Consider this equilibrium system:

${\mathbf{\text{CO(g)+Fe}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}_{\mathbf{\text{4}}}\mathbf{\text{(s)}}\mathbf{\rightleftharpoons }{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)+3FeO(s)}}$

How does the equilibrium position shift as a result of each of the following disturbances? (a) $\mathit{C}\mathit{O}$is added.

(b) $\mathit{C}{\mathit{O}}_{\mathbf{2}}$is removed by adding solid $\mathit{N}\mathit{a}\mathit{O}\mathit{H}$.

(c) Additional $\mathit{F}{\mathit{e}}_{\mathbf{3}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$is added to the system.

(d) Dry ice is added at constant temperature.

Le-Chatelier's principle states that, whenever a system at equilibrium is disturbed, the system will undergo reactions and try to cancel that effect and retain equilibrium. Changes in concentration of any component, temperature, pressure, or volume are all examples of disturbances.

(a)

Considering the given information,

The given equilibrium reaction is,

${\mathbf{CO}}_{\left(g\right)}{\mathbf{Fe}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{}{\mathbf{CO}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}\mathbf{3}{\mathbf{FeO}}_{\left(s\right)}$

Condition: CO is added:

For the above reaction, the reaction quotient expression is,

${\mathit{Q}}_{\mathbf{c}}\mathbf{=}\frac{\left|C{O}_2\right|}{\left|CO\right|}$

The addition of CO reduces the value of ${\mathit{Q}}_{\mathbf{c}}$. As a result, in order to maintain equilibrium, the reaction will proceed in the product side in order to consume the extra CO.

Therefore, adding CO shifts the equilibrium toward product formation.

(b)

Considering the given information,

The given equilibrium reaction is,

$\mathit{C}{\mathit{O}}_{\left(g\right)}\mathit{F}{\mathit{e}}_{\mathbf{3}}{\mathit{O}}_{\mathbf{4}\mathbf{\left(}\mathit{s}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathit{C}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathit{g}\mathbf{\right)}}\mathbf{+}\mathbf{3}\mathit{F}\mathit{e}{\mathit{O}}_{\left(s\right)}$

Condition: $\mathit{C}{\mathit{O}}_{\mathbf{2}}$is removed by adding solid $\mathit{N}\mathit{a}\mathit{O}\mathit{H}$:

For the above reaction, the reaction quotient expression is,

${\mathit{Q}}_{\mathbf{c}}\mathbf{=}\frac{\left[C{O}_2\right]}{\left[CO\right]}$

Whenis removed, the value of ${\mathit{Q}}_{\mathbf{c}}$is decreased. As a result, in order to maintain equilibrium, the reaction will proceed in the product side, producing more $\mathit{C}{\mathit{O}}_{\mathbf{2}}$.

Therefore, when solid $\mathit{N}\mathit{a}\mathit{O}\mathit{H}$is added to remove $\mathit{C}{\mathit{O}}_{\mathbf{2}}$, the equilibrium shifts toward the formation of product.

(c)

Considering the given information,

The given equilibrium reaction is,

$\mathit{C}{\mathit{O}}_{\left(g\right)}\mathit{F}{\mathit{e}}_{\mathbf{3}}{\mathit{O}}_{\mathbf{4}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathit{C}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}\mathbf{3}\mathit{F}\mathit{e}{\mathit{O}}_{\left(s\right)}$

Condition: when $\mathit{F}\mathit{e}{\mathit{O}}_{\left(s\right)}$is added:

For the above reaction, the reaction quotient expression is,

${\mathit{Q}}_{\mathbf{c}}\mathbf{=}\frac{\left[C{O}_2\right]}{\left[CO\right]}$

Addition of solid $\mathit{F}{\mathit{e}}_{\mathbf{3}}{\mathit{O}}_{\mathbf{4}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}$does show any impact on equilibrium.

Therefore, there is no change in the equilibrium position.

(d)

Considering the given information,

The given equilibrium reaction is,

$\mathit{C}{\mathit{O}}_{\left(g\right)}\mathit{F}{\mathit{e}}_{\mathbf{3}}{\mathit{O}}_{\mathbf{4}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathit{C}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}\mathbf{3}\mathit{F}\mathit{e}{\mathit{O}}_{\left(s\right)}$

Condition: When dry ice is added at constant temperature:

For the above reaction, the reaction quotient expression is,

${\mathit{Q}}_{\mathbf{c}}\mathbf{=}\frac{\left[C{O}_2\right]}{\left[CO\right]}$

Dry ice is a solid carbon dioxide that sublimes to carbon dioxide gas. Since the reaction has $\mathit{C}{\mathit{O}}_{\mathbf{2}}$gas and the reaction takes place at same temperature, it is considered that sublimation is taking place here.

The reaction will shift toward the reactant side when dry ice is added. When the product concentration rises, the reaction proceeds in such a way that the product concentration falls, consuming more $\mathit{C}{\mathit{O}}_{\mathbf{2}}$.

Therefore, the reaction will go in the direction of the reactant.