Question

Heating solid sodium bicarbonate in a closed vessel established this equilibrium: \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system, (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system, (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.

Short answer

(a) The equilibrium will shift to the right. (b) Adding \(Na_{2}CO_{3}\) will not affect the equilibrium position. (c) Removing \(NaHCO_{3}\) will not affect the equilibrium position.

Step-by-step solution

Effect of Removing \(CO_{2}\) from the system

According to Le Chatelier's Principle, if \(CO_{2}\) is removed from the system, the reaction will shift to the right to replace the quantity of \(CO_{2}\) that was removed. This is because the system will want to achieve its equilibrium state again by increasing the concentration of \(CO_{2}\) to its original level. This means the reaction will lean towards the products.

Effect of Adding \(Na_{2}CO_{3}\) to the system

Adding more of \(Na_{2}CO_{3}\) to the system does not affect the equilibrium position as it is a solid. In terms of equilibrium, the concentration of a solid or liquid does not change as it doesn’t have a volume in which to be spread.

Effect of Removing \(NaHCO_{3}\) from the system

If some of the solid \(NaHCO_{3}\) were removed from the system, it would not affect the equilibrium position. This is because in the equilibrium expression, solids are not considered as their concentrations do not change.

Key concepts

Le Chatelier's Principle

Imagine you're trying to keep your room at a comfortable temperature. If it gets too hot, you might open a window to let cooler air in, and if it gets too cold, you might close it to keep warmth inside. Le Chatelier's Principle acts similarly for chemical reactions trying to maintain balance, or equilibrium.

When a change occurs in a system at equilibrium, this principle predicts how the system will react to restore balance. For example, if a product or reactant is added or removed, the system will shift to counteract that change.

• Adding a reactant or removing a product will shift the reaction towards the products, encouraging more substances to be made.
• Conversely, adding a product or removing a reactant shifts the reaction towards the reactants, discouraging excess formation.

Le Chatelier's Principle helps us understand and predict the behavior of a reaction when exposed to external changes.

sodium bicarbonate decomposition

Sodium bicarbonate, commonly known as baking soda, breaks down in a unique way. When heated, it undergoes decomposition, resulting in a fascinating chemical reaction:
\[2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2}\mathrm{CO}_{3}(s) + \mathrm{H}_{2}\mathrm{O}(g) + \mathrm{CO}_{2}(g)\]
Here, solid sodium bicarbonate decomposes into solid sodium carbonate, water vapor, and carbon dioxide gas. This process is called chemical equilibrium because the reaction doesn’t proceed to completion but instead reaches a state where the rates of the forward and reverse reactions are equal, thus maintaining a balance of reactants and products.

• Sodium bicarbonate is the key starting material in this process.
• Heat causes the breaking down of bicarbonate into carbonate, water, and carbon dioxide.
• Equilibrium is reached when the production and decomposition rates match.

This decomposition is important not only in chemistry but also in practical applications like baking, where it contributes to the rising of dough.

equilibrium shift effects

Equilibrium systems can be delicate and respond significantly to changes in conditions. When you alter a component of the chemical reaction, the reaction tries to re-establish equilibrium by shifting in a particular direction.

Consider the reaction: \[2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2}\mathrm{CO}_{3}(s) + \mathrm{H}_{2}\mathrm{O}(g) + \mathrm{CO}_{2}(g)\]Here are some equilibrium shift effects:

• **Removing \(CO_2\)**: The equation will shift right, striving to produce more \(CO_2\) to replace what was removed. This results in more products being formed.
• **Adding \(Na_2CO_3\)**: Since \(Na_2CO_3\) is a solid, its addition doesn't affect the equilibrium position because solids don't change concentration.
• **Removing \(NaHCO_3\)**: This also doesn't change the equilibrium position due to the same logic applied to adding solids.

By understanding these effects, we can predict and manipulate chemical reactions in various settings, making the process an invaluable tool in chemistry.