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Sodium bicarbonate undergoes thermal decomposition according to the reaction $$ 2 \mathrm{NaHCO}_{3}(s) \Longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ How does the equilibrium position shift as a result of each of the following disturbances? (a) 0.20 atm of argon gas is added. (b) \(\mathrm{NaHCO}_{3}(s)\) is added. (c) \(\mathrm{Mg}\left(\mathrm{ClO}_{4}\right)_{2}(s)\) is added as a drying agent to remove \(\mathrm{H}_{2} \mathrm{O}\). (d) Dry ice is added at constant \(T\)

Short Answer

Expert verified
(a) No shift, (b) No shift, (c) Shift to the right, (d) Shift to the left.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
02

Examine the effect of adding argon gas

Adding 0.20 atm of argon gas increases the total pressure but does not affect the partial pressures of the reactants or products since argon is an inert gas and does not participate in the reaction. Therefore, the equilibrium position does not shift.
03

Examine the effect of adding more \(\text{NaHCO}_{3}(s)\)

Adding solid \(\text{NaHCO}_{3}(s)\) increases its amount, but since \(\text{NaHCO}_{3}(s)\) is a solid, it does not affect the concentration of any gaseous components. Therefore, the equilibrium position does not shift.
04

Examine the effect of removing \(\text{H}_{2} \text{O}\)

Using \(\text{Mg}(\text{ClO}_{4})_{2}(s)\) as a drying agent to remove \(\text{H}_{2} \text{O}(g)\) reduces its concentration. According to Le Chatelier's Principle, the equilibrium will shift to the right to produce more \(\text{H}_{2} \text{O}\).
05

Examine the effect of adding dry ice

Adding dry ice (solid \(\text{CO}_{2}\)) at constant temperature increases the concentration of \(\text{CO}_{2} \text{(g)}\). According to Le Chatelier's Principle, the equilibrium will shift to the left to counteract the increase in \(\text{CO}_{2} \text{(g)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction reaches a state where the concentrations of reactants and products remain constant over time. This does not mean that the reactions stop, but rather that the forward and reverse reactions continue at the same rate. At equilibrium, the system is in a state of balance.

To understand this better, consider a reversible reaction:
and the forward reaction converts A to B, while the reverse reaction converts B back to A. When the rate of the forward reaction equals the rate of the reverse reaction, the reaction has reached equilibrium.

Here are key points to remember about chemical equilibrium:
  • At equilibrium, the concentration of reactants and products remains unchanged.
  • The system must be closed, meaning no reactants or products can escape.
  • Dynamic process: reactions are still occurring but at equal rates.
Thermal Decomposition
Thermal decomposition is a chemical reaction where a compound breaks down into simpler compounds when heated. In the context of sodium bicarbonate (\text{NaHCO}_{3}), the thermal decomposition can be represented as:

$$2 \text{NaHCO}_{3}(s) \rightarrow \text{Na}_{2} \text{CO}_{3}(s) + \text{CO}_{2}(g) + \text{H}_{2} \text{O}(g)$$

In this reaction, solid sodium bicarbonate breaks down into solid sodium carbonate, carbon dioxide gas, and water vapor when heated.

Some important aspects of thermal decomposition include:
  • It requires heat to proceed.
  • Produces at least one gas and/or liquid product.
  • Usually results in a significant change in the state of matter of the substances involved (solid to gas, etc.).
Thermal decomposition reactions are common in various fields, including industry and environmental science. Understanding how these reactions work is crucial for controlling processes that involve heating substances.
Equilibrium Shift
An equilibrium shift occurs when the position of equilibrium is disturbed by a change in concentration, pressure, or temperature. Le Chatelier's Principle helps predict how the system will respond to these changes.

According to this principle, if a system at equilibrium is disturbed, it will shift in the direction that counteracts the disturbance. Let's explore how different disturbances affect equilibrium:
  • Adding a reactant or product: Adding more reactant will shift the equilibrium to the right (toward products), while adding more product will shift it to the left (toward reactants).
  • Removing a reactant or product: Removing a reactant shifts the equilibrium to the left, and removing a product shifts it to the right.
  • Changing pressure: For reactions involving gases, increasing pressure by reducing volume will shift the equilibrium toward the side with fewer moles of gas, while decreasing pressure by increasing volume shifts it towards the side with more moles of gas.
  • Changing temperature: Increasing temperature favors endothermic reactions (absorbs heat), while decreasing temperature favors exothermic reactions (releases heat).
Understanding how equilibrium shifts in response to changes is essential for controlling chemical processes in the laboratory and industry. By manipulating conditions, chemists can drive reactions to produce more desired products.

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Most popular questions from this chapter

For the following equilibrium system, which of the changes will form more \(\mathrm{CaCO}_{3} ?\) $$ \begin{array}{r} \mathrm{CO}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-113 \mathrm{~kJ} \end{array} $$ (a) Decrease temperature at constant pressure (no phase change). (b) Increase volume at constant temperature. (c) Increase partial pressure of \(\mathrm{CO}_{2}\). (d) Remove one-half of the initial \(\mathrm{CaCO}_{3}\).

Predict the effect of decreasing the container volume on the amounts of each reactant and product in the following reactions: (a) \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

Hydrogen iodide decomposes according to the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ A sealed 1.50-L container initially holds \(0.00623 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), \(0.00414 \mathrm{~mol}\) of \(\mathrm{I}_{2}\), and \(0.0244 \mathrm{~mol}\) of \(\mathrm{HI}\) at \(703 \mathrm{~K}\). When equilibrium is reached, the concentration of \(\mathrm{H}_{2}(g)\) is \(0.00467 \mathrm{M}\). What are the concentrations of \(\mathrm{HI}(g)\) and \(\mathrm{I}_{2}(g) ?\)

Does \(Q\) for the formation of \(1 \mathrm{~mol}\) of NO from its elements differ from \(Q\) for the decomposition of \(1 \mathrm{~mol}\) of \(\mathrm{NO}\) to its elements? Explain and give the relationship between the two \(Q\) 's.

The two most abundant atmospheric gases react to a tiny extent at \(298 \mathrm{~K}\) in the presence of a catalyst: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad K_{\mathrm{p}}=4.35 \times 10^{-31} $$ (a) What are the equilibrium pressures of the three gases when the atmospheric partial pressures of \(\mathrm{O}_{2}(0.210 \mathrm{~atm})\) and of \(\mathrm{N}_{2}(0.780 \mathrm{~atm})\) are put into an evacuated \(1.00-\mathrm{L}\) flask at \(298 \mathrm{~K}\) with the catalyst? (b) What is \(P_{\text {total }}\) in the container? (c) Find \(K_{\text {c }}\) at \(298 \mathrm{~K}\).

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