Chapter U6.122, Problem 1E

Interpretation Introduction

Interpretation: Two things needs to be named that could form more reactants in an equilibrium mixture.

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 1E

Two things that could form more reactants in an equilibrium mixture are:

• Increasing the product concentration.

Decreasing the reactant concentration.

Explanation of Solution

Two things that could form more reactants in an equilibrium mixture are discussed in detail by taking an example of equilibrium mixture described by equation as:

  ${\text{SO}}_{\text{2}}{\text{(g) + NO}}_{\text{2}}\text{(g) }\rightleftharpoons {\text{ SO}}_{\text{3}}\text{(g) + NO(g)}$

• Increasing the product concentration:

The addition of $\text{NO(g)}$ 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 $\text{NO(g)}$ 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.

Decreasing the reactant concentration.

The removal of ${\text{SO}}_2\text{(g)}$ from the above equilibrium mixture means that the concentration of the reactant decreases. Now, according to LeChatlier's principle, the equilibrium will shift in backward direction (reactant side as the concentration of reactant is decreasing) in order to nullify the change. Thus, on removing ${\text{SO}}_2\text{(g)}$ from 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.