Chapter U6, Problem SII5RE

Interpretation Introduction

Interpretation: The effect of adding ${\text{N}}_{\text{2}}\text{O}$ to the following equilibrium mixture needs to be describe which exothermic in nature:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g)}$

Concept introduction: According to LeChatlier's principle, when a dynamic equilibrium is subjected to a change in either concentration or temperature or pressure, then the position of equilibrium will shift in such a direction that it nullifies the change.

For example:

  $\text{A + B }\rightleftharpoons \text{ C}$ is a system at equilibrium.

If concentration of A is increased by adding some more amount of A at equilibrium, then left side concentration is increasing, so to nullify this change, equilibrium shifts towards right side. Hence, forward reaction or formation of C is favoured in this case.

Answer to Problem SII5RE

The concentration of reactants will increase and that of products will decrease and hence, resulting in decrease in ${\text{K}}_{\text{eq}}$ value.

Explanation of Solution

The given equilibrium mixture described by exothermic equation is:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g)}$

The addition of ${\text{N}}_{\text{2}}\text{O}$ to the above equilibrium mixture means that the concentration of the product increases. Now, according to LeChatlier's principle, the equilibrium will shift in backward direction (reactant side as the concentration of product is increasing) in order to nullify the change. Thus, on adding ${\text{N}}_{\text{2}}\text{O}$ to the given equilibrium equation, the concentration of reactants will increase and that of products will decrease and hence, resulting in decrease in ${\text{K}}_{\text{eq}}$ value.

Interpretation Introduction

Interpretation: The effect of removing ${\text{N}}_{\text{2}}\text{O}$ to the following equilibrium mixture needs to be describe which exothermic in nature:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g)}$

Concept introduction: According to LeChatlier's principle, when a dynamic equilibrium is subjected to a change in either concentration or temperature or pressure, then the position of equilibrium will shift in such a direction that it nullifies the change.

For example:

  $\text{A + B }\rightleftharpoons \text{ C}$ is a system at equilibrium.

If concentration of A is increased by adding some more amount of A at equilibrium, then left side concentration is increasing, so to nullify this change, equilibrium shifts towards right side. Hence, forward reaction or formation of C is favoured in this case.

Answer to Problem SII5RE

The concentration of reactants will decrease and that of products will increase and hence, resulting in increase in ${\text{K}}_{\text{eq}}$ value.

Explanation of Solution

The given equilibrium mixture described by exothermic equation is:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g)}$

The removal of ${\text{N}}_{\text{2}}\text{O}$ from the above equilibrium mixture means that the concentration of the product decreases. Now, according to LeChatlier's principle, the equilibrium will shift in forward direction (product side as the product is decreasing) in order to nullify the change. Thus, on removing ${\text{N}}_{\text{2}}\text{O}$ from the given equilibrium equation, the concentration of reactants will decrease and that of products will increase and hence, resulting in increase in ${\text{K}}_{\text{eq}}$ value.

Interpretation Introduction

Interpretation: The effect of raising the temperature on the following equilibrium mixture needs to be describe which exothermic in nature:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g)}$

Concept introduction: According to LeChatlier's principle, when a dynamic equilibrium is subjected to a change in either concentration or temperature or pressure, then the position of equilibrium will shift in such a direction that it nullifies the change.

For example:

  $\text{A + B }\rightleftharpoons \text{ C}$ is a system at equilibrium.

If concentration of A is increased by adding some more amount of A at equilibrium, then left side concentration is increasing, so to nullify this change, equilibrium shifts towards right side. Hence, forward reaction or formation of C is favoured in this case.

Answer to Problem SII5RE

The concentration of reactants will increase and that of products will decrease and hence, resulting in decrease in ${\text{K}}_{\text{eq}}$ value.

Explanation of Solution

The given equilibrium mixture described by exothermic equation is:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g)}$

The exothermic equation represents that the thermal energy is removedfrom the reaction. So, the reaction can be rewritten as:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g) + heat}$

From the above equation it can be interpreted that heat is acting as a product so, heating the equilibrium mixture means adding more amount of heat to the system. Now, according to LeChatlier's principle, the equilibrium will shift in backward direction (reactant side as the product is increasing) in order to nullify the change. Thus, on raising the temperatureof the given exothermic equation, the concentration of reactants will increase and that of products will decrease and hence, resulting in decrease in ${\text{K}}_{\text{eq}}$ value.

Interpretation Introduction

Interpretation: The effect of lowering the temperature on the following equilibrium mixture needs to be describe which exothermic in nature:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g)}$

Concept introduction: According to LeChatlier's principle, when a dynamic equilibrium is subjected to a change in either concentration or temperature or pressure, then the position of equilibrium will shift in such a direction that it nullifies the change.

For example:

  $\text{A + B }\rightleftharpoons \text{ C}$ is a system at equilibrium.

If concentration of A is increased by adding some more amount of A at equilibrium, then left side concentration is increasing, so to nullify this change, equilibrium shifts towards right side. Hence, forward reaction or formation of C is favoured in this case.

Answer to Problem SII5RE

The concentration of reactants will decrease and that of products will increase and hence, resulting in increase in ${\text{K}}_{\text{eq}}$ value.

Explanation of Solution

The given equilibrium mixture described by exothermic equation is:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g)}$

The exothermic equation represents that the thermal energy is removed from the reaction. So, the reaction can be rewritten as:

  $\text{3NO(g) }\rightleftharpoons {\text{ N}}_2{\text{O(g) + NO}}_{\text{2}}\text{(g) + heat}$

From the above equation it can be interpreted that heat is acting as a product so, lowering the temperature means cooling and the equilibrium mixture representsremoving amount of heat from the system. Now, according to LeChatlier's principle, the equilibrium will shift in forward direction (product side as the product is decreasing) in order to nullify the change. Thus, on lowering the temperature of the given exothermic equation, the concentration of reactants will decrease and that of products will increase and hence, resulting in increase in ${\text{K}}_{\text{eq}}$ value.