Chapter U6.6, Problem 4E

Interpretation Introduction

Interpretation: The effect of removing ammonia, ${\text{NH}}_{\text{3}}\text{(g)}$ from following equilibrium mixture needs to be explained:

  ${\text{N}}_2{\text{(g) + 3H}}_2\text{(g) }\rightleftharpoons {\text{ 2NH}}_{\text{3}}\text{(g)}$

Concept introduction: According to Le Chatlier'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 4E

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 equation is:

  ${\text{N}}_2{\text{(g) + 3H}}_2\text{(g) }\rightleftharpoons {\text{ 2NH}}_{\text{3}}\text{(g)}$

The removal of ammonia, ${\text{NH}}_{\text{3}}\text{(g)}$ 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 ammonia, ${\text{NH}}_{\text{3}}\text{(g)}$ 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.