Sodium bicarbonate undergoes thermal decomposition according to the reaction

$\mathbf{2}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{}\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

How does the equilibrium position shift as a result of each of the following disturbances? (a) $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$of argon gas is added.

(b) $\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$is added.

(c) $\mathit{M}\mathit{g}{\left(CI{O}_4\right)}_{\mathbf{2}}\left(s\right)$is added as a drying agent to remove${\mathit{H}}_{\mathbf{2}}\mathit{O}$.

(d) Dry ice is added at constant T.

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 equation is,

$\mathbf{2}{\mathbf{NaHCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Condition: if $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$of argon gas is added:

Argon is a non-combustible gas that is not involved in the reaction. As a result, there will be no change in the system and no effect on the reaction.

Therefore, there will be no change in the system and no effect on the reaction.

(b)

Considering the given information,

The given equilibrium reaction is,

$\mathbf{2}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Condition: if $\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$is added:

There is no effect when $\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$is added or removed so the system doesn't shift .

(c)

Considering the given information,

The given equilibrium reaction is,

$\mathbf{2}\mathit{N}\mathit{a}\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Condition: if $\mathit{M}\mathit{g}{\left(CI{O}_4\right)}_{\mathbf{2}}\left(s\right)$is added as a drying agent to remove ${\mathit{H}}_{\mathbf{2}}\mathit{O}$:

The product side includes ${\mathit{H}}_{\mathbf{2}}\mathit{O}$. According to the Le Chatelier's principle, lowering the product concentration causes the reaction to shift to the right, favouring its formation.

The product side is now missing its product, causing the equilibrium to be disrupted. To compensate for the loss, the reaction will produce more of the product.

(d)

Considering the given information,

The given equilibrium reaction is,

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

Condition: if Dry ice is added at constant T :

$\mathit{C}{\mathit{O}}_{\mathbf{2}}$is solid dry ice. It, on the other hand, easily transforms from a solid to a gaseous state. The addition of dry ice will increase the concentration of $\mathit{C}{\mathit{O}}_{\mathbf{2}}$as it sublimes. As a result, there is a shift in equilibrium to the left, favoring the formation of reactants.

There are now excess components on the product side. The reaction will consume the excess concentration in order to restore equilibrium on both sides.