}}_{\text{2}}(g)\rightleftharpoons {\text{CH}}_{\text{4}}(g)+\text{18}\text{kcal}$

Addition of catalyst increases both rate of forward reaction and reverse reaction.  Therefore, there is no change in concentration of the reactant or product.  There will not be any shift in the equilibrium also.  Addition of catalyst increases the rate at which the equilibrium can be attained only.  Therefore, there will not be any effect on the equilibrium.